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2 years ago

What is the molar volume (of a gas)?

Chemistry

1 answer:

Nookie1986 [14]2 years ago

7 0

Answer:

The molecular weight (molar mass) of any gas is the mass of one particle of that gas multiplied by Avogadro's number (6.02 x 1023).

Explanation:

How do you find the molar mass of a gas?

First the ideal gas law will be used to solve for the moles of unknown gas (n). Then the mass of the gas divided by the moles will give the molar mass.

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Alpha, Beta, and Gamma Decay

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g Consider an ideal atomic gas in a cylinder. The upper part of the cylinder is a moveable piston of negligible weight. The heig

kumpel [21]

A cylindrical weight with a mass of 3 kg is dropped onto the piston from a height of 10 m. The entropy of the gas is 1.18 J/K and the change in the entropy of the environment is -1.18 J/K.

A cylindrical weight with a mass (m) of 3 kg is dropped, that is, its initial velocity (u) is 0 m/s and travels 10 m (s). Assuming the acceleration (a) is that of gravity (9.8 m/s²). We can calculate the velocity (v) of the weight in the instant prior to the collision with the piston using the following kinematic equation.

$v^{2} = u^{2} + 2as = 2 (9.8 m/s^{2} ) (10m) \\\\v = 14 m/s$

The object with a mass of 3 kg collides with the piston at 14 m/s, The kinetic energy (K) of the object at that moment is:

$K = \frac{1}{2} m v^{2} = \frac{1}{2} (3kg) (14m/s)^{2} = 294 J$

The kinetic energy of the weight is completely converted into heat transferred into the gas cylinder. Thus, Q = 294 J.

Given all the process is at 250 K (T), we can calculate the change of entropy of the gas using the following expression.

$\Delta S_{gas} = \frac{Q}{T} = \frac{294 J}{250K} = 1.18 J/K$

The change in the entropy of the environment, has the same value but opposite sign than the change in the entropy of the gas. Thus, $\Delta S_{env} = -1.18 J/K$

Learn more: brainly.com/question/22655760

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2 years ago

What is the Ka of a weak acid (HA) if the initial concentration of weak acid is 4.5 x 10-4 M and the pH is 6.87? (pick one)

rewona [7]

Answer:

Ka = $4.04 \times 10^{-11}$

Explanation:

Initial concentration of weak acid = $4.5 \times 10^{-4}\ M$

pH = 6.87

$pH = -log[H^+]$

$[H^+]=10^{-pH}$

$[H^+]=10^{-6.87}=1.35 \times 10^{-7}\ M$

HA dissociated as:

$HA \leftrightharpoons H^+ + A^{-}$

(0.00045 - x) x x

[HA] at equilibrium = (0.00045 - x) M

x = $1.35 \times 10^{-7}\ M$

$Ka = \frac{[H^+][A^{-}]}{[HA]}$

$Ka = \frac{(1.35 \times 10^{-7})^2}{0.00045 - 0.000000135}$

0.000000135 <<< 0.00045

$Therefore, Ka = \frac{(1.35 \times 10^{-7})^2}{0.00045 } = 4.04 \times 10^{-11}$

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2 years ago

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