A) For balanced chemical equation: 2HgO(s) → 2Hg(l) + O₂(g).
1) Mole ratio 1: n(HgO) : n(Hg) = 2 : 2 (1 : 1).
2) Mole ratio 2: n(HgO) : n(O₂) = 2 : 1.
3) Mole ratio 3: n(Hg) : n(O₂) = 2 : 1.
B) Balanced chemical equation: 4NH₃(g) + 6NO(g) → 5N₂(g) + 6H₂O(l).
1) Mole ratio 1: n(NH₃) : n(NO) = 4 : 6 (2 : 3).
2) Mole ratio 2: n(NH₃) : n(N₂) = 4 : 5.
3) Mole ratio 3: n(NH₃) : n(H₂O) = 4 : 6 (2 : 3).
4) Mole ratio 4: n(NO) : n(N₂) = 6 : 5.
5) Mole ratio 5: n(NO) : n(H₂O) = 6 : 6 (1 :1).
6) Mole ratio 6: n(N₂) : n(H₂O) = 5 : 6.
Answer:
Theoretical yield of C6H10 = 3.2 g.
Explanation:
Defining Theoretical yield as the quantity of product obtained from the complete conversion of the limiting reactant in a chemical reaction. It can be expressed as grams or moles.
Equation of the reaction
C6H11OH --> C6H10 + H2O
Moles of C6H11OH:
Molar mass of C6H110H = (12*6) + (1*12) + 16
= 100 g/mol
Mass of C6H10 = 3.8 g
number of moles = mass/molar mass
=3.8/100
= 0.038 mol.
Using stoichoimetry, 1 moles of C6H110H was dehydrated to form 1 mole of C6H10 and 1 mole of water.
Therefore, 0.038 moles of C6H10 was produced.
Mass of C6H10 = molar mass * number of moles
Molar mass of C6H10 = (12*6) + (1*10)
= 82 g/mol.
Mass = 82 * 0.038
= 3.116 g of C6H10.
Theoretical yield of C6H10 = 3.2 g
Half life is the time taken for a radioactive isotope to decay by half its original mass. In this case the half life of carbon-14 is 5.730 years.
Using the formula;
New mass = original mass × (1/2)^n; where n is the number of half lives (in this case n=1 )
New mass = 2 g × (1/2)^1
= 1 g
Therefore; the mass of carbon-14 that remains will be 1 g
Answer:
Answers are bellow.
Explanation:
The element with electron configuration 1s22s22p63s1 belong I A group in the periodic table and it is sodium because it loses one electron.
We have periodic table in attachments.
Sodium is a chemical element with the symbol Na and atomic number 11. It is a soft, silvery-white, highly reactive metal. Sodium is an alkali metal, being in group 1 of the periodic table, because it has a single electron in its outer shell, which it readily donates, creating a positively charged ion—the Na+ cation. Its only stable isotope is 23Na.
The free metal does not occur in nature, and must be prepared from compounds.
It is soft metal, reactive and with a low melting point.
