The crust
1. divergent (moves away from each other)
2. convergent (moves towards each other)
3. transform (slides past each other)
<h3>
Answer:</h3>
1.93 g
<h3>
Explanation:</h3>
<u>We are given;</u>
The chemical equation;
2C₂H₆(g) + 7O₂(g) → 4CO₂(g) + 6H₂O(l) ΔH = -3120 kJ
We are required to calculate the mass of ethane that would produce 100 kJ of heat.
- 2 moles of ethane burns to produce 3120 Kilo joules of heat
Number of moles that will produce 100 kJ will be;
= (2 × 100 kJ) ÷ 3120 kJ)
= 0.0641 moles
- But, molar mass of ethane is 30.07 g/mol
Therefore;
Mass of ethane = 0.0641 moles × 30.07 g/mol
= 1.927 g
= 1.93 g
Thus, the mass of ethane that would produce 100 kJ of heat is 1.93 g
The reaction between mercury (Hg) and sulfur (S) to form HgS is:
Hg + S ------------- HgS
Therefore: 1 mole of Hg reacts with 1 mole of S to form 1 mole of HgS
The given mass of Hg = 246 g
Atomic mass of Hg = 200.59 g/mol
# moles of Hg = 246 g/ 200.59 gmol-1 = 1.226 moles
Based on the reaction stoichiometry,
# moles of S that would react = 1.226 moles
Atomic mass of S = 32 g/mol
Therefore, mass of S = 1.226 moles*32 g/mole = 39.23 g
39.2 g of sulfur would be needed to react completely with 246 g of Hg to produce HgS
<h3>Answer:</h3>
Limiting reactant is Lithium
<h3>
Explanation:</h3>
<u>We are given;</u>
- Mass of Lithium as 1.50 g
- Mass of nitrogen is 1.50 g
We are required to determine the rate limiting reagent.
- First, we write the balanced equation for the reaction
6Li(s) + N₂(g) → 2Li₃N
From the equation, 6 moles of Lithium reacts with 1 mole of nitrogen.
- Second, we determine moles of Lithium and nitrogen given.
Moles = Mass ÷ Molar mass
Moles of Lithium
Molar mass of Li = 6.941 g/mol
Moles of Li = 1.50 g ÷ 6.941 g/mol
= 0.216 moles
Moles of nitrogen gas
Molar mass of Nitrogen gas is 28.0 g/mol
Moles of nitrogen gas = 1.50 g ÷ 28.0 g/mol
= 0.054 moles
- According to the equation, 6 moles of Lithium reacts with 1 mole of nitrogen.
- Therefore, 0.216 moles of lithium will require 0.036 moles (0.216 moles ÷6) of nitrogen gas.
- On the other hand, 0.054 moles of nitrogen, would require 0.324 moles of Lithium.
Thus, Lithium is the limiting reagent while nitrogen is in excess.
The metallic elements located anywhere between 3-12 of the periodic table
