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Kipish [7]
3 years ago
14

Calculate the molarity of 80.0 ml of a solution that is 0.92 % by mass nacl. assume the density of the solution is the same as p

ure water.
Chemistry
1 answer:
Sav [38]3 years ago
6 0

<span>We are given the volume of the solution which is 80 mL or 0.08 L and the density which is equal to water, density = 1 kg / L.  Therefore calculate total mass:</span>

total mass = (1 kg / L) * (0.08 L) = 0.08 kg = 80 g

 

So the mass of NaCl is:

mass NaCl = 0.0092 * 80 g = 0.736 g

The molar mass of NaCl is 58.44 g/mol, so the number of moles is:

moles NaCl = 0.736 g / (58.44 g/mol) = 0.0126 mol

 

SO the molarity is mol / L:

<span>Molarity = 0.0126 mol / 0.08 L = 0.157 mol/L = 0.157 M</span>

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5 0
3 years ago
1. Gases are made up of molecules which are relatively far apart.2. The molecules are in motion at high speeds.3. The molecular
Katen [24]

Answer:

A

Explanation:

Molecules of a gas are relatively more compressible than those of liquids and solids because they are relatively far apart without any intermolecular forces between them. However, at lower temperature and higher pressure, there is now a significant intermolecular interaction between the gas molecules and they are no longer relatively far apart. Hence they are more compressible than liquids and solids which already possess significant intermolecular interaction and thus a definite volume.

4 0
3 years ago
When salt is added to water, the entropy, S,
Dovator [93]

Answer:

C. number of particles

Explanation:

Entropy is the measure of disorderliness of a system. Remember that when you dissolve salt in water, you increase the number of particles in the solution. The greater the number of particles in solution, the greater the entropy of the  solution system.

Hence dissolution of a salt in water increases the entropy by increasing the number of particles in solution leading to the inequality; Ssolution > Swater + Ssalt.

5 0
3 years ago
On a hot summer day, air rises most quickly when it is above a
NNADVOKAT [17]

Answer:

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6 0
3 years ago
Read 2 more answers
Butane (C4 H10(g), mc031-1.jpgHf = –125.6 kJ/mol) reacts with oxygen to produce carbon dioxide (CO2 , mc031-2.jpgHf = –393.5 kJ/
ankoles [38]

The balanced chemical equation for the combustion of butane is:

2C_{4}H_{10}(g) +13 O_{2}(g)-->8CO_{2}(g)+10H_{2}O(g)

ΔH_{reaction}^{0} = Σn_{products}ΔH_{f}^{0}_{(products)}-Σn_{reactants}ΔH_{f}^{0}_{(reactants)}

                         = [{8*(-393.5kJ/mol)}+{10*(-241.82kJ/mol)}]-[{2*(-125.6kJ/mol)}+13*(0 kJ/mol)}]=[-3148kJ/mol+(-2418.2kJ/mol)]-[(-251.2kJ/mol)+0]

                      = -5315 kJ/mol

Calculating the enthalpy of combustion per mole of butane:

1mol C_{4}H_{10}*( \frac{-5315kJ}{2mol C_{4}H_{10} })=-2657.5 \frac{kJ}{molC_{4}H_{10}}

Therefore the heat of combustion per one mole butane is -2657.5 kJ/mol

Correct answer: -2657.5 kJ/mol

6 0
3 years ago
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