Answer:
0.1110 mol
Explanation:
Mass = 2g
Molar mass = 18.016 g/mol
moles = ?
These quantities are realted by the following equation;
Moles = Mass / Molar mass
Substituting the values of the quantities and solving for moles, we have;
Moles = 2 / 18.016 = 0.1110 mol
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
What 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.
I hope it helps!
Two moles of neon gas enclosed in a constant volume system receive 4250 J of heat. If the gas was initially at 293 K, what is the final temperature of the neon
Answer:
=355.5K
Explanation:
Specific heat, Q = mcΔT
where
Q= 4250JΔT= change in temp = final temp - initial tempc = specific heat capacity = 1.7m = mass of substance in grams[1 mole of Ne = 20g; 2 moles of Ne = 2 × 20 = 40g]
4250 = 40 × 1.7 × (final - 293K)
final - 293k = 4250 / ( 40 × 1.7)
Final temp = 62.5 + 293
=355.5K
I hope this steps are simple to follow and understand.
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.
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
The combustion of propane (C 3H 8) in the presence of excess oxygen yields CO 2 and H 2O: C 3H 8 (g) + 5O 2 (g) → 3CO 2 (g) + 4H 2O (g) When 2.5 mol of O 2 are consumed in their reaction, ________ mol of CO 2 are produced.
Answer:
1.5 mol of CO₂
Explanation:
Use the mole ratio to find how many moles of CO₂ are produced from the reaction.
For every 5 moles of O₂, three moles of CO₂ is produced.
2.5 mol O₂ × 3 mol CO₂ ÷ 5 mol O₂
= 2.5 mol O₂ × 0.6
= 1.5 mol CO₂
When 2.5 mol of O₂ is consumed in the reaction, 1.5 mol of CO₂ is produced.
Hope that helps.
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.
Match each property of a liquid to what it indicates about the relative strength of the intermolecular forces in that liquid.
Strong intermolecular forces
Weak intermolecular forces
Answer:
Strong intermolecular forces: an increase in viscosity of the liquid, increase in surface tension, decrease in vapor pressure, and an increase in the boiling point.
Weak intermolecular forces: a decrease in viscosity, a decrease in surface tension, an increase in vapor pressure and an increase in boiling point.
Explanation:
Intermolecular forces are forces of attraction or repulsion between neighboring molecules in a substance. These intermolecular forces inclde dispersion forces, dipole-dipole interactions, hydrogen bonding, and ion-dipole forces.
The strength of the intermolecular forces in a liquid usually affects the various properties of the liquid such as viscosity, surface tension, vapour pressure and boiling point.
Strong intermolecular forces in a liquid results in the following; an increase in viscosity of the liquid, increase in surface tension, decrease in vapor pressure, and an increase in the boiling point of the liquid.
Weak intermolecular forces in a liquid results in the following; a decrease in viscosity, a decrease in surface tension, an increase in vapor pressure and an increase in boiling point of that liquid.
Strong intermolecular force is defined as the increase in viscosity of the liquid, increase in surface tension, decrease in vapor pressure, and an increase in the boiling point while weak intermolecular forces define as the decrease in viscosity, a decrease in surface tension, an increase in vapor pressure, and an increase in boiling point.
Intermolecular forces are forces of attraction or repulsion between neighboring molecules in a substance. These intermolecular forces include as follows:-
Dispersion forcesDipole-dipole interactionsHydrogen bondingion-dipole forces.
Strong intermolecular forces in a liquid result in the following; an increase in viscosity of the liquid, increase in surface tension, decrease in vapor pressure, and an increase in the boiling point of the liquid.
Weak intermolecular forces in a liquid result in the following; a decrease in viscosity, a decrease in surface tension, an increase in vapor pressure, and an increase in the boiling point of that liquid.
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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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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
Use bond energies provided to estimate 2Br2
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 = -338.8kJHow 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:
Identify 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
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
“Denitrifying” bacteria return molecular nitrogen gas (N2) back into the biosystem by a series of reductions. Identify the correct sequence. Select the correct answer below: A. NO−3→NO−2→N2O→N2 B. N2O→NO−3→NO−2→N2 C. N2O→NO−2→NO−3→N2 D. NO−3→N2O→NO−2→N2
Answer:
NO−3→NO−2→N2O→N2
Explanation:
Denitrification is the process by which nitrogen is returned to the atmosphere by denitrifying bacteria. The process of denitrification involves a sequence of reduction reactions in the sequence; NO3−→NO2−→N2O→N2.
