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meriva
2 years ago
13

What name should be used for the ionic compound Co2 (SO4)3?

Chemistry
1 answer:
Pie2 years ago
5 0
Bonjour,

it’s cobalt(III) sulfate
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Is it kinetic energy
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What is the mass in grams of a single formula unit of silver chloride, AgCI? A) 4.21 x 1021 g B) 8.61 x 10258 C) 1.66 x 10-248 D
bekas [8.4K]

Answer: D) 2.38*10^-^2^2 g

Explanation: The question asks to convert formula unit to grams. It is a unit conversion problem.

1 mole equals to Avogadro number of formula units. So, to convert the given number of formula units to moles, we need to divide by the Avogadro number. After this, we do moles to grams conversion and for this the moles are multiplied by the molar mass of the compound. Molar mass of AgCl is 143.32 gram per mol.

1FormulaUnitAgCl(\frac{1mol}{6.022*10^2^3formulaUnits})(\frac{143.32g}{1mol})

= 2.38*10^-^2^2 g

So, the correct option is D) 2.38*10^-^2^2 g

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What common item can detect the presence of radiation? A. Bleach B. A roll of photographic film C. LED watch D. Baking soda
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Answer: B- a roll of photographic film

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Founders education answer

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Which is a factor in determining the average atomic mass of an element
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The mass numbers of the different isotopes of that element are averaged according to their respective abundances in nature.

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Determine the empirical formula of a compound containing 48.38 grams of carbon, 6.74 grams of hydrogen, and 53.5 grams of oxygen
Darya [45]

Answer: The empirical formula of the compound becomes CH_2O

<u>Explanation:</u>

The empirical formula is the chemical formula of the simplest ratio of the number of atoms of each element present in a compound.

We are given:

Mass of C = 48.38 g

Mass of H = 6.74 g

Mass of O = 53.5 g

The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}} ......(1)

To formulate the empirical formula, we need to follow some steps:

  • <u>Step 1:</u> Converting the given masses into moles.

Molar mass of C = 12 g/mol

Molar mass of H = 1 g/mol

Molar mass of O = 16 g/mol

Putting values in equation 1, we get:

\text{Moles of C}=\frac{48.38g}{12g/mol}=3.023 mol

\text{Moles of H}=\frac{6.74g}{1g/mol}=6.74 mol

\text{Moles of O}=\frac{53.5g}{1g/mol}=3.34 mol

  • <u>Step 2:</u> Calculating the mole ratio of the given elements.

Calculating the mole fraction of each element by dividing the calculated moles by the least calculated number of moles that is 3.023 moles

\text{Mole fraction of C}=\frac{3.023}{3.023}=1

\text{Mole fraction of H}=\frac{6.74}{3.023}=2.23\approx 2

\text{Mole fraction of O}=\frac{3.34}{3.023}=1.105\approx 1

  • <u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : H : O = 1 : 2 : 1

Hence, the empirical formula of the compound becomes CH_2O

5 0
3 years ago
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