Answer:
They are called coefficients.
Answer:
it has six significant figures
<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.
Air on a hot day translates to hot air and air on a cold day translates to cold air. The hot air usually have more energy due to particles being more excited by the heat.
D) otherwise we wouldn't be able to have water, oxygen gas, and all other compounds that have oxygen. We need at least 1 free slot to make a molecule
