Answer: (a) K *[A][B]^2
(b) The answer is B
Explanation:
A)
Step1:A+B<--> C (fast)
Step2: B+C→D(slow)
Rate depends on slowest step.
so,
rate = k' [B][C] ...eqn 1
But C is intermediate.so use step 1
Since 1st step is an equilibrium,
Kc = [C] /[A][B]
so,
[C] = Kc [A][B] ...eqn 2
put eqn 2 in eqn 1
rate = k' *[B] * Kc [A][B]
= k'Kc*[A][B]^2
= K *[A][B]^2 {writing k'Kc = K}
Answer: K *[A][B]^2
B)
Answer is B
Since rate depends on slowest step.
if slowest step is:
X2Y2+Z2→X2Y2Z+Z
then only,
rate= k[X2Y2][Z2]
Answer: B
They are the elements from scandium to copernicium. The elements between group 2 and group 3
Use a proportion ...
<span>100.0g - 38.67g - 13.86g = 47.47g Oxygen </span>
<span>285.0 mg = 0.285g </span>
<span>47.47/100 = x/0.285g </span>
<span>x = ( 47.47/100) X 0.285g </span>
The two ways to make a saturated solution are 1 reducing the temperature of the solution and Adding more solute.<span> </span>
Answer:
It is equal to the number of moles of acid that reacted. When Oxalic acid is your limiting reactant it is the # of moles of oxalic acid used. When NaOH is your limiting reactant it is equal to the number of moles of NaOH used.
