Answer:
I believe your answer would be grams.
The given example is a chemical reaction.
The contents (separated as reactants and products) :

The written reaction is :

<em>I hope it helped you solve the problem.</em>
<em>Good luck on your studies!</em>
Answer:
C Group 11
Explanation:
Group 11, by modern IUPAC numbering, is a group of chemical elements in the periodic table, consisting of copper (Cu), silver (Ag), and gold (Au).
These elements show highest electrical conductivity.
Answer:
<h3>25.0 grams is the mass of the steel bar.</h3>
Explanation:
Heat gained by steel bar will be equal to heat lost by the water

Mass of steel=
Specific heat capacity of steel =
Initial temperature of the steel = 
Final temperature of the steel = 

Mass of water= 
Specific heat capacity of water=
Initial temperature of the water = 
Final temperature of water = 

On substituting all values:

<h3>25.0 grams is the mass of the steel bar.</h3>
increases my factor of 10