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2 years ago

14

Consider Example 12.10. Suppose the experiment is repeated with 0.032 mol of helium instead with everything else staying the sam

e, what is the amount of heat required now to achieve that process

1 answer:

Lemur [1.5K]2 years ago

5 0

Answer:

120 extracted from the gas

Explanation:

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Which of the following acts as a catalyst in catalytic converters?

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Answer: B. Metal

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2 years ago

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8. What is the mass of 4.50 x 1022 Cu atoms?

vladimir2022 [97]

Answer:

4.7485 g

Explanation:

4.50 x 10^22 Cu atoms * (1 mol Cu / 6.022 x 10^23 Cu atoms) * 63.546 g Cu/(mol Cu) = 4.7485 g

In every mole of Cu, there are 6.022 x 10^23 atoms (Avogadro's number). The molecular weight of copper is 63.546 g/mol.

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2 years ago

At what point would a chemical bond form between two atoms.

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2 years ago

An ethylene glycol solution contains 21.4 g of ethylene glycol (C2H6O2) in 97.6 mL of water. (Assume a density of 1.00 g/mL for

8090 [49]

Answer: The freezing point and boiling point of the solution are $-6.6^0C$ and $101.8^0C$ respectively.

Explanation:

Depression in freezing point:

$T_f^0-T^f=i\times k_f\times \frac{w_2\times 1000}{M_2\times w_1}$

where,

$T_f$ = freezing point of solution = ?

$T^o_f$ = freezing point of water = $0^0C$

$k_f$ = freezing point constant of water = $1.86^0C/m$

i = vant hoff factor = 1 ( for non electrolytes)

m = molality

$w_2$ = mass of solute (ethylene glycol) = 21.4 g

$w_1$ = mass of solvent (water) = $density\times volume=1.00g/ml\times 97.6ml=97.6g$

$M_2$ = molar mass of solute (ethylene glycol) = 62g/mol

Now put all the given values in the above formula, we get:

$(0-T_f)^0C=1\times (1.86^0C/m)\times \frac{(21.4g)\times 1000}{97.6g\times (62g/mol)}$

$T_f=-6.6^0C$

Therefore,the freezing point of the solution is $-6.6^0C$

Elevation in boiling point :

$T_b-T^b^0=i\times k_b\times \frac{w_2\times 1000}{M_2\times w_1}$

where,

$T_b$ = boiling point of solution = ?

$T^o_b$ = boiling point of water = $100^0C$

$k_b$ = boiling point constant of water = $0.52^0C/m$

i = vant hoff factor = 1 ( for non electrolytes)

m = molality

$w_2$ = mass of solute (ethylene glycol) = 21.4 g

$w_1$ = mass of solvent (water) = $density\times volume=1.00g/ml\times 97.6ml=97.6g$

$M_2$ = molar mass of solute (ethylene glycol) = 62g/mol

Now put all the given values in the above formula, we get:

$(T_b-100)^0C=1\times (0.52^0C/m)\times \frac{(21.4g)\times 1000}{97.6g\times (62g/mol)}$

$T_b=101.8^0C$

Thus the boiling point of the solution is $101.8^0C$

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3 years ago

What is the mass number of chlorine?

Anastasy [175]

Answer:

Explanation:

35.453 u

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2 years ago