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3 years ago

14

A sample of Oxygen effuses through a porous container about 2 times faster than an unknown gas. Estimate the molar mass of the u

nknown gas.

1 answer:

sdas [7]3 years ago

5 0

Answer:

The molar mass of the unknown gas is 128g/mol

Explanation:

The rate of effusion of 2 different gases follows the equation:

Rate gas A / Rate gas B = √MM gas B / √MM gas A

<em>Where rate is the speed of effusion of each gas and MM the molar mass of each gas</em>

<em />

If gas A is oxygen and gas B the unknown:

Rate gas A / Rate gas B = 2 = √MM gas B / √32g/mol

11.134 = √MM gas B

128g/mol = MM gas B

The molar mass of the unknown gas is 128g/mol

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3 years ago

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3 years ago

The total volume of seawater is 1.5 x 10²¹ L. Seawater contains approximately 3.5% sodium chloride by mass. At that high of a co

garri49 [273]

Answer:

There are $5.408\times 10^{22}$ grams contained in all the seawater in the world.

Explanation:

At first let is determinate the total mass of seawater ( $m_{sw}$ ), measured in grams, in the world by definition of density and considering that mass is distributed uniformly:

$m_{sw} = \rho_{sw}\cdot V_{sw}$

Where:

$\rho_{sw}$ - Density of seawater, measured in grams per liters.

$V_{sw}$ - Volume of seawater, measured in liters.

If $V_{sw} = 1.5\times 10^{21}\,L$ and $\rho_{sw} = 1030\,\frac{g}{L}$ , then:

$m_{sw}=\left(1030\,\frac{g}{L} \right)\cdot (1.5\times 10^{21}\,L)$

$m_{sw} = 1.545\times 10^{24}\,g$

The total mass of sodium chloride is determined by the following ratio:

$r = \frac{m_{NaCl}}{m_{sw}}$

$m_{NaCl} = r\cdot m_{sw}$

Given that $m_{sw} = 1.545\times 10^{24}\,g$ and $r = 0.035$ , the total mass of sodium chloride in all the seawater in the world is:

$m_{NaCl} = 0.035\cdot (1.545\times 10^{24}\,g)$

$m_{NaCl} = 5.408\times 10^{22}\,g$

There are $5.408\times 10^{22}$ grams contained in all the seawater in the world.

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4 years ago

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fredd [130]

Answer:

Explanation:

combustion

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4 years ago

How many grams of CuSO4 • 5H2O do you need to make 0.8 L of a 0.6 M solution

Bumek [7]

<h3><u>Unders</u><u>tanding</u><u> </u><u>the</u><u> </u><u>conc</u><u>ept</u><u>:</u><u>-</u></h3>

Here the concept of Molarity used .We will put the formula of Molarity then we find moles of copper sulphate .Then we can find the.Given weight of Copper sulphate.So let's do

<h3>Solution:-</h3>

Volume of Solvent=0.8L
Molarity=0.6M

We know

$\boxed{\sf Molarity=\dfrac{Moles\:of\: solute}{Volume\:of\: Solvent \:in\;L}}$

$\\ \rm\longmapsto Moles\:of\: solute=Molarity\times Volume\:of\: Solvent\:in\:L$

$\\ \rm\longmapsto Moles\:of\: Solute=0.8(0.6)$

$\\ \rm\longmapsto Moles\:of\: Solute=0.48mol$

Now

Molar Mass of Solute:-

$\\ \rm\longmapsto CuSO_4.5H_2O$

$\\ \rm\longmapsto 65u+32u+4(16u)+5(2\times 1u+16u)$

$\\ \rm\longmapsto 96u+64u+5(2+16)$

$\\ \rm\longmapsto 160u+90u$

$\\ \rm\longmapsto 250u$

$\\ \rm\longmapsto 250g/mol$

Now

$\boxed{\sf No\:of\:moles=\dfrac{Given\:Mass}{Molar\:Mass}}$

$\\ \rm\longmapsto Given\:Mass=No\:of\:Moles\times Molar\:Mass$

$\\ \rm\longmapsto Given\:Mass=250(0.48)$

$\\ \rm\longmapsto Given\:Mass=120g$

8 0

3 years ago