Answer: The initial temperature of the iron was 
Explanation:

As we know that,

.................(1)
where,
q = heat absorbed or released
= mass of iron = 360 g
= mass of water = 750 g
= final temperature = 
= temperature of iron = ?
= temperature of water = 
= specific heat of iron = 
= specific heat of water= 
Now put all the given values in equation (1), we get
![-360\times 0.450\times (46.7-x)=[750\times 4.184\times (46.7-22.5)]](https://tex.z-dn.net/?f=-360%5Ctimes%200.450%5Ctimes%20%2846.7-x%29%3D%5B750%5Ctimes%204.184%5Ctimes%20%2846.7-22.5%29%5D)

Therefore, the initial temperature of the iron was 
Answer:

Explanation:
Any gas at standard temperature and pressure (STP) has a volume of 22.4 liters per mole or 22.4 L/mol. We can create a proportion with this value.

Multiply both sides of the equation by 6.8 moles of krypton.

The units of moles of krypton will cancel.

The denominator of 1 can be ignored, so this becomes a simple multiplication problem.


If we round to the nearest whole number, the 3 in the tenths place tells us to leave the 2 in the ones place.

6.8 moles of krypton gas at standard temperature and pressure is equal to <u>152 liters</u>.
The density of it would be 57. Do the math and apply the formula to it and you would get this answer.
soluble substances dissolves in solvents