All categories

4 years ago

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

Chemistry

2 answers:

STatiana [176]4 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

You might be interested in

The subatomic particles found in the nucleus are

sattari [20]

C.) protons , the nucleus contains 2 subatomic particles , one protons & neutrons ( protons have a + charge & neutrons have no electrical charge

4 0

3 years ago

Read 2 more answers

2NO (g) + O2 (g) →2NO2 (g) At equilibrium [NO] = 2.4 × 10 -3 M, [O2] = 1.4 × 10 -4 M, and [NO2] = 0.95 M.

azamat

Answer:

$K=1.12x10^9$

Explanation:

Hello there!

Unfortunately, the question is not given in the question; however, it is possible for us to compute the equilibrium constant as the problem is providing the concentrations at equilibrium. Thus, we first set up the equilibrium expression as products/reactants:

$K=\frac{[NO_2]^2}{[NO]^2[O_2]}$

Then, we plug in the concentrations at equilibrium to obtain the equilibrium constant as follows:

$K=\frac{(0.95)^2}{(0.0024)^2(0.00014)}\\\\K=1.12x10^9$

In addition, we can infer this is a reaction that predominantly tends to the product (NO2) as K>>>>1.

Best regards!

4 0

3 years ago

In an accident, a solution containing 2.5 kg of nitric acid was spilled. Two kilograms of Na2CO3 was quickly spread on the area

storchak [24]

First we must write a balanced chemical equation for this reaction

$Na_2CO_3 _(_a_q_)+ 2HNO_3_(_a_q_) \implies 2NaNO_3_(_s_) + CO_2_(_g_) + H_2O_(_l_)$

The mole ratio for the reaction between $HNO_3$ and $Na_2CO_3$ is 1:2. This means 1 moles of $Na_2CO_3$ will neutralize 2 moles $HNO_3$ . Now we find the moles of each reactant based on the mass and molar mass.

$2500g HNO_3 \times \frac{mol}{63.01g\ HNO_3} = 39.67 mol\ HNO_3$

$2000g\ Na_2CO_3 \times \frac{mol}{105.99g \ Na_2CO_3} = 18.87 mol\ Na_2CO_3$

$\frac{18.87 mol Na_2CO_3}{39.67\ HNO_3} = \frac{1 molNa_2CO_3}{2 mol HNO_3}$

The $Na_2CO_3$ was enough to neutralize the acid because 18.87:39.67 is the same as 1:2 mol ratio.

4 0

4 years ago

Please help I will give extra points

Svetllana [295]

I think that it would be 12

7 0

3 years ago

Read 2 more answers

12. How many grams of glucose (C6H1206) is produced if 17.3 mol of H20 is reacted according to this

iren2701 [21]

Answer:

1. 17.3 MOLES OF WATER WILL PRODUCE 518.4 G OF GLUCOSE.

2. 87.4 G OF HNO3 WILL PRODUCE 7.86 G OF NITROGLYCERIN WHEN REACTED WITH EXCESS GLYCEROL.

Explanation:

1. How many grams of glucose is produced from 17.3 mole of water?

Equation:

6CO2 + 6H20 ------> C6H12O6 + 6O2

From the reaction, 6 moles of carbon dioxide reacts with 6 moles of water to produce 1 mole of glucose

So therefore,

6 moles of water = 1 mole of glucose

Since 17.3 mole of water was used, we can first calculate the number of moles of glucose produced:

Then, we have:

6 moles of water = 1 mole of glucose

17.3 moles of water = ( 17.3 * 1/ 6) moles of glucose

= 2.883 moles of glucose

So we say 17.3 moles of water produces 2.883 moles of glucose

At standard conditions, 1 mole of a substance is its molar mass

Molar mass of water = 18 g/mol

Molar mass of glucose = 180 g/mol

From the reaction:

17.3 moles of water produces 2.883 moles of glucose

17.3 * 18 g of water produces 2.833 * 180 g of glucose

= 518.94 g of glucose.

From 17.3 mole of water, 518.4 g of glucose will be produced.

C3H5(OH)3 + 3HNO3 -------> C3H5(ONO2)3 + 3H20

3 moles of HNO3 reacts to produce 1 mole of nitroglycerin

Molar mass of HNO3 = ( 1 + 14 + 16*3) = 63 g/mol

Molar mass of nitroglycerin = ( 12 *3 + 1 *5 + 16*6 + 14*3) = 179 g/mol

3 moles of HNO3 = 1 mole of nitroglycerin

3 * 63 g of HNO3 = 179 g of nitroglycerin

if 87.4 g of HNO3 were to be reacted, we have:

189 g of HNO3 = 179 g of nitroglycerin

87.4 g of HNO3 = ( 87.4 * 179 / 189) of nitroglycerin

= 7.86 g of nitroglycerin

So therefore, 87.4 g of HNO3 will produce 7.86 g of nitroglycerin when reacted with excess glycerol.

4 0

4 years ago