Answer:
1. False
2. True
3. True
4. True
5. True
6. True
7. True
8. False
9. True
Explanation:
Density is a physical property since its measurement does not involve any chemical process.
Since transition elements exhibit variable oxidation states, the actual oxidation state of the transition element must be specified in the compound.
Due to the fact that neutron has no charge, it was discovered by Chadwick long after the electron and proton were discovered.
The balancing of chemical reaction equations is a demonstration that atoms are neither created no destroyed. It also shows that mass is neither created nor destroyed in chemical reactions.
When a gas is heated, it expands. Its volume and its kinetic energy increases. Since volume and pressure are inversely proportional (Boyle's law) the pressure decreases.
Enthalpy is said to be an extensive property. This implies that the magnitude of change in enthalpy is known to depend on the amount of reactants that is actually reacted.
The combined gas law is applicable to all ideal gas systems irrespective of their individual chemical formulas.
10g of CO2 contains 0.227 moles of CO2 while 10g of NO contains 0.33 moles of NO hence 10.0 g of NO will contain more molecules than 10.0g of CO2.
If a sample is not at absolute zero, the particles are known to possess kinetic energy which decreases continuously until absolute zero is attained.
what is the concentration in ppm of a solution which is prepared by dissolving in 15mg of nacl in 200ml water
Answer:
Explanation:
In weight/volume (w/v) terms,
1 ppm = 1g m-3 = 1 mg L-1 = 1 μg mL-1
200 mL = 0.2 L
15 / 0.2 mg L-1 =75 ppm
The concentration in ppm of a solution which is prepared by dissolving in 15mg of NaCl in 200ml water is 75 mg/.,
What is ppm?ppm stand for 'part per million' and it is used to define the concentration of any substance as mass of any substance present in per liter of volume of solution, its unit for measurement is mg/L.
Given that, mass of NaCl = 15mg
Volume of solution = 200mL = 0.2L
Concentration in ppm will be calculated as:
ppm = 15/0.2 = 75mg/L
Hence ppm concentration of NaCl is 75 mg/L.
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Which types of electron orbitals will have higher energy than a 4d orbital?
A) 4p
B) 3s
C) 5s
D) 4f
Answer:
D) 4f
Explanation:
To determine which electron orbital that will have higher energy than a 4d orbital, we write the electron configuration starting with s-orbital.
1s
2s 2p
3s 3p 3d 3f
4s 4p 4d 4f
5s 5p 5d 5f
6s 6p 6d 6f
7s 7p 7d 7f
In ascending order, 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 3f, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p, 6f, 7d, 7f.
From the electronic configuration formula above, the electron orbitals that have higher energy than a 4d orbital are 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p, 6f, 7d, 7f.
Therefore, 4f is the correct answer.
Answer:
D) 4f
Explanation:
A 27.9 mL sample of 0.289 M dimethylamine, (CH3)2NH, is titrated with 0.286 M hydrobromic acid.
(1) Before the addition of any hydrobromic acid, the pH is___________.
(2) After adding 12.0 mL of hydrobromic acid, the pH is__________.
(3) At the titration midpoint, the pH is___________.
(4) At the equivalence point, the pH is________.
(5) After adding 45.1 mL of hydrobromic acid, the pH is_________.
Answer:
(1) Before the addition of any HBr, the pH is 12.02
(2) After adding 12.0 mL of HBr, the pH is 10.86
(3) At the titration midpoint, the pH is 10.73
(4) At the equivalence point, the pH is 5.79
(5) After adding 45.1 mL of HBr, the pH is 1.18
Explanation:
First of all, we have a weak base:
0 mL of HBr is added(CH₃)₂NH + H₂O ⇄ (CH₃)₂NH₂⁺ + OH⁻ Kb = 5.4×10⁻⁴
0.289 - x x x
Kb = x² / 0.289-x
Kb . 0.289 - Kbx - x²
1.56×10⁻⁴ - 5.4×10⁻⁴x - x²
After the quadratic equation is solved x = 0.01222 → [OH⁻]
- log [OH⁻] = pOH → 1.91
pH = 12.02 (14 - pOH)
After adding 12 mL of HBrWe determine the mmoles of H⁺, we add:
0.286 M . 12 mL = 3.432 mmol
We determine the mmoles of base⁻, we have
27.9 mL . 0.289 M = 8.0631 mmol
When the base, react to the protons, we have the protonated base plus water (neutralization reaction)
(CH₃)₂NH + H₃O⁺ ⇄ (CH₃)₂NH₂⁺ + H₂O
8.0631 mm 3.432 mm -
4.6311 mm 3.432 mm
We substract to the dimethylamine mmoles, the protons which are the same amount of protonated base.
[(CH₃)₂NH] → 4.6311 mm / Total volume (27.9 mL + 12 mL) = 0.116 M
[(CH₃)₂NH₂⁺] → 3.432 mm / 39.9 mL = 0.0860 M
We have just made a buffer.
pH = pKa + log (CH₃)₂NH / (CH₃)₂NH₂⁺
pH = 10.73 + log (0.116/0.0860) = 10.86
Equivalence pointmmoles of base = mmoles of acid
Let's find out the volume
0.289 M . 27.9 mL = 0.286 M . volume
volume in Eq. point = 28.2 mL
(CH₃)₂NH + H₃O⁺ ⇄ (CH₃)₂NH₂⁺ + H₂O
8.0631 mm 8.0631mm -
8.0631 mm
We do not have base and protons, we only have the conjugate acid
We calculate the new concentration:
mmoles of conjugated acid / Total volume (initial + eq. point)
[(CH₃)₂NH₂⁺] = 8.0631 mm /(27.9 mL + 28.2 mL) = 0.144 M
(CH₃)₂NH₂⁺ + H₂O ⇄ (CH₃)₂NH + H₃O⁻ Ka = 1.85×10⁻¹¹
0.144 - x x x
[H₃O⁺] = √ (Ka . 0.144) → 1.63×10⁻⁶ M
pH = - log [H₃O⁺] = 5.79
Titration midpoint (28.2 mL/2)This is the point where we add, the half of acid. (14.1 mL)
This is still a buffer area.
