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AURORKA [14]
3 years ago
5

The molecular weight of NaCl is 58.44 grams/mole. If you had a 1.0 molar solution (1.0 M), you would have to put 58.44 g of salt

in 1.0 liter of solution.
How many moles of NaCl would you have in 100 mL of this solution?

a. 0.010
b. 0.10
c. 0.58
d. 5.8
Chemistry
2 answers:
Rzqust [24]3 years ago
7 0
Molarity is expressed as the number of moles of solute per volume of the solution. For example, we are given a solution of 2M NaOH this describes a solution that has 2 moles of NaOH per 1 L volume of the solution. To calculate the moles of NaCl in 1.0 M of solution, we simply multiply the volume given of the solution.

moles NaCl = 1.0 M (0.100 L ) = 0.10 mol NaCl --------> OPTION B
Anna11 [10]3 years ago
4 0

Answer:

0.10    ...............................

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If 1/4 cups are in 1 serving jow many servings are in 4 cups​
mote1985 [20]

Answer:

One serving is 1/4 cup therefore 1 cup would be 4 servings. Since we have 9 cups we would then multiply 9 x 4 and the answer would be 36 servings. You have a ratio of cups to serving, so 1/4 cup : 1 serving, or .

Explanation:

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3 years ago
The density of liquid mercury is 13.6 g/mL. What is its density in units of ? (2.54 cm = 1 in., 2.205 lb = 1 kg)
nalin [4]

Correct question

The density of liquid mercury is 13.6 g/mL. What is its density in units of lb/in​3​? (2.5 cm = 1 in., 2.205 lbs= 1 kg., 1000 g =1 kg, 1 mL = 1 cm³)

Answer:

\rho0.4916\ lb/in^3

Explanation:

Given that;-

The density = 13.6 g/mL

Also, 1 kg = 2.205 lb

1 kg = 1000 g

So, 1000 g = 2.205 lb

1 g = 0.002205 lb

Also,

1 in = 2.54 cm

1 in³ = 16.39 cm³

1 cm³ = 1 mL

So,  1 in³ = 16.39 mL

1 mL = 0.061 in³

The expression for the calculation of density is shown below as:-

\rho=\frac{m}{V}

Thus,

\rho=\frac{13.6\ g}{1\ mL}=\frac{13.6\times 0.002205\ lb}{0.061\ in^3}=0.4916\ lb/in^3

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Molar mass of 4Fe(OH)2
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3 0
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The first known telescopes, used around 1600, consisted of a lens and an eyepiece. Today, telescopes are often computer controll
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Answer:

B.

Explanation:

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