Answer:
Number of moles = 0.0005 mol.
Explanation:
Given data:
pH = 3
Volume of solution = 500 mL
Number of moles = ?
Solution:
HCl dissociate to gives H⁺ and Cl⁻
HCl → H⁺ + Cl⁻
It is known that,
pH = -log [H⁺]
3 = -log [H⁺]
[H⁺] = 10⁻³ M
[H⁺] = 0.001 M
Number of moles of HCl:
Molarity = number of moles / Volume in litter
Number of moles = Molarity × Volume in litter
Number of moles = 0.001 mol/L × 0.5 L
Number of moles = 0.0005 mol
Answer:
False
Explanation:
The urine is produced in the kidneys, each one of the kidneys is connected to a ureter. The ureter is a tube that propels the excreted urine to the urinary bladder. The urinary bladder is a deposit for the urine, here is collected and stored before disposal. After the urinary bladder, the urine goes through the urethra, which is a tube, to exit the body.
Answer:
34.3 g NH3
Explanation:
M(H2) = 2*1 = 2 g/mol
M(N2) = 2*14 = 28 g/mol
M(NH3) = 14 + 3*1 = 17 g/mol
23.6 g H2* 1 mol/2 g = 11.8 mol H2
28.3 g N2 * 1 mol/28 g = 1.01 mol N2
3H2 + N2 ------> 2NH3
from reaction 3 mol 1 mol
given 11.8 mol 1.01 mol
We can see that H2 is given in excess, N2 is limiting reactant.
3H2 + N2 ------> 2NH3
from reaction 1 mol 2 mol
given 1.01 mol x
x = 2*1.01/1= 2.02 mol NH3
2.02 mol * 17g/1 mol ≈ 34.3 g NH3
According to the reaction equation:
CH3COO- + H+ → CH3COOH
initial 0.25 0.15
change - 0.025 + 0.025
Equ (0.25-0.025) (0.15 + 0.025)
first, we have to get moles acetate and moles acetic acid:
moles of acetate = 0.25 - 0.025 = 0.225 moles
∴ [CH3COO-] = 0.225 mol / 1 L = 0.225 M
moles of acetic acid = 0.15 + 0.025 = 0.175 moles
∴ [ CH3COOH] = 0.175 mol / 1L = 0.175 M
Pka = -㏒ Ka
= -㏒ 1.8 x 10^-5
= 4.74
from H-H equation we can get the PH value:
PH = Pka + ㏒ [acetate / acetic acid]
PH = 4.74 + ㏒[0.225/0.175]
∴ PH = 4.8
