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3 years ago

A solution is prepared by dissolving 15 grams of sugar in 85 grams of water. How many percent of sugar is in the solution?

Chemistry

1 answer:

trasher [3.6K]3 years ago

7 0

Answer:

15%

Explanation:

From the question given above, the following data were obtained:

Mass of sugar = 15 g

Mass of water = 85 g

Percentage of sugar in the solution =?

Next, we shall determine the mass of the solution. This can be obtained as follow:

Mass of solute (sugar) = 15 g

Mass of solvent (water) = 85 g

Mass of solution =?

Mass of solution = mass of solute + mass of solvent

Mass of solution = 15 + 85

Mass of solution = 100 g

Finally, we shall determine the percentage of the sugar in the solution. This can be obtained as follow:

Mass of solute (sugar) = 15 g

Mass of solution = 100 g

Percentage of sugar in the solution =?

Percentage = solute /solution × 100

Percentage = 15 / 100 × 100

Percentage of sugar in the solution = 15%

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The answer
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3 years ago

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Answer:

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As the reaction is endothermic, when the reaction occurs,

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72.8g are:

72.8g * (1mol / 216.59g) = 0.3361 moles HgO.

that absorb:

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4 years ago

In a lab, a student dissolves 6.9 g of sodium chloride (NaCl) in 125 g of water (H2O).

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3 years ago

A volume of 125 mL of H2O is initially at room temperature (22.00 ∘C). A chilled steel rod at 2.00 ∘C is placed in the water. If

VMariaS [17]

Answer:

41.9 g

Explanation:

We can calculate the heat released by the water and the heat absorbed by the steel rod using the following expression.

Q = c × m × ΔT

where,

c: specific heat capacity

m: mass

ΔT: change in temperature

If we consider the density of water is 1.00 g/mol, the mass of water is 125 g.

According to the law of conservation of energy, the sum of the heat released by the water (Qw) and the heat absorbed by the steel (Qs) is zero.

Qw + Qs = 0

Qw = -Qs

cw × mw × ΔTw = -cs × ms × ΔTs

(4.18 J/g.°C) × 125 g × (21.30°C-22.00°C) = -(0.452J/g.°C) × ms × (21.30°C-2.00°C)

ms = 41.9 g

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