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goldfiish [28.3K]
3 years ago
8

How much heat is required to raise the temperature of 81.0 g of water from its melting point to its boiling point?

Chemistry
1 answer:
Dovator [93]3 years ago
6 0

Answer:

Specific heat of water = 33.89 KJ

Explanation:

Given:

mass of water = 81 gram

Initial temperature = 0°C

Final temperature = 100°C

Specific heat of water = 4.184

Find:

Required heat Q

Computation:

Q = Mass x Specific heat of water x (Final temperature - Initial temperature)

Q = (81)(4.184)(100-0)

Q = 33,890.4

Specific heat of water = 33.89 KJ

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What is the molarity of a sodium chloride solution made by dissolving 4.512 moles to make 2.0 L?
Svet_ta [14]

Answer:

2.40 M

Explanation:

The molarity of a solution tells you how many moles of solute you get per liter of solution.

Notice that the problem provides you with the volume of the solution expressed in milliliters,

mL

. Right from the start, you should remember that you must convert this volume to liters by using the conversion factor

1 L

=

10

3

mL

Now, in order to get the number of moles of solute, you must use its molar mass. Now, molar masses are listed in grams per mol,

g mol

−

1

, which means that you're going to have to convert the mass of the sample from milligrams to grams

1 g

=

10

3

mg

Sodium chloride,

NaCl

, has a molar mass of

58.44 g mol

−

1

, which means that your sample will contain

unit conversion



280.0

mg

⋅

1

g

10

3

mg

⋅

molar mass



1 mole NaCl

58.44

g

=

0.004791 moles NaCl

This means that the molarity of the solution will be

c

=

n

solute

V

solution

c

=

0.004791 moles

2.00

⋅

10

−

3

L

=

2.40 M

The answer is rounded to three sig figs, the number of sig figs you have for the volume of the solution.

6 0
3 years ago
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Arlecino [84]
Am letting the picture doing the talk.

8 0
3 years ago
How many calories of heat are necessary to raise the temperature of 319.5 g of water from 35.7 °C
rjkz [21]

20600Cal              

Explanation:

Given parameters:

Mass of water = 319.5g

Initial temperature = 35.7°C

Final temperature = 100°C

Unknown:

Calories needed to heat the water = ?

Solution:

The calories is the amount of heat added to the water. This can be determined using;

     H  =   m  c Ф

c  = specific heat capacity of water = 4.186J/g°C

   H is the amount of heat

    Ф is the change in temperature

    H = m c (Ф₂ - Ф₁)

    H = 319.5 x 4.186 x (100 - 35.7) = 85996.56J

Now;

     1kilocalorie = 4184J

     

85996.56J to kCal; \frac{85996.56}{4184}   = 20.6kCal  = 20600Cal

               

learn more:

Specific heat brainly.com/question/3032746

#learnwithBrainly

6 0
3 years ago
If it takes 20757 J to heat 96.5 g of a substance from 18.76°C to 78.62 °C, what is the specific heat of the substance?
Free_Kalibri [48]

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5 0
3 years ago
Reaction type for this equation <br> 2 AgNO3 + Cu → Cu(NO3)2 + 2 Ag
horrorfan [7]

Answer:

single displacement reaction

6 0
3 years ago
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