Answer:
The mass of ice required to melt to lower the temperature of 353 mL of water from 26 ⁰C to 6 ⁰C is 85.4197 kg
Explanation:
Heat gain by ice = Heat lost by water
Thus,
Heat of fusion +
Where, negative sign signifies heat loss
Or,
Heat of fusion +
Heat of fusion = 334 J/g
Heat of fusion of ice with mass x = 334x J/g
For ice:
Mass = x g
Initial temperature = 0 °C
Final temperature = 6 °C
Specific heat of ice = 1.996 J/g°C
For water:
Volume = 353 mL
Density of water = 1.0 g/mL
So, mass of water = 353 g
Initial temperature = 26 °C
Final temperature = 6 °C
Specific heat of water = 4.186 J/g°C
So,
345.976x = 29553.16
x = 85.4197 kg
Thus,
<u>The mass of ice required to melt to lower the temperature of 353 mL of water from 26 ⁰C to 6 ⁰C is 85.4197 kg</u>
Answer:
pH of 7.86, then the [OH-] is equal to 10^(14-7.86) = 10^6.14 M
Explanation:
Seval ozhabes Seval your dad will get you the first time in the next few months for the first time I am so happy for you and your family will have a great time together for you and
Answer: The SI unit for measuring distance is meter
Explanation:
Answer:
0 587 mL
Explanation:
First we convert 180 grams of HCl into moles, using its molar mass:
- 180 g ÷ 36.46 g/mol = 4.94 mol HCl
Now we can <u>use the number of moles and the given concentration to calculate the required volume</u>, applying the <em>definition of molarity</em>:
- Molarity = moles / liters
- Liters = Moles / molarity
- 4.94 mol / 8.40 M = 0.588 L
Finally we <u>convert liters into milliliters</u>:
The closest answer is option C, 587 mL.