Answer:
(a) rate = 4.82 x 10⁻³s⁻¹ [N2O5]
(b) rate = 1.16 x 10⁻⁴ M/s
(c) rate = 2.32 x 10⁻⁴ M/s
(d) rate = 5.80 x 10⁻⁵ M/s
Explanation:
We are told the rate law is first order in N₂O₅, and its rate constant is 4.82 x 10⁻³s⁻¹ . This means the rate is proportional to the molar concentration of N₂O₅, so
(a) rate = k [N2O5] = 4.82 x 10⁻³s⁻¹ x [N2O5]
(b) rate = 4.82×10⁻³s⁻¹ x 0.0240 M = 1.16 x 10⁻⁴ M/s
(c) Since the reaction is first order if the concentration of N₂O₅ is double the rate will double too: 2 x 1.16 x 10⁻⁴ M/s = 2.32 x 10⁻⁴ M/s
(d) Again since the reaction is halved to 0.0120 M, the rate will be halved to
1.16 x 10⁻⁴ M/s / 2 = 5.80 x 10⁻⁵ M/s
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I believe it is 6ml because you do the doseage times the ml and mutiply it by 1
Answer: m = 50 g ZnSO4
Explanation: First is convert the moles of Zn to the moles of ZnSO4 by having their mole ratio which is 2:2 based from the balanced equation. Next is convert the moles of ZnSO4 to mass using its molar mass.
0.311 mole Zn x 2 moles ZnSO4 / 2 moles Zn
= 0.311 moles ZnSO4
0.311 moles ZnSO4 x 161 g ZnSO4 / 1 mole ZnSO4
= 50 ZnSO4
Product are favored at ...
example 2A + 3B = 5C + D if reaction is exothermic ....delta H = NEGATIVE)
A decrease in temperature favors the forward reaction (more product formed)
as heat is considered as a product
For the exam I have given: A decrease in pressure (volume increases as pressure decreases) .... there are more number of moles on the product side (6 in all) .....so according to LCP decreasing pressure will revert back to increase pressure and to do that equilibrium position shifts to the right (product)
decreasing concentration of product will cause a disturbance in equilibrium position....and reaction will restore its equilibrium by shifting to the right
(I believe its like this)
