Answer:
Explanation:
Hello,
In this case, the undergoing chemical reaction is:
Thus, given the 1.00-kg of 12.5% ore, we can compute the theoretical yield of copper (II) oxide via stoichiometry:
Whereas the third factor accounts for the percent purity of the covellite. Then, given the percent yield, we can compute the actual yield by:
Regards.
Answer:
108.9g of Silver can be produced from 125g of Ag2S
Explanation:
The compound Ag2S shows that two atoms of Silver Ag, combined with an atom of Sulphur S to form Ag2S. We can as well say the combination ration of Silver to Sulphur is 2:1
•Now we need to calculate the molecular weight of this compound by summing up the molar masses of each element in the compound.
•Molar mass of Silver Ag= 107.9g/mol
•Molar mass of Sulphur S= 32g/mol
•Molecular weight of Ag2S= (2×107.9g/mol) + 32g/mol
•Molecular weight of Ag2S= 215.8g/mol + 32g/mol= 247.8g/mol
•From our calculations, we know that 215.8g/mol of Ag is present in 247.8g/mol of Ag2S
If 247.8g Ag2S produced 215.8g Ag
125g Ag2S will produce xg Ag
cross multiplying we have
xg= 215.8g × 125g / 247.8g
xg= 26975g/247.8
xg= 108.85g
Therefore, 108.9g of Silver can be produced from 125g of Ag2S
Answer:
16.0%.
Explanation:
Volume percent of a substance is the ratio of the substance volume to the solution volume multiplied by 100.
V % of ethanol = (volume of ethanol / volume of the solution) x 100.
volume of ethanol = 90.0 mL, volume of the solution = 550.0 mL.
∴ V % of ethanol = (90.0 mL / 550.0 mL) x 100 = 16.36% ≅ 16.0%.
Answer:
Distance = 200 km
Distance = 204 km
Speed = 77 km/h
Time = 21.42 h
Explanation:
Given:
A.
Speed = 100 km/h , Time = 2 h
Find:
Distance
B.
Speed = 68 km/h , Time = 3 h
Find:
Distance
C.
Distance = 154 km , Time = 2 h
Find:
Speed
D.
Distance = 1500 km speed = 70 km/h
Find:
Time
Computation:
Speed = distance / time
A.
Distance = 100 x 2
Distance = 200 km
B.
Distance = 68 x 3
Distance = 204 km
C.
Speed = 154 / 2
Speed = 77 km/h
D.
Time = 1500 / 70
Time = 21.42 h
<span>70.4 mg CO2 x 1.0 g /1000 mg x 1 mole CO2/ 44 gCO2 x 1 mole C/1 mole CO2 = 0.0016 moles C
14.4 mg H2O x 1.0 g/1000 mg x 1 mole H2O/18 g H2O x 2 moles H/ 1 mole H2O = 0.0016 moles O
molar mass of C=12 g/mole
molar mass of H=1 g/mole
0.0016 moles C x 12 g C/ 1 mole C = 0.0192 g C or 19.2 mg C
0.00156 moles H x 1 g H/1 mole H = 0.00156 g H or 1.56 mg H
mg O= 30.4 mg vanillin - 19.2 mg C – 1.56 mg H = 9.64 mg O
molar mass of O=16 g/mole
9.64 mg O x 1 g/1000 mg x 1 mole O/16.0 g = 0.000602
C.0016 H.0016 O.000602; divide all the moles by the smallest value of0.000602
C2.66H2.66O1 is the empirical formula;
to obtain whole numbers multiply by 3
3[C2.66H2.66O1] = C8H8O3
above formula weight: 8(C) + 8(H) + 3(O) = 8(12) + 8(1) + 3(16) = 152 amu
The empirical formula weight and the molecular formula weight are the same .
Molecular formula is C8H8O3.</span>
