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3 years ago

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If measurements of a gas are 100L and 300 kilopascals and then the gas is measured a second time and found to be 75L, describe w

hat had to happen to the pressure (if temperature remained constant). Include which law supports this observation.

1 answer:

Tanya [424]3 years ago

4 0

Answer:

The pressure of the gas increased (if temperature remained constant).

The Boyle's law supports this observation.

Explanation:

The initial measurements of the gas are given as;

volume = 100 L

Pressure = 300 kpa

The second measurement is given as;

Volume = 75 L

The second reading implies that the volume of the gas has decreased. If the temperature of the gas remained constant, then the pressure must have increased according to the Boyle's law;

At constant temperature, the pressure of a given mass of an ideal gas is inversely proportional to its volume.

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4 years ago

Which describes the polarity in a water molecule?

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4 years ago

Read 2 more answers

What volume of water vapor would be produced from the combustion of 815.74 grams of propane (C3H8) with 1,006.29 grams of oxygen

d1i1m1o1n [39]

3940.2 is the volume of water vapour that would be produced from the combustion of 815.74 grams of propane ( $C_3H_8$ ) with 1,006.29 grams of oxygen gas, under a pressure of 1.05 atm and a temperature of 350. degrees C.

<h3>What is an ideal gas equation?</h3>

The ideal gas law (PV = nRT) relates the macroscopic properties of ideal gases. An ideal gas is a gas in which the particles (a) do not attract or repel one another and (b) take up no space (have no volume).

Stoichiometric calculations:

$C_3H_8(g) + 5 O_2(g)$ → $3 CO_2(g) + 4 H_2O(g)$

From the equation of the reaction, the mole ratio of propane to oxygen is 1:5.

Mole of 815.74 grams of propane = $\frac{ 815.74}{44.1 }$

Mole of 815.74 grams of propane = 18.49750567 moles

Mole of 1,006.29 grams of oxygen = $\frac{ 1,006.29}{32 }$

Mole of 1,006.29 grams of oxygen = 31.4465625 moles

Going by the mole ratio, it appears propane is limiting while oxygen is in excess.

From the equation, 1 mole of propane produces 4 moles of water vapour. Thus, the equivalent mole of water vapour will be:

18.49750567 moles x 4 = 73.99 moles.

Using the ideal gas equation:

PV = nRT

v = (73.99 x 0.08206 x 623) ÷ 0.96

v = 3940.2

Hence, 3940.2 is the volume of water vapour that would be produced from the combustion of 815.74 grams of propane ( $C_3H_8$ ) with 1,006.29 grams of oxygen gas, under a pressure of 1.05 atm and a temperature of 350. degrees C.

Learn more about the ideal gas here:

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2 years ago

Carbon dioxide dissolves in water to form carbonic acid, which is primarily dissolved CO2. Dissolved CO2 satisfies the equilibri

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Explanation:

The reaction equation will be as follows.

$CO_{2}(aq) + H_{2}O \rightleftharpoons H^{+}(aq) + HCO^{-}_{3}(aq)$

Calculate the amount of $CO_{2}$ dissolved as follows.

$CO_{2}(aq) = K_{CO_{2}} \times P_{CO_{2}}$

It is given that $K_{CO_{2}}$ = 0.032 M/atm and $P_{CO_{2}}$ = $1.9 \times 10^{-4}$ atm.

Hence, $[CO_{2}]$ will be calculated as follows.

$[CO_{2}]$ = $K_{CO_{2}} \times P_{CO_{2}}$

= $0.032 M/atm \times 1.9 \times 10^{-4}atm$

= $0.0608 \times 10^{-4}$

or, = $0.608 \times 10^{-5}$

It is given that $K_{a} = 4.46 \times 10^{-7}$

As, $K_{a} = \frac{[H^{+}]^{2}}{[CO_{2}]}$

$4.46 \times 10^{-7} = \frac{[H^{+}]^{2}}{0.608 \times 10^{-5}}$

$[H^{+}]^{2}$ = $2.71 \times 10^{-12}$

$[H^{+}]$ = $1.64 \times 10^{-6}$

Since, we know that pH = $-log [H^{+}]$

So, pH = $-log (1.64 \times 10^{-6})$

= 5.7

Therefore, we can conclude that pH of water in equilibrium with the atmosphere is 5.7.

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C).dd offspring have a nice day

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