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Bingel [31]
3 years ago
11

The concentration of hydrogen peroxide (H2O2) can be determined by titrating it with an acidified MnO4− solution. The following

is an unbalanced equation.
MnO4− (aq) + H2O2 (aq) → O2 (g) + Mn2+ (aq)

(a) Balance the above redox reaction.
(b) Determine the concentration of H2O2 solution in molarity if 10.00 mL of this solution requires 20.00 mL of 1.5 M MnO4− for a complete reaction.
Chemistry
1 answer:
Vaselesa [24]3 years ago
5 0

Answer:

A) {MnO_4}^- + 3 H^+ + \frac{5}{2} H_2O_2 \rightarrow Mn^{2+} + 4 H_2O + \frac{5}{2} O_2

B) 7.5 molar

Explanation:

A) <u>Reduction</u>

{MnO_4}^- + 8 H^+ + 5e^- \rightarrow Mn^{2+} + 4 H_2O

<u>Oxidation</u>

H_2O_2 \rightarrow O_2 + 2 H^+ + 2e^-

Multiplying the oxidation reaction by 5/2 and adding it to the reduction equation:

{MnO_4}^- + 8 H^+ + 5e^- \rightarrow Mn^{2+} + 4 H_2O

+

\frac{5}{2} H_2O_2 \rightarrow \frac{5}{2} O_2 + 5 H^+ + 5e^-

-----------------------------------------------------------------------------------------------------

{MnO_4}^- + 3 H^+ + \frac{5}{2} H_2O_2 \rightarrow Mn^{2+} + 4 H_2O + \frac{5}{2} O_2

B) 10 ml = 0.01 L

20 ml = 0.02 L

mol of MnO4− = molarity*volume = 1.5*0.02 = 0.03

1 mol of MnO4− reacts with 5/2 mol of H2O2, then:

mol of H2O2 = 0.03*5/2 = 0.075

molarity = mol/volume = 0.075/0.01 = 7.5 molar

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er solutions can be produced by mixing a weak acid with its conjugate base or by mixing a weak base with its conjugate acid. The
liubo4ka [24]

Answer:

You need to add 19,5 mmol of acetates

Explanation:

Using the Henderson-Hasselbalch equation:

pH = pKa + log₁₀ [base]/[acid]

For the buffer of acetates:

pH = pKa + log₁₀ [CH₃COO⁻]/[CH₃COOH]

As pH you want is 5,03, pka is 4,74 and milimoles of acetic acid are 10:

5,03 = 4,74 + log₁₀ [CH₃COO⁻]/[10]

1,95 = [CH₃COO⁻]/[10]

<em>[CH₃COO⁻] = 19,5 milimoles</em>

Thus, to produce an acetate buffer of 5,03 having 10 mmol of acetic acid, you need to add 19,5 mmol of acetates.

I hope it helps!

7 0
4 years ago
What is the mass of 2.35 x 10^21 moles of sodium
lora16 [44]

Answer:

5.41 x 10²² g

Explanation:

Data Given:

no. of moles of sodium = 2.35 x 10²¹ moles

mass of sodium = ?

Solution:

Formula used

         no. of moles = mass in grams / molar mass

To find mass rearrange the above equation:

   mass in grams = no. of moles x molar mass. . . . . . (1)

  • molar mass of Sodium (Na) = 23 g/mol

Put values in equation 1

          mass in grams = 2.35 x 10²¹ moles x 23 g/mol

          mass in grams =  5.41 x 10²² g

So,

mass of Sodium (Na) =  5.41 x 10²² g

8 0
4 years ago
How many moles are in a sample of Sodium that contains 1.806 x 1024 atoms?
Ad libitum [116K]

Answer: 3 moles Na

Explanation: To find the number of moles of Na, divide the number of atoms of Na with the Avogadro's Number.

1.806x10²⁴ atoms Na x 1 mole Na / 6.022x10²³ atoms Na

= 2.99 or 3 moles Na

3 0
4 years ago
Student follow up
sleet_krkn [62]

Answer:

Magnesium and calcium belong to the second group i. e. alkaline earth metals. They are known as earth metals because they are extracted from the earth. They are very reactive elements. Their reactivity increases when we go from top to bottom because the outermost electrons goes farther from the nucleus i. e. atomic radius increases so less energy is needed for its removal.

8 0
3 years ago
Please help :)<br> I don’t know what to do
3241004551 [841]

Answer:

1. MgCl2 + Zn -> Mg + ZnCl2

2. 2Al + 6H2O(g) -> 2Al(OH)3 + 3H2

3. 2Cd + O2 -> 2CdO

4. I2 + KF -> it's not going to react

5. Zn + H2SO4 -> ZnSO4 + H2

6.2KBr + Cl2 -> 2KCl + Br2

7. AgNO3 + Na -> NaNO3 + Ag

8. 2NaCl + F2 -> 2NaF + Cl2

9. AgNO3 + Mg(NO3)2 + Ag

10. Ni + H2SO4 -> NiSO4 + H2

11. Al + K2SO4

12. FeCl3 + 3Mn-> 3MnCl + Fe

13. 2Na + 2H2O -> 2NaOH + H2

14. 2K + MgBr2 -> 2KBr + Mg

15. Zn + Pb(NO3)2 -> Pb + Zn(NO3)2

16. 2AlBr3+ 3Cl2 -> 2AlCl3 + 3Br2

3 0
3 years ago
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