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zheka24 [161]
1 year ago
7

What is the molarity of a solution that was prepared by dissolving 82 g of cacl2?

Chemistry
1 answer:
ExtremeBDS [4]1 year ago
7 0

The molarity of a solution is 0.909 M.

m(CaCl₂) = 82 g; mass of calcium chloride

M(CaCl₂) = 111 g/mol; molar mass of calcium chloride

n(CaCl₂) = m(CaCl₂) ÷ M(CaCl₂)

n(CaCl₂) = 82 g ÷ 111 g/mol

n(CaCl₂) = 0.74 mol; amount of calcium chloride

V(CaCl₂) = 812 mL = 0.812 L; volume of calcium chloride

c(CaCl₂) = n(CaCl₂) ÷ V(CaCl₂)

c(CaCl₂) = 0.74 mol ÷ 0.812 L

c(CaCl₂) = 0.909 mol/L = 0.909 M; molarity of calcium chloride

Calcium chloride is a inorganic salt with ionic bonds between calcium and chlorine. It is a colorless crystalline solid at room temperature, highly soluble in water.

More about molarity: brainly.com/question/26873446

#SPJ4

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Margarita [4]

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I believe the answer would be A. equal temperature

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6 0
3 years ago
Read 2 more answers
How does a sample of hydrogen at 10 °C compare to a sample of hydrogen at 350 K?
Aleks04 [339]

Answer: -

The hydrogen at 10 °C has slower-moving molecules than the sample at 350 K.

Explanation: -

Temperature of the hydrogen gas first sample = 10 °C.

Temperature in kelvin scale of the first sample = 10 + 273 = 283 K

For the second sample, the temperature is 350 K.

Thus we see the second sample of the hydrogen gas more temperature than the first sample.

We know from the kinetic theory of gases that

The kinetic energy of gas molecules increases with the increase in temperature of the gas. The speed of the movement of gas molecules also increase with the increase in kinetic energy.

So higher the temperature of a gas, more is the kinetic energy and more is the movement speed of the gas molecules.

Thus the hydrogen at 10 °C has slower-moving molecules than the sample at 350 K.

8 0
3 years ago
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Helpppp pleaseee ill give brainliest
Studentka2010 [4]

Answer:

The answers are in the explanation.

Explanation:

The energy required to convert 10g of ice at -10°C to water vapor at 120°C is obtained per stages as follows:

Increasing temperature of ice from -10°C - 0°C:

Q = S*ΔT*m

Q is energy, S specific heat of ice = 2.06J/g°C, ΔT is change in temperature = 0°C - -10°C = 10°C and m is mass of ice = 10g

Q = 2.06J/g°C*10°C*10g

Q = 206J

Change from solid to liquid:

The heat of fusion of water is 333.55J/g. That means 1g of ice requires 333.55J to be converted in liquid. 10g requires:

Q = 333.55J/g*10g

Q = 3335.5J

Increasing temperature of liquid water from 0°C - 100°C:

Q = S*ΔT*m

Q is energy, S specific heat of ice = 4.18J/g°C, ΔT is change in temperature = 100°C - 0°C = 100°C and m is mass of water = 10g

Q = 4.18J/g°C*100°C*10g

Q = 4180J

Change from liquid to gas:

The heat of vaporization of water is 2260J/g. That means 1g of liquid water requires 2260J to be converted in gas. 10g requires:

Q = 2260J/g*10g

Q = 22600J

Increasing temperature of gas water from 100°C - 120°C:

Q = S*ΔT*m

Q is energy, S specific heat of gaseous water = 1.87J/g°C, ΔT is change in temperature = 20°C and m is mass of water = 10g

Q = 1.87J/g°C*20°C*10g

Q = 374J

Total Energy:

206J + 3335.5 J + 4180J + 22600J + 374J =

30695.5J =

30.7kJ

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krok68 [10]

B,speedof light>speed of sound

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