<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.
I think it might be lead? I'm not quite too sure though. You might have to research more into it.
Chromium (Cr): 2, 8, 8, 6
Copper (Cu): 2, 8, 8, 11
Chromium+2 (Cr+2): 2, 8, 8, 4 (if the ion retains a positive charge then the amount of electrons will decrease)
Copper+2 (Cu+2): 2, 8, 8, 9
Manganese+2 (Mn+2): 2, 8, 8, 5
Option A. Health claim
Health claim must be supported by scientifically valid evidence to be approved in using for food label.
This is also a food guide to all consumers the food content implied by manufacturers to their products.
