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1 year ago

The density of liquid Z is 0.9237 g/mL. A student masses a cup

Chemistry

1 answer:

Ulleksa [173]1 year ago

3 0

Answer:

37.65mL

Explanation:

Given parameters:

density of liquid Z = 0.9237g/mL

Mass of liquidZ + mass of cup = 50.7g

Mass of cup= 15.92g

Volume of liquid Z in cup=?

Solution:

Density is the mass per unit volume of a substance. It is mathematically expressed as shown below:

Density = $\frac{mass}{volume}$

To find the volume of liquid Z, we know the density of the liquid but we dont know the mass yet.

Mass of liquidZ = 50.7g - mass of cup = 50.7g - 15.92g = 34.98g

Therefore:

Volume of liquidZ = $\frac{mass of liquidZ}{density of liquidZ}$

= $\frac{34.98}{0.9237}$

= 37.65mL

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Answer : The complete chemical equation is,

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As we know that, in a chemical equation the reacting species present on left side and the product formed present on right side and a right arrow inserted between the reactants and product that show a chemical reaction taking place.

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Q1) A vapor-compression refrigeration system operates on the cycle of Fig. 9.1. The refrigerant is 1,1,1,2-Tetrafluoroethane. Gi

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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

n ( compressor efficiency ) = 0.76

refrigeration rate = 1200 kJ.s^-1

i) determine the circulation rate of the refrigerant

m = $\frac{Q}{H2 - H1}$ = $\frac{Q}{H2 - H4\\}$ ------- 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 = $\frac{H'3 - H2 }{0.76}$ = $\frac{2814.7-2508.9}{0.76}$ = 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

$w_{carnot} = \frac{T2}{T4 - T2}$

= 277.15 / ( 307.15 - 277.15 )

= 9.238

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