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3 years ago

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Greenland ice sheets

1 answer:

olga nikolaevna [1]3 years ago

8 0

<span>The Greenland ice sheet is a vast body of ice covering 1,710,000 square kilometers, roughly 80% of the surface of Greenland. It is the second largest ice body in the world, after the Antarctic ice sheet.</span><span />

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Deep sea divers use a mixture of helium and oxygen to breathe. Assume that a diver is going to a depth of 150 feet where the tot

Svetlanka [38]

Answer:

4.525% is the percentage by volume of oxygen in the gas mixture.

Explanation:

Total pressure of the mixture = p = 4.42 atm

Partial pressure of the oxygen = $p_1=0.20 atm$

Partial pressure of the helium = $p_2$

$p_1=p\times \chi_1$ (Dalton law of partial pressure)

$0.20 atm=4.42 atm\times \chi_1$

$\chi_1=\frac{0.20 atm}{4.42 atm}=0.04525$

$\chi_2=1-\chi_1=1-0.04525=0.95475$

$chi_1+chi_2=1$

$n_1=0.04525 mol,n_2=0.95475 mol$

According Avogadro law:

$Moles\propto Volume$ (At temperature and pressure)

Volume occupied by oxygen gas = $V_1$

Total moles of gases = n = 1 mol

Total Volume of the gases = V

$\frac{n_1}{V_1}=\frac{n}{V}$

$\frac{V_1}{V}=\frac{n_1}{n}=\frac{0.04525 mol}{1 mol}$

Percent by volume of oxygen in the gas mixture:

$\frac{V_1}{V}\times 100=\frac{0.04525 mol}{1 mol}\times 100=4.525\%$

6 0

3 years ago

A rigid tank contains 0.66 mol of oxygen (O2). Find the mass of oxygen that must be withdrawn from the tank to lower the pressur

dsp73

Answer:

12.8 g of $O_{2}$ must be withdrawn from tank

Explanation:

Let's assume $O_{2}$ gas inside tank behaves ideally.

According to ideal gas equation- $PV=nRT$

where P is pressure of $O_{2}$ , V is volume of $O_{2}$ , n is number of moles of $O_{2}$ , R is gas constant and T is temperature in kelvin scale.

We can also write, $\frac{V}{RT}=\frac{n}{P}$

Here V, T and R are constants.

So, $\frac{n}{P}$ ratio will also be constant before and after removal of $O_{2}$ from tank

Hence, $\frac{n_{before}}{P_{before}}=\frac{n_{after}}{P_{after}}$

Here, $\frac{n_{before}}{P_{before}}=\frac{0.66mol}{43atm}$ and $P_{after}=17atm$

So, $n_{after}=\frac{n_{before}}{P_{before}}\times P_{after}=\frac{0.66mol}{43atm}\times 17atm=0.26mol$

So, moles of $O_{2}$ must be withdrawn = (0.66 - 0.26) mol = 0.40 mol

Molar mass of $O_{2}$ = 32 g/mol

So, mass of $O_{2}$ must be withdrawn = $(32\times 0.40)g=12.8g$

7 0

3 years ago

I need a answer quick. PLEASE<br><br>What is the volume of 2.43 x 1023 molecules of N2 gas at STP?

Readme [11.4K]

I think it is 2485.89.

6 0

3 years ago

What is the density of a sample of a substance with a volume of 120ml<br> and a mass of 90g?

tigry1 [53]

Answer:

<h3>The answer is 0.75 g/mL</h3>

Explanation:

The density of a substance can be found by using the formula

$density = \frac{mass}{volume} \\$

From the question

mass = 90 g

volume = 120 mL

We have

$density = \frac{90}{120} = \frac{9}{12} = \frac{3}{4} \\$

We have the final answer as

<h3>0.75 g/mL</h3>

Hope this helps you

6 0

3 years ago

3. What is the mass of 5 moles of Hydrogen sulfate?I

solong [7]

Answer:

490

98 for 1 mole, Hence for 5 moles 5 X 98 =490.

Explanation:

Brainliest please?

3 0

3 years ago