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liubo4ka [24]
3 years ago
5

What is the concentration of a solution that has 15.0 g NaCl dissolved to a total of 750 ml?

Chemistry
1 answer:
Oksana_A [137]3 years ago
6 0

<u>Answer:</u> The concentration of solution is 0.342 M

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}

We are given:

Mass of solute (Sodium chloride) = 15 g

Molar mass of sodium chloride = 58.5 g/mol

Volume of solution = 750 mL

Putting values in above equation, we get:

\text{Molarity of solution}=\frac{15g\times 1000}{58.5g/mol\times 750mL}\\\\\text{Molarity of solution}=0.342M

Hence, the concentration of solution is 0.342 M

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Consider the following reaction representing the combustion of propane: ????????3HH8 + O2 → CO2 + H2O a. Balance the equation b.
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Answer:

a) C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O

b) You need 5 moles of O₂ per mole of propane.

c) 181 g of O₂

d) 254 L of O₂ and 1210 L of air

e) 152 L of carbon dioxide

f) The gross heat released is 2354 kJ

Explanation:

C₃H₈ + O₂ → CO₂ + H₂O

a) To balance a combustion reaction you must add CO₂ as carbons of hydrocarbons you have. Then, you should add waters as half of hydrogens of the hydrocarbon and, in the last, balance oxygen with O₂, thus:

C₃H₈ + O₂ → 3 CO₂ + 4 H₂O

10 oxygens, so you sholud add 5 O₂:

C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O

b) The equation balanced says that you need 5 moles of O₂ per mole of propane.

c) To burn 100g of propane you need:

100 g C₃H₈×\frac{1 mol}{44,1g}×\frac{5 molO_2}{1mol C_{3}H_{8}}×\frac{16g}{1mol O_2}= 181 g of O₂

d) 181g of O₂ are 11,34 moles. The volume you require is:

V =nRT/P

where:

n are moles (11,34 moles)

R is gas constant (0,082 atmL/molK)

T is temperature (273 K at STP)

P is pressure (1 atm at STP)

V is 254 L of oxygen.

The liters of air are:

254L O₂ ₓ \frac{100 air}{21 O_2} = 1210 L of air

e) The volume of CO₂ produced is:

100 g C₃H₈×\frac{1 mol}{44,1g}×\frac{3 molCO_2}{1mol C_{3}H_{8}}= 6,80 moles of CO₂

V =nRT/P

where:

n are moles (6,80 moles)

R is gas constant (0,082 atmL/molK)

T is temperature (273 K at STP)

P is pressure (1 atm at STP)

V is 152 L of carbon dioxide.

f) C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O ΔH = -103,8 kJ/mol

1 kg of C₃H₈ are:

1000 g × \frac{1mol}{44,1 g} = 22,68 moles

Thus, the gross heat released is:

103,8 kJ/mol × 22,68 moles = 2354 kJ

I hope it helps!

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