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3 years ago

Approximately what will be the total solute molarity for a 0.47 m solution of c6h8o7

1 answer:

Colt1911 [192]3 years ago

8 0

C(C₆H₈O₇) = 0,47 M = 0,47 mol/L.
If we have one liter of solution:
V(solution) = 1 L.
n(solution) = 0,47 mol.
According to molecular formula of compound this can be citric acid.
Citric acid dissociation in water: C₆H₈O₇ → C₆H₅O₇³⁻ + 3H⁺.
There is four ions from one citric acid. Molarity is than c = 4 · 0,47 M = 1,88 M.

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igomit [66]

<h2>Giant impact and metalcore.</h2>

Explanation :

Mercury has a large core of liquid metal.
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3 years ago

Calculate the following quantity: molarity of a solution prepared by diluting 45.45 mL of 0.0404 M ammonium sulfate to 550.00 mL

dybincka [34]

Answer:

$M_2=3.34x10^{-3}M$

Explanation:

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In this case, since a dilution process implies that the moles of the solute remain the same before and after the addition of diluting water, we can write:

$M_1V_1=M_2V_2$

Thus, since we know the volume and concentration of the initial sample, we compute the resulting concentration as shown below:

$M_2=\frac{M_2V_2}{V_1} =\frac{45.45mL*0.0404M}{550.00mL}\\\\M_2=3.34x10^{-3}M$

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2 years ago

For the reaction of hydrogen with iodine

fenix001 [56]

Answer:

$r_{H_2} = \frac{-1}{2} r_{HI}$

Explanation:

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In this case, considering the given chemical reaction:

$H_2(g) + I_2(g) \rightarrow 2HI(g)$

Thus, by applying the law of rate proportions, we can write:

$\frac{1}{-1} r_{H_2} = \frac{1}{-1}r_{i_2} = \frac{1}{2} r_{HI}$

Whereas the stoichiometric coefficients of reactants are negative due their disappearance and that of the product is positive due to its appearance. In such a way, when we relate the rate of disappearance of hydrogen gas to the rate of formation of hydrogen iodide, we obtain:

$r_{H_2} = \frac{-1}{2} r_{HI}$

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2 years ago

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