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mina [271]
4 years ago
7

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]4 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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Explanation:

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The oxidation state of the oxygen is -2.

As the compound is neutral, the sum of the oxidation states of all atoms must be 0.

Oxidation State Ca + Oxidation State C + (Oxidation State O)×3 = 0

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2 + x - 6 = 0

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Covert 12.64 g NaOH to moles
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3 0
3 years ago
Read 2 more answers
A 35.0 g sample of a mixture of kcl and nacl required 199 ml of a 2.54 m agno3 solution to precipitate all the chloride as agcl.
notsponge [240]

Mass of the sample (NaCl + KCl ) = 35.0 g

Concentration of AgNO3 = 2.54 M

Volume of AgNo3 = 199 ml = 0.199 L

Moles of AgNO3 = 2.54 moleL-1 * 0.199 L = 0.5055 moles

Since all the Ag in AgCl has to come from AgNO3, then

Moles of AgNO3 = moles of AgCl = 0.5055

Now, since all the Cl- comes from NaCl and KCl, we can write:

# Moles NaCl + # moles KCl = 0.5055

Therefore,  # moles NaCl = 0.5055 - moles KCl--------(1)

It is given that

Mass NaCl + Mass KCl = 35.0 g

i.e.

(58 g/mole)NaCl * #Moles NaCl + (75 g/mol)KCl * #Moles KCl = 35.0 g---(2)

from eq(1) we have:

58 (0.5055-moles KCl) + 75 *moles KCl = 35.0

Therefore,

moles KCl = 5.681/133 = 0.0427 moles

Now, molar mass of KCl = 75 g/mol

Thus, the mass of KCl in the mixture = 75 g/mole * 0.0427 moles = 3.203 g



8 0
3 years ago
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