_Mg + _HCL = _MgCl2 + H2
Separate the terms on each side:
_Mg + _HCl = _MgCl2 + H2
Mg- 1 Mg-1
H-1 H-2
Cl-1 Cl-2
Mg is balanced on both sides so move on to the next (put a 1 in the space).
1Mg
There are two H's and two Cl's on the results side, so to balance the equation put a 2 as a coefficient for HCl and it'll all balance out.
2HCl
Balamced equation will be:
1Mg + 2HCL = 1MgCl2 + H2
Answer:
C6H12O6 —> 2C2H5OH + 2CO2
Explanation:
The equation for the reaction is given below:
C6H12O6 —> C2H5OH + CO2
We can balance the equation above as follow:
There are 12 atoms of H on the left side and 6 atoms of the right side. It can be balance by putting 2 in front of C2H5OH as shown below:
C6H12O6 —> 2C2H5OH + CO2
There are 6 atoms of C on the left side and 5 atoms on the right side. It can be balance by putting 2 in front of CO2 as shown below:
C6H12O6 —> 2C2H5OH + 2CO2
Now the equation is balanced.
Answer:
Rate = k [OCl] [I]
Explanation:
OCI+r → or +CI
Experiment [OCI] M I(-M) Rate (M/s)2
1 3.48 x 10-3 5.05 x 10-3 1.34 x 10-3
2 3.48 x 10-3 1.01 x 10-2 2.68 x 10-3
3 6.97 x 10-3 5.05 x 10-3 2.68 x 10-3
4 6.97 x 10-3 1.01 x 10-2 5.36 x 10-3
The table above able shows how the rate of the reaction is affected by changes in concentrations of the reactants.
In experiments 1 and 3, the conc of iodine is constant, however the rate is doubled and so is the conc of OCl. This means that the reaction is in first order with OCl.
In experiments 3 and 4, the conc of OCl is constant, however the rate is doubled and so is the conc of lodine. This means that the reaction is in first order with I.
The rate law is given as;
Rate = k [OCl] [I]
All are CORRECT except (d)
