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3 years ago

Which gas is least likely to obey the ideal gas laws at very high pressures and very low temperatures?

Chemistry

1 answer:

dem82 [27]3 years ago

3 0

Xe !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

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Gases and particles which are put into the air or emitted by various sources are called __________. photochemical smog emissions

Leto [7]

I would say the answer is emissions. These are the particles that are not supposed to be present in air but due to the production of different substances from humans daily activities these substances go with the air we breath. Hope this helped.

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3 years ago

What negative impact did chemistry have on society through the production and use of chlorofluorocarbons (CFCs)?

andrey2020 [161]

<span>The answer to your question is the 3rd option </span>

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3 years ago

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Can some answer this please?

marysya [2.9K]

Answer:

Explanation:

it's b because I just went over that frome my class and got it correct

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3 years ago

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3mL of cyclohexanol (density = 0.9624 g/mL, Molecular weight = 100.158 g/mol) reacts with excess sulfuric acid to produce cycloh

TiliK225 [7]

Answer:

$n_{C_6H_{10}}=0.03molC_6H_{10}$

Explanation:

Hello,

In this case the undergoing chemical reaction is shown on the attached picture whereas cyclohexanol is converted into cyclohexene and water by the dehydrating effect of the sulfuric acid. Thus, for the starting 3 mL of cyclohexanol, the following stoichiometric proportional factor is applied in order to find the theoretical yield of cyclohexene in moles:

$n_{C_6H_{10}}=3mLC_6H_{12}O*\frac{0.9624gC_6H_{12}O}{1mLC_6H_{12}O}*\frac{1molC_6H_{12}O}{100.158gC_6H_{12}O}*\frac{1molC_6H_{10}}{1molC_6H_{12}O} \\n_{C_6H_{10}}=0.03molC_6H_{10}$

Besides, the mass could be computed as well by using the molar mass of cyclohexene:

$m_{C_6H_{10}}=0.03molC_6H_{10}*\frac{82.143gC_6H_{10}}{1molC_6H_{10}} \\\\m_{C_6H_{10}}=2.4gC_6H_{10}$

Even thought, the volume could be also computed by using its density:

$V_{C_6H_{10}}=2.4gC_6H_{10}*\frac{1mLC_6H_{10}}{0.811gC_6H_{10}} \\V_{C_6H_{10}}=3mLC_6H_{10}$

Best regards.

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3 years ago

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If the volume occupied by 0.500 mol of nitrogen gas at 0°C is 11.2 L, then the volume occupied by 2.00 mol of nitrogen gas at th

Paraphin [41]

Avagadros law states that volume of gas is directly proportional to number of moles of gas at constant pressure and temperature.
$\frac{V1}{n1} = \frac{V2}{n2}$
where V -volume , n - number of moles
parameters for the first instance are on the left side and parameters for the second instance are on the right side of the equation
substituting these values in the equation
$\frac{11.2 L}{0.500 mol} = \frac{V}{2.00 mol}$
V = 44.8 L
answer is C. 44.8 L

5 0

3 years ago