mmoles of H₃O⁺ = 4.0326 mmol (0.286M . 14.1mL)
mmoles of base = 8.0631 mmol - 4.0326 mmol
[(CH₃)₂NH] = 4.0305 mm / (27.9 mL + 14.1 mL) = 0.096 M
[(CH₃)₂NH₂⁺] = 4.0326 mm (27.9 mL + 14.1 mL) = 0.096 M
pH = pKa + log (0.096M / 0.096 M)
pH = 10.73 + log 1 = 10.73
Both concentrations are the same, so pH = pKa. This is the maximum buffering capacity.
When we add 45.1 mL of HBrmmoles of acid = 45.1 mL . 0.286 M = 12.8986 mmol
mmoles of base = 8.0631 mmoles
This is an excess of H⁺, so, the new [H⁺] = 12.8986 - 8.0631 / Total vol.
(CH₃)₂NH + H₃O⁺ ⇄ (CH₃)₂NH₂⁺ + H₂O
8.0631 mm 12.8986 mm -
- 4.8355 mm
[H₃O⁺] = 4.8355 mm / (27.9 ml + 45.1 ml)
[H₃O⁺] = 4.8355 mm / 73 mL → 0.0662 M
- log [H₃O⁺] = pH
- log 0.0662 = 1.18 → pH
Mass of the condensed unknown liquid: 0.3175 g Temperature of the water bath: 99.00 oC Pressure of the gas: 748.2 mmHg Volume of the flask (volume of the gas): 145.0 mL Given : Kelvin = t oC + 273.15 1 L = 1000 mL 1 atm = 760 mmHg Gas constant: R = 0.08206 atm L / mole K; Ideal Gas Law: PV = nRT 1. What is the pressure of the gas in atm? (1 points) 2.
Answer:
1. 0.98 atm
Explanation:
The following data were obtained from the question:
Mass of unknown liquid (m) = 0.3175 g
Temperature (T) = 99 °C
Pressure (P) = 748.2 mmHg
Volume (V) = 145.0 mL
Gas constant (R) = 0.08206 atm.L/Kmol
1. Determination of the pressure in atm.
760 mmHg = 1 atm
Therefore,
748.2 mmHg = 748.2/760 = 0.98 atm
Therefore, the pressure in atm is 0.98 atm.
Use bond energies provided to estimate 2Br2
How many mL of 2.5M HCl would be needed to completely neutralize a standard solution of 0.53M NaOH in a titration
Answer:
Amount of HCL = 0.00318 L of 3.18 ml
Explanation:
Given:
HCL = 2.5 M
NaOH = 0.53 M
Amount of NaOH = 15 ml = 0.015 L
Find:
Amount of HCL
Computation:
HCL react with NaOH
HCl + NaOH ⇒ NaCl + H₂O
So,
Number of moles = Molarity × volume
Number of moles of NaOH = 0.53 × 0.015
Number of moles of NaOH = 0.00795 moles
So,
Number of moles of HCl needed = 0.00795 mol es
So,
Volume = No. of moles / Molarity
Amount of HCL = 0.00795 / 2.5
Amount of HCL = 0.00318 L of 3.18 ml
Methanol is produced industrially by catalytic hydrogenation of carbon monoxide according to the following equation: CO(g) + 2 H2(g) → CH3OH(l) If the yield of the reaction is 40%, what volume of CO (measured at STP) would be needed to produce 1.0 × 106 kg CH3OH?
Answer:
1.7 × 10⁹ L
Explanation:
Step 1: Write the balanced equation
CO(g) + 2 H₂(g) → CH₃OH(l)
Step 2: Calculate the moles corresponding to 1.0 × 10⁶ kg CH₃OH
The molar mass of CH₃OH is 32.04 g/mol.
[tex]1.0 \times 10^{6} kg \times \frac{10^{3}g }{1kg} \times \frac{1mol}{32.04g} = 3.1 \times 10^{7} mol[/tex]
Step 3: Calculate the theoretical yield of CH₃OH
The real yield of CH₃OH is 3.1 × 10⁷ mol and the percent yield is 40%. The theoretical yield is:
[tex]3.1 \times 10^{7} mol (R) \times \frac{100mol(T)}{40mol(R)} = 7.8 \times 10^{7}mol(T)[/tex]
Step 4: Calculate the moles of CO required to produce 7.8 × 10⁷ mol of CH₃OH
The molar ratio of CO to CH₃OH is 1:1. The moles of CO required are 1/1 × 7.8 × 10⁷ mol = 7.8 × 10⁷ mol
Step 5: Calculate the volume of 7.8 × 10⁷ mol of CO at STP
The volume of 1 mole of CO at STP is 22.4 L.
[tex]7.8 \times 10^{7}mol \times \frac{22.4L}{mol} = 1.7 \times 10^{9}L[/tex]
Fructose-2,6-bisphosphate is a regulator of both glycolysis and gluconeogenesis for the phosphofructokinase reaction of glycolysis and the fructose-1,6-bisphosphatase reaction of gluconeogenesis. In turn, the concentration of fructose-2,6-bisphosphate is regulated by many hormones, second messengers, and enzymes.
How do the following affect glycolysis and gluconeogenesis?
