Answer:
A
Explanation:
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)
chromium (iii) or chromium(ii) are frequently used to apply chrome finish to sink fixtures such as faucets. if 45.2 Amps flow through a solution of chromium (iii) for 2 hours, how many grams of chromium can be deposited on a fixture A...175.35g B...58.45g c...0.016g d....0.974g
Answer:
58.45g is the answer
Explanation:
took the test
The mass of chromium that can be deposited is equal to 58.45 g. Therefore, option B is correct.
What is electric current?For a steady flow of charge through a conductor, the current can be determined with the following equation:
[tex]{\displaystyle I={Q \over t}[/tex]
where Q is the electric charge while time t. If Q and t are measured in coulombs (C) and seconds then I will be in amperes.
Electric charge flows by electrons, from lower potential to higher electrical potential. Any stream of charged objects can constitute an electric current.
Given, the amount of electric current flowing through the solution:
I = 45.2 A
The time for which the current flows, t = 2 hrs = 2 × 60 ×60 = 7200 sec
The charge flowing through the solution, Q = I × t
Q = 45.2 × 7200
Q = 325440 C
The number of moles of electrons in 325440 C charge = 325440/96500 = 3.37 mol
We know Cr³⁺ + 3e⁻ → Cr (s)
3 moles of electrons deposit of chromium = 1 mol
3.37 mol of electrons deposit of chromium = 3.37/3 = 1.12 mol
The mass of chromium in 1.12 mol = 1.12 × 52 = 58.45 g
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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
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]
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].
Balance the following
Na+02-→ Na20
Al+O2 ->Al2O3
H2+12+ ->HI
Mg+H2O → Mg(OH)2+H2
Ca+O2 -> Cao
Answer:
1. Na + O2 → Na2O (Balanced)
2. 4Al + 3O2 → 2(Al2O3) (Balanced)
3. H2 + i2 → 2HI (Balanced)
4. Mg + 2H2O → Mg(OH)2+ H2 (Balanced)
5. 2Ca +O2 → 2CaO (Balanced)
In the following reaction: Mg + 2HCl → MgCl2 + H2 How many liters of H2 would be produced if you started with 24.3 g of Mg?
Answer:
22.4 L H2
Explanation:
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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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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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Which is the electron configuration for bromine?
Answer:
The answer below would be written in a straight line from left to right but I wrote it as a list to make it easier to read.
Explanation:
1s^2
2s^2
2p^6
3s^2
3p^6
4s^2
3d^10
4p^5
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
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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An organic acid is composed of carbon (68.84%), hydrogen (4.96%), and oxygen (26.20%). Its molar mass is 122.12 g/mol. Determine the molecular formula of the compound.
Answer:
The molecular formula of the compound is [tex]C_{7}H_{6}O_{2}[/tex].
Explanation:
Let consider that given percentages are mass percentages, so that mass of each element are determined by multiplying molar massof the organic acid by respective proportion. That is:
Carbon
[tex]m_{C} = \frac{68.84}{100}\times \left(122.12\,\frac{g}{mol} \right)[/tex]
[tex]m_{C} = 84.067\,g[/tex]
Hydrogen
[tex]m_{H} = \frac{4.96}{100}\times \left(122.12\,\frac{g}{mol} \right)[/tex]
[tex]m_{H} = 6.057\,g[/tex]
Oxygen
[tex]m_{O} = \frac{26.20}{100}\times \left(122.12\,\frac{g}{mol} \right)[/tex]
[tex]m_{O} = 31.995\,g[/tex]
Now, the number of moles ([tex]n[/tex]), measured in moles, of each element are calculated by the following expression:
[tex]n = \frac{m}{M}[/tex]
Where:
[tex]m[/tex] - Mass of the element, measured in grams.
[tex]M[/tex]- Molar mass of the element, measured in grams per mol.
