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Yakvenalex [24]
3 years ago
13

URGENT HELP NEEDED! PLEASE ANSWER WITH EXPLANATION*

Chemistry
1 answer:
VladimirAG [237]3 years ago
6 0

Answer:

1) 94.0° C, 2) 34.7°C

Explanation:

1)

Q=mcΔt° =mc(t2 - t1)

2350 J = 85.0 g* 0.385 J/(g*°C) *(t2 - 22.2°C)

t2 - 22.2°C = 71.81°C

t2 = 94.0° C

2)

Q=mcΔt° =mc(t2 - t1)

- 2070 J = 75.0 g * 0.903 J/(g*°C) *(t2 -65.3°C)

t2 - 65.3°C =  - 30.56°C

t2 = 34.7°C

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What mass of solid that has a molar mass of 46.0 g/mol should be added to 150.0 g of benzene to raise the boiling point of benze
enyata [817]

Answer : 17.12 g

Explanation:\Delta T =k_b\times m

\Delta T = elevation in boiling point

k_b = boiling point elevation constant

m= molality

molality=\frac{mass of solute}{molecular mass of solute\times weight of the solvent in kg}

given \Delta T=6.28^{\circ}C

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Weight of the solvent = 150.0 g = 0.15 kg

Putting in the values

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3 0
3 years ago
chemistry A basketball is inflated to a pressure of 1.10 atm in a 28.0°C garage. What is the pressure of the basketball outside
butalik [34]

Answer : The final pressure of the basketball is, 0.990 atm

Explanation :

Gay-Lussac's Law : It is defined as the pressure of the gas is directly proportional to the temperature of the gas at constant volume and number of moles.

P\propto T

or,

\frac{P_1}{P_2}=\frac{T_1}{T_2}

where,

P_1 = initial pressure = 1.10 atm

P_2 = final pressure = ?

T_1 = initial temperature = 28.0^oC=273+28.0=301.0K

T_2 = initial temperature = -2.00^oC=273+(-2.00)=271.0K

Now put all the given values in the above equation, we get:

\frac{1.10atm}{P_2}=\frac{301.0K}{271.0K}

P_2=0.990atm

Thus, the final pressure of the basketball is, 0.990 atm

7 0
3 years ago
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