Answer:
Zr (Zirconium)
Explanation:
1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d2
Answer:
the heat rate required to cool down the gas from 535°C until 215°C is -2.5 kW.
Explanation:
assuming ideal gas behaviour:
PV=nRT
therefore
P= 109 Kpa= 1.07575 atm
V= 67 m3/hr = 18.6111 L/s
T= 215 °C = 488 K
R = 0.082 atm L /mol K
n = PV/RT = 109 Kpa = 1.07575 atm * 18.611 L/s /(0.082 atm L/mol K * 488 K)
n= 0.5 mol/s
since the changes in kinetic and potencial energy are negligible, the heat required is equal to the enthalpy change of the gas:
Q= n* Δh = 0.5 mol/s * (- 5 kJ/mol) =2.5 kW
Answer:
energy required=qnet=87.75kJ
Explanation:
we will do it in three seperate step and then add up those value.
first step is to heat the sample of water upto 100C i.e upto boiling pont. because just after this sample of water started vaporization.
q 1= m c (T2-T1)
q1 = 36.0 g (4.18 J/gC) (100 - 65 C)
q1 = 5267 J
=5.267kJ
next is to vaporize the sample at 100C
q2 = 36.0 g / 18.0 g/mol X 40.7 kJ/mol
q2= 81.4 kJ
Finally, heat the steam upto 115C
q3 = m c (T2-T1)
q 3= 36.0 g (2.01 J/gC)(115-100C)
q3 = 1085 J
=1.085kJ
qnet=q1 +q2 +q3
energy required=qnet=87.75kJ
Answer:
Calcium
Explanation:
Calcium is the only metal here
hope this helps...
