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geniusboy [140]
3 years ago
6

How many calories of heat are necessary to raise the temperature of 319.5 g of water from 35.7 °C

Chemistry
1 answer:
rjkz [21]3 years ago
6 0

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

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4 0
3 years ago
A 25.00 mL sample of the ammonia solution
musickatia [10]

Answer:

1.634 molL-1

Explanation:

The mol ration between NH3 and HCl is 1 : 1

Using Ca Va / Cb Vb = Na / Nb   where a = acid and b = base

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Ca = 0.208 molL-1

Cb = ?

Va = 19.64 mL