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3 years ago

Determine the number of formula units of Cr(ClO2)3 and moles of Oxygen contained in 6.85 moles of Cr(ClO2)3

Chemistry

1 answer:

Advocard [28]3 years ago

4 0

The formula units of Cr(ClO2)3 that are in 6.85 moles of Cr(Clo203 is <u>4.1237 x10^24 formula units</u>

calculation

the formula units are calculated using of Avogadro's law constant

that is according to Avogadro's law 1 mole = 6.02 x10 ^23 formula units

6.85 moles= ? formula units

by cross multiplication.

= ( 6.85 moles x 6.02 x10^23 formula units ) / 1 mole = <u>4.1237 x 10 ^24 formula units</u>

<h3>The moles of Oxygen contained in 6.85 moles of cr(ClO2)3 is <em><u>41.1 moles</u></em></h3>

<em><u> </u></em><u><em>calculation</em></u>

<em>moles of oxygen = number of oxygen atom in Cr(ClO2)3 x 6.85(moles of Cr(ClO2)3</em>

calculate the number of oxygen atoms that are in Cr(ClO2)3

that is 2 x 3 = 6 atoms of Oxygen.

moles is therefore = 6 x 6.85 moles= 41.1 moles

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Answer:

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Explanation:

The unit of concentration $\rm M$ is the same as $\rm mol \cdot L^{-1}$ (moles per liter.) On the other hand, the volume of both the $\rm NaOH$ solution and the original $\rm HCl$ solution here are in milliliters. Convert these two volumes to liters:

$V(\mathrm{NaOH}) = 18.2\; \rm mL = 18.2 \times 10^{-3}\; \rm L = 0.0182\; \rm L$ .
$V(\text{$\mathrm{HCl}$, original}) = 10.0\; \rm mL = 10.0\times 10^{-3}\; \rm L = 0.0100\; \rm L$ .

Calculate the number of moles of $\rm NaOH$ in that $0.0182\; \rm L$ of $0.160\; \rm M$ solution:

$\begin{aligned} n(\mathrm{NaOH}) &= c(\mathrm{NaOH})\cdot V(\mathrm{NaOH})\\ &= 0.160\; \rm mol \cdot L^{-1} \times 0.0182\; \rm L \approx 0.00291\; \rm mol\end{aligned}$ .

$\rm HCl$ reacts with $\rm NaOH$ at a one-to-one ratio:

$\rm HCl\; (aq) + NaOH\; (aq) \to NaCl\; (aq) + H_2O\; (l)$ .

Coefficient ratio:

$\displaystyle \frac{n(\mathrm{HCl})}{n(\mathrm{NaOH})} = 1$ .

In other words, one mole of $\rm NaOH$ would neutralize exactly one mole of $\rm HCl$ . In this titration, $0.291\; \rm mol$ of $\rm NaOH\!$ was required. Therefore, the same amount of $\rm HC$ should be present in the original solution:

$\begin{aligned}&n(\text{$\mathrm{HCl}$, original})\\ &= n(\mathrm{NaOH})\cdot \frac{n(\mathrm{HCl})}{n(\mathrm{NaOH})} \\ &\approx 0.00291\; \rm mol \times 1 = 0.00291\; \rm mol\end{aligned}$ .

Calculate the concentration of the original $\rm HCl$ solution:

$\displaystyle c(\text{$\mathrm{HCl}$, original}) = \frac{n(\text{$\mathrm{HCl}$, original})}{V(\text{$\mathrm{HCl}$, original})} \approx \frac{0.00291\; \rm mol}{0.0100\; \rm L} \approx 0.291\; \rm M$ .

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A) It's correctly written

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Explanation:

A) From online sources, we have number of calories as follows;

Fats: 9 calories per gram

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First of all, you have to translate the words into an equation.

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