Answer:
1. 15.71 g CO2
2. 38.19 % of efficiency
Explanation:
According to the balanced reaction (2 CO(g) + O2(g) → 2 CO2(g)), it is clear that the CO is the limitant reagent, because for every 2 moles of CO we are using only 1 mole of O2, so even if we have the same quantity for both reagents, not all of the O2 will be consumed. This means that we can just use the stoichiometric ratios of the CO and the CO2 to solve this question, and for that we need to convert the gram units into moles:
For CO:
C = 12.01 g/mol
O = 16 g/mol
CO = 28.01 g/mol
(10.0g CO) x (1 mol CO/28.01 g) = 0.3570 mol CO
For CO2:
C = 12.01 g/mol
O = 16 x 2 = 32 g/mol
CO2 = 44.01 g/mol
We now that for every 2 moles of CO we are going to get 2 moles of CO2, so we resolve as follows:
(0.3570 mol CO) x (2 mol CO2/2 mol CO) = 0.3570 moles CO2
We are obtaining 0.3570 moles of CO2 with the 10g of CO, now lets convert the CO2 moles into grams:
(0.3570 moles CO2) x (44.01 g/1 mol CO2) = 15.71 g CO2
Now for the efficiency question:
From the previous result, we know that if we produce 15.71 CO2 with all the 10g of CO used, we would have an efficiency of 100%. So to know what would that efficiency be if we would only produce 6g of CO2, we resolve as follows,
(6g / 15.71g) x 100 = 38.19 % of efficiency
<u>Given:</u>
Concentration of Cr2+ = 0.892 M
Concentration of Fe2+ = 0.0150 M
<u>To determine:</u>
The cell potential, Ecell
<u>Explanation:</u>
The half cell reactions for the given cell are:
Anode: Oxidation
Cr(s) ↔ Cr2+(aq) + 2e⁻ E⁰ = -0.91 V
Cathode: Reduction
Fe2+ (aq) + 2e⁻ ↔ Fe (s) E⁰ = -0.44 V
------------------------------------------
Net reaction: Cr(s) + Fe2+(aq) ↔ Cr2+(aq) + Fe(s)
E°cell = E°cathode - E°anode = -0.44 - (-0.91) = 0.47 V
The cell potential can be deduced from the Nernst equation as follows:
Ecell = E°cell - (0.0591/n)log[Cr2+]/[Fe2+]
Here, n = number of electrons = 2
Ecell = 0.47 - 0.0591/2 * log[0.892]/[0.0150] = 0.418 V
Ans: The cell potential is 0.418 V
Answer:
Molarity of Na₂CO₃ = 0.25M
% mass = 2.69
Explanation:
Molarity means mole of solute in 1L of solution
Molar mass of solute (Na₂CO₃) = 105,98 g/m
Moles = mass / molar mass → 6.73 g / 105.98 g/m = 0.0635 m
Mol/L = [M]
0.0635 mol/0.250L = 0.25M
Density of solution = Solution mass / Solution volume
1 g/ml = Solution mass / 250 mL → Solution mass is 250g
% mass will be:
In 250 g of solution we have 6.73 g of solute
in 100 g of solution we have (100 . 6.73)/250 = 2.69
Answer:
i dont know, i wish i could help
Explanation:
1.Products of most combustion reactions are:
Carbon dioxide and water
2.Alloys are made by melting and combining metals together.
Chemical changes involve electrons,in this process no electron is used/shared or given.
Thus,its not a chemical change.
