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julia-pushkina [17]
3 years ago
5

Can manganese be a catalyst

Chemistry
1 answer:
Genrish500 [490]3 years ago
7 0
Manganese alone cannot be a catalyst. However, its oxides can work as a catalyst. Manganes (II, III) oxide has found some applications in certain reactions as a catalyst. These reactions are the oxidation of methane, carbon monoxide, decomposition of NO and the catalytic combustion of organic compounds.
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What is the equilibrium expression for the reatcion bellow<br><br> 2SO3(g)&lt;=&gt;O2(g)+2SO2(g)
Luda [366]

Answer:

[O2(g)][SO2(g)]^2/[SO3(g)]^2

8 0
3 years ago
An analytical chemist is titrating of a solution of nitrous acid with a solution of . The of nitrous acid is . Calculate the pH
Burka [1]

Answer:

pH = 2.69

Explanation:

The complete question is:<em> An analytical chemist is titrating 182.2 mL of a 1.200 M solution of nitrous acid (HNO2) with a solution of 0.8400 M KOH. The pKa of nitrous acid is 3.35. Calculate the pH of the acid solution after the chemist has added 46.44 mL of the KOH solution to it.</em>

<em />

The reaction of HNO₂ with KOH is:

HNO₂ + KOH → NO₂⁻ + H₂O + K⁺

Moles of HNO₂ and KOH that react are:

HNO₂ = 0.1822L × (1.200mol / L) = <em>0.21864 moles HNO₂</em>

KOH = 0.04644L × (0.8400mol / L) = <em>0.0390 moles KOH</em>

That means after the reaction, moles of HNO₂ and NO₂⁻ after the reaction are:

NO₂⁻ = 0.03900 moles KOH = moles NO₂⁻

HNO₂ = 0.21864 moles HNO₂ - 0.03900 moles = 0.17964 moles HNO₂

It is possible to find the pH of this buffer (<em>Mixture of a weak acid, HNO₂ with the conjugate base, NO₂⁻), </em>using H-H equation for this system:

pH = pKa + log₁₀ [NO₂⁻] / [HNO₂]

pH = 3.35 + log₁₀ [0.03900mol] / [0.17964mol]

<h3>pH = 2.69</h3>
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3 years ago
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I hope this helps

5 0
2 years ago
What is the ratio of hydrogen atoms to oxygen atoms in a water molecule?
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2 hydrogen molecules to one oxygen molecule
Remember by H2O since O doesn’t have a number after it you know there is only one oxygen molecule
6 0
3 years ago
Read 2 more answers
A 36.4-l volume of methane gas is heated from 25°c to 88°c at constant pressure. what is the final volume of the gas?
nignag [31]
<span>using the law pv=nrT and equating these you get the equation v1/t1 = v2/t2 since pressure is constant it also cancels with n and r. show that v1=36.4, t1 = 25 + 273.15 and t2 = 88 +273.15. 273.15 is the Kelvin conversion. then solve for v2. This is 44.1 L.</span>
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