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3 years ago

13

What is the final temperature of a system if 10.0 grams of cold water, originally at 10°C is mixed with 120.0 grams of hot water

, originally at 77.0°C?
A. 83.1°C
B. 100.0°C
C. 71.8 °C
D. 95.1°C

1 answer:

Aleonysh [2.5K]3 years ago

3 0

D because I did it on my test

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Let the mixture is X% by mass of CuSO

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2 years ago

Which type of intermolecular attractive force operates between (part A) all molecules, (part B) polar molecules, (part C) the hy

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Answer:

PART A: The LDF occurs between all molecules. Dispersion forces result from shifting electron clouds, which cause weak, temporary dipole.

PART B: Dipole dipole operates only between polar molecules. This is when two polar molecules get near each other and the positively charged portion of the molecule is attracted to the negatively charged portion of another molecule.

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Near the surface of a liquid, fast-moving particles can break free and become a gas.

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3 years ago

An 80 kg long jumper in flight accelerates at a rate of 10 m/s^2. What is the force of the long jumper

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Answer:

F = 800 N

Explanation:

Given data:

Mass = 80 Kg

Acceleration = 10 m/s²

Force = ?

Solution:

Formula:

<em>F = m × a </em>

F = force

m = mass

a = acceleration

Now we will put the values in formula:

<em>F = m × a </em>

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3 years ago

A 1.800-g sample of solid phenol (C6H5OH(s)) was burned in a bomb calorimeter whose total heat capacity is 11.66 kJ/?C. The temp

vichka [17]

Answer:

The balanced chemical equation:

$C_6H_5OH(s)+7O_2(g)\rightarrow 6CO_2(g)+3H_2O(g)$

Heat of combustion per gram of phenol is 32.454 kJ/g

Heat of combustion per gram of phenol is 3,050 kJ/mol

Explanation:

$C_6H_5OH(s)+7O_2(g)\rightarrow 6CO_2(g)+3H_2O(g)$

Heat capacity of calorimeter = C = 11.66 kJ/°C

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Final temperature of the calorimeter = $T_2= 26.37^oC$

Heat absorbed by calorimeter = Q

$Q=C\times \Delta T$

Heat released during reaction = Q'

Q' = -Q ( law of conservation of energy)

Energy released on combustion of 1.800 grams of phenol = Q' = -(58.4166 kJ)

Heat of combustion per gram of phenol:

$\frac{Q'}{1.800 g}=\frac{-58.4166 kJ}{1.800 g}=32.454 kJ/g$

Molar mass of phenol = 94 g/mol

Heat of combustion per gram of phenol:

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