Activate glycolysis Inhibit gluconeogenesis Activate gluconeogenesis Inhibit glycolysis
1. increased levels of fructose-2,6-bisphosphatase
2. activation of fructose-2,6-bisphosphate (FBPase-2)
3. increased glucagon levels
4. activation of PFK-2
5. increased levels of CAMP
Answer:
1. Increased levels of fructose-2,6-bisphosphatase : Activate gluconeogenesis Inhibit glycolysis
2. Activation of fructose-2,6-bisphosphate (FBPase-2) : Activate glycolysis Inhibit gluconeogenesis
3. Increased glucagon levels : Activate gluconeogenesis Inhibit glycolysis
4. Activation of PFK-2 : Activate glycolysis Inhibit gluconeogenesis
5. Increased levels of CAMP : Activate gluconeogenesis Inhibit glycolysis
Explanation:
Glycolysis is the breakdown of glucose molecules in order to release energy in the form of ATP in response to the energy needs of the cells of an organism.
Gluconeogenesis is the process by which cells make glucose from other molecules for other metabolic needs of the cell other than energy production.
Glycolysis and gluconeogenesis are metabolically regulated in the cell by various enzymes and molecules.
The following shows the various regulatory methods and their effects on both processes:
1. The enzyme fructose-2,6-bisphosphatase functions in the regulation of both processes. It catalyzes the breakdown of the molecule fructose-2,6-bisphosphate which is an allosteric effector of two enzymes phosphofructokinasse-1, PFK-1 and fructose-1,6-bisphosphatase, FBPase-1 which fuction in glycolysis and gluconeogenesis respectively.
Increased levels of fructose-2,6-bisphosphatase activates gluconeogenesis and inhibits glycolysis by its breakdown of fructose-2,6-bisphosphate.
2. Fructose-2,6-bisphosphate increases the activity of PFK-1 and inhibits the the activity of FBPase-1. The effect is that glycolysis is activated while gluconeogenesis is inhibited.
3. Glucagon is a hormone that stimulates the synthesis of cAMP. It fuctions to activate gluconeogenesis and inhibit glycolysis.
4. Phosphosfructikinase-2, PFK-2 is an enzyme that catalyzes the formation of fructose-2,6-bisphosphate. Activation of PFK-2 results the activation of glycolysis and inhibition of gluconeogenesis.
5. Cyclic-AMP (cAMP) synthesis in response to glucagon release serves to activate a cAMP-dependent protein kinase which phosphorylates the bifunctional protein PFK-2/FBPase-2. This phosphorylation enhances the activity of FBPase-2 while inhibiting the activity of PFK-2, resulting in the activation of gluconeogenesis and inhibition of glycolysis.
Permanganate ion reacts in basic solution with oxalate ion to form carbonate ion and solid manganese dioxide. Balance the skeleton ionic equation for the reaction between NaMnO4 and Na2C2O4 in basic solution: Fill in all blanks with numbers so if the term is not in the equation make it 0.
Mno4^- (aq)+ C204^2- (aq)+
H^+(aq) + OH^-(aq)
H2O(l) MnO2(s)+
CO3^2 (aq)+ H^+(aq)+
OH^- (aq) + H2O(l)
Answer:
2MnO4^- (aq) + 3C2O4^2- (aq) + 2H2O (l) --> 2MnO2(s) +6CO3^2 -(aq) + 4H^+ (aq)
Explanation:
First, write the half equations for the reduction of MnO4^- and the oxidation of C2O4^2- respectively. Balance it.
Reduction requires H+ ions and e- and gives out water, vice versa for oxidation.
Reduction:
MnO4^- (aq) + 4H^+ (aq) + 3e- ---> MnO2(s) + 2H2O (l)
Oxidation:
C2O4^2- (aq) + 2H2O (l) ---> 2CO3^2 -(aq) + 4H^+ (aq) + 2e-
Balance the no. of electrons on both equations so that electrons can be eliminated. we can do so by multiplying the reduction eq by 2, and oxidation eq by 3.
2MnO4^- (aq) + 8H^+ (aq) + 6e- ---> 2MnO2(s) + 4H2O (l)
3C2O4^2- (aq) + 6H2O (l) ---> 6CO3^2 -(aq) + 12H^+ (aq) + 6e-
Now combine both equations and eliminate repeating H+ and H2O.
2MnO4^- (aq) + 8H^+ (aq) + 3C2O4^2- (aq) + 6H2O (l) --> 2MnO2(s) + 4H2O (l) +6CO3^2 -(aq) + 12H^+ (aq)
turns into:
2MnO4^- (aq) + 3C2O4^2- (aq) + 2H2O (l) --> 2MnO2(s) +6CO3^2 -(aq) + 4H^+ (aq)
Complete ionic equation K2CO3(aq)+2CuF(aq) → Cu2CO3(s)+2KF(aq) Examine each of the chemical species involved to determine the ions that would be present in solution. Be sure to consider both the coefficients and subscripts of the molecular equation, and then write this precipitation reaction in the form of a balanced complete ionic equation. Express your answer as a chemical equation including phases.
Answer:
2K+(aq) + CO3²¯(aq) + Ca^2+(aq) + 2F¯(aq) —› Cu2CO3(s) + 2K+(aq) + 2F¯(aq)
Explanation:
K2CO3(aq) + 2CuF(aq) → Cu2CO3(s) + 2KF(aq)
The complete ionic equation for the above equation can be written as follow:
In solution, K2CO3 and CuF will dissociate as follow:
K2CO3(aq) —› 2K+(aq) + CO3²¯(aq)
CuF(aq) —› Ca^2+(aq) + 2F¯(aq)
Thus, we can write the complete ionic equation for the reaction as shown below:
K2CO3(aq) + 2CuF(aq) —›
2K+(aq) + CO3²¯(aq) + Ca^2+(aq) + 2F¯(aq) —› Cu2CO3(s) + 2K+(aq) + 2F¯(aq)
The second-order decomposition of HI has a rate constant of 1.80 · 10-3 M-1s-1. How much HI remains after 27.3 s if the initial concentration of HI is 4.78 M?
