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gizmo_the_mogwai [7]
1 year ago
7

how many formula units of CaCl2 are in 111 g CaCl2? the molar mass of calcium chloride is about 111 g/mol

Chemistry
1 answer:
d1i1m1o1n [39]1 year ago
5 0

There are 6.02 × 10²³ formula units of CaCl2 in 111 g CaCl2. Details about formula units can be found below.

<h3>What is a formula unit?</h3>

Formula unit refers to the empirical formula of an ionic compound for use in stoichiometric calculations.

According to this question, there are 111g of CaCl2. The formula units can be calculated by multiplying the number of moles by Avogadro's number.

no of moles in CaCl2 = 111g ÷ 111g/mol = 1mol

Formula units of CaCl2 = 1mol × 6.02 × 10²³ = 6.02 × 10²³ formula units.

Therefore, there are 6.02 × 10²³ formula units of CaCl2 in 111 g CaCl2.

Learn more about formula units at: brainly.com/question/21494857

#SPJ1

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Answer:

The Henry's law constant for argon is k=2.11*10^{-3}\frac{ M}{atm}

Explanation:

Henry's Law indicates that the solubility of a gas in a liquid at a certain temperature is proportional to the partial pressure of the gas on the liquid.

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So, being the concentration C=\frac{ngas}{V}  

where ngas is the number of moles of gas and V is the volume of the solution, you must calculate the number of moles ngas. This is determined by the Ideal Gas Law: P*V=n*R*T where P is the gas pressure, V is the volume that occupies, T is its temperature, R is the ideal gas constant, and n is the number of moles of the gas. So n=\frac{P*V}{R*T}

In this case:

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Then:

n=\frac{1 atm*5.16*10^{-2} L}{0.082 \frac{atm*L}{mol*K} *298K}

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n= 2.11 *10⁻³ moles

So: C=\frac{ngas}{V}=\frac{2.11*10^{-3} moles}{1 L} =2.11*10^{-3} \frac{moles}{L}= 2.11*10^{-3} M

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2.11*10⁻³ M= k* 1 atm

Solving:

k=\frac{2.11*10^{-3} M}{1 atm}

You get:

k=2.11*10^{-3}\frac{ M}{atm}

<u><em>The Henry's law constant for argon is </em></u>k=2.11*10^{-3}\frac{ M}{atm}<u><em></em></u>

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