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shepuryov [24]
2 years ago
7

Complete the sentences to explain whether molar volumes in the solid state are as similar as they are in the gaseous state and w

hether you would you expect the molar volumes in the liquid state to be closer to those in the solid state or the gaseous state. Match the words in the left column to the appropriate blanks in the sentences on the right. Make certain each sentence is complete before submitting your answer.
less than In the solid state, it is expected that the amount gaseous state of empty space is that of the gaseous state, and molar volumes greater than equivalent to influenced by molecular characteristics. are Thus, solid state molar volumes solid state similar to those in the gaseous state. are not When comparing empty space in the liquid state, it is expected that space in the liquid phase is space in the gaseous phase and space to the solid phase. Thus, the molar volumes of the liquids are expected to be closer to those in the
Chemistry
1 answer:
Inessa05 [86]2 years ago
8 0

Answer:

Thus, molar volumes of the liquids are expected to be closer to those in the "solid state."

Explanation:

Anything which occupies mass is known as the matter and we know that matter is existed in three states i.e. solid, liquid and gas. The atoms in the solid are closely packed and therefore, they posses very less kinetic energy. The atoms in liquid state are less closely packed and possess more kinetic energy in comparison to the solid state. The atoms in the gaseous state are loose and possess maximum kinetic energy. The molar volume is the volume occupied by 1 mole of a substance. Solid being compressed will have lesser volume in comparison to the liquid and liquid will have lesser volume in comparison to the gas.

In the solid state, it is expected that the amount of empty space is less than that of the gaseous state, and molar volumes are influenced by molecular characteristics. Thus, solid state molar volumes are not similar to those in the gaseous state. When, comparing empty space in the liquid state, it is expected that space in the liquid phases is less than space in the gaseous phase and equivalent to space to the solid state. Thus, molar volumes of the liquids are expected to be closer to those in the solid state.

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A student builds a model of a race car. The scale is 1:15. In the scale model, the car is 8 cm tall.How tall is the actual car?
Xelga [282]

let the actual height of car be x

now, according to question,

  • \dfrac{1}{15}  =  \dfrac{8}{x}

  • x = 15 \times 8

  • x = 120 \:  \: cm

  • height \:  \: of \:  \: car   =   120 \:   cm \:  \: or \:  \: 1.2 \:  \: m
5 0
1 year ago
Q1) A vapor-compression refrigeration system operates on the cycle of Fig. 9.1. The refrigerant is 1,1,1,2-Tetrafluoroethane. Gi
hoa [83]

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 = \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

4 0
2 years ago
If 25 g of NH3, and 96 g of H2S react according to the following reaction, what is the
jeyben [28]

25 g of NH₃ will produce 47.8 g of (NH₄)₂S​

<u>Explanation:</u>

2 NH₃ + H₂S ----> (NH₄)₂S​

Molecular weight of NH₃ = 17 g/mol

Molecular weight of (NH₄)₂S​ = 68 g/mol

According to the balanced reaction:

2 X 17 g of NH₃ produces 68 g of (NH₄)₂S​

1 g of NH₃ will produce \frac{68}{34} g of (NH₄)₂S​

25g of NH₃ will produce \frac{65}{34} X 25 g of (NH₄)₂S​

                                     = 47.8 g of (NH₄)₂S​

Therefore, 25 g of NH₃ will produce 47.8 g of (NH₄)₂S​

4 0
2 years ago
Consider the reaction. X ( g ) + Y ( g ) − ⇀ ↽ − Z ( g ) K p = 1.00 at 300 K In which direction will the net reaction proceed fo
marta [7]

Answer:

Explanation:

We have in this question the equilibrium

X ( g ) + Y ( g ) ⇆  Z ( g )

With the equilibrium contant Kp = pZ/(pX x pY)

The moment we change the concentration of Y, we are changing effectively the partial pressure of Y since pressure and concentration are directly proportional

pV = nRT ⇒ p = nRT/V and n/V is molarity.

Therefore we can calculate the reaction quotient Q

Qp = pZ/(pX x pY) = 1/ 1  x 0.5 atm = 2

Since Qp is greater than Kp the system proceeds from right to left.

We could also arrive to the same conclusion by applying LeChatelier´s principle which states that any disturbance in the equilibrium, the system will react in such a way to counteract the change to restore the equilibrium. Therefore, by having reduced the pressure of Y the system will react favoring the reactants side increasing some of the y pressure until restoring the equilibrium Kp = 1.

4 0
2 years ago
Hydrogen chloride gas can be prepared by the following reaction: 2NaCl(s) + H2SO4(aq) → 2HCl(g) + Na2SO4(s) How many grams of HC
SVETLANKA909090 [29]

Answer:

The correct answer is is option B

b. 93.3 g

Explanation:

SEE COMPLETE QUESTION BELOW

Hydrogen chloride gas can be prepared by the following reaction: 2NaCl(s) + H2SO4(aq) → 2HCl(g) + Na2SO4(s)

How many grams of HCl can be prepared from 2.00 mol H2SO4 and 2.56 mol NaCl?

a. 7.30 g

b. 93.3 g

c. 146 g

d. 150 g

e. 196 g

CHECK THE ATTACHMENT FOR STEP BY STEP EXPLANATION

7 0
2 years ago
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