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3 years ago

Someone help1!1!1!1!!!!!

Chemistry

1 answer:

alisha [4.7K]3 years ago

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The question is cut off in the picture

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Consider the following equilibrium: H2CO3+H2O = H3O+HCO3^-1. What is the correct equilibrium expression?

mylen [45]

Equilibrium expression is $Keq = \frac{[H3O+][HCO3^-]}{[H2CO3]}\\$

<u>Explanation:</u>

Equilibrium expression is denoted by Keq.

Keq is the equilibrium constant that is defined as the ratio of concentration of products to the concentration of reactants each raised to the power its stoichiometric coefficients.

Example -

aA + bB = cC + dD

So, Keq = conc of product/ conc of reactant

$Keq = \frac{[C]^c [D]^d}{[A]^a [B]^b}$

So from the equation, H₂CO₃+H₂O = H₃O+HCO₃⁻¹

$Keq = \frac{[H3O^+]^1 [HCO3^-]^1}{[H2CO3]^1 [H2O]^1}$

The concentration of pure solid and liquid is considered as 1. Therefore, concentration of H2O is 1.

Thus,

$Keq = \frac{[H3O+][HCO3^-]}{[H2CO3]}\\$

Therefore, Equilibrium expression is $Keq = \frac{[H3O+][HCO3^-]}{[H2CO3]}\\$

4 0

2 years ago

Which of these formulas is NOT consistent with the guidelines for creating ionic compounds? Mark all that apply.

iVinArrow [24]

Answer:

Explanation:

because boron and fluorine are both nonmetals and don't fit the guidlelines for creating ionic compounds

8 0

3 years ago

(GIVING BRAINLIEST) do what the following image says.

S_A_V [24]

Answer:

hope it helps.

<h3>stay safe healthy and happy.<u>.</u><u>.</u></h3>

7 0

2 years ago

Read 2 more answers

Are you for or against transgender? why?

EleoNora [17]

Explanation:

that u know right .........................

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3 years ago

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The density of gold is 19.3 g /cm cubed The density of iron pyrite is 5.0 g /cm cubed. Is a nugget of iron pyrite and a nugget o

telo118 [61]

If you clear volume in the density equation:

$\rho = \frac{m}{V}\ \to\ V = \frac{m}{\rho}$

The greater the density the lower the volume. This means, the volume of gold nugget will be smaller than the volume of iron pyrite nugget.

$V_{gold} = \frac{m}{\rho} = \frac{50\ g}{19.3\ g/cm^3} = \bf 2.59\ cm^3$

$V_{iron} = \frac{m}{\rho} = \frac{50\ g}{5.0\ g/cm^3} = \bf 10\ cm^3$

6 0

3 years ago