Carbon ([tex]m_{C} = 84.067\,g[/tex], [tex]M_{C} = 12.011\,\frac{g}{mol}[/tex])
[tex]n = \frac{84.067\,g}{12.011\,\frac{g}{mol} }[/tex]
[tex]n = 7[/tex]
Hydrogen ([tex]m_{H} = 6.057\,g[/tex], [tex]M_{H} = 1.008\,\frac{g}{mol}[/tex])
[tex]n = \frac{6.057\,g}{1.008\,\frac{g}{mol} }[/tex]
[tex]n = 6[/tex]
Oxygen ([tex]m_{O} = 31.995\,g[/tex], [tex]M_{O} = 15.999\,\frac{g}{mol}[/tex])
[tex]n = \frac{31.995\,g}{15.999\,\frac{g}{mol} }[/tex]
[tex]n = 2[/tex]
For each mole of organic acid, there are 7 moles of carbon, 6 moles of hydrogen and 2 moles of oxygen. Hence, the molecular formula of the compound is:
[tex]C_{7}H_{6}O_{2}[/tex]
Which of the following is an example of a formula of a compound?
A. NH3
B. Nitrogen trihydride
C. Nitrogen + Hydrogen = Nitrogen trihydride
D. Ammonia
Answer:
A) NH3
It is the only one that is a formula!!
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.
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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What is the product of the unbalanced combustion reaction below?
C4H10(g) + O2(g) →
Answer:
Option C . CO2(g) + H2O(g)
Explanation:
When hydrocarbon undergoes combustion, carbon dioxide (CO2) and water (H2O) are produced.
C2H4(g) + O2(g) —› CO2(g) + H2O(g)
Thus, the product of the unbalanced combustion reaction is:
CO2(g) + H2O(g)
Thus, we can balance the equation as follow:
C2H4(g) + O2(g) —› CO2(g) + H2O(g)
There are 2 atoms of C on the left side and 1 atom on the right side. It can be balanced by putting 2 in front of CO2 as shown below:
C2H4(g) + O2(g) —› 2CO2(g) + H2O(g)
There are 4 atoms of H on the left side and 2 atoms on the right side. It can be balanced by putting 2 in front of H2O as shown below:
C2H4(g) + O2(g) —› 2CO2(g) + 2H2O(g)
There are a total of 6 atoms of O on the right side and 2 atom on the left side. It can be balanced by putting 3 in front of O2 as shown below:
C2H4(g) + 3O2(g) —› 2CO2(g) + 2H2O(g)
Thus, the equation is balanced.
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.
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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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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:
Calculate the maximum wavelength of light that will cause the photoelectric
effect for potassium. Potassium has work function 2.29 eV = 3.67 x 10-19 J.
Answer:
Explanation:
Work function of potassium = 2.29 eV = 3.67 X 10⁻¹⁹ J
So the minimum energy of photon must be equal to 3.67 X 10⁻¹⁹ J .
energy of photon of wavelength λ = hc / λ
where h = 6.67 x 10⁻³⁴
c = 3 x 10⁸
Putting the values in the equation above
6.67 x 10⁻³⁴ x 3 x 10⁸ / λ = 3.67 X 10⁻¹⁹
λ = 6.67 x 10⁻³⁴ x 3 x 10⁸ / 3.67 X 10⁻¹⁹
= 5.452 x 10⁻⁷
= 5452 x 10⁻¹⁰ m
= 5452 A .
Classify each of these reactions.
1) Ba(ClO3)2(s)--->BaCl2(s)+3O2(g)
2) 2NaCl(aq)+K2S(aq)--->Na2S(aq)+2KCl(aq)
3) CaO(s)+CO2(g)--->CaCO3(s)
4) KOH(aq)+AgCl(aq)---->KCl(aq)+AgOH(s)
5) Ba(OH)2(aq)+2HNO2(aq)--->Ba(NO2)2(aq)+2H2O(l)
Each classify reaction should be either one of this.
a. acid-base neutralization
b. precipitation
c. redox
d. none of the above
Answer:
1. REDOX
2. None of the above
3. Precipitation
4. Preicipitation
5. Acid base neutralization
Explanation:
Reactions where a solid is formed, are named as precipitation. This solid is called precipitated.
