<h3>
Answer:</h3>
= 19.712 kJoules
<h3>
Explanation:</h3>
- Heat of vaporization refers to the amount of heat required to change a unit mass of a substance from liquid to gaseous state without change in temperature.
To calculate the amount of heat, we use,
Amount of heat = Mass × Heat of vaporization
Q = m×Lv
Given;
Mass of liquid Zinc = 11.2 g
Lv of liquid Zinc = 1.76 kJ/g
Therefore;
Q = 11.2 g × 1.76 kJ/g
= 19.712 kJ
Thus, the amount of heat needed to boil 11.2 g of zinc is 19.712 kilo-joules.
50/35 = √x/ √31.9988 g
50 x (√31.9988/ 35) = √x
√x = 8.081...
x= 65.30...
x= 70 g/ mol
Answer:
Option (2) At equilibrium, there is a much higher concentration of products than reactants.
Explanation:
The equilibrium constant for a reaction is simply the ratio of the concentration products raised to their mole ratio divided by the concentration of the reactants raised to their mole ratio.
If the equilibrium constant is close to 1 or 1, it means the concentration of the reactants and products are almost the same. But if the equilibrium constant is large as in the case of the question given above, it means that at equilibrium, the concentration of the products are higher than that of the reactants.
Answer:
8 to 1.
Explanation:
- Oxygen combines with hydrogen atoms to form water according to the balanced equation:
<em>O₂ + 2H₂ → 2H₂O.</em>
It is clear that one mole of oxygen combines with two moles of hydrogen atoms to form 2 moles of water.
So, the molar ratio of oxygen to hydrogen is (1 to 2).
- The mass of 1 mole of oxygen = (no. of moles)(molar mass) = (1 mol)(32.0 g/mol) = 32.0 g.
- The mass of 2 moles of hydrogen = (no. of moles)(molar mass) = (2 mol)(2.0 g/mol) = 4.0 g.
<em>So, the mass ratio of oxygen to hydrogen (32.0 g/4.0 g) = (8: 1).</em>
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