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3 years ago

What is the molarity of a NaCl solution containing 9.0 moles of NaCl in 3.0 L of solution?

Chemistry

1 answer:

ss7ja [257]3 years ago

4 0

Answer:

3 M

Explanation:

Molarity equation: M = n/v

n = moles of solute

v = liters of solution

9 moles of NaCl / 3 L

9/3 = 3 M

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A certain substance X has a normal freezing point of -6.4 C and a molal freezing point depression constant Kf= 3.96 degrees C.kg

Brut [27]

Answer: $1.0\times 10^2g$

Explanation:

Depression in freezing point is given by:

$\Delta T_f=i\times K_f\times m$

$\Delta T_f=T_f^0-T_f=(-6.4-(13.6))^0C=7.2^0C$ = Depression in freezing point

i= vant hoff factor = 1 (for non electrolyte like urea)

$K_f$ = freezing point constant = $3.96^0C/m$

m= molality

$\Delta T_f=i\times K_f\times \frac{\text{mass of solute}}{\text{molar mass of solute}}\times \text{weight of solvent in kg}}$

Weight of solvent (X)= 950 g = 0.95 kg

Molar mass of non electrolyte (urea) = 60.06 g/mol

Mass of non electrolyte (urea) added = ?

$7.2=1\times 3.96\times \frac{xg}{60.06 g/mol\times 0.95kg}$

$x=1.0\times 10^2g$

Thus $1.0\times 10^2g$ urea was dissolved.

8 0

3 years ago

What piece of equipment is used to measure: length

avanturin [10]

Answer:

Ruler

Explanation:

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3 years ago

Read 2 more answers

Calculate the missing variables in each experiment below using Avogadro’s law.

blagie [28]

Answer:

The answer to your question is: letter c

Explanation:

Data

V1 = 612 ml n1 = 9.11 mol

V2 = 123 ml n2 = ?

Formula

$\frac{V1}{n1} = \frac{V2}{n2}$

$n2 = \frac{n1V2}{V1}$

$n2 = \frac{(9.11)((123)}{(612)}$

n2 = 1.83 mol

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3 years ago

What Is the Connection Between Sulfuric Acid and Hydrochloric Acid?

d1i1m1o1n [39]

They are both strong, corrosive acids. Hydrochloric acid can be synthesize using table salt and sulfuric acid.

7 0

3 years ago

What is the mass of 8.0 x 10^26 UF6 molecules?

gulaghasi [49]

First, find the number of moles of UF6
Avagadro's number = 6.023 x 10^23

Number of moles = 8.0 x 10^26 / Avagadro's number = 8.0 x 10^26 / 6.023 x 10^23 = 1.328 x 10³ moles

Molecular weight of UF6 = Molecular weight of U (238.02891) + Molecular weight of F6 (6 x 18.9984032) = 238.02891 + 113.9904192 = 352.0193292 g/mol

Therefore mass of 8.0 x 10^26 UF6 molecules = 352.0193292 g/mol x 1.328 x 10³ moles = 467.481669 x 10³ grams

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3 years ago