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ohaa [14]
3 years ago
7

Redox titrations are used to determine the amounts of oxidizing and reducing agents in solution. For example, a solution of hydr

ogen peroxide, H2O2, can be titrated against a solution of potassium permanganate, KMnO4. The following equation represents the reaction: 2 KMnO4 (aq) + H2O2 (aq) + 3 H2SO4 (aq) → 3 O2 (g) + 2 MnSO4 (aq) + K2SO4 (aq) + 4 H2O (l) A certain amount of hydrogen peroxide was dissolved in 100. mL of water and then titrated with 1.66 M KMnO4. What mass of H2O2 was dissolved if the titration required 13.2 mL of the KMnO4 solution?

Chemistry
1 answer:
xxTIMURxx [149]3 years ago
6 0

Answer:

0.374 g

Explanation:

Hello,

Since both the molarity and the volume allows us to know the moles of potassium permanganate, and we already have the balanced chemical reaction, the stoichiometric procedure that is attached in the picture, is developed to substantiate the the titrated mass of hydrogen peroxide was 0.374 g.

Best regards.

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Which of these is an example of a chemical change?
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Answer:

D. Burning a peice of wood

Explanation:

Because when you burn wood a chemical reaction happenes between the flames and the wood making the wood into ashes.

Hope this helps you :)

3 0
3 years ago
\The specific heat of aluminum is 0.21 cal g°C . How much heat is released when a 10 gram piece of aluminum foil is taken out of
atroni [7]

<u>Answer: </u>The amount of heat released is 84 calories.

<u>Explanation: </u>

The equation used to calculate the amount of heat released or absorbed, we use the equation:

Q= m\times c\times \Delta T

where,

Q = heat gained  or released = ? Cal

m = mass of the substance = 10g

c = specific heat of aluminium = 0.21 Cal/g ° C

Putting values in above equation, we get:

\Delta T={\text{Change in temperature}}=(10-50)^oC=-40^oC  

Q=10g\times 0.21Cal/g^oC\times-40^oC

Q = -84 Calories

Hence, the amount of heat released is 84 calories.

8 0
3 years ago
How many elements are in Li2SO4?how many elements are in li2s 04 how many elements are in li2s 04 ​
Free_Kalibri [48]

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3

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6 0
3 years ago
Empirical formula of a compound composed of 32.1 g potassium (k) and 6.57 g oxygen (o)?
tester [92]

The empirical formula is K₂O.

The empirical formula is the <em>simplest whole-number ratio</em> of atoms in a compound.

The <em>ratio of atom</em>s is the same as the <em>ratio of moles</em>.

So, our job is to calculate the <em>molar ratio</em> of K to O.

Step 1. Calculate the <em>moles of each element </em>

Moles of K = 32.1 g K × (1 mol K/(39.10 g K =) = 0.8210 mol K

Moles of O = 6.57 g O × (1 mol O/16.00 g O) = 0.4106 mol 0

Step 2. Calculate the <em>molar ratio of each elemen</em>t

Divide each number by the smallest number of moles and round off to an integer

K:O = 0.8210:0.4106 = 1.999:1 ≈ 2:1

Step 3: Write the <em>empirical formula </em>

EF = K₂O

5 0
3 years ago
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morpeh [17]

Answer:

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8 0
3 years ago
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