Option 4 and 3.
3) CaO (s) + CO₂ (g) → CaCO₃(s)
4) KOH (aq) + AgCl (aq) → KCl (aq) + AgOH(s)
Reactions where water is produced, and you have an acid and a base as reactants, are named as neutralization. You called them acid-base because, the products.
5) Ba(OH)₂ (aq) + 2HNO₂(aq) → Ba(NO₂)₂ (aq) + 2H₂O(l)
Redox, are the reactions where one of the reactans can be oxidized and reduced, when a mole of electrons is released, or gained.
1) Ba(ClO₃)₂ (s) → BaCl₂ (s) + 3O₂(g)
Oxygen from the chlorate is oxidized (increases the oxidation state from -2 to 0) and the chlorine is reduced (decreases the oxidation state from +5 to -1).
2. 2NaCl(aq) + K₂S(aq) Na₂S (aq) + 2KCl (aq)
None of the above
A student found the mass of an object to be 26.5 g. To find the volume, the student submerged the object in a graduated cylinder of water. Submerging the object in the water in the graduated cylinder increased the water level by 24.1 mL. The density of the object is
Question 17 options:
A) 0.909 g/mL.
B) 1.1 g/mL.
C) 1.10 g/mL.
D) 1.0906 g/mL.
Answer: The density of the object is 1.10 g/ml
Explanation: Density of object = ?
Mass of object = 26.5 g
Volume of object = volume of water displaced = 24.1 ml
Putting values in above equation, we get:
[tex]Density: \frac{26.5g}{24.1ml} = 1.10g/ml[/tex]
Thus density of the object is 1.10 g/ml
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 free energy obtained from the oxidation (reaction with oxygen) of glucose (C6H12O6) to form carbon dioxide and water can be used to re-form ATP by driving the above reaction in reverse. Calculate the standard free energy change for the oxidation of glucose.
Answer:
The correct answer is -2878 kJ/mol.
Explanation:
The reaction that takes place at the time of the oxidation of glucose is,
C₆H₁₂O₆ (s) + 6O₂ (g) ⇒ 6CO₂ (g) + 6H₂O (l)
The standard free energy change for the oxidation of glucose can be determined by using the formula,
ΔG°rxn = ∑nΔG°f (products) - ∑nΔG°f (reactants)
The ΔG°f for glucose is -910.56 kJ/mol, for oxygen is 0 kJ/mol, for H2O -237.14 kJ/mol and for CO2 is -394.39 kJ/mol.
Therefore, ΔG°rxn = 6 (-237.14) + 6 (-394.39) - (-910.56)
ΔG°rxn = -2878 kJ/mol
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 amount of space an object takes up is called _____. gravity weight mass volume
What is the mass of a sample of water that takes 2000 kJ of energy to boil into steam at 373 K. The latent heat of vaporization of water is 2.25 x 10^6 J kg-1
Answer:
0.89kg
Explanation:
Q=mL L=specific latent heat
Q=energy required in J
m=mass in Kg
Q=mL
m=Q/L
m=2000000J/2.25 x 10^6 J kg-1
m=0.89kg
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.
2,4-Dimethylpent-2-ene undergoes an electrophilic addition reaction in the presence of HBr to form 2-bromo-2,4-dimethylpentane. Complete the mechanism of this addition and draw the intermediates formed as the reaction proceeds.
Answer:
See figure 1
Explanation:
In this case, we have to start with the ionization reaction of HBr to produce the hydronium ion ([tex]H^+[/tex]) and the bromide ion ([tex]Br^-[/tex]). Then the double bond in the alkene can attack the hydronium ion to produce a carbocation. The most stable carbocation would be the tertiary one, therefore we have to put the positive charge in the tertiary carbon. Then, the bromide attacks the carbocation to produce the final halide.
See figure 1
I hope it helps!
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 produced