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krok68 [10]
3 years ago
6

How many joules of heat are given off when 5.00g of water cool from 348.0K to 298.0K?

Chemistry
1 answer:
forsale [732]3 years ago
6 0

Answer:

Q =  -1045 J

Explanation:

Given data:

Mass of water = 5.00 g

Initial temperature = 348.0 K

Final temperature = 298.0 K

Heat given off = ?

Solution:

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Specific heat capacity of water is 4.18 J/g.K

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 = 298.0 K - 348.0 K

ΔT = - 50 K

Q = 5.0 g ×4.18 J/g.K× - 50 K

Q =  -1045 J

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5 0
1 year ago
We heavier the object the less friction that exist is this true or false
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4 0
3 years ago
What is the molarity of the solution resulting from the dissolution of 239 g glucose (C6H12O6) in 250
kifflom [539]

Answer:

Molarity =5.32 M

Explanation:

Given data:

Mass of glucose = 239 g

Volume = 250 mL (250 /1000 = 0.25 L)

Molarity = ?

Solution;

Formula:

Molarity = number of moles / volume in litter

Number of moles:

Number of moles = mass/ molar mass

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Number of moles = 1.33 mol

Molarity:

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6 0
3 years ago
What is the percent by mass of carbon in acetone, c 3 h 6 o?
Tatiana [17]
In the problem, we are tasked to solved for the amount of carbon (C) in the acetone having a molecular formula of C 3 H 6 O. We need to find first the molecular weight if Carbon (C), Hydrogen (H), Oxygen (O).

Molecular Weight:
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H=1 g/mol
O=16 g/mol

To calculate for the percent by mass of acetone, we assume 1 mol of acetone.
%C=( \frac{3(12 g/mol)}{3(12 g/mol)+6(1 g/mol)+16 g/mol} ) x 100%
%C=62.07%
Therefore, the percent by mass of carbon in acetone is 62.07%
5 0
3 years ago
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