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slamgirl [31]
3 years ago
9

Hydrogen peroxide is a versatile chemical, its uses include bleaching wood pulp and fabrics and substituting for chlorine in wat

er purification. One reason for its versatility is that it can be either an oxidizing or reducing agent. For the following reactions, identify whether hydrogen peroxide is an oxidizing or reducing agent.
a) H2O2 (aq) + 2 Fe2+ (aq) + 2 H+ + 2H20 (1) + 2 Fe3+(aq)
b) 5 H2O2 (aq) + 2 MnO4 (aq) + 6 H+ (aq) → 8 H20 (1) + 2 Mn2+ (aq) + 5 O2 (g)
Chemistry
1 answer:
Tasya [4]3 years ago
6 0

Answer:

Explanation:

a )

H₂O₂ (aq) + 2 Fe²⁺ (aq) + 2 H⁺  = 2H20 (l) + 2 Fe³⁺(aq)

Here oxidation number  of Fe is increasing from + 2 to + 3 so it is being oxidized . Hence H₂O₂ is acting as oxidizing agent here .

b )

5 H₂O₂ (aq) + 2 MnO₄⁻¹ (aq) + 6 H⁺ (aq) → 8 H20 (l) + 2 Mn⁺² (aq) + 5O₂ (g)

In this reaction,  oxidation number of Mn is reducing from + 7 to + 2 so it is being reduced . Here H₂O₂ is acting as reducing agent .

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Interaction of Orchids on A tree? ​
ziro4ka [17]

Answers:

<h2>The symbiotic relationship that occurs between an orchid and a tree would be classified as commensalism. Most orchids are epiphytes, which mean that that they grow on other plants. This benefits the orchids because they can grow on top of the canopy, which prevents the orchids from being walked on or eaten by ground-dwelling organisms.</h2><h3 /><h3>I HOPE TO IT'S HELP YOU:)</h3>

3 0
2 years ago
A chemistry student needs of acetic acid for an experiment. He has available of a w/w solution of acetic acid in acetone. Calcul
Volgvan

Answer:

49.4 g Solution

Explanation:

There is some info missing. I think this is the original question.

<em>A chemistry student needs 20.0g of acetic acid for an experiment. He has 400.g  available of a 40.5 %  w/w solution of acetic acid in acetone.  </em>

<em> Calculate the mass of solution the student should use. If there's not enough solution, press the "No solution" button. Round your answer to 3 significant digits.</em>

<em />

We have 400 g of solution and there are 40.5 g of solute (acetic acid) per 100 grams of solution. We can use this info to find the mass of acetic acid in the solution.

400gSolution \times \frac{40.5gSolute}{100gSolution} = 162 g Solute

Since we only need 20.0 g of acetic acid, there is enough of it in the solution. The mass of solution that contains 20.0 g of solute is:

20.0gSolute \times \frac{100gSolution}{40.5gSolute} = 49.4g Solution

4 0
3 years ago
2H2 + O2 → 2H2O
pantera1 [17]
Okay
Mr (H2O)= 18g
therefore moles of H2O
is 720.8/18= 40.04mol
the ratio of H2 to O2 to H2O is
2 : 1 : 2
so moles of H2 is same as H2O here
H2= 40.04moles

moles of O2 is half
so 40.04 x 0.5
20.02moles

grams of O2 is
its moles into Mr of O2
that's 20.02 x 32 = 640.64g

6 0
3 years ago
Consider the following reaction: CH3OH(g)⇌CO(g)+2H2(g) Part A Calculate ΔG for this reaction at 25 ∘C under the following condit
kati45 [8]

<u>Answer:</u> The \Delta G of the reaction at given temperature is -12.964 kJ/mol.

<u>Explanation:</u>

For the given chemical reaction:

CH_3OH(g)\rightleftharpoons CO(g)+2H_2(g)

The expression of K_p for the given reaction:

K_p=\frac{(p_{CO})\times (p_{H_2}^2)}{p_{CH_3OH}}

We are given:

p_{CO}=0.140atm\\p_{H_2}=0.180atm\\p_{CH_3OH}=0.850atm

Putting values in above equation, we get:

K_p=\frac{(0.140)\times (0.180)^2}{0.850}\\\\K_p=5.34\times 10^{-3}

To calculate the Gibbs free energy of the reaction, we use the equation:

\Delta G=\Delta G^o+RT\ln K_p

where,

\Delta G = Gibbs' free energy of the reaction = ?

\Delta G^o = Standard gibbs' free energy change of the reaction = 0 J (at equilibrium)

R = Gas constant = 8.314J/K mol

T = Temperature = 25^oC=[25+273]K=298K

K_p = equilibrium constant in terms of partial pressure = 5.34\times 10^{-3}

Putting values in above equation, we get:

\Delta G=0+(8.314J/K.mol\times 298K\times \ln(5.34\times 10^{-3}))\\\\\Delta G=-12963.96J/mol=-12.964kJ/mol

Hence, the \Delta G of the reaction at given temperature is -12.964 kJ/mol.

5 0
3 years ago
Why is it important for scientists to review and repeat the work of other scientists
guapka [62]

Answer:

Accuracy

Explanation:

It is important for scientists to review the work of other scientists, so they can be sure there are no mistakes or lack of judgement. They repeat to compare results to make hypotheses.

3 0
2 years ago
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