Answer: 3.87M of HI remains after 27.3 s
Explanation:
Using the Second order decomposition equation of
1/[H]t =K x t +1/[A]o
Given initial concentration ,[A]o = 4.78M
time, t = 27.3 s
rate of constant , k= 1.80 x 10^-3 M-1s-1
1/[H] t= 1/[A] t= concentration after time, t=?
SOLUTION
1/[A] t =kt +1/[A]o
1/[A] t =(1.80 x 10^-3 (27.3)+1/4.78
0.04914+0.2092=0.2583
1/[A] t =0.2583
[A] t =1/0.2583= 3.87M
An aqueous solution is 40.0 % by mass hydrochloric acid, HCl, and has a density of 1.20 g/mL. The mole fraction of hydrochloric acid in the solution is
Answer:
The molar concentration of HCl in the aqueous solution is 0.0131 mol/dm3
Explanation:
To get the molar concentration of a solution we will use the formula:
Molar concentration = mass of HCl/ molar mass of HCl
Mass of HCl in the aqueous solution will be 40% of the total mass of the solution.
We can extract the mass of the solution from its density which is 1.2g/mL
We will further perform our analysis by considering only 1 ml of this aqueous solution.
The mass of the substance present in this solution is 1.2g.
The mass of HCl Present is 40% of 1.2 = 0.48 g.
The molar mass of HCl can be obtained from standard tables or by adding the masses of Hydrogen (1 g) and Chlorine (35.46 g) = 36.46g/mol
Therefore, the molar concentration of HCl in the aqueous solution is 0.48/36.46 = 0.0131 mol/dm3
Draw the Lewis structure for methane (CH4) and ethane (C2H6) in the box below. Then predict which would have the higher boiling point. Finally, explain how you came to that conclusion.
Answer:
Ethane would have a higher boiling point.
Explanation:
In this case, for the lewis structures, we have to keep in mind that all atoms must have 8 electrons (except hydrogen). Additionally, each carbon would have 4 valence electrons, with this in mind, for methane we have to put the hydrogens around the carbon, and with this structure, we will have 8 electrons for the carbon. In ethane, we will have a bond between the carbons, therefore we have to put three hydrogens around each carbon to obtain 8 electrons for each carbon.
Now, the main difference between methane and ethane is an additional carbon. In ethane, we have an additional carbon, therefore due to this additional carbon, we will have more area of interaction for ethane. If we have more area of interaction we have to give more energy to the molecule to convert from liquid to gas, so, the ethane will have a higher boiling point.
I hope it helps!
The Lewis structure shows the valence electrons in a molecule. Ethane will have a higher boiling point than methane.
We can deduce the number of valence electrons in a molecule by drawing the Lewis structure of the molecule. The Lewis structure consists of the symbols of elements in the compound and the valence electrons in the compound.
We know that the higher the molar mass of a compound the greater its boiling point. Looking at the Lewis structures of methane and ethane, we cam see that ethane has a higher molecular mass (more atoms) and consequently a higher boiling point than methane.
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11. (2 pts) Sodium Hydroxide, is also known as lye and was a critical component in
homemade soap. Now it is a commonly used drain cleaner because it chemically reacts
with fats (the typical cause of a clog) to form a soap that can be swept down the drain.
What is the molarity of 5.00 g Sodium Hydroxide in 750.0 mL of solution?
Answer:
0.167M
Explanation:
Molarity, M, is an unit of concentration in chemistry defined as the ratio between moles of solute (NaOH in this case) and volume of the solution in liters.
To find molarity of 5.00 g Sodium Hydroxide in 750.0 mL of solution we need to convert mass of NaOH to moles (Using its molar mass: 40g/mol) and the mililiters of solution to liters (1L = 1000mL), thus:
Moles NaOH = 5.00g × (1mol/ 40g) = 0.125 moles NaOH = Moles solute
Liters solution = 750.0mL × (1L / 1000mL) = 0.7500L solution
And molariy is:
0.125 moles NaOH / 0.7500L solution =
0.167M
How many moles of aqueous magnesium ions and chloride ions are formed when 0.250 mol of magnesium chloride dissolves in water
Answer:
0.250 mol Mg²⁺
0.500 mol Cl⁻
Explanation:
Magnesium chloride (MgCl₂) dissociates into ions according to the following equilibrium:
MgCl₂ ⇒ Mg²⁺ + 2 Cl⁻
1 mol 1 mol 2 mol
1 mol of Mg²⁺ and 2 moles of Cl⁻ are formed per mole of MgCl₂. If we have 0.250 mol of MgCl₂, the following amounts of ions will be formed:
0.250 mol MgCl₂ x 1 mol Mg²⁺/mol MgCl₂= 0.250 mol Mg²⁺
0.250 mol MgCl₂ x 2 mol Cl⁻/mol MgCl₂= 0.500 mol Cl⁻
Answer:
HEY THE ANSWER ABOVE ME IS RIGHT!! i defientely misclicked my rating :/
5/5 all the way.
Explanation:
A saturated solution of lead(II) iodide, PbI2 has an iodide concentration of 3.0 x 10^-3 mol/L.
a) What is the molar solubility of PbI2?
b) Determine the solubility constant, Ksp, for lead(II) iodide.
c) Does the molar solubility of lead (II) iodide increase, decrease, or remain unchanged with the addition of potassium iodide to the solution? EXPLAIN.
