Answer:
The substance that remained on the filter paper is Al(OH).
Explanation:
- Filter paper is the substance that is used in laboratory to separate the solid objects. It doesn't filter aqueous and gaseous products.
- So in our experiment the product formed is aluminum hydroxide Al(OH) and sodium chloride (NaCl).
- Between two products, sodium chloride is in aqueous form as indicated in the question. So it wont remain in the filter paper.
- Hence aluminum hydroxide being only solid product remains on the filter paper.
I think the first one applys. I may be wrong.
Answer:
k = 100 mol⁻² L² s⁻¹, r= k[A][B]²
Explanation:
A + B + C --> D
[A] [B] [C] IRR
0.20 0.10 0.40 .20
0.40 0.20 0.20 1.60
0.20 0.10 0.20 .20
0.20 0.20 0.20 .80
Comparing the third and fourth reaction, the concentrations of A and C are constant. Doubling the concentration of B causes a change in the rate of the reaction by a factor of 4.
This means the rate of reaction is second order with respect to B.
Comparing reactions 2 and 3, the concentrations of B and C are constant. Halving the concentration of A causes a change in the rate of the reaction by a factor of 2.
This means the rate of reaction is first order with respect to A.
Comparing reactions 1 and 3, the concentrations of A and B are constant. Halving the concentration of A causes no change in the rate of the reaction.
This means the rate of reaction is zero order with respect to C.
The rate expression for this reaction is given as;
r = k [A]¹[B]²[C]⁰
r= k[A][B]²
In order to obtain the value of the rate constant, let's work with the first reaction.
r = 0.20
[A] = 0.20 [B] = 0.10
k = r / [A][B]²
k = 0.20 / (0.20)(0.10)²
k = 100 mol⁻² L² s⁻¹
It would be 12.4 moles of HCL because you multiply 6.2 by how many moles of HCL in te equation then divide it by how many moles of h2 in the equation so you multiple 6.2 by 2 moles because there’s two moles of HCL which gives you 12.4 then you divide it by one because in the equation there is only 1 mole of H2
