The gases in air dissolve in water. What would be one way to reduce the amount of a gas dissolved in water?

Darya [45]

Answer:

Coulomb's law, mathematical description of the electric force between charged objects. Formulated by the 18th-century French physicist Charles-Augustin de Coulomb, it is analogous to Isaac Newton's law of gravity.

Explanation:

Soo yeah I think this can help you

3 0

3 years ago

Select the answer which shows the correct order, slowest to fastest, of the relative rate of diffusion of the gases.

IrinaK [193]

Diffusion is the process of a substance spreading out to evenly fill its container or environment. Rate of diffusion of a gas is inversely proportional to the molar mass of the gas.

$Rate of diffusion \propto \frac{1}{Molar mass}$

Lighter(lower) the molar mass of the gas , faster will be its rate of diffusion and heavier (higher) the molar mass of the gas , slower will be its rate of diffusion.

We have to arrange the given gases from slowest rate of diffusion to fastest rate of diffusion that means we need to arrange gases from higher molar mass to lower molar mass.

Molar mass of given gases are :

Cl = 35.5 g/mol

Xe = 131.29 g/mol

He = 4.00 g/mol

N = 14.00 g/mol

So correct order for slowest rate of diffusion (highest molar mass) to fastest rate of diffusion (lowest molar mass) is :

Xe , Cl , N , He

Xe having the highest molar mass will have the slowest rate of diffusion and He with lowest molar mass will have the fastest rate of diffusion, so option 'c' is correct.

Note : Slowest rate of diffusion = High Molar Mass

Fastest rate of diffusion = Low Molar Mass

7 0

4 years ago

What is the pH of 4.3x10^-7 M solution of H2CO3?

Rzqust [24]

Answer:

pH = -log[H⁺] = -log(4.75 x 10⁻⁷) = -(- 6.32 ) = 6.32

Explanation:

Given 4.3 x 10⁻⁷M H₂CO₃ (Ka1 = 4.2 x 10⁻⁷ & Ka2 = 4.8 x 10⁻¹¹)

Note: The Ka2 value for the 2nd ionization step is so small (Ka2 = 4.8 x 10⁻¹¹) It will be assumed all of the hydronium ions (H⁺) come from the 1st ionization step.

1st Ionization step

H₂CO₃ ⇄ H⁺ + HCO₃⁻

C(initial) 4.3 x 10⁻⁷ 0 0

ΔC -x +x +x

C(final) 4.3 x 10⁻⁷ - x x x

Note: the 'x' value in this analysis can not be dropped as the Conc/Ka value is less than 10². In this case he C/Ka ratio* (4.3E-7/4.2E-7 ≈ 1) is far below 10².

So, one sets up the equilibrium equation to be quadric and the x-value can be determined.

Ka1 = [H⁺][HSO⁻]/[H₂SO₃] = (x)(x)/(4.3 x 10⁻ - x) = x²/(4.3 x 10⁻⁷ - x) = 4.2 x 10⁻⁷

=> x² = 4.2 x 10⁻⁷(4.3 x 10⁻⁷ - x)

=> x² + 4.2 x 10⁻⁷x - 1.8 x 10⁻¹³ = 0

a = 1, b = 4.2 x 10⁻⁷, c = - 1.8 x 10⁻¹³

x = b² ± SqrRt(b² - 4(1)(-1.8 x 10⁻¹³ / 2(1) = 4.75 x 10⁻⁷

x = [H⁺] = 4.75 x 10⁻⁷

pH = -log[H⁺] = -log(4.75 x 10⁻⁷) = -(- 6.4 ) = 6.4

______________________

* The Concentration/Ka-value is the simplification test for quadratic equations used in Equilibrium studies. If the C/Ka > 100 then one can simplify the C(final) by dropping the 'x' if used in this type analysis. However, if the C/Ka value is < 100 then the x-value must be retained and the solution is determined using the quadratic equation formula.

for ax² + bx + c = 0

x = b² ± SqrRt(b² - 4ac) / 2a

4 0

3 years ago

What is the name for the prevailing global wind pattern that comes from the poles and blows from east to west?

Fittoniya [83]

Answer: Trade winds :)

Explanation: just took the test

7 0

3 years ago

What is produced when LiOH neutralizes HCl

iren [92.7K]

LiOH+HCl ---->LiCl+H2O
Base...Acid......Salt...

6 0

3 years ago