An aqueous solution of na2co3 saturates at 300k to form na2co3.h2o crystals. 10kg/h of 35 mass % aqueous solution enters a continuous evaporative crystallizer at 335k and 15 kpa.
additional information:
a) the boiling point rise of the solution is 12k
b) the heat of crytallization is 175 kj/kg
c) the mean heat capacity of the solution is 4.2 kj/kg
d) at 1.5kpa water boils at 286k and the latent heat of vaporization is 2470 kj/kg
e) over the temperatue range 270 k and 350 k the solubility of na2co3 in water is given by: solubility = 0.55t - 127.6 kg/100kg water
1.1) draw a schematic diagram that depicts the crystallization process showing all the entering and leaving streams with the given information. (4)
1.2) determine the actual evaporation rate (8)
1.3) determine the yield of crystals and solution formed using the mass balance method

the ideal gas law equation is as follows

pv = nrt

where p - pressure

v - volume

n - number of moles

r - universal gas constant

t - temperature

so if the pressure, volume and temperature are already known

we are left with n and r

since r is the universal gas constant that has a known fixed value then r too is known

so we are left with 'n'

once we know temperature volume and pressure

we can find the number of moles of gas present using the ideal gas law equation

this is not my own answer, this is an answer for the same question by the user blahblahmali. i do not claim ownership of this account.

(newtons first law)

conservation of momentum law states :

momentum before collision = momentum after collision

momentum p (kg-m/s)= mass * velocity

say moving car=1 000 kg, velocity = 10 m/s then p = 1 000*10= 10 000 kg-m/s

say still car = 2 000 kg, velocity = 0 m/s then p = 2 000* 0 = 0 kg-m/s

total momentum prior to collision = 10 000 + 0 = 10 000 kg-m/s

momentum after collision = 10 000 kg-m/s

mass now = 1 000 kg+2 000 kg = 3 000 kg

10 000 = 3 000 kg* velocity m/s

10 000/3 000 = velocity

3.33 m/s = velocity after collision

an impulse is a force applied over time , momentum is exclusive of external forces and a perfect collision is implied

option c: 1605.1 kj

the balanced chemical reaction for combustion of methane is as follows:

here, enthalpy formation of is -74.7 kj/mol, 0 kj/mol, -393.5 kj/mol and -241.82 kj/mol respectively.

the heat released or can be calculated as follow:

putting the values,

thus, heat released by combustion of 1 mol of methane is 802.54 kj/mol. thus, for 2 mol, heat released will be:

therefore, heat released by the combustion of 2 mol of methane is 1605.1 kj.