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12345 [234]
2 years ago
10

How many liters of 0.37 M solution can be made with 29.53 grams of lithium fluoride. (LiF)?

Chemistry
1 answer:
dsp732 years ago
4 0

Answer:

V = 3.1 L      

Explanation:

Given data:

Molarity of solution = 0.37 M

Mass of LiF = 29.53 g

Volume of solution = ?

Solution:

Number of moles of LiF:

Number of moles = mass/molar mass

Number of moles = 29.53 g/ 25.94g/mol

Number of moles = 1.14 mol

Volume:

Molarity = number of moles of solute / Volume in L

0.37 M = 1.14 mol / V

V = 1.14 mol / 0.37 M

V = 3.1 L         (M = mol/L)

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1. Name the three particles of the atom and their respective charges are:
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a protons

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c electrons

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1 year ago
Read 2 more answers
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3 0
2 years ago
KCl and KBr are both ionic solids. A mixture of KCl and KBr has a mass of 3.595 g. When this mixture is heated in the presence o
monitta

Answer:

The percentage (by mass) of KBr in the original mixture was 33.1%.

Explanation:

The mixture of KCl and KBr has a mass of 3.595g, thus the sum of the moles of KCl (<em>x</em>) multiplied by it molar mass (74.5g/mol) and the moles of KBr (<em>y</em>) multiplied by it molar mass (119g/mol) is the total mass of the mixture:

x.74.5g/mol + y.119g/mol = 3.595g

Also, after the conversion of KBr into KCl, the total mass of 3.129 g is only from KCl moles, hence

\frac{3.129g}{74.5g/mol} = 0.042 moles

But the 0.042 moles came from the originals KCl and KBr moles, thus

x + y = 0.042moles

Now it is possible to propose a system of equations:

x.74.5g/mol + y.119g/mol = 3.595g

x + y = 0.042moles

Solving the system of equations,

x=0.032moles\\y=0.010 moles

0.010 moles of KBr multiplied it molar mass is

0.010molesx119g/mol = 1.19g

Therefore, the percentage (by mass) of KBr in the original mixture was:

\frac{1.19g}{3.595g}x100% = 33.1%%

4 0
3 years ago
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