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lions [1.4K]
3 years ago
14

Which represents the overall (net) reaction of this mechanism: step 1: H2O2 + I- → H2O + OI- step 2: H2O2 + OI- → H2O + O2 + I-

Note that I- acts as a catalyst in this reaction.
A. 2H2O2 → 2H2O + O2

B. H2O2 → H2 + O2

C. H2O2 + I- → OI- + H2O

D. H2O2 + OI- → H2O + O2 + I-
Chemistry
1 answer:
11111nata11111 [884]3 years ago
4 0

Answer:

A. 2H₂O₂  → 2H₂O + O₂

Explanation:

  • 1. H₂O₂ + I⁻ → H₂O + OI⁻
  • 2. H₂O₂ + OI⁻ → H₂O + O₂ + I⁻

If we <u>make a net sum of both reactions</u>, we're left with:

  • H₂O₂ + I⁻ + H₂O₂ + OI⁻ → H₂O + OI⁻ + H₂O + O₂ + I⁻

Grouping species:

  • 2H₂O₂ + OI⁻ + I⁻ → 2H₂O + OI⁻ + O₂ + I⁻

There is OI⁻ at both sides, so it is eliminated -same goes for the catalyst, I⁻-.

  • 2H₂O₂  → 2H₂O + O₂

Thus the answer is option A.

You might be interested in
Which statement describes the masses of the particles that make up an atom?
Novay_Z [31]
The mass of the atom is equal to the sum of the number of protons and the number of neutrons. In a neutral atom, the number of protons is equal to the number of electrons.  The atomic number meanwhile of an atom is equal to the number of protons of the atom.
6 0
3 years ago
A compound contains 6.0 g of carbon and 1.0 g of hydrogen and has a molar mass of 42.0 g/mol.
makvit [3.9K]

Answer:

%C = 85.71 wt%; %H = 14.29 wt%; Empirical Formula => CH₂; Molecular Formula => C₃H₆

Explanation:

%Composition

Wt C = 6 g

Wt H = 1 g

TTL Wt = 6g + 1g = 7g

%C per 100wt = (6/7)100% = 85.71 wt%

%H per 100wt = (1/7)100% = 14.29 wt % or, %H = 100% - %C = 100% - 85.71% = 14.29 wt% H

What you should know when working empirical formula and molecular formula problems.

Empirical Formula=> <u>smallest</u> whole number ratio of elements in a compound

Molecular Formula => <u>actual</u> whole number ratio of elements in a compound

Empirical Formula Weight x Whole Number Multiple = Molecular Weight

From elemental %composition values given (or, determined as above), the empirical formula type problem follows a very repeatable pattern. This is ...

% => grams => moles => ratio => reduce ratio => empirical ratio

for determination of molecular formula one uses the empirical weight - molecular weight relationship above to determine the whole number multiple for the molecular ratios.

Caution => In some 'textbook' empirical formula problems, the empirical ratio may contain a fraction in the amount of 0.25, 0.50 or 0.75. If such an issue arises, multiply all empirical ratio numbers containing 0.25 and/or 0.75 by '4'  to get the empirical ratio and multiply all empirical ration numbers containing 0.50 by '2' to get the final empirical ratio.

This problem:

Empirical Formula:

Using the % per 100wt values in part 'a' ...

              %     =>         grams                 =>                 moles

%C => 85.71% => 85.71 g* / 100 g Cpd => (85.71 / 12) = 7.14 mol C

%H => 14.29% => 14.29 g / 100 g Cpd => (14.29 / 1) = 14.29 mol H

=> Set up mole Ratio and Reduce to Empirical Ratio:

mole ratio C:H =>  7.14 : 14.29

<u>To reduce mole values to the smallest whole number ratio,  divide all mole values by the smaller mole value of the set.</u>

=> 7.14/7.14 : 14.29/7.14 => Empirical Ration=> 1 : 2

∴ Empirical Formula => CH₂

Molecular Formula:

(Empirical Formula Wt)·N = Molecular Wt => N = Molecular Wt / Empirical Wt

N = 42 / 14 = 3 => multiply subscripts of empirical formula by '3'.

Therefore, the molecular formula is C₃H₆

3 0
3 years ago
How many moles of O2 are needed to react with 2.35 mol of C2H2?
Mrrafil [7]
2 C2H2 + 5 O2 --> 4 CO2 + 2 H2O

2.35 mol C2H2 - x mol O2
2 mol C2H2 - 5 mol O2

x =  \frac{2.35 \times 5}{2}  = 5.875 \: mol
answer: 5.875 mol
5 0
3 years ago
Read 2 more answers
Fluorine has 7 valence electrons what type of bond is likely to form between two atoms?
Semmy [17]
I'm not quiet sure...possibly an ionic bond.
4 0
3 years ago
Imagine you need to explain to a friend how to convert a value on a food label to one that is measured in grams. Assume the pack
Mariulka [41]

Answer:

453.592 grams

Explanation:

Given

Mass = 1 lb

Required.

Convert to grams using dimensional analysis

Represent 1 lb with x g

In unit conversion, we have that.

1 lb = 453.592 g

So:

Getting the equivalent of lb in g, we have:

x g = 1 lb * (453.592 g/ 1 lb)

x g = 1 * 453.592 g

x g = 453.592 grams

Hence:

The equivalent of 1 lb in grams is 453.592 grams

4 0
3 years ago
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