Answer:
a) 1.5 x 10^-3 mol/L
b) 1.35×10^-8
c) decrease
Explanation:
The solubility of lead II iodide is given by the equation;
PbI2(s) -----> Pb^2+(aq) + 2I^-
By looking at the ICE table, I^-=2x= 3.0 x 10^-3 mol/L/2 = 1.5×10^-3 mol/L
Hence molar solubility of PbI2 = 1.5 x 10^-3 mol/L
Ksp= [Pb^2+] [2I^-]^2 =
Let the molar solubility of each ion be x, therefore;
Ksp= 4x^3
Ksp= 4(1.5 x 10^-3 mol/L)^3= 1.35×10^-8
Addition of kI to the saturated solution will shift the equilibrium position to the left thereby decreasing the solubility of the PbI2 in the system due to common ion effect. The concentration of the iodide ion is now excess in the system leading to the reverse reaction being favoured according to Le Chateliers principle.
a) The molar solubility of PbI₂ is [tex]1.5 * 10^{-3} mol/L[/tex]
b) The solubility constant is [tex]1.35*10^{-8}[/tex]
c) The molar solubility of lead (II) will decrease.
Molar Solubility:The solubility of lead II iodide is given by the equation;
[tex]PbI_2(s) ----- > Pb^{2+}(aq) + 2I^-[/tex]
By looking at the ICE table,
[tex]I^-=2x= 3.0 * 10^{-3} mol/L/2 =[/tex] [tex]1.5 * 10^{-3} mol/L[/tex]
Hence, molar solubility of PbI2 = [tex]1.5 * 10^{-3} mol/L[/tex]
[tex]Ksp= [Pb^{2+}] [2I^-]^2[/tex]
Let the molar solubility of each ion be x, therefore;
[tex]Ksp= 4x^3\\\\Ksp= 4(1.5 * 10^{-3} mol/L)^3\\\\Ksp= 1.35*10^{-8}[/tex]
The addition of KI to the saturated solution will shift the equilibrium position to the left thereby decreasing the solubility of the PbI₂ in the system due to common ion effect. The concentration of the iodide ion is now excess in the system leading to the reverse reaction being favoured according to Le- Ch-ateliers principle.
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What is an anode? Explain.
Answer:
Anode is the positively charged electrode which has the following characteristics:
1) Electrons leave anode to enter to the cathode by the battery.
2) Negatively charged ions are attracted towards cathode.
3) It is connected to the positive terminal of the battery.
The initial concentrations of I2 and I− in the reaction below are each 0.0401 M. If the initial concentration of I−3 is 0.0 M and the equilibrium constant is Kc=0.25 under certain conditions, what is the equilibrium concentration (in molarity) of I−? I−3(aq)↽−−⇀I2(aq)+I−(aq)
Answer:
[I⁻] = 0.0352M
Explanation:
Based on the equilibrium:
I₃⁻(aq) ⇄ I₂(aq) + I⁻(aq)
Kc is defined as:
Kc = 0.25 = [I₂] [I⁻] / [I₃⁻]
The system reaches the equilbrium when the ratio [I₂] [I⁻] / [I₃⁻] is equal to 0.25
In the beginning, you add 0.0401M of both [I₂] [I⁻]. When the reaction reach the equilibrium, xM of both [I₂] [I⁻] is consumed producing xM of [I₃⁻]. That is written as:
[I₃⁻] = X
[I₂] = 0.0401M - X
[I⁻] = 0.0401M - X
X is known as reaction coordinate.
Replacing in Kc:
0.25 = [I₂] [I⁻] / [I₃⁻]
0.25 = [0.0401M - X] [0.0401M - X] / [X]
0.25X = 0.00160801 - 0.0802X + X²
0 = 0.00160801 - 0.3302X + X²
Solving for X:
X = 0.0049M → Right solution
X = 0.3252M → False solution. Produce negative concentrations
Replacing, equilibrium concentrations will be:
[I₃⁻] = X
[I₂] = 0.0401M - X
[I⁻] = 0.0401M - X
[I₃⁻] = 0.0049M
[I₂] = 0.0352M
[I⁻] = 0.0352M
The equilibrium concentration (in molarity) of [I⁻] should be considered as the 0.0352M.
Calculation of the equilibrium concentration:Since
I₃⁻(aq) ⇄ I₂(aq) + I⁻(aq)
Here Kc should be defined
Kc = 0.25 = [I₂] [I⁻] / [I₃⁻]
Also, The system finished the equilibrium at the time when the ratio [I₂] [I⁻] / [I₃⁻] is equivalent to 0.25.
Also,
[I₃⁻] = X
[I₂] = 0.0401M - X
[I⁻] = 0.0401M - X
Also,
0.25 = [I₂] [I⁻] / [I₃⁻]
0.25 = [0.0401M - X] [0.0401M - X] / [X]
0.25X = 0.00160801 - 0.0802X + X²
0 = 0.00160801 - 0.3302X + X²
Now
X = 0.0049M → Right solution
X = 0.3252M → False solution
Now equilibrium concentrations will be:
[I₃⁻] = X
[I₂] = 0.0401M - X
[I⁻] = 0.0401M - X
[I₃⁻] = 0.0049M
[I₂] = 0.0352M
[I⁻] = 0.0352M
Hence, The equilibrium concentration (in molarity) of [I⁻] should be considered as the 0.0352M.
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assume that amonia can be prepared by the folowing reaction in the gas phase at STP. If the reaction conditions are maintainted at STP, how many liters of NH3 can be produced by the reaction of 12.0 L of H2 and the exact required volumen of N2
Answer:
8.00L of ammonia can be produced
Explanation:
The reaction is:
N₂(g) + 3H₂(g) → 2NH₃(g)
Where 1 mole of nitrogen reacts with 3 moles of hydrogen to produce 2 moles of ammonia.
Avogadro's law states that, under constant pressure and temperature, equal volumes of gases contains equal number of moles.
