Answer:
Explanation:
we know that specific heat is the amount of heat required to raise the temperature of substance by one degree mathmeticaly
Q=mcΔT
ΔT=T2-T1
ΔT=26.8-10.2=16.6
C for water is 4.184
therefore
Q=1.00*4.184*16.6
Q=69.4 j
now we have to covert joule into calorie
1 calorie =4.2 j
x calorie=69.4 j/2
so 69.4 j =34.7 calorie thats why 34.7 calorie heat is required to raise the temperature of water from 10.2 to 26.8 degree celsius
2NH₃(g) + CO₂(g) → CO(NH₂)₂(s) + H₂O(l)
is the balanced equation for the synthesis of urea.
What amount of heat absorbs 50 g of steel (ce = 0.115 cal / g. ° C) that
does its temperature vary by 25 ° C?
Answer:
143.75cal
Explanation:
Given parameters:
Mass of steel = 50g
Specific heat capacity of the steel = 0.115cal/g°C
Temperature = 25°C
Unknown:
Amount of heat = ?
Solution:
The amount of heat to cause this temperature change is dependent on mass and specific heat capacity of the substance.
Amount of heat = m C (ΔT)
m is the mass
c is the specific heat capacity
ΔT is the temperature change
Now insert the parameters and solve;
Amount of heat = 50 x 0.115 x 25
Amount of heat = 143.75cal
Answer:
-1
Explanation:
The relation between Kp and Kc is given below:
Where,
Kp is the pressure equilibrium constant
Kc is the molar equilibrium constant
R is gas constant
T is the temperature in Kelvins
Δn = (No. of moles of gaseous products)-(No. of moles of gaseous reactants)
For the first equilibrium reaction:
<u>Δn = (2)-(2+1) = -1 </u>
Thus, Kp is: