Answer:
Ka = 1.39x10⁻⁶
Explanation:
A monoprotic acid, HX, will be in equilibrium in an aqueous medium such as:
HX(aq) ⇄ H⁺(aq) + X⁻(aq)
<em>Where Ka is:</em>
Ka = [H⁺] [X⁻] / [HX]
<em>Where [] is the molar concentration in equilibrium of each specie.
</em>
The equilibrium is reached when some HX reacts producing H+ and X-, that is:
[HX] = 1.64M - X
[H⁺] = X
[X⁻] = X
As pH is 2.82 = -log [H⁺]:
[H⁺] = 1.51x10⁻³M:
[HX] = 1.64M - 1.51x10⁻³M = 1.638M
[H⁺] = 1.51x10⁻³M
[X⁻] = 1.51x10⁻³M
And Ka is:
Ka = [1.51x10⁻³M] [1.51x10⁻³M] / [1.638M]
<h3>Ka = 1.39x10⁻⁶</h3>
The answer for the following problem is mentioned below.
<u><em>Therefore volume occupied by methane gas is 184.78 × 10^-3 liters </em></u>
Explanation:
Given:
mass of methane(
) = 272 grams
pressure (P) = 250 k Pa =250×10^3 Pa
temperature(t) = 54°C =54 + 273 = 327 K
Also given:
R = 8.31JK-1 mol-1 ,
Molar mass of methane() = 16.0 grams
We know;
According to the ideal gas equation,
<u><em>P × V = n × R × T</em></u>
here,
n = m÷M
n =272 ÷ 16
<u><em>n = 17 moles</em></u>
Therefore,
250×10^3 × V = 17 × 8.31 × 327
V = ( 17 × 8.31 × 327 ) ÷ ( 250×10^3 )
V = 184.78 × 10^-3 liters
<u><em>Therefore volume occupied by methane gas is 184.78 × 10^-3 liters </em></u>
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The answer is b. 4.2 mole. The balanced reaction formula is 2LiOH + H2SO4 -->Li2SO4 + 2H2O. And the ratio of mole number of the reactants is the same as the ratio of coefficients.
To calculate the average mass of the element, we take the summation of the product of the isotope and the percent abundance. In this case, the equation becomes 186.207=187*0.626+185*x where x is the percent abundance of 185. The answer is 0.374 or 37.4%. This can also be obtained by 100%-62.6%= 37.4%.