As in the reaction conditions are mantained at STP (Pressure and temperature are constant) you can say of the reaction that:
1 liter of nitrogen reacts with 3 liters of hydrogen to produce 2 liters of ammonia
Thus, if 12.0L of hydrogen reacts and 3L of hydrogen produce 2L of ammonia, liters of ammonia produced are:
12L H₂(g) ₓ (2L NH₃(g) / 3L H₂(g)) =
8.00L of ammonia can be producedIdentify a homogeneous catalyst:
a. SO2 over vanadium (V) oxide
b. H2SO4 with concentrated HCl
c. Pd in H2 gas
d. N2 and H2 catalyzed by Fe
e. Pt with methane
Answer:
b, H2SO4 with HCl, as they are both liquid acids
Calculate the pH of a buffer solution that contains 0.25 M benzoic acid (C 6H 5CO 2H) and 0.15M sodium benzoate (C 6H 5COONa). [K a = 6.5 × 10 –5 for benzoic acid]
Answer:
3.97
Explanation:
pH of buffer solution = pKa+Log(Cb/Ca)
pH of buffer solution = -log(Ka)+log(Cb/Ca)............... Equation 1
Where Ca = concentration of acid, Cb = concentration of base.
Given: Ka = 6.5×10⁻⁵, Ca = 0.25 M, Cb = 0.15 M
Substitute into equation 1
pH of buffer solution = -log(6.5×10⁻⁵)+log(0.15/0.25)
pH of buffer solution = 4.19+(0.22)
pH of buffer solution = 3.97.
A compound, C11H12O2, has an IR spectrum showing a peak at 1710 cm-1. Its 1H NMR spectrum has peaks at delta 1.3 (3 H, triplet), 4.3 (2 H, quartet), 6.5 (1 H, doublet), 7.4-7.6 (5 H, multiplet), and 7.7 (1 H, doublet).
Required:
Draw its structure below.
Answer:
Ethyl cinnamate
Explanation:
For this question, we have to start with the IR info. If we have a peak at 1710 this indicates the presence of a carbonyl group in the molecule (C=O). Additionally, if we calculate the I.H.D (index of hydrogen deficiency), we will have a value of "6". We already know that we have a C=O group, so, this counts for 1 of the 6 additionally, we can have a benzene ring so, this counts for 4, so far we have 5. Finally, we will have a double bond outside of benzene and we will have a total of 6, so:
Benzene: 4
Carbonyl group: 1
Double bond: 1
For a total of six (that fits with the I.H.D calculation). So, so far we know that we have a benzene ring, a double bond, and a carbonyl group. In the formula we have 2 oxygens, therefore we can have a carboxylic acid or an ester. In this case, the IR info doesn't give any additional info, so our best option is the ester group.
The 1H NMR info give is:
Signal A= 1.3 (3 H, triplet)
Signal B= 4.3 (2 H, quartet)
Singal C= 6.5 (1 H, doublet)
Signal D= 7.4-7.6 (5 H, multiplet)
Signal E= 7.7 (1 H, doublet)
The molecule that fits with this NMH spectrum and the info given by the I.H.D is "ethyl cinnamate".
See figure 1
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D-Fructose is the sweetest monosaccharide. How does the Fischer projection of D-fructose differ from that of D-glucose? Match the words in the left column to the appropriate blanks in the sentences on the right. Fill in the blanks.
a ketone
carbon 3
carbon 2
carbon 1
an aldehyde
carbon 4
In D-glucose, there is__________ functional group, and the carbonyl group is at___________ when looking at the Fischer projection.
In D-tructose. there is functional group, and the carbonyl group is at when looking at______ the Fischer projection.
Answer:
aldehyde
carbon-1
ketone
carbon-2
Explanation:
Monosaccharides are colorless crystalline solids that are very soluble in water. Moat have a swwet taste. D-Fructose is the sweetest monosaccharide.
In the open chain form, monosaaccharides have a carbonuyl group in one of their chains. If the carbonyl group is in the form of an aldehyde group, the monosaccharide is an aldose; if the carbonyl group is in the form of a ketone group, the monosaccharide is known as a ketose. glucose is an aldose while fructose is a ketose.
In D-glucose, there is an aldehyde functional group, and the carbonyl group is at carbon-1 when looking at the Fischer projection.
In D-fructose, there is a ketone functional group, and the carbonyl group is at carbon-2 when looking at the Fischer projection.
Determine whether each of the following salts will form a solution that is acidic, basic, or pH-neutral. Drag the appropriate items to their respective bins.
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AI(NO3)3 CH3NH3CN NaCIO
CH3NH3CI NaNO3
Acidic Basic pH-neutral
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Answer:
AI(NO₃)₃ → Acidic pH < 7
CH₃NH₃CN → Neutral pH = 7
NaCIO → Basic pH > 7
CH₃NH₃CI → Acidic pH < 7
NaNO₃ → Neutral pH = 7
Explanation:
First of all we dissociate the salts:
Al(NO₃)₃ → Al³⁺ + 3NO₃⁻Nitrate anion comes from the nitric acid which is strong, so the anion is the conjugate weak base. It does not react to water, but the Al is an special case. Aluminum as a cathion comes from the Al(OH)₃ which is a base but this compound can also react as an acid, it is called amphoterous.
Al³⁺ + H₂O ⇄ Al(OH)²⁺ + H⁺
Aluminium cathion reacts to water in order to produce a complex and to give protons to the medium, so the salt is acid.
CH₃NH₃CN → CH₃NH₃⁺ + CN⁻Both ions come from a weak base and a strong acid, so both ions are the conjugate strong base and acid, respectively. They can make hydrolysis to water so the salt is neutral.
