Answer:
The molarity of the solution is 0,12 M.
Explanation:
We calculate the molarity, which is a concentration measure that indicates the moles of solute (in this case KOH) in 1000ml of solution (1 liter):
0,25 L solution----- 0,030 moles of KOH
1 L solution----x= (1 L solution x 0,030 moles of KOH)/0,25 L solution
x= 0,12 moles of KOH ---> <em>The solution is 0,12 M</em>
Answer:
1. [OH⁻] = 2.4 x 10⁻¹⁴ M.
2. [H₃O⁺] = 4.167 x 10⁻¹ M.
Explanation:
Firstly, we need to calculate the molarity of Mg(OH)₂:
- Molarity (M) is the no. of moles of solute that dissolved in 1.0 L of the solution.
M = (no. of moles of solute)/(V of the solution (L).
∴ M of MgCl₂ = (no. of moles of solute)/(V of the solution (L) = (1.2 x 10⁻¹⁴ mol)/(1.0 L) = 1.2 x 10⁻¹⁴ M.
<u><em>1. [OH⁻]:</em></u>
- Mg(OH)₂ is dissociated in the solution according to the equation:
<em>Mg(OH)₂ → Mg²⁺ + 2OH⁻.</em>
<em></em>
- It is clear that every 1.0 mol of Mg(OH)₂ produces 2.0 moles of (OH⁻).
∴ [OH⁻] = 2(1.2 x 10⁻¹⁴) = 2.4 x 10⁻¹⁴ M.
<u><em>2. [H₃O⁺]:</em></u>
<u><em></em></u>
<em>∵ [H₃O⁺][OH⁻] = 10⁻¹⁴.</em>
∴ [H₃O⁺] = 10⁻¹⁴/[OH⁻] = 10⁻¹⁴/(2.4 x 10⁻¹⁴ M) = 4.167 x 10⁻¹ M.
Answer:
+1
Explanation:
A Potassium atom is represented with the sign "K" in chemistry and have atomic number 19.
The charge of electrons causes a matter to experience a force and it can be positive or negative.
In Potassium atom, the electric charge is +1 to enter a stable electron configuration as there is only one valence electron in the outermost shell of potassium atom. As potassium atom will lose electron it will become positively charged.
Hence, the correct answer is "+1".
Answer:
Explanation
Let, Wavelength = W,
Given, frequency (f) = 50 Hz, wave velocity (V) = 342 m/s
We know, V = f × W
=> 342 = 50 × W
=> W = 342/50 metre
=> W = 6.84 metre
Explanation:
This energy comes from the food we eat. Our bodies digest the food we eat by mixing it with fluids (acids and enzymes) in the stomach. When the stomach digests food, the carbohydrate (sugars and starches) in the food breaks down into another type of sugar, called glucose.