Answer:
CO(g) + H2(g) + H2O(g) ==> CO2(g) + 2H2(g)
Explanation:
In the industry, hydrogen is prepared from water and hydrocarbons. Water gas being the major method of preparation of hydrogen industrially.
The water-gas reaction is an industrial process in which steam is passed over red-hot coke giving a gaseous mixture of carbon monoxide and hydrogen:
C + H2O(g) → CO + H2.
The mixture of CO and H2 is Futher passed through steam according to the equation:
CO(g) + H2(g) + H2O(g) ==> CO2(g) + 2H2(g) to give hydrogen and carbon dioxide.
Answer:
B: Adding water, then adding solute
Explanation:
This is because, say you have a solution with a certain concentration.
If you add more water, it will become more diluted (less concentrated)
If you add more solute, it will become more concentrated.
Therefore if you add water and solute, it could cancel out, and the concentration would remain the same.
Hope this helps! Let me know if you have any questions/ would like anything further explained :)
That is correct…….. i think
Answer: 147 mL
Explanation:
<u>Given:</u>
Molarity of the sodium bromide (NaBr) solution (M1) = 1.75 M
Volume of the solution (V1) = 84 mL
Molarity of the diluted NaBr solution (M2) = 1 M
Using the dilution formula to solve for V2:
Therefore, the new volume of the solution is 147 mL
Missing question: What is the vapor pressure of the solution at 25°<span>C?
n(NaCl) = 100 g </span>÷ 58,4 g/mol.
n(NaCl) = 1,71 mol.
NaCl → Na⁺ + Cl⁻, amount of ions are 2 · 1,71 mol = 3,42 mol.
n(CaCl₂) = 100 g ÷ 111 g/mol = 0,9 mol.
CaCl₂ → Ca²⁺ + 2Cl⁻, amount of ions 3 · 0,9 mol = 2,7 mol.
m(solution) = 1000 ml (1,00 L) · 1,15 g/ml = 1150 g.
m(H₂O) = 1150 g - 100 g - 100 g = 950 g.
n(H₂O) = 950 g ÷ 18 g/mol = 118,75 mol.
<span>water's mole fraction = 118,75 mol </span>÷ (118,75 mol + 2,7 mol + 3,42 mol).
water's mole fraction = 0,95.
p(solution) = 0,95 · 23 mmHg = 21,85 mmHg.
