Missing question: What is the rate constant for the reaction?
<span>[RS2](mol L-1) Rate (mol/(L·s))
0.150 0.0394
0.250 0.109
0.350 0.214
0.500 0.438</span>
Chemical reaction: 3RS₂ → 3R + 6S.
Compare second and fourth experiment, when concentration is doubled, rate of concentration is increaced by four. So rate is:
rate = k·[RS₂]².
k = 0,438 ÷ (0,500)².
k = 1,75 L/mol·s.
Answer:
E. Q < K and reaction shifts right
Explanation:
Step 1: Write the balanced equation
A(s) + 3 B(l) ⇄ 2(aq) + D(aq)
Step 2: Calculate the reaction quotient (Q)
The reaction quotient, as the equilibrium constant (K), only includes aqueous and gaseous species.
Q = [C]² × [D]
Q = 0.64² × 0.38
Q = 0.15
Step 3: Compare Q with K and determine in which direction will shift the reaction
Since Q < K, the reaction will shift to the right to attain the equilibrium.
Answer:
i think the answer is c hoped this helped
Explanation:
1) 0.89% m/v = 0.89 grams of NaCl / 100 ml of solution
=> 8.9 grams of NaCl in 1000 ml of solution = 8.9 grams of NaCl in 1 liter of solution
2) Molarity = M = number of moles of solute / liters of solution
=> calculate the number of moles of 8.9 grams of NaCl
3) molar mass of NaCl = 23.0 g /mol + 35.5 g/mol = 58.5 g / mol
4) number of moles of NaCl = mass / molar mass = 8.9 g / 58.5 g / mol = 0.152 mol
5) M = 0.152 mol NaCl / 1 liter solution = 0.152 M
Answer: 0.152 M