Nitrogen is usually present in soil in the form of soil nitrates which are soluble in water and can be absorbed by plant roots. These denitrifying bacteria reduce soil nitrates to nitrites, then to nitrogen I oxide and finally to molecular nitrogen as shown in the sequence above.
Denitrification can release N2O, is an ozone-depleting substance and
greenhouse gas into the atmosphere with its attendant consequence on global warming.
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.
A crystal lattice formed by positive and negative ions is called a
Answer:
Ionic Crystal
Explanation:
If a diatomic molecule has a vibrational force constant of k=240 kg s-2 and a reduced mass of 1.627x10-27 kg, its vibrational frequency should be (in cm-1):
A. 2040
B. 4079
C. 2885
D. 5770
E. 1443
Answer:
2040 cm-1
Explanation:
The vibrations frequency is obtained from;
v=1/2πc √k/μ
Where;
k= force constant = 240kgs-2
μ= reduced mass = 1.627×10^-27 kg
c= speed of light= 3×10^10cms-1
v= 1/2×3.142×3×10^10√240/1.627×10^-27
v= 5.3×10^-12 × 3.84×10^14
v= 20.4×10^2
v= 2040 cm-1
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 producedFor each reaction, write the chemical formulae of the oxidized reactants in the space provided. Write the chemical formulae of the reduced reactants.
reactants oxidized _____
reactants reduced _____
a. 2Fe(s)+3Pb(NO3)2(aq)→3Pb(s)+2Fe(NO3)3(aq)
b. AgNO3(aq)+Cu(s)→2Ag(s)+CuNO)2(a)
c. 3AgNO(aq)+Al()→3Ags)+Al(NO3)3(aq)
Answer:
a. Oxidized: Fe(s)
Reduced: Pb(NO3)2
b.Oxidized: Cu(s)
Reduced: AgNO3
c. Oxidized: Al(s)
Reduced: AgNO3
Explanation:
In a redox reaction, one reactant is been oxidized whereas the other is reduced.. The reduced reactant is the one that is gaining electrons and the oxidized one is loosing electrons.
In the reactions:
a. 2Fe(s)+3Pb(NO3)2(aq)→3Pb(s)+2Fe(NO3)3(aq)
The Fe is as reactant as Fe(s) (Oxidation state 0) and the product is +3 (Because NO3, nitrate ion, is always -1). That means Fe is oxidized. The Pb as reactant is +2 and as product 0 (Gaining 2 electrons). Pb(NO3)2 is reduced
b. 2AgNO3(aq)+Cu(s)→2Ag(s)+Cu(NO3)2(a)
AgNO3 is +1 and Ag(s) is 0. AgNO3 is reduced. Cu(s) is 0 as reactant and +2 as product. Cu(s) is been oxidized
c. 3AgNO3(aq)+Al(s)→3Ag(s)+Al(NO3)3(aq)
Here, in the same way, AgNO3 is +1 as reactant and 0 as product. AgNO3 is reduced. And Al(s) is 0 as reactant but + 3 as product. Al(s) is oxidized.
formic acid buffer containing 0.50 M HCOOH and 0.50 M HCOONa has a pH of 3.77. What will the pH be after 0.010 mol of NaOH has been added to 100.0 mL of the buffer
Answer:
pH = 3.95
Explanation:
It is possible to calculate the pH of a buffer using H-H equation.
pH = pka + log₁₀ [HCOONa] / [HCOOH]
If concentration of [HCOONa] = [HCOOH] = 0.50M and pH = 3.77:
3.77 = pka + log₁₀ [0.50] / [0.50]
3.77 = pka
Knowing pKa, the NaOH reacts with HCOOH, thus:
HCOOH + NaOH → HCOONa + H₂O
That means the NaOH you add reacts with HCOOH producing more HCOONa.
