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2 years ago

How do we think that Mars lost atmospheric gas?

Chemistry

1 answer:

grin007 [14]2 years ago

7 0

the rocky planet was exposed to harsh and forceful solar winds, which blew the Martian atmosphere away.

Another option for thickening the atmosphere of Mars, and, in turn, raising the temperature of the planet, would be to set up solar-powered, greenhouse-gas producing factories.

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What does Le Châtelier's principle state?

bonufazy [111]

When a system experiences a disturbance ( such as concentration, temperature, or pressure changes), it will respond to restore a new equilibrium state.

5 0

3 years ago

The partial pressure of CO2 gas in a bottle of carbonated water is 4.60 atm at 25 ºC. How much CO2 gas (in g) will be released f

frutty [35]

If the partial pressure of CO₂ in a bottle of carbonated water decreases from 4.60 atm to 1.28 atm, the mass of CO₂ released is 0.265 g.

The partial pressure of CO₂ gas in a bottle of carbonated water is 4.60 atm at 25 ºC. We can calculate the concentration of CO₂ using Henry's law.

$C = k \times P = \frac{1.65 \times 10^{-3} M }{atm} \times 4.60 atm = 7.59 \times 10^{-3} M$

We can calculate the mass of CO₂ in 1.1 L considering its molar mass is 44.01 g/mol.

$\frac{7.59 \times 10^{-3} mol}{L} \times 1.1 L \times \frac{44.01 g}{mol} = 0.367 g$

Now, we will repeat the same procedure for a partial pressure of 1.28 atm.

$C = k \times P = \frac{1.65 \times 10^{-3} M }{atm} \times 1.28 atm = 2.11 \times 10^{-3} M$

$\frac{2.11 \times 10^{-3} mol}{L} \times 1.1 L \times \frac{44.01 g}{mol} = 0.102 g$

The mass of CO₂ released will be equal to the difference in the masses at the different pressures.

$m = 0.367 g - 0.102 g = 0.265 g$

If the partial pressure of CO₂ in a bottle of carbonated water decreases from 4.60 atm to 1.28 atm, the mass of CO₂ released is 0.265 g.

Learn more: brainly.com/question/18987224

<em>The partial pressure of CO₂ gas in a bottle of carbonated water is 4.60 atm at 25 ºC. How much CO₂ gas (in g) will be released from 1.1 L of the carbonated water when the partial pressure of CO2 is lowered to 1.28 atm? At 25 ºC, the Henry’s law constant for CO₂ dissolved in water is 1.65 x 10⁻³ M/atm, and the density of water is 1.0 g/cm³.</em>

5 0

2 years ago

How much energy is required to heat 36.0 g H2O from a liquid at 65°C to a gas at 115°C? The following physical data may be usefu

In-s [12.5K]

Answer:

energy required=qnet=87.75kJ

Explanation:

we will do it in three seperate step and then add up those value.

first step is to heat the sample of water upto 100C i.e upto boiling pont. because just after this sample of water started vaporization.

q 1= m c (T2-T1)

q1 = 36.0 g (4.18 J/gC) (100 - 65 C)

q1 = 5267 J =5.267kJ

next is to vaporize the sample at 100C

q2 = 36.0 g / 18.0 g/mol X 40.7 kJ/mol

q2= 81.4 kJ

Finally, heat the steam upto 115C

q3 = m c (T2-T1)

q 3= 36.0 g (2.01 J/gC)(115-100C)

q3 = 1085 J =1.085kJ

qnet=q1 +q2 +q3

energy required=qnet=87.75kJ

8 0

3 years ago

How would you prepare 1.00 L of a 0.50 M solution of each of the following?

Lunna [17]

Answer:

Take approx 41.7 mL of 12-M HCl in a 1.00-L flask and fill the rest of the volume with distilled water.

Explanation:

Hello,

In this case, for the dilution process from concentrated 12-M hydrochloric acid to 1.00 L of the diluted 0.50M hydrochloric acid, the volume of concentrated HCl you must take is computed by considering that the moles remain constant for all dilution processes as shown below:

$n_1=n_2$