18. c
19. A
20. decomposition reaction
21. 4
22. 0.04%
Answer:
0.74 moles of oxygen are required for combustion
Explanation:
Given data:
Number of moles of oxygen required = ?
Mass of Mg = 35.5 g
Solution:
Chemical equation:
2Mg + O₂ → 2MgO
Number of moles of Mg:
Number of moles = mass/molar mass
Number of moles = 35.5 g/ 24 g/mol
Number of moles = 1.48 mol
Now we will compare the moles of Mg with oxygen.
Mg : O₂
2 : 1
1.48 : 1/2×1.48 = 0.74 mol
0.74 moles of oxygen are required for combustion.
Answer:
Volume O₂ at STP = 50.4 Liters
Explanation:
4Al(s) + 3O₂(g) => 2Al₂O₃(s) at STP conditions
81g Al(s) = 81g/27g/mole = 3mole Al
moles O₂ consumed = 4/3(3)moles O₂ = 2.25 moles O₂ consumed
Volume O₂ at STP = 2.25moles x 22.4L/mole = 50.4 Liters O₂
Answer:
A-10
Explanation:
In the SI, designations of multiples and subdivision of any unit may be arrived at by combining with the name of the unit the prefixes deka, hecto, and kilo meaning, respectively, 10, 100, and 1000, and deci, centi, and milli, meaning, respectively, one-tenth, one-hundredth, and one-thousandth.
IM NOT SURE PO
Answer: 7.6 g of water are required to make 12.6 g of glucose
Explanation:
Moles is calculated by using the formula:
The balanced chemical equation is:
According to stoichiometry:
1 mole of glucose is produced by = 6 moles of water
Thus 0.07 moles of glucose is produced by = moles of water
Mass of water required =
Thus 7.6 g of water are required to make 12.6 g of glucose