Answer:
its 0.163 g
Explanation:
From the total pressure and the vapour pressure of water we can calculate the partial pressure of O2
PO 2 =P t −P H 2 O
= 760 − 22.4
= 737.6 mmHg
From the ideal gas equation we write.
W= RT/PVM = (0.0821Latm/Kmol)(273+24)K(0.974atm)(0.128L)(32.0g/mol/) =0.163g
KH₂PO₄ hydrolyzes as;
H₂PO₄⁻ + H₂O ↔ H₃PO₄ + OH⁻
Let x amount of H₂PO₄⁻ has reacted with water then,
Kb₁ = [H₃PO₄][OH⁻] / [H₂PO₄⁻]
[H₂PO₄⁻] = 0.8-x M
Kb₁ = x² / (0.8 - x)
Given Ka₁ = 7.5 x 10⁻³
so Kb₁ = 1 x 10⁻¹⁴ / (7.5 x 10⁻³) = 1.33 x 10⁻¹²
From this information:
1.33 x 10⁻¹² = x² / 0.8
x = [OH⁻] = 1.03 x 10⁻⁶ M
pOH = - log (1.03 x 10⁻⁶) = 5.99
pH = 14 - pOH = 14 - 5.99 = 8.01
Answer:
Explanation:
Bromine >Tellurium > Phosphorus > Helium > Sodium
Electron affinity of Bromine , Tellurium , Phosphorus are positive , of helium is zero and of sodium is negative .
Answer:
it must be testable I think that's the answer
We will assume that the solvent is water. So, if we have 100 grams of the solution, 19 grams will be sodium hydroxide, while the remaining 81 grams will be water.
The molar weight of sodium hydroxide, NaOH, is 40. The molar weight of water is 18. Finding the moles of each:
NaOH:
19 / 40 = 0.475
Water:
81 / 18 = 4.5
Total moles present:
4.5 + 0.475 = 4.975 moles
The mole fraction of NaOH is:
0.475 / 4.975 = 0.0955
The mole fraction of NaOH is 0.0955
