<span>2.28 kg x 213 kJ/kg = 486 kJ = 4.86E+05 J</span>
<u>Answer:</u> The amount of heat required to warm given amount of water is 470.9 kJ
<u>Explanation:</u>
To calculate the mass of water, we use the equation:

Density of water = 1 g/mL
Volume of water = 1.50 L = 1500 mL (Conversion factor: 1 L = 1000 mL)
Putting values in above equation, we get:

To calculate the heat absorbed by the water, we use the equation:

where,
q = heat absorbed
m = mass of water = 1500 g
c = heat capacity of water = 4.186 J/g°C
= change in temperature = 
Putting values in above equation, we get:

Hence, the amount of heat required to warm given amount of water is 470.9 kJ
Given :
A 250 ml beaker weighs 13.473 g .
The same beaker plus 2.2 ml of water weighs 15.346 g.
To Find :
How much does the 2.2 ml of water, alone, weigh .
Solution :
Now, mass of water is given by :

Therefore , mass of 2.2 ml of water alone is 1.873 g .
Hence , this is the required solution .
<h2>
Answer:</h2>
Average atomic mass of an element is the sum of the masses of its isotopes each multiplied by its natural abundance




