Answer: fourth option, 10.8 kJ
Explanation:
The <em>heat of fusion</em>, also named latent heat of fusion, is the amount of heat energy required to change the state of a substance from solid to liquid (at constant pressure).
The data of the <em>heat of fusions</em> of the substances are reported in tables and they can be shown either per mole or per gram of substance.
In this case we have that the<em> heat of fusion for water </em>is reported per mole: <em>6.02 kJ/mole</em>.
The formula to calculate <em>how many kJ of heat (total heat) are needed to completely melt 32.3 g of water, given that the water is at its melting point</em> is:
- Heat = number of moles × heat of fusion
The calculations are:
- number of moles = mass / molar mass
number of moles = 32.3 g / 18.015 g/mol = 1.79 mol
- Heat = 1.79 mol × 6.02 kJ / mol = 10.8 kJ ← answer
Answer:
The ratio of f at the higher temperature to f at the lower temperature is 5.356
Explanation:
Given;
activation energy, Ea = 185 kJ/mol = 185,000 J/mol
final temperature, T₂ = 525 K
initial temperature, T₁ = 505 k
Apply Arrhenius equation;
![Log(\frac{f_2}{f_1} ) = \frac{E_a}{2.303 \times R} [\frac{1}{T_1} -\frac{1}{T_2} ]](https://tex.z-dn.net/?f=Log%28%5Cfrac%7Bf_2%7D%7Bf_1%7D%20%29%20%3D%20%5Cfrac%7BE_a%7D%7B2.303%20%5Ctimes%20R%7D%20%5B%5Cfrac%7B1%7D%7BT_1%7D%20-%5Cfrac%7B1%7D%7BT_2%7D%20%5D)
Where;
is the ratio of f at the higher temperature to f at the lower temperature
R is gas constant = 8.314 J/mole.K
![Log(\frac{f_2}{f_1} ) = \frac{E_a}{2.303 \times R} [\frac{1}{T_1} -\frac{1}{T_2} ]\\\\Log(\frac{f_2}{f_1} ) = \frac{185,000}{2.303 \times 8.314} [\frac{1}{505} -\frac{1}{525} ]\\\\Log(\frac{f_2}{f_1} ) = 0.7289\\\\\frac{f_2}{f_1} = 10^{0.7289}\\\\\frac{f_2}{f_1} = 5.356](https://tex.z-dn.net/?f=Log%28%5Cfrac%7Bf_2%7D%7Bf_1%7D%20%29%20%3D%20%5Cfrac%7BE_a%7D%7B2.303%20%5Ctimes%20R%7D%20%5B%5Cfrac%7B1%7D%7BT_1%7D%20-%5Cfrac%7B1%7D%7BT_2%7D%20%5D%5C%5C%5C%5CLog%28%5Cfrac%7Bf_2%7D%7Bf_1%7D%20%29%20%3D%20%5Cfrac%7B185%2C000%7D%7B2.303%20%5Ctimes%208.314%7D%20%5B%5Cfrac%7B1%7D%7B505%7D%20-%5Cfrac%7B1%7D%7B525%7D%20%5D%5C%5C%5C%5CLog%28%5Cfrac%7Bf_2%7D%7Bf_1%7D%20%29%20%3D%200.7289%5C%5C%5C%5C%5Cfrac%7Bf_2%7D%7Bf_1%7D%20%20%3D%2010%5E%7B0.7289%7D%5C%5C%5C%5C%5Cfrac%7Bf_2%7D%7Bf_1%7D%20%20%3D%205.356)
Therefore, the ratio of f at the higher temperature to f at the lower temperature is 5.356
Answer:
X(Cl-35) = 75.95% => Answer 'A'
Explanation:
34.9689·X(Cl-35) + 36.9695·X(Cl-37) = 35.45; X = fractional abundance
X(Cl-35) + X(Cl-37) = 1 ⇒ X(Cl-37) = 1 - X(Cl-25)
34.9689·X(Cl-35) + 36.9695(1 - X(Cl-35)) = 35.45
34.9689·X(Cl-35) + 36.9695 - 36.9695·X(Cl-35) = 35.45
Rearrange ...
36.9695·X(Cl-35) - 34.9689·X(Cl-35) = 36.9689 - 35.45
2.0006·X(Cl-35) = 1.5195
X(Cl-35) = 1.5195/2.0006 = 0.7595 fractional abundance
⇒ % abundance = 75.95%
Answer:
None of these cause the correct answer is 9/-5