Answer:
Mode
Explanation:
The mode is the number that appears most frequently in a data set. A set of numbers may have one mode, more than one mode, or no mode at all. Other popular measures of central tendency include the mean, or the average (mean) of a set, and the median, the middle value in a set.
<span>Si is reduced/Oxidizing reagent Mg is oxidized/reducing reagent</span>
Explanation:
a. CGS system: centimetre as the unit of length, the gram as the unit of mass, and the second as the unit of time
b. SI system: metre as the unit of length, the kilogram as the unit of mass, and the second as the unit of time
Answer : 21.12 g
explanation :
- A limiting reactant is the substance that is totally consumed when the chemical reaction is complete and the reaction cannot continue without it.
- This is a limiting reactant problem because the amount of product (CO2) formed is limited by this substance (either O2 or C3H8).
- We can predict the limiting reactant by calculating number of mole for all reactants;
n (C3H8) = mass/ molar mass = 7/[(12 × 3)+(1 ×8)] = 0.16 mol
n (O2) = mass/ molar mass = 98/(16 × 2) = 3 mol
But we know from this balanced equation that for the reaction to continue, for (n) of propane there must be (5n) of oxygen. Clearly we have more oxygen than required for the reaction to continue as
[ 3 O2 mol > (0.16 propane mol × 5) . Hence, the limiting reactant is propane.
- Using cross multiplication,
0.16 mol propane → 1 mol propane
? → 3 mol CO2
So (n) of CO2 produced = 0.16 × 3 / 1 = 0.48 mol ,
And mass of CO2 produced = n × molar mass = 0.48 × [12 + (16 ×2)] = 21.12 g..
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
