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3 years ago

How many formula units make up 30.4 g of magnesium chloride (mgcl2)?

Chemistry

2 answers:

STatiana [176]3 years ago

4 0

<u>Answer:</u> The number of formula units in the given amount of magnesium chloride is $1.93\times 10^{23}$

<u>Explanation:</u>

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

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

Given mass of magnesium chloride = 30.4 g

Molar mass of magnesium chloride = 95.2 g/mol

Putting values in above equation, we get:

$\text{Moles of magnesium chloride}=\frac{30.4g}{95.2g/mol}=0.32mol$

Formula units is defined as lowest whole number ratio of ions in an ionic compound. It is calculate by multiplying the number of moles by Avogadro's number which is $6.022\times 10^{23}$

We are given:

Number of moles of magnesium chloride = 0.32 moles

Number of formula units = $0.32\times 6.022\times 10^{23}=1.93\times 10^{23}$

Hence, the number of formula units in the given amount of magnesium chloride is $1.93\times 10^{23}$

ahrayia [7]3 years ago

3 0

There are 1.923×10^23 formula units in 30.4g mgcl2

molar mass of mgcl2= 95.211g/mol
magnesium= 24.305g/mol
chlorine= 35.45g/mol times 2
24.305+(35.45*2)= 95.211g/mol

sample divided by molar mass times Avogadro's number gives u the formula units

30.4÷ 95.211= .31929
.3193*6.022=1.9288

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3 years ago

BARSIC [14]

The number of C atoms in 0.524 moles of C is 3.15 atoms.

The number of $SO_2$ molecules in 9.87 moles $SO_2$ is 59.43 molecules.

The moles of Fe in 1.40 x $10^{22}$ atoms of Fe is 0.23 x $10^{-1}$

The moles of $C_2H_6O$ in 2.30x $10^{24}$ molecules of $C_2H_6O$ is 3.81.

<h3>What are moles?</h3>

A mole is defined as 6.02214076 × $10^{23}$ of some chemical unit, be it atoms, molecules, ions, or others. The mole is a convenient unit to use because of the great number of atoms, molecules, or others in any substance.

A. The number of C atoms in 0.524 mole of C:

6.02214076 × $10^{23}$ x 0.524 mole

3.155601758 atoms =3.155 atoms

B. The number of $SO_2$ molecules in 9.87 moles of $SO_2$ :

6.02214076 × $10^{23}$ x 9.87

59.4385293 molecules= 59.43 molecules

C. The moles of Fe in 1.40 x $10^{22}$ atoms of Fe:

1.40 x $10^{22}$ ÷ 6.02214076 × $10^{23}$

0.2324754694 x $10^{-1}$ moles.

0.23 x $10^{-1}$ moles.

D. The moles of $C_2H_6O$ in 2.30x $10^{24}$ molecules of $C_2H_6O$ :

2.30x $10^{24}$ ÷ 6.02214076 × $10^{23}$

3.819239854 moles=3.81 moles

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1 year ago

Calculate the molar solubility of CaF2 at 25°C in a solution that is 0.010 M in Ca(NO3)2. The Ksp for CaF2 is 3.9 x 10-11.

madreJ [45]

Answer:

$Molar \ solubility=3.12x10^{-5}M$

Explanation:

Hello,

In this case, for the dissociation of calcium fluoride:

$CaF_2(s)\rightleftharpoons Ca^{2+}+2F^-$

The equilibrium expression is:

$Ksp=[Ca^{2+}][F^-]^2$

In such a way, via the ICE procedure, including an initial concentration of calcium of 0.01 M (due to the calcium nitrate solution), the reaction extent $x$ is computed as follows: