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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 4.1237 x10^24 formula units

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 = 4.1237 x 10 ^24 formula units

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

calculation

moles of oxygen = number of oxygen atom in Cr(ClO2)3 x 6.85(moles of Cr(ClO2)3

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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4 years ago

6. How many grams are present in 13.2 moles of O₂?

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Answer:

422.3868 Grams

Explanation:

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2 years ago

In a laboratory experiment, students synthesized a new compound and found that when 7.229 grams of the compound were dissolved i

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The question is incomplete, here is the complete question:

The common laboratory solvent chloroform is often used to purify substances dissolved in it. The vapor pressure of chloroform, CHCl₃, is 173.11 mm Hg at 25°C.

In a laboratory experiment, students synthesized a new compound and found that when 7.229 grams of the compound were dissolved in 207.8 grams of chloroform, the vapor pressure of the solution was 170.51 mm Hg. The compound was also found to be nonvolatile and a non-electrolyte. What is the molecular weight of this compound?

Answer: The molecular mass of the compound is 481.9 g/mol

Explanation:

The equation used to calculate relative lowering of vapor pressure follows:

$\frac{p^o-p_s}{p^o}=i\times \chi_{solute}$

where,

$\frac{p^o-p_s}{p^o}$ = relative lowering in vapor pressure

i = Van't Hoff factor = 1 (for non electrolytes)

$\chi_{solute}$ = mole fraction of solute = ?

$p^o$ = vapor pressure of pure chloroform = 173.11 mmHg

$p_s$ = vapor pressure of solution = 170.51 mmHg

Putting values in above equation, we get:

$\frac{173.11-170.51}{173.11}=1\times \chi_A\\\\\chi_A=0.0150$

This means that 0.0150 moles of compound is present in the solution

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Moles of compound = 0.0150 moles

Given mass of compound = 7.229 g

Putting values in above equation, we get:

$0.0150mol=\frac{7.229g}{\text{Molar mass of compound}}\\\\\text{Molar mass of compound}=\frac{7.229g}{0.0150mol}=481.9g/mol$

Hence, the molecular mass of the compound is 481.9 g/mol

5 0

3 years ago

Calculate ΔH⁰298 (in kJ) for the process P(s) + 5/2 Cl2(g) → PCl5(g) from the following information.

Viefleur [7K]

Answer:

$\Delta H_{298}^{0}$ for the process is -375 kJ

Explanation:

Given reaction is a combination of the two given elementary steps.

Summation of change in standard enthalpy ( $\Delta H_{298}^{0}$ )of the two elementary reactions give $\Delta H_{298}^{0}$ of the reaction $P(s)+\frac{5}{2}Cl_{2}(g)\rightarrow PCl_{5}(g)$ .