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3 years ago

What pressure in atmospheres(atm) is equal to 45.6 kPa?

Chemistry

1 answer:

postnew [5]3 years ago

4 0

1 kpa = 0.0098692327 atm so just multiply that by 45.6

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A gas that exerts a pressure of

spayn [35]

Answer:

3.089 L

Explanation:

From the given information, provided that the no of moles and the temperature remains constant;

$P_1$ = 15.6 psi

$V_1$ = ???

$P_2$ = 25.43 psi

$V_2$ = 1.895 L

Using Boyle's law:

$P_1V_1 =P_2V_2 \\ \\ V_1 = \dfrac{P_2V_2}{P_1} \\ \\ V_1 = \dfrac{25.43 \times 1.895}{15.6} \\ \\ \mathbf{ V_1 = 3.089 \ L}$

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3 years ago

The atom as a whole<br> doesn’t carry any electric charge. It’s called a ____________ _________.

olchik [2.2K]

Answer:

Neutron has no charge while electron has a negative charge and proton has a positive charge

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3 years ago

A colligative property is when the addition of a solute to a solvent alters the boiling point or The freezing point of a solvent

Mrrafil [7]

Answer:

this cp, is probably vapor pressure.

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3 years ago

What is the total number of atoms in 1.0 mole of CO2?

worty [1.4K]

18 x 10^23. This is the answer because there are 6.022 x 10^23 atoms of carbon and 12.044 x 10^23 atoms of oxygen in one mole of CO2, if you combine them, you get 18 x 10^23

5 0

2 years ago

One mole of an ideal gas doubles its volume in a reversible isothermal expansion. (a) What is the change in entropy of the gas?

Andre45 [30]

Explanation:

The given data is as follows.

n = 1 mol, $V_{f} = 2V_{i}$

Q = 1500 J, R = 8.314 J/mol k

(a) $\Delta S = \frac{dQ}{dT}$

And, according to the first law of thermodynamics

$\Delta E_{int} = Q - W$

And, in an isothermal process the change in internal energy of the gas is zero.

Hence, 0 = Q - W

or, W = Q

Expression for work done in an isothermal process is as follows.

W = $nRT ln \frac{V_{f}}{V_{i}}$

As W = Q, Hence expression for Q will also be given as follows.

Q = $nRT ln \frac{V_{f}}{V_{i}}$

Now,

$\Delta S = \frac{nRT ln \frac{V_{f}}{V_{i}}}{T}$

[/tex]\Delta S = nR ln \frac{V_{f}}{V_{i}}[/tex]

= $nR ln \frac{2V_{i}}{V_{i}}$

= nR ln 2

= $1 \times 8.314 \times 0.693$

= 5.76 J/K

Therefore, change in entropy is 5.76 J/K.

(b) As, Q = $nRT ln \frac{V_{f}}{V_{i}}$

= $nRT ln \frac{2V_{i}}{V_{i}}$

= nRT ln 2

T = $\frac{Q}{nR ln 2}$

= $\frac{1500}{1 \times 8.314 ln 2}$

= 260.4 K

Therefore, temperature of the gas is 260.4 K.

7 0

3 years ago