Answer:
[HF]₀ = 0.125M
Explanation:
NaOH + HF => NaF + H₂O
Adding 20ml of 0.200M NaOH into 25ml of HF solution neutralizes 0.004 mole of HF leaving 0.004 mole NaF in 0.045L with 0.001M H⁺ at pH = 3. This is 0.089M NaF and 0.001M HF remaining.
=> 45ml of solution with pH = 3 and contains 0.089M NaF from titration becomes a common ion problem.
HF ⇄ H⁺ + F⁻
C(eq) [HF] 10⁻³M 0.089M (<= soln after adding 20ml 0.200M NaOH)
Ka = [H⁺][F⁻]/[HF]₀ => [HF]₀ = [H⁺][F⁻]/Ka
[HF]₀ = (0.001)(0.089)/(7.1 x 10⁻⁴) M = 0.125M
Ice has a lower density than the density of water.
Answer:
The answer is SiO2
Explanation:
Silocon dioxide is written without a 1 after the silocon and with a 2 after the oxygen.
Answer:
2.77 mol N
Explanation:
M(N2O) = 2*14 + 16 = 44 g/mol
61.0 g * 1 mol/44g = (61/44) mol N2O
N2O ---- 2N
1 mol 2 mol
(61/44) mol x mol
x = (61/44)*2/1 = 2.77 mol N
