Answer:
373.1 mL of AgCN (aq) must be poured into your electrolysis vat to ensure you have sufficient Ag to plate all of the forks.
Explanation:
Mass of silver to be precipitated on ecah spoon = 0.500 g
Number of silver spoons = 250
Total mass of silver = 250 × 0.500 g = 125 g
Moles of AgCN = n =
Volume of AgCN solution =V
Molarity of the AgCN = 2.50 M
(1 L = 1000 mL)
373.1 mL of AgCN (aq) must be poured into your electrolysis vat to ensure you have sufficient Ag to plate all of the forks.
Answer: 770 g water are needed to dissolve 27.8 g of ammonium nitrate in order to prepare a 0.452 m solution
Explanation:
Molality : It is defined as the number of moles of solute present per kg of solvent
Formula used :
where,
n= moles of solute
Moles of
= weight of the solvent in g = ?
Thus 770 g water are needed to dissolve 27.8 g of ammonium nitrate in order to prepare a 0.452 m solution
Answer:
i) 0.5071 (kg/s)
ii) -1407.1 kj/kg
iii) 204.05 Kw
iv) 5.881
v) 9.238
Explanation:
Given Data:
evaporation temperature ( T ) = 4°c = 277.15 K
Condensation Temperature ( T ) = 34°c = 307.15 K
<em>n</em> ( compressor efficiency ) = 0.76
refrigeration rate = 1200 kJ.s^-1
i) determine the circulation rate of the refrigerant
m = =
------- 1
Q = 1200 Kj.s^-1
H2 = entropy at step 2 = 2508.9 (kJ / kg ) ( gotten from Table F )
H4 = entropy at step 4 = 142.4 ( kJ/ kg )
back to equation 1
m ( circulation rate of refrigerant ) = 0.5071 (kg/s)
ii) heat transfer rate in the condenser
Q = m ( H4 - H3 )
= 0.5071 ( 142.4 - 2911.27 )
= -1407.1 kj/kg
where H3 = H2 + ΔH23 = 2911.27 (kj/kg) ( as calculated )
iii) power requirement
w = m * ΔH23
= 0.5071 (kg/s) * 402.37 (kj/kg) = 204.05 Kw
where: ΔH23 = =
= 402.37 (kj/kg)
iv) coefficient of performance of a cycle
W = Qc / w
= 1200 Kj.s^-1/ 204.05 kw
= 5.881
v) coefficient of performance of a Carnot refrigeration cycle
= 277.15 / ( 307.15 - 277.15 )
= 9.238