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3 years ago

What is the process of gas changes to a solid

Chemistry

2 answers:

Bogdan [553]3 years ago

5 0

Answer:

Deposition is the phase transition in which gas transforms into solid without passing through the liquid phase. Deposition is a thermodynamic process. The reverse of deposition is sublimation and hence sometimes deposition is called desublimation

Explanation:

i hope this help

igor_vitrenko [27]3 years ago

3 0

Answer:

Deposition is the phase transition in which gas transforms into solid

Explanation:

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3 years ago

What does the pH scale for acids and bases range from?

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Answer:

Anything below 7.0 is acidic, and anything above 7.0 is alkaline, or basic. pH scale, ranging from 0 (very acidic) to 14 (very basic/alkaline) and listing the pH values of common substances.

Explanation:

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3 years ago

Two Carnot engines are operated in series with the exhaust (heat output) of the first engine being the input of the second engin

OlgaM077 [116]

Answer:

(a) 140 F

(b) The temperature rise at the point where the heat is dumped is 2.51 degC

Explanation:

(a) Considering T1 the temperature of input of the first engine, T2 the temperature of the exhaust of the first engine (and input of the second engine) and T3 the exhaust of the second engine, if both engines have the same efficiency we have:

$\eta=1-\frac{T_1}{T2}=1-\frac{T_2}{T_3}$

The temperatures have to be expressed in Rankine (or Kelvin) degrees

$1-\frac{T_1}{T2}=1-\frac{T_2}{T_3}\\\\\frac{T_1}{T2}=\frac{T_2}{T_3}\\\\(T_2)^{2} =T_1*T_3\\\\T_2=\sqrt{T_1*T_3} =\sqrt{(459.67+260)*(459.67+40)}= \sqrt{719.67*499.67}\\\\ T_2=599 \, R= (599-459.67) ^{\circ} F=140^{\circ} F$

(b) The Carnot efficiency of the cycle is

$\eta_{c}=1-Th/Ts=1-(273+20)/(273+315)=0.502$

If the efficiency of the plant is 60% of the Carnot efficiency, we have

$\eta=0.6*\eta_{c}=0.6*0.502=0.302$

The heat used in the plant can be calculated as

$Q_i=W/\eta=750MW/0.302=2483MW$

And the heat removed to the heat sink is

$Q_o=Qi-W=2483-750=1733MW$

If the flow of the river is 165 m3/s, the heat per volume in the sink is

$\frac{Q_o}{f} =\frac{1733 MJ/s}{165 m3/s}= 10.5MJ/m3$

Considering a heat capacity of water C=4.1796 kJ/(kg*K) and a density ρ of 1000 kg/m3, the temperature rise of the water is

$\Delta Q=C*\Delta T\\\Delta T=(1/C)*\Delta Q\\\Delta T=(\frac{1}{4.1796\frac{kJ}{kgK} } )*10,500\frac{kJ}{m3}*\frac{1m3}{1000kg}\\\Delta T= 2.51 ^{\circ}C$

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lianna [129]

Anything with a triple bond ends with -yne so that would be C

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3 years ago

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