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olasank [31]
2 years ago
14

The average kinetic energy of water molecules is greatest in which of these samples?(1) 10 g of water at 35°C(2) 10 g of water a

t 55°C(3) 100 g of water at 25°C(4) 100 g of water at 45°C
Chemistry
2 answers:
liberstina [14]2 years ago
5 0

Answer : The correct option is, (2) 10 g of water at 55°C

Explanation :

Average kinetic energy of the gas particle is directly proportional to the temperature of the gas particle.

Formula used :

K.E=\frac{3}{2}\frac{RT}{N_A}

where,

R = Gas constant

T = temperature

N_A = Avogadro's number

From this we conclude that the kinetic energy is directly proportional to the temperature where 'R' and N_A are constant. That means kinetic energy depends only on the temperature not on the mass.

(Higher the temperature, higher will be the kinetic energy)

Hence, the average kinetic energy of water molecules is greatest in 10 g of water at 55°C.

Ipatiy [6.2K]2 years ago
3 0
The question asks about the average kinetic energy so it is not related with mass. We only need to compare the temperature. The higher temperature is, the higher kinetic energy is. So the answer is (2).
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How calculate how many milliliters of glycerin (specific gravity=1.26) will have a mass of 0.75 lbs?
Dimas [21]
Specific gravity is the ratio between the density of the material to the density of water at 4 degrees celcius (density of water at this temperature is 1000 kg/m^3)

1.26 = density of glycerin / 1000
density of glycerin = 1260 kg/m^3

1 Kg is equivalent to 2.2 lbs, therefore:
0.75 lbs = 0.34 kg

density = mass/volume
1260 = 0.34/volume
volume = 0.34/1260 = 2.857 x 10^-4 m^3

now, 1 m^3 is equivalent to 1000 liters
2.857 x 10^-4 m^3 = 0.2857 liters

one liter is equivalent to 1000 ml
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3 years ago
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The answer to this question will be C
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2 years ago
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24
Firlakuza [10]

Answer:

Q = 52668 J

Explanation:

Given data:

Amount of heat required = ?

Mass of water = 350 g

Initial temperature = 20°C

Final temperature = 56°C

Specific heat capacity of water = 4.18 J/g°C

Solution:

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = 56°C - 20°C

ΔT = 36°C

Q = 350 g× 4.18 J/g°C ×36°C

Q = 52668 J

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