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DIA [1.3K]
3 years ago
8

You add 100.0 g of water at 52.0 °C to 100.0 g of ice at 0.00 °C. Some of the ice melts and cools the water to 0.00 °C. When the

ice and water mixture reaches thermal equilibrium at 0 °C, how much ice has melted? (The specific heat capacity of liquid water is 4.184 J/g ⋅ K. The enthalpy of fusion of ice at 0 °C is 333 J/g.) Mass of ice = g
Chemistry
1 answer:
lara31 [8.8K]3 years ago
5 0

Answer:

m_{ice} = 65.336\,g

Explanation:

Accoding to the First Law of Thermodynamics, the heat released by the water melts a portion of ice. That is to say:

Q_{water} = Q_{ice}

(100\,g)\cdot \left(4.184\,\frac{J}{kg\cdot ^{\textdegree}C}\right)\cdot (52\,^{\textdegree}C - 0\,^{\textdegree}C) = m_{ice}\cdot \left(333\,\frac{J}{g} \right)

The amount of ice that is melt is:

m_{ice} = 65.336\,g

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3 years ago
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At 25.0°c, a solution has a concentration of 3.179 m and a density of 1.260 g/ml. the density of the solution at 50.0°c is 1.249
oksano4ka [1.4K]

Answer: -

3.151 M

Explanation: -

Let the volume of the solution be 1000 mL.

At 25.0 °C, Density = 1.260 g/ mL

Mass of the solution = Density x volume

= 1.260 g / mL x 1000 mL

= 1260 g

At 25.0 °C, the molarity = 3.179 M

Number of moles present per 1000 mL = 3.179 mol

Strength of the solution in g / mol

= 1260 g / 3.179 mol = 396.35 g / mol (at 25.0 °C)

Now at 50.0 °C

The density is 1.249 g/ mL

Mass of the solution = density x volume = 1.249 g / mL x 1000 mL

= 1249 g.

Number of moles present in 1249 g = Mass of the solution / Strength in g /mol

= \frac{1249 g}{396.35 g/mol}

= 3.151 moles.

So 3.151 moles is present in 1000 mL at 50.0 °C

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7 0
2 years ago
A solution of ammonia has a pH of 11.8. What is the concentration of OH– ions in the solution?
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Answer B

hope this helps!





3 0
3 years ago
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