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3 years ago

Aqueous acetic acid is neutralized by aqueous potassium hydroxide. True or False

Chemistry

2 answers:

Stels [109]3 years ago

8 0

Answer:

i think its true

Explanation:

liberstina [14]3 years ago

4 0

Answer:

i think its true but I’m not sure

Explanation:

I know that they can. Be mixed

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If a material conducts heat easily, its a good _____. It has to be a 18 letter word

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Answer:

a thermal conductor

Explanation:

thats what its called

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3 years ago

If you were to place the beaker containing the products of the reaction between copper ii chloride and aluminum on a balance, ho

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3 years ago

?Al + ?H2SO4 → ?Al2(SO4)3 +3H2

AfilCa [17]

The answer is: 27 grams of aluminium.

Balanced chemical reaction: 2Al + 3H₂SO₄ → Al₂(SO₄)₃ + 3H₂.

n(H₂) = 1.5 mol; amount of hydrogen.

Form chemical reaction: n(Al) : n(H₂) = 2 : 3.

n(Al) = 2 · 1.5 mol ÷ 3.

n(Al) = 1.0 mol; amount of aluminium.

m(Al) = n(Al) · M(Al).

m(Al) = 1 mol · 27 g/mol.

m(Al) = 27 g; mass of aluminium.

5 0

3 years ago

Assume the hydrolysis of ATP proceeds with ΔG′° = –30 kJ/mol. ATP + H2O → ADP + Pi Which expression gives the ratio of ADP to AT

Andru [333]

Answer:

$6.14\cdot 10^{-6}$

Explanation:

Firstly, write the expression for the equilibrium constant of this reaction:

$K_{eq} = \frac{[ADP][Pi]}{ATP}$

Secondly, we may relate the change in Gibbs free energy to the equilibrium constant using the equation below:

$\Delta G^o = -RT ln K_{eq}$

From here, rearrange the equation to solve for K:

$K_{eq} = e^{-\frac{\Delta G^o}{RT}}$

Now we know from the initial equation that:

$K_{eq} = \frac{[ADP][Pi]}{ATP}$

Let's express the ratio of ADP to ATP:

$\frac{[ADP]}{[ATP]} = \frac{[Pi]}{K_{eq}}$

Substitute the expression for K:

$\frac{[ADP]}{[ATP]} = \frac{[Pi]}{K_{eq}} = \frac{[Pi]}{e^{-\frac{\Delta G^o}{RT}}}$

Now we may use the values given to solve:

$\frac{[ADP]}{[ATP]} = \frac{[Pi]}{K_{eq}} = \frac{[Pi]}{e^{-\frac{\Delta G^o}{RT}}} = [Pi]e^{\frac{\Delta G^o}{RT}} = 1.0 M\cdot e^{\frac{-30 kJ/mol}{2.5 kJ/mol}} = 6.14\cdot 10^{-6}$

7 0

3 years ago