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2 years ago

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A solution contains cr3+ ions. the addition of 0.063 l of 1.50 m naf solution was needed to completely precipitate the chromium

ions as crf3(s). what was the mass of cr3+ ions in the original solution?

1 answer:

steposvetlana [31]2 years ago

6 0

Cr{3+} + 3 NaF → CrF3 + 3 Na{+} <span>

First calculate the total mols of NaF.

(0.063 L) x (1.50 mol/L NaF) = 0.0945 mol NaF total </span>

Using stoichiometric ratio:

<span>0.0945 mol NaF * (1 mol Cr3+ / 3 mol NaF) * (51.9961 g Cr3+/mol) = 1.6379 g Cr3+</span>

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2 years ago

A sample of 0.53 g of carbon dioxide was obtained by heating 1.31 g of calcium carbonate. what is the percent yield for this rea

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2 years ago

Read 2 more answers

What would be the composition and ph of an ideal buffer prepared from lactic acid (ch3chohco2h), where the hydrogen atom highlig

Mashutka [201]

Answer:

$P_H =2.86$

$c=1.4\times 10^{-4}$

Explanation:

first write the equilibrium equaion ,

$C_3H_6O_3$ ⇄ $C_3H_5O_3^{-} +H^{+}$

assuming degree of dissociation $\alpha$ =1/10;

and initial concentraion of $C_3H_6O_3$ =c;

At equlibrium ;

concentration of $C_3H_6O_3 = c-c\alpha$

$[C_3H_5O_3^{-} ]= c\alpha$

$[H^{+}] = c\alpha$

$K_a = \frac{c\alpha \times c\alpha}{c-c\alpha}$

$\alpha$ is very small so $1-\alpha$ can be neglected

and equation is;

$K_a = {c\alpha \times \alpha}$

$[H^{+}] = c\alpha$ = $\frac{K_a}{\alpha}$

$P_H =- log[H^{+} ]$

$P_H =-logK_a + log\alpha$

$K_a =1.38\times10^{-4}$

$\alpha = \frac{1}{10}$

$P_H= 3.86-1$

$P_H =2.86$

composiion ;

$c=\frac{1}{\alpha} \times [H^{+}]$

$[H^{+}] =antilog(-P_H)$

$[H^{+} ] =0.0014$

$c=0.0014\times \frac{1}{10}$

$c=1.4\times 10^{-4}$

6 0

2 years ago