Answer:
0.1988 J/g°C
Explanation:
-Qmetal = Qwater
Q = mc∆T
Where;
Q = amount of heat
m = mass of substance
c = specific heat of substance
∆T = change in temperature
Hence;
-{mc∆T} of metal = {mc∆T} of water
From the information provided in this question, For water; m= 22.0g, ∆T = (24°C-19°C), c = 4.18J/g°C.
For metal; m= 34.0g, ∆T = (24°C-92°C), c = ?
Note that, the final temperature of water and the metal = 24°C
-{34 × c × (24°C-92°C)} = 22 × 4.18 × (24°C-19°C)
-{34 × c × (-68°C)} = 459.8
-{34 × c × -68} = 459.8
-{-2312c} = 459.8
+2312c = 459.8
c = 459.8/2312
c = 0.1988
The specific heat capacity of the metal is 0.1988 J/g°C
Answer:
if u question my last question i will answer yours please
Explanation:
the volume of 2.5% m/v koh which can prepared from 125 ml of a 5% koh solution is calculated using the following formula
m1v1= m2 v2
M1= 5/100= 0.05
v1= 125
m2=2.5/100=0.025
V2=?
v2= m1v1/m2
=0.05 x125 /0.025=250 ml
Answer:
38.7 g is the answer to your question. Your welcome:)
Answer:
When water melts from ice to liquid, or boils and becomes a gas, it is still water.
Explanation:
