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3 years ago

Strong root systems and long flexable stems allow plants to survive on the strong currents of which ecosystem?

Chemistry

1 answer:

V125BC [204]3 years ago

4 0

Answer:

its B. rivers

Explanation:

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Three gases (8.00 g of methane, CH4, 18.0 g of ethane, C2H6, and an unknown amount of propane, C3H8) were added to the same 10.0

klemol [59]

Answer:

Partial pressure of methane: 1.18 atm

Partial pressure of ethane: 1.45 atm

Partial pressure of propane: 2.35 atm

Explanation:

Let the total moles of gases in a container be n.

Total pressure of the gases in a container =P = 5.0 atm

Temperature of the gases in a container =T = 23°C = 296.15 K

Volume of the container = V = 10.0 L

$PV=nRT$ (Ideal gas equation)

$n=\frac{PV}{RT}=\frac{5.0 atm\times 10.0 L}{0.0821 atm L/mol K\times 296.15 K}=2.0564 mol$

Moles of methane gas = $n_1=\frac{8.00 g}{16.04 g/mol}=0.4878 mol$

Moles of ethane gas = $n_2=\frac{18.00 g}{30.07 g/mol}=0.5986 mol$

Moles of propane gas = $n_3=?$

$n=n_1+n_2+n_3$

$n_3=n-n_1-n_2=2.0564 mol-0.4878 mol-0.5986 mol= 0.9700 mol$

Partial pressure of all the gases can be calculated by using Raoult's law:

$p_i=P\times \chi_i$

$p_i$ = partial pressure of 'i' component.

$\chi_1$ = mole fraction of 'i' component in mixture

P = total pressure of the mixture

Partial pressure of methane:

$p_1=P\times \chi_1=P\times \frac{n_1}{n_1+n+2+n_3}=P\times \frac{n_1}{n}$

$p_1=5.00 atm\times \frac{0.4878 mol}{2.0564 mol}=1.18 atm$

Partial pressure of ethane:

$p_2=P\times \chi_2=P\times \frac{n_2}{n_1+n+2+n_3}=P\times \frac{n_2}{n}$

$p_2=5.00 atm\times \frac{0.5986 mol}{2.0564 mol}=1.45 atm$

Partial pressure of propane:

$p_3=P\times \chi_3=P\times \frac{n_3}{n_1+n+2+n_3}=P\times \frac{n_3}{n}$

$p_3=5.00 atm\times \frac{0.9700 mol}{2.0564 mol}=2.35 atm$

5 0

4 years ago

The mathematical relationship between gas solubility and pressure is called Henry's Law, solubility = kHPgas where kH is the Hen

Diano4ka-milaya [45]

Answer : The unit of $k_H$ in mol/L.mm Hg is, $1.8\times 10^{-6}mol/L.mmHg$

Explanation :

As we know that the $k_H$ is the Henry's Law constant for argon at $25^oC$ is, $1.4\times 10^{-3}mol/L.atm$

Now we have to determine the unit of $k_H$ in mol/L.mm Hg

Conversion used for pressure from atm to mmHg is:

1 atm = 760 mmHg

So,

$k_H=1.4\times 10^{-3}mol/L.atm\times \frac{1atm}{760mmHg}$

$k_H=1.8\times 10^{-6}mol/L.mmHg$

Thus, the unit of $k_H$ in mol/L.mm Hg is, $1.8\times 10^{-6}mol/L.mmHg$

5 0

3 years ago

What force must act on a 30 kg mass object to give it an acceleration of 0.45 m/s²?

Degger [83]

Explanation:

Force=mass×acceleration

F=m×a

F=30×0.45= 13.5N

hope this helps you.

5 0

3 years ago

Plz help me with this one question

Lunna [17]

Answer:

Cl2(g) + H2O <–> HOCl + H+ + Cl-

8 0

3 years ago

Read 2 more answers

How many elements are in 2CaCO3 and C8H10N402

Ilia_Sergeevich [38]

Answer:

In 2CaCO3 there is 3 elements.

In C8H10N402 there is 3 elements.

3 0

4 years ago