Answer : The specific heat of the metal is,
Explanation
In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.
where,
= specific heat of metal = ?
= specific heat of water =
= mass of metal = 95.0g
= mass of water = 50.0 g
= final temperature of mixture =
= initial temperature of metal =
= initial temperature of water =
Now put all the given values in the above formula, we get
Therefore, the specific heat of the metal is,
Answer:
3.2 x 10²¹molecules
Explanation:
Given parameters:
Volume of nitrogen gas = 120cm³
Unknown:
Mass of nitrogen gas = ?
Number of molecules = ?
Solution:
To solve this problem, note that;
1 mole of gas occupies a volume of 22.4L at STP
Now;
convert 120cm³ to L ;
1000cm³ = 1L
120cm³ gives 0.12L
Since;
22.4L of gas has 1 mole at STP
0.12L of gas will have = 0.0054mole at STP
So;
Mass of N₂ = number of moles x molar mass
Molar mass of N₂ = 2(14) = 28g/mol
Mass of N₂ = 0.0054 x 28 = 0.15g
Now;
1 mole of a gas will have 6.02 x 10²³ molecules
0.0054 mole of N₂ will contain 0.0054 x 6.02 x 10²³ =
3.2 x 10²¹molecules