Answer is: <span>the percent ionization is 0,19%.
</span>Chemical reaction: HA(aq) ⇄ H⁺(aq) + A⁻(aq).
Ka(HA) = 3,6·10⁻⁷.
c(HA) = 0,1 M.
[H⁺] = [A⁻] = x; equilibrium concentration.
[HA] = 0,1 M - x.
Ka = [H⁺] · [A⁻] / [HA].
0,00000036 = x² / 0,1 M - x.
Solve quadratic equation: x = 0,00019 M.
α = 0,00019 M ÷ 0,1 M · 100% = 0,19%.
Explanation:
The term reproductive strategy is loosely used to refer to the way an animal mates and/or raises offspring. ... It could also refer to semelparous reproduction, one reproductive episode in a lifetime, or iteroparous reproduction, multiple reproductive episodes over the course of an organism's life.
Elements with the same properties would be in the same group (Up and down row). Look at a periodic table and see which of the elements are in the same group. That will be your answer.<span />
Answer:
B. 0.55 M
Explanation:
Step 1: Given data
- Moles of calcium chloride (solute): 1.5 mol
- Volume of solution: 2.75 M
Step 2: Calculate the molarity of the aqueous solution
The molarity is a way to express the concentration of a solution. It is equal to the quotient between the moles of solute and the liters of solution.
M = moles of solute / liters of solution
M = 1.5 mol / 2.75 L = 0.55 mol/L = 0.55 M
<span>6.38x10^-2 moles
First, let's determine how many moles of gas particles are in the two-liter container. The molar volume for 1 mole at 25C and 1 atmosphere is 24.465 liters/mole. So
2 L / 24.465 L/mol = 0.081749438 mol
Now air doesn't just consist of nitrogen. It also has oxygen, carbon dioxide, argon, water vapor, etc. and the total number of moles includes all of those other gasses. So let's multiply by the percentage of nitrogen in the atmosphere which is 78%
0.081749438 mol * 0.78 = 0.063764562 mol.
Rounding to 3 significant figures gives 6.38x10^-2 moles</span>
