A. Because you have to simplify
Answer: -
3.3° C
Explanation: -
Mass of water m = 180.5 g
Energy released as heat Q = 2494 J
Specific heat is defined as the heat required to raise the temperature of the unit mass of a given substance by 1 C.
Specific heat of water Cp = 4.184 (J/g)⋅∘C
Using the formula
Q = m x Cp x ΔT
We get temperature change ΔT = Q / (m x Cp)
= 2494 J / ( 180.5 g x 4.184 (J/g)⋅∘C
= 3.3° C
Thus the temprature change, (ΔT), of the wateris 3.3 °C if 180.5 g of water sat in the copper pipe from part A, releasing 2494 J of energy to the pipe
Answer:
32.00 g. Hope this helps! PLEASE GIVE ME BRAINLIEST!!!!! =)
Answer is B. gas formation
Answer:
D
Explanation:
If the pressure remains constant then the temperature and Volume are all that you have to consider.
Givens
T1 = 19oC = 19 + 273 = 292o K
T2 = 60oC = 60 + 273 = 333oK
V1 = 250 mL
V2 = x
Formula
V1/T1 = V2/T2
250/292 = x/333
Solution.
The solves rather neatly. Multiplly both sides by 333
250*333 / 292 = 333 *x / 333
Do the multiplication
250 * 333 / 292 = x
83250 / 292 = x
Divide by 292
x = 285.1 mL
The answer is D
