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SSSSS [86.1K]
2 years ago
9

A sample of oxygen gas (O2) has a volume of 7.51 L at a temperature of 19°C and a pressure of 1.38 atm. Calculate the moles of O

2 molecules present in this gas sample.
Chemistry
1 answer:
Maslowich2 years ago
8 0

Answer: 0.43molO₂

Explanation:

The ideal gas law for moles is n=\frac{PV}{RT}. If you did not know, R is the ideal gas constant, 0.0821\frac{L*atm}{mol*K}.

Since the temperature must be in Kelvin, we must convert our given temperature from ℃ to K. 19C+273=292K

Now that we have the temperature converted to Kelvin, we can plug in our information to the ideal gas law for the number of moles.

n=\frac{PV}{RT}

n=\frac{1.38atm*7.51L}{0.0821\frac{L*atm}{mol*K} *292K} =0.43molO_{2}

n=0.43molO_{2}

Therefore, the number of moles is 0.43molO₂.

I hope this helps! Pls mark brainliest!! :)

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A. Wax drips down the side of a lot candle.

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The chemical change from solid to liquid. This is a combustion reaction, so carbon dioxide gas and water vapour is also produced but you can't see them

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Is barium hydroxide Ba(OH)2 or is BaOH?
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Answer:

What is the charge on the barium ion and what is the charge of the hydroxide ion.

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To get the correct formula they have to add to zero over all.

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3 years ago
3. If you start with 8x1025 molecules of Cl, and 25 grams of KI, how many grams of KCl would
miskamm [114]

Answer:

Percent yield = 89.1%

Explanation:

Based on the equation:

Cl₂ + 2KI → 2KCl + I₂

<em>1 mole of Cl₂ reacts with 2 moles of KI to produce to moles of KCl</em>

<em />

To solve this quesiton we must find the moles of each reactant in order to find the limiting reactant. With the limiting reactant we can find the moles of KCl and the mass:

<em>Moles Cl₂:</em>

8x10²⁵ molecules * (1mol / 6.022x10²³ molecules) = 133 moles

<em>Moles KI -Molar mass: 166.0028g/mol-</em>

25g * (1mol / 166.0028g) = 0.15 moles

Here, clarely, the KI is the limiting reactant

As 2 moles of KI produce 2 moles of KCl, the moles of KCl produced are 0.15 moles. The theoretical mass is:

0.15 moles * (74.5513g / mol) =

11.2g KCl

Percent yield is: Actual yield (10.0g) / Theoretical yield (11.2g) * 100

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3 0
3 years ago
20 ML of a gas at 200 K is heated until the new volume is 55ML what is the final temperature of the gas
MrRa [10]

Answer:

T2 = 550K

Explanation:

From Charles law;

V1/T1 = V2/T2

Where;

V1 is initial volume

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T1 is initial temperature

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We are given;

V1 = 20 mL

V2 = 55 mL

T1 = 200 K

Thus from V1/T1 = V2/T2, making T2 the subject;

T2 = (V2 × T1)/V1

T2 = (55 × 200)/20

T2 = 550K

8 0
3 years ago
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