Answer:
Explanation:
V1= 600.0 mL
T1= 293 K
V2 unknown
T2= 333K
Charles law V1*T2=V2*T1
where V1 and T1 are the initial volume and temperature. V2 and T2 are the final temperature and volume.
V2=(V1*T2)/T1 = ( 600.0*333)/293= 682 mL (with significant figures)(681.911 mL before signifcant figures.)
Velocity vs. Time shows an objects acceleration because... in Position vs. Time the line does not go through the origin and it’s very uneven, But Velocitu vs. time goes through the origin and has straight lines.
Answer:
(185 g Ca) / (40.0784 g Ca/mol) × (6.022 × 10^23 atoms/mol) = 2.78 × 10^24 atoms Ca
Explanation:
Answer:
The answer to the question above is
The energy required to heat 87.1 g acetone from a solid at -154.0°C to a liquid at -42.0°C = 29.36 kJ
Explanation:
The given variables are
ΔHfus = 7.27 kJ/mol
Cliq = 2.16 J/g°C
Cgas = 1.29 J/g°C
Csol = 1.65 J/g°C
Tmelting = -95.0°C.
Initial temperature = -154.0°C
Final temperature = -42.0°C?
Mass of acetone = 87.1 g
Molar mass of acetone = 58.08 g/mol
Solution
Heat required to raise the temperature of solid acetone from -154 °C to -95 °C or 59 °C is given by
H = mCsolT = 87.1 g* 1.65 J/g°C* 59 °C = 8479.185 J
Heat required to melt the acetone at -95 °C = ΔHfus*number of moles =
But number of moles = mass÷(molar mass) = 87.1÷58.08 = 1.5
Heat required to melt the acetone at -95 °C =1.5 moles*7.27 kJ/mol = 10.905 kJ
The heat required to raise the temperature to -42 degrees is
H = m*Cliq*T = 87.1 g* 2.16 J/g°C * 53 °C = 9971.21 J
Total heat = 9971.21 J + 10.905 kJ + 8479.185 J = 29355.393 J = 29.36 kJ
The energy required to heat 87.1 g acetone from a solid at -154.0°C to a liquid at -42.0°C is 29.36 kJ
