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lianna [129]
3 years ago
13

Which property of manganese dioxide helps it turn hydrogen peroxide into water faster than it would without it? Manganese dioxid

e is an acid. Manganese dioxide is a catalyst. Manganese dioxide is highly reactive. Manganese dioxide is used up very fast.
Chemistry
1 answer:
MakcuM [25]3 years ago
6 0

Answer:

Manganese dioxide is a catalyst.

Explanation:

The decomposition of hydrogen peroxide into water and oxygen is a slow reaction and MnO₂ is used as a catalyst to speed up the reaction.

The role of MnO₂ (catalyst):

  • Chemical reactions occur faster in the presence of a catalyst because the catalyst provides an alternative reaction pathway with a lower activation energy than the non-catalyzed mechanism.
  • In catalyzed mechanisms, the catalyst usually reacts to form a temporary intermediate, which then regenerates the original catalyst in a cyclic process.
  • A substance which provides a mechanism with a higher activation energy does not decrease the rate because the reaction can still occur by the non-catalyzed route.
  • The catalyst does not used up in the reaction.

So, the right choice is:

Manganese dioxide is a catalyst.

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Water is a pure substance. Which of the following is true about water? (4 points) Its compounds can only be physically separated
jonny [76]

Answer:

Water/H20 is a compound made of hydrogen and oxygen. Although water is the most abundant substance on earth, it is rarely found naturally in its pure form. Most of the time, pure water has to be created. Pure water is called distilled water or deionized water.

Explanation:

7 0
2 years ago
A method used by the U.S. Environmental Protection Agency (EPA) for determining the concentration of ozone in air is to pass the
baherus [9]

Answer: 1. 9.08\times 10^{-6} moles

2. 90 mg

Explanation:

O_3(g)+2NaI(aq)+H_2O(l) \rightarrow O_2(g)+I_2(s)+2NaOH(aq)

According to stoichiometry:

1 mole of ozone is removed by 2 moles of sodium iodide.

Thus 4.54 \times 10^{-6} moles of ozone is removed by =\frac{2}{1}\times 4.54 \times 10^{-6}=9.08\times 10^{-6} moles of sodium iodide.

Thus 9.08\times 10^{-6} moles of sodium iodide are needed to remove 4.54\times 10^{-6} moles of O_3

2. \text{Number of moles of ozone}=\frac{0.01331g}{48g/mol}=0.0003moles

According to stoichiometry:

1 mole of ozone is removed by 2 moles of sodium iodide.

Thus 0.0003 moles of ozone is removed by =\frac{2}{1}\times 0.0003=0.0006 moles of sodium iodide.

Mass of sodium iodide= moles\times {\text {molar mass}}=0.0006\times 150g/mol=0.09g=90mg    (1g=1000mg)

Thus 90 mg of sodium iodide are needed to remove 13.31 mg of O_3.

3 0
2 years ago
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PIT_PIT [208]

Answer:

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Explanation:

7 0
3 years ago
What mass of lead (II) chloride is produced when 200.0 mL of a 0.250 M solution of sodium chloride is mixed with 200.0 mL of a 0
kow [346]

Answer:

Option d. 6.95 g

Explanation:

First of all, we state the reaction:

2NaCl + Pb(NO₃)₂ → PbCl₂ +  2NaNO₃

We determine the moles of each reactant, to state the limiting

Firstly we convert volume frm mL to L

0.200 L . 0.250M = 0.05 moles of NaCl

0.200L . 0.250M = 0.05 moles of Pb(NO₃)₂

Acording to stoichiometry we know that relation is 1:2, so the limiting reagent is the NaCl.

For 1 mol of Pb(NO₃)₂ I need 2 moles of NaCl

For 0.05 moles of Pb(NO₃)₂ I would need, the double → 0.1 moles

(We only have, 0.05 moles of NaCl)

Stoichiometry to the formed product is 2:1

From 2 moles of NaCl I produce 1 mol of PbCl₂

From 0.05 moles I would produce, the half → 0.025 moles

Let's convert the moles to mass → 0.025 mol . 278.1 g / 1mol = 6.95 g

8 0
3 years ago
Buffers resist change in pH in a system when?
Sedbober [7]
The answer to this question is <span>b) hydrogen chloride (HCl) is added to the system. This is the only acid/base on the list. Only acids and bases have the potential to directly change pH as they contribute hydronium and hydroxide ions. Glucose, sodium chloride, and sodium bromide do not affect pH in the first place.</span>
5 0
2 years ago
Read 2 more answers
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