Explanation:
Mg(s) + Cr(C2H3O2)3 (aq)
Overall, balanced molecular equation
Mg(s) + Cr(C2H3O2)3(aq) --> Mg(C2H3O2)3(aq) + Cr(s)
To identify if an element has been reduced or oxidized, the oxidation number is observed in both the reactant and product phase.
An increase in oxidation number denotes that the element has been oxidized.
A decrease in oxidation number denotes that the element has been reduced.
Oxidation number of Mg:
Reactant - 0
Product - +3
Oxidation number of Cr:
Reactant - +3
Product - 0
Note: C2H3O2 is actually acetate ion; CH3COO- The oxidatioon number of C, H and O do not change.
Oxidized : Mg
Reduced : Cr
Answer:
464.1 J absorbed.
Explanation:
Given data:
Specific heat of zinc = 0.39 J/g°C
Mass of zinc = 34 g
Temperature changes = 22°C to 57°C
Energy absorbed or released = ?
Solution:
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = 57°C - 22°C
ΔT = 35°C
Q = m.c. ΔT
Q = 34 g. 0.39 J/g°C. 35°C
Q = 464.1 J
Answer:
Mass of NH₃ produced = 34 g
Explanation:
Given data:
Mass of nitrogen = 28 g
Mass of Hydrogen = 12 g
Mass of NH₃ produced = ?
Solution:
Chemical equation:
N₂ + 3H₂ → 2NH₃
Moles of nitrogen:
Number of moles = mass/molar mass
Number of moles = 28 g/ 28 g/mol
Number of moles = 1 mol
Moles of hydrogen:
Number of moles = mass/molar mass
Number of moles = 12 g/ 2 g/mol
Number of moles = 6 mol
Now we will compare the moles of hydrogen and nitrogen with ammonia.
H₂ : NH₃
3 : 2
6 : 2/3×6 = 4 mol
N₂ : NH₃
1 : 2
Number of moles of ammonia produced by nitrogen are less thus it will act as limiting reactant.
Mass of ammonia produced:
Mass = number of moles × molar mass
Mass = 2 mol × 17 g/mol
Mass = 34 g
<h3>
Answer:</h3>
0.35 M
<h3>
Explanation:</h3>
<u>We are given;</u>
- Initial volume as 35.0 mL or 0.035 L
- Initial molarity as 12.0 M
- Final volume is 1.20 L
We are required to determine the final molarity of the solution;
- Dilution involves adding solvent to a solution to make it more dilute which reduces the concentration and increases the solvent while maintaining solute constant.
- Using dilution formula we can determine the final molarity.
M1V1 = M2V2
M2 = M1V1 ÷ V2
= (12.0 M × 0.035 L) ÷ 1.2 L
= 0.35 M
Thus, the final concentration of the solution is 0.35 M