Initial moles of 100.0mL = 0.1000L:
[HCOOH] = (0.50mol / L) ₓ 0.1000L = 0.0500moles HCOOH
[HCOONa] = (0.50mol / L) ₓ 0.1000L = 0.0500moles HCOONa
After the reaction, moles of each species is:
0.0500moles HCOOH - 0.010 moles NaOH (Moles added of NaOH) = 0.0400 moles HCOOH
0.0500moles HCOONa + 0.010 moles NaOH (Moles added of NaOH) = 0.0600 moles HCOONa
With these moles of the buffer, you can calculate pH:
pH = 3.77 + log₁₀ [0.0600] / [0.0400]
pH = 3.95When the pH be after 0.010 mol of NaOH has been added to 100.0 mL of the buffer pH is = 3.77 + log₁₀ [0.0600] / [0.0400] = 3.95
What is Formic Acid?It is possible to Computation the pH of a buffer using H-H equation.
Then pH is = pka + log₁₀ [HCOONa] / [HCOOH]
Then If concentration of [HCOONa] is = [HCOOH] then = 0.50M and pH = 3.77:
3.77 is = pka + log₁₀ [0.50] / [0.50]
After that, 3.77 = pka
Then, Knowing pKa, the NaOH reacts with HCOOH, thus:
After that,[tex]HCOOH + NaOH \rightarrow HCOONa + H2O[/tex]
Now, That means the NaOH you add reacts with HCOOH producing more HCOONa.
Then, Initial moles of 100.0mL = 0.1000L:
After that, [HCOOH] = (0.50mol / L) ₓ 0.1000L = 0.0500moles HCOOH
Then, [HCOONa] = (0.50mol / L) ₓ 0.1000L = 0.0500moles HCOONa
After that, when the reaction, moles of each species is:
Then, 0.0500moles HCOOH - 0.010 moles NaOH (Moles added of NaOH) = 0.0400 moles HCOOH
Now, 0.0500moles HCOONa + 0.010 moles NaOH (Moles added of NaOH) = 0.0600 moles HCOONa
Then, With these moles of the buffer, you can calculate pH:
pH = 3.77 + log₁₀ [0.0600] / [0.0400]
Therefore, pH = 3.95
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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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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
which statement describes the use of a flowchart?
Answer:
A flowchart is a type of diagram that represents a workflow or process
Explanation:
Answer: orders in which steps in a process happen
Explanation:
The rate law for the reaction 2NO2 + O3 → N2O5 + O2 is rate = K[NO2][O3].
Which one of the following mechanisms Is consistent with this rate law?
A. NO2 + NO2 → N2O2 (fast)
N2O4 + O3 → N2O5 + O2 (slow)
B. NO2 + O3 → NO5 (fast)
NO5 + NO5 → N2O5 + (5/2)O2 (slow)
C. NO2 + O3 → NO3 + O2 (slow)
NO3 + NO2 → N2O5 (fast)
D. NO2 + NO2 → N2O2 + O2 (slow)
N2O2 + O3 → N2O5 (fast)
Answer:
C. NO2 + O3 → NO3 + O2 (slow)
NO3 + NO2 → N2O5 (fast)
Explanation:
A reaction mechanism represents an amount of elementary steps that explain how a reaction proceeds. The mechanism must explain the experimental rate law. Also, the slow step is the rate determining step.
This rate law is obtained from the multiplication of the reactants in the slow step, thus:
A. NO2 + NO2 → N2O2 (fast)
N2O4 + O3 → N2O5 + O2 (slow)
Rate law:
rate = k [N2O4] [O3]
This mechanism is not consistent with rate law.
B. NO2 + O3 → NO5 (fast)
NO5 + NO5 → N2O5 + (5/2)O2 (slow)
Rate law:
rate = k [NO5]²
This mechanism is not consistent with rate law.
C. NO2 + O3 → NO3 + O2 (slow)
NO3 + NO2 → N2O5 (fast)
Rate law:
rate = k [NO2] [O3]
This mechanism is consistent with rate law.D. NO2 + NO2 → N2O2 + O2 (slow)
N2O2 + O3 → N2O5 (fast)
Rate law:
rate = k [NO2]²
This mechanism is not consistent with rate law.
Thus, right solution is:
C. NO2 + O3 → NO3 + O2 (slow)
NO3 + NO2 → N2O5 (fast)
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]
Regards.
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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