Answer:
6.15.3 k
Explanation:
From the question we can see that
q = 0, Δu = w
Then,
putting values wet
=
T_f = 615.3 K
Answer choice B 259 kelvins
Answer:
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Answer:
0.15 l of 4.0 m stock KCl solution should betaken
Explanation:
N1V1=N2V2
6*0.1=V2*4
V2=0.15L
Answer:
6.2 calories
Explanation:
Data Given:
change in temperature = 20 °C
specific heat of gold = 0.031 calories/gram °C
mass of gold = 10.0 grams
Amount of Heat = ?
Solution:
Formula used
Q = Cs.m.ΔT
Where:
Q = amount of heat
Cs = specific heat of gold = 0.031 calories/gram °C
m = mass
ΔT = Change in temperature
Put values in above equation
Q = 0.031 calories/gram °C x 10.0 g x 20 °C
Q = 6.2 calories
So option A is correct = 6.2 calories
