Answer:
= 8350 joules
Solution and explanation:
- The heat of fusion refers to the quantity of heat released when a given amount of water freezes.
- For example, 1 g of water releases 334 J when it freezes at 0°C.
Therefore; For 25.0 g of water.
Heat released = Mass of water × heat of fusion
= 25 g × 334 J/g
= 8350 Joules
Hence, the amount of heat released when 25.0 g of water freezes at 0°C is 8350 J.
Answer:
x= 138.24 g
Explanation:
We use the avogradro's number
6.023 x 10^23 molecules -> 1 mol C2H8
26.02 x 10^23 molecules -> x
x= (26.02 x 10^23 molecules * 1 mol C2H8 )/6.023 x 10^23 molecules
x= 4.32 mol C2H8
1 mol C2H8 -> 32 g
4.32 mol C2H8 -> x
x= (4.32 mol C2H8 * 32 g)/ 1 mol C2H8
x= 138.24 g
Upon a constant pressure (P), volume (V) of a gas will vary in direct proportion to changes in temperature (T). So V1/T1 = V2/T2
V2 = V1T2/T1 = (500)(300)/150
V2 = 150000/150 = 1000 mL
Answer:
it would go back to being deflated
the ion version of potassium is k+
