The third option because the elements mentioned in that selection are in the same group, they must have similar properties.
Answer:
The equilibrium temperature of the coffee is 72.4 °C
Explanation:
Step 1: Data given
Mass of cream = 15.0 grams
Temperature of the cream = 10.0°C
Mass of the coffee = 150.0 grams
Temperature of the coffee = 78.6 °C
C = respective specific heat of the substances( same as water) = 4.184 J/g°C
Step 2: Calculate the equilibrium temperature
m(cream)*C*(T2-T1) = -m(coffee)*c*(T2-T1)
15.0 g* 4.184 J/g°C *(T2 - 10.0°C) = -150.0g *4.184 J/g°C*(T2-78.6°C)
62.76T2 - 627.6 = -627.6T2 + 49329.36
690.36T2 = 49956.96
T2 = 72.4 °C
The equilibrium temperature of the coffee is 72.4 °C
All of the above because you didn't send us a image for reference..
<span>PV=nRT= a universal constant
For any condition
P1V1/n1T1=R
and
P2V2/n2T2=R
i.e
P1V1/n1T1=P2V2/n2T2
Becomes
V1/n1=V2/n2
rearranging and solving
V2=V1X(n2/n1)= 750 mLx((0.65+0.35)/(0.65))=1200ml=1.2L...2 sig figs</span>