CH₃NH₃⁺ + H₂O ⇄ CH₃NH₂ + H₃O⁺ Ka
CN⁻ + H₂O ⇄ HCN + OH⁻ Kb
NaCIO → Na⁺ + ClO⁻Sodium cathion, comes from the strong base NaOH so it is does not react to water. It is the conjugate weak acid. Hypochlorite comes from the weak acid, so it can hydrolyse to water.
ClO⁻ + H₂O ⇄ HClO + OH⁻ Kb
Hypochlorous acid is formed giving OH⁻ to medium, so the salt is basic.
CH₃NH₃CI → CH₃NH₃⁺ + Cl⁻Chloride comes from the strong acid HCl. It does not react to water.
Methylammonium comes from the weak base, methylamine so it can react to water in order to make hydrolysis. The salt will be acidic.
CH₃NH₃⁺ + H₂O ⇄ CH₃NH₂ + H₃O⁺ Ka
NaNO₃ → Na⁺ + NO₃⁻
Both ions come from a strong base and acid, so they are the conjugate base and acid, respectively. As they do not make hydrolisis in water, the salt will be neutral.
What mass of product, in grams, can be made by reacting 5.0g of aluminum and 22g of bromine?
Answer:
Approximately [tex]24\; \rm g[/tex] (at most.)
Explanation:
Aluminum [tex]\rm Al[/tex] reacts with bromine [tex]\rm Br_2[/tex] at a [tex]2:3[/tex] ratio:
[tex]\rm 2\; Al\, (s) + 3\; Br_2\, (g) \to 2\; AlBr_3\, (s)[/tex].
Look up the relative atomic mass of [tex]\rm Al[/tex] and [tex]\rm Br[/tex]. From a modern periodic table:
[tex]\rm Al[/tex]: [tex]26.982[/tex].[tex]\rm Br[/tex]: [tex]79.904[/tex].Calculate the formula mass of the reactants and of the product:
[tex]M(\mathrm{Al}) = 26.986\; \rm g \cdot mol^{-1}[/tex].[tex]M(\mathrm{Br_2}) = 2\times 79.904 = 159.808\; \rm g \cdot mol^{-1}[/tex].[tex]M(\mathrm{AlBr_3}) = 26.986 + 3 \times 79.904 = 266.698\; \rm g \cdot mol^{-1}[/tex].Calculate the quantity (in number of moles of formula units) of each reactant:
[tex]\displaystyle n(\mathrm{Al}) = \frac{m(\mathrm{Al})}{M(\mathrm{Al})} = \frac{5.0\; \rm g}{26.986\; \rm g \cdot mol^{-1}} \approx 0.18528\; \rm mol[/tex].[tex]\displaystyle n(\mathrm{Br_2}) = \frac{m(\mathrm{Br_2})}{M(\mathrm{Br_2})} = \frac{22\; \rm g}{159.808\; \rm g \cdot mol^{-1}} \approx 0.13767\; \rm mol[/tex].Assume that [tex]\rm Al\, (s)[/tex] is the limiting reactant. From the coefficients:
[tex]\displaystyle \frac{n(\mathrm{AlBr_3})}{n(\mathrm{Al})} = 1[/tex].
Based on the assumption that [tex]\rm Al\, (s)[/tex] is the limiting reactant:
[tex]\begin{aligned}n(\mathrm{AlBr_3}) &= \frac{n(\mathrm{AlBr_3})}{n(\mathrm{Al})} \cdot n(\mathrm{Al}) \\ &=1\times 0.18528\; \rm mol \approx 0.185\; \rm mol\end{aligned}[/tex].
In other words, if [tex]\rm Al[/tex] is the limiting reactant (meaning that [tex]\rm Br_2[/tex] is in excess,) then approximately [tex]0.556\; \rm mol[/tex] of [tex]\rm AlBr_3[/tex] will be produced.
On the other hand, assume that [tex]\rm Br_2\; (g)[/tex] is the limiting reactant. Similarly, from the coefficients:
[tex]\displaystyle \frac{n(\mathrm{AlBr_3})}{n(\mathrm{Br_2})} = \frac{2}{3}[/tex].
Based on the assumption that [tex]\rm Br_2\, (g)[/tex] is the limiting reactant:
[tex]\begin{aligned}n(\mathrm{AlBr_3}) &= \frac{n(\mathrm{AlBr_3})}{n(\mathrm{Br_2})} \cdot n(\mathrm{Br_2}) \\ &= \frac{2}{3}\times 0.13767\; \rm mol \approx 0.0918\; \rm mol\end{aligned}[/tex].
Compare the [tex]n(\mathrm{AlBr_3})[/tex] value based on the two assumptions. Only the smallest value, [tex]n(\mathrm{AlBr_3}) \approx 0.0918\; \rm mol[/tex] (under the assumption that [tex]\rm Br_2\, (g)[/tex] is the limiting reactant,) would resemble the theoretical yield. The reason is that [tex]\rm Br_2\, (g)[/tex] would run out before all that [tex]\rm 5.0\; g[/tex] of [tex]\rm Al\, (s)[/tex] was converted to [tex]\rm AlBr_3\, (g)[/tex].
Apply the formula mass of [tex]\rm AlBr_3[/tex] to find the mass of that (approximately) [tex]0.0918\; \rm mol[/tex] of [tex]\rm AlBr_3[/tex] formula units:
[tex]\begin{aligned}m(\mathrm{AlBr_3}) &= n(\mathrm{AlBr_3}) \cdot M(\mathrm{AlBr_3}) \\ &= 0.0918\; \rm mol \times 266.698\; g \cdot mol^{-1} \approx 24\; \rm g\end{aligned}[/tex].
Consider the equilibrium system: N2O4 (g) = 2 NO2 (g) for which the Kp = 0.1134 at 25 C and deltaH rx is 58.03 kJ/mol. Assume that 1 mole of N2O4 and 2 moles of NO2 are introduced into a 5 L contains. What will be the equilibrium value of [N204]?
