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Maurinko [17]
3 years ago
5

What is the pOH of a solution with a [OH^ - ] of 10^ -11?

Chemistry
1 answer:
castortr0y [4]3 years ago
7 0

Answer:

pOH= 11

Explanation:

pOH= -log[10^ -11]= 11

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Select the correct answer.
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Well, this is somewhat difficult, because an electron already creates an electric field.  However, I know that when an electron moves it then creates a magnetic field.  So, I'm going to safely assume that when an electron moves, it creates an electric, a magnetic, and a gravitational field.  

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3 years ago
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If an equal number of moles of reactants are used, do the following equilibrium mixtures contain primarily reactants or products
puteri [66]

<u>Answer:</u>

<u>For a:</u> The equilibrium mixture contains primarily reactants.

<u>For b:</u> The equilibrium mixture contains primarily products.

<u>Explanation:</u>

There are 3 conditions:

  • When K_{eq}>1; the reaction is product favored.
  • When K_{eq}; the reaction is reactant favored.
  • When K_{e}=1; the reaction is in equilibrium.

For the given chemical reactions:

  • <u>For a:</u>

The chemical equation follows:

HCN(aq.)+H_2O(l)\rightleftharpoons CN^-(aq.)+H_3O^+(aq.);K_{eq}=6.2\times 10^{-10}

The expression of K_{eq} for above reaction follows:

K_{eq}=\frac{[CN^-][H_3O^+]}{[HCN][H_2O]}=6.2\times 10^{-10}

As, K_{eq}, the reaction will be favored on the reactant side.

Hence, the equilibrium mixture contains primarily reactants.

  • <u>For b:</u>

The chemical equation follows:

H_2(g)+Cl_2(g)\rightleftharpoons 2HCl(g);K_{eq}=2.51\times 10^{4}

The expression of K_{eq} for above reaction follows:

K_{eq}=\frac{[HCl]^2}{[H_2][Cl_2]}=2.51\times 10^{4}

As, K_{eq}>1, the reaction will be favored on the product side.

Hence, the equilibrium mixture contains primarily products.

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3 years ago
Which one of the following polymer is prepared by condensation polymerization?
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The answer is Darcon my friend, have a fantastic day

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<span>6.05×10^3 kg H2SO4 ( 1 kmol / 98.08 kg) ( 1 kmol Na2CO3 / 1 kmol H2SO4 ) ( 105.99 kg / kmol ) = 6537.92 kg Na2CO3 needed</span>
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Idk cause I’m not completely sure I apologize
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