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Tanzania [10]
3 years ago
12

You are making homemade ice cream and can use either 200 g Ice Melt (CaCl2) or 200 g rock salt (NaCl) to lower the freezing poin

t of 7.00 kg of ice (assume the ice is pure water). Which salt would help you eat your ice cream faster (i.e. lower the freezing temperature of the ice)?
Chemistry
1 answer:
mash [69]3 years ago
5 0

Answer:

Calcium chloride (CaCl2) would help me eat my ice cream faster

Explanation:

Addition of salt to ice melts the lowers the freezing temperature of the ice thus melting the ice easily

Adding calcium chloride to my ice cream would help me eat it faster because the melting point of calcium chloride (772°C) is lower than that of sodium chloride (melting point is 801°C). Calcium chloride would melt my ice cream faster because of its lower melting point

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Ammonia and oxygen react to form nitrogen and water.
Nata [24]

Answer:

A. 19.2 g of O2.

B. 3.79 g of N2.

C. 54 g of H2O.

Explanation:

The balanced equation for the reaction is given below:

4NH3(g) + 3O2(g) → 2N2+ 6H2O(g)

Next, we shall determine the masses of NH3 and O2 that reacted and the masses of N2 and H2O produced from the balanced equation.

This is illustrated below:

Molar mass of NH3 = 14 + (3x1) = 17 g/mol

Mass of NH3 from the balanced equation = 4 x 17 = 68 g

Molar mass of O2 = 16x2 = 32 g/mol

Mass of O2 from the balanced equation = 3 x 32 = 96 g

Molar mass of N2 = 2x14 = 28 g/mol

Mass of N2 from the balanced equation = 2 x 28 = 56 g

Molar mass of H2O = (2x1) + 16 = 18 g/mol

Mass of H2O from the balanced equation = 6 x 18 = 108 g

Summary:

From the balanced equation above,

68 g of NH3 reacted with 96 g of O2 to produce 56 g of N2 and 108 g of H2O.

A. Determination of the mass of O2 needed to react with 13.6 g of NH3.

This is illustrated below:

From the balanced equation above,

68 g of NH3 reacted with 96 g of O2.

Therefore, 13.6 g of NH3 will react with = (13.6 x 96)/68 = 19.2 g of O2.

Therefore, 19.2 g of O2 are needed for the reaction.

B. Determination of the mass of N2 produced when 6.50 g of O2 react.

This is illustrated below:

From the balanced equation above,

96 g of O2 reacted to produce 56 g of N2.

Therefore, 6.5 g of O2 will react to produce = (6.5 x 56)/96 = 3.79 g of N2.

Therefore, 3.79 g of N2 were produced from the reaction.

C. Determination of the mass of H2O formed from the reaction of 34 g of NH3.

This is illustrated below:

From the balanced equation above,

68 g of NH3 reacted to 108 g of H2O.

Therefore, 34 g of NH3 will react to produce = (34 x 108)/68 = 54 g of H2O.

Therefore, 54 g of H2O were obtained from the reaction.

4 0
3 years ago
No spam or links! Thanks!
Monica [59]

Answer:

364 K or 91°C

Explanation:

Applying,

V₁/T₁ = V₂/T₂................ Equation 1

Where V₁ = Initial Volume, V₂ = Final volume, T₁ = initial Temperature, T₂ = final Temperature.

make T₂ the subject of the equation,

T₂ = V₂T₁/V₁................. Equation 2

From the question,

Given: V₁ = 375 mL, V₂ = 500 mL, T₁ = 0.0°C = (273+0) K = 273 K

Substitute these values into equation 2

T₂ = (500×273)/375

T₂ = 364 K

T₂ = (364-273) °C = 91 °C

5 0
3 years ago
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