A) 0.358 M
B) 0.042 M
C) 0.0822 M
D) 0.928 M
E) 0.379 M
Answer: The equilibrium value of [tex]N_2O_4[/tex] is 0.379 M
Explanation:
Equilibrium constant is the ratio of the concentration of products to the concentration of reactants each term raised to its stochiometric coefficients.
Using ideal gas equation : [tex]PV=nRT[/tex]
P = pressure of gas
V = volume of gas
n = no of moles
R = gas constant
T = Temperature
pressure of [tex]N_2O_4[/tex] = [tex]\frac{1\times 0.0821Latm/Kmol\times 298}{5L}=5atm[/tex]
pressure of [tex]NO_2[/tex] = [tex]\frac{2\times 0.0821Latm/Kmol\times 298}{5L}=10atm[/tex]
[tex]N_2O_4(g)\rightleftharpoons 2NO_2(g)[/tex]
at t= 0 5 atm 10 atm
at eqm (5-x) atm (10+2x) atm
[tex]K_p=\frac{[p_NO_2]^2}{[p_N_2O_4]}[/tex]
[tex]0.1134=\frac{(10+2x)^2}{(5-x)}[/tex]
[tex]x=-4.48[/tex]
pressure of [tex]N_2O_4[/tex] at equilibrium = (5-(-4.48))= 9.48 atm
pressure of [tex]N_2O_4[/tex] = [tex]\frac{n\times 0.0821Latm/Kmol\times 298}{V}[/tex]
9.48 = [tex]{M\times 0.0821Latm/Kmol\times 298}[/tex]
[tex]M=0.379[/tex]
Thus the equilibrium value of [tex]N_2O_4[/tex] is 0.379 M
What was one idea Dalton taught about atoms?
A. Atoms contained negatively charged particles scattered inside.
B. Atoms of one type would not react with atoms of another type.
C. All atoms of one type were identical in mass and properties.
D. Atoms changed into new elements when they formed compounds.
Answer:
C
Explanation:
I had this question and C is the right answer
One idea that Dalton taught about atoms was that all atoms of one type were identical in mass and properties.
What is an atom?
An atom is defined as the smallest unit of matter which forms an element. Every form of matter whether solid,liquid , gas consists of atoms . Each atom has a nucleus which is composed of protons and neutrons and shells in which the electrons revolve.
The protons are positively charged and neutrons are neutral and hence the nucleus is positively charged. The electrons which revolve around the nucleus are negatively charged and hence the atom as a whole is neutral and stable due to presence of oppositely charged particles.
Atoms of the same element are similar as they have number of sub- atomic particles which on combination do not alter the chemical properties of the substances.
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below are three reactions showing how chlorine from CFCs (chlorofluorocarbons) destroy ozone (O3) in the stratosphere. Ozone blocks harmful ultraviolet radiation from reaching earth’s surface. Show how these 3 equations sum to produce the net equation for the decomposition of two moles of ozone to make three moles of diatomic oxygen (2 O3→ 3 O2), and calculate the enthalpy change. (6 points) R1 O2 (g) → 2 O (g) ΔH1°= 449.2 kJ R2 O3 (g) + Cl (g) → O2 (g) + ClO (g) ΔH2° = -126 kJ R3 ClO (g) + O (g) → O2 (g) + Cl (g) ΔH3°= -268 kJ
Answer:
ΔH = -338.8kJ
Explanation:
it is possible to sum the enthalpy changes of some reactions to obtain the enthalpy change of the whole reaction (Hess's law).
Using the reactions:
R₁ O₂(g) → 2O(g) ΔH₁°= 449.2 kJ
R₂ O₃(g) + Cl(g) → O₂(g) + ClO(g) ΔH₂° = -126 kJ
R₃ ClO (g) + O (g) → O₂ (g) + Cl (g) ΔH₃°= -268 kJ
By the sum 2R₂ + 2R₃:
(2R₂ + 2R₃) = 2O(g) + 2O₃(g) → 4O₂(g)
ΔH = 2ₓ(-126kJ) + (2ₓ-268kJ) = -788kJ
Now, this reaction + R₁
2O₃(g) → 3O₂(g)
ΔH = -768kJ + 449.2kJ
ΔH = -338.8kJWhat is a major product of the reaction in the box?
Answer:
Molecule C
Explanation:
In this case, on the first reaction, we will have the production of a Grignard reagent. This molecule will react with [tex]D_2O[/tex] and a deuterium atom will be transferrred to the benzene ring. Then at the top of the molecule, we will have an acetal structure. This acetal can be broken by the action of the acid [tex]DCl[/tex], In the mechanism at the end, we will obtain a carbonyl group bonded to a hydrogen atom. Therefore we will have in the final product the aldehyde group. See figure 1 to further explanations.
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An aqueous solution of cobalt(II) fluoride, , is made by dissolving 6.04 grams of cobalt(II) fluoride in sufficient water in a 200. mL volumetric flask, and then adding enough water to fill the flask to the mark. What is the weight/volume percentage of cobalt(II) fluoride in the solution
Answer:
[tex]w/v\%=3.02\frac{g}{mL} \%[/tex]
Explanation:
Hello,
In this case, we first define the formula for the calculation of weight/volume percentage considering cobalt (II) fluoride as the solute, water the solvent and the both of them as the solution:
[tex]w/v\%=\frac{mass_{solute}}{V_{solution}}*100\%[/tex]
In such a way, since the mass of the solute is given as 6.04 g and the final volume of the solution 200 mL, the weight/volume percentage turns out:
[tex]w/v\%=\frac{6.04g}{200mL}*100\%\\\\w/v\%=3.02\frac{g}{mL} \%[/tex]
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