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vfiekz [6]
3 years ago
5

How many moles of MgCl2 are there in 319 g of the compound?

Chemistry
1 answer:
Leokris [45]3 years ago
3 0

Hey there!:

Molar mass MgCl2 = 95.2110 g/mol

So:

1 mole MgCl2 -------------- 95.2110 g

moles MgCl2 ---------------- 319 g

moles MgCl2 = 319 * 1 / 95.2110

moles MgCl2 = 319 / 95.2110

=> 3.350 moles of MgCl2


Hope that helps!

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What hydrocarbons burn completely in an excess of oxygen, the products are
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The right answer for the question that is being asked and shown above is that: "C) carbon monoxide and carbon dioxide" hydrocarbons burn completely in an excess of oxygen, the products are <span>C) carbon monoxide and carbon dioxide</span>
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3 years ago
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what volume of a 0.149 m potassium hydroxide solution is required to neutralize 17.0 ml of a 0.112 m hydrobromic acid solution?
IgorLugansk [536]

Answer: 12.78ml

Explanation:

Given that:

Volume of KOH Vb = ?

Concentration of KOH Cb = 0.149 m

Volume of HBr Va = 17.0 ml

Concentration of HBr Ca = 0.112 m

The equation is as follows

HBr(aq) + KOH(aq) --> KBr(aq) + H2O(l)

and the mole ratio of HBr to KOH is 1:1 (Na, Number of moles of HBr is 1; while Nb, number of moles of KOH is 1)

Then, to get the volume of a 0.149 m potassium hydroxide solution Vb, apply the formula (Ca x Va)/(Cb x Vb) = Na/Nb

(0.112 x 17.0)/(0.149 x Vb) = 1/1

(1.904)/(0.149Vb) = 1/1

cross multiply

1.904 x 1 = 0.149Vb x 1

1.904 = 0.149Vb

divide both sides by 0.149

1.904/0.149 = 0.149Vb/0.149

12.78ml = Vb

Thus, 12.78 ml of potassium hydroxide solution is required.

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2 years ago
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Answer:

All the carbon will combust with atmospheric oxygen to form carbon dioxide. This means that the charcoal present will become consumed to form colorless carbon dioxide gas.

Explanation:

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Na2S2O3 + AgBr NaBr + Na3ſAg(S203)2] What is
gtnhenbr [62]

Answer:

Mass of NaBr produced  = 23.67 g

Explanation:

Given data:

Mass of AgBr = 42.7 g

Mass of NaBr produced = ?

Solution:

Chemical equation:

2Na₂S₂O₃ + AgBr    →    NaBr + Na₃(Ag(S₂O₃)₂

Number of moles of AgBr:

Number of moles = mass/molar mass

Number of moles = 42.7 g/ 187.7 g/mol

Number of moles = 0.23 mol

now we will compare the moles of AgBr with NaBr.

             AgBr        :         NaBr

                1            :           1

              0.23       :         0.23

Mass of NaBr:

Mass = number of moles × molar mass

Mass = 0.23 mol × 102.89 g/mol

Mass = 23.67 g

8 0
2 years ago
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