Answer:
Al + MnO4- + 2H2O → Al(OH)4- + MnO2
Explanation:
First of all, we out down the skeleton equation;
Al + MnO4- → MnO2 + Al(OH)4-
Secondly, we write the oxidation and reduction equation in basic medium;
Oxidation half equation:Al + 4H2O + 4OH- → Al(OH)4- + 4H2O + 3e-
Reduction half equation:MnO4- + 4H2O + 3e- → MnO2 + 2H2O + 4OH-
Thirdly, we add the two half reactions together to obtain:
Al + MnO4- + 8H2O + 4OH- + 3e- → Al(OH)4- + MnO2 + 6H2O + 3e- + 4OH-
Lastly, cancel out species that occur on both sides of the reaction equation;
Al + MnO4- + 8H2O→ Al(OH)4- + MnO2 + 6H2O
The simplified equation now becomes;
Al + MnO4- + 2H2O → Al(OH)4- + MnO2
Do you have a picture of the graph?
Answer:

Explanation:
Given that,
Mass of sample, m = 996.9 g
The change in temperature of the sample, 
Heat produced, Q = 62.9 calories = 263173.6 J
The heat released by a sample due to change in temperature is given by :

Where
c is the specific heat capacity
So,

So, the specific heat of ethanol is equal to
.
1 mole C₆H₁₂O₆ ------------- 6 moles oxygen
3 moles C₆H₁₂O₆ ----------- X
X = (3×6)/1
X = 18 moles
Answer:
1.5
Explanation:
Given that :
Compound A and B are formed from Sulfur + Oxygen.
Compound A :
6g sulfur + 5.99g Oxygen
Compound B:
8.6g sulfur + 12.88g oxygen
Comparing the ratios :
Compound A:
S : O = 6.00 : 5.99
S/0 = 6.0g S / 5.99g O
Compound B :
S : O = 8.60 : 12.88
S / O = 8.60g S / 12.88g O
Mass Ratio of A / mass Ratio of B
(6.0g S / 5.99g O) ÷ (8.60g S / 12.88g O)
(6.0 g S / 5.99g O) × (12.88g O / 8.60g S)
(6 × 12.88) / (5.99 × 8.60)
= 77.28 / 51.514
= 1.50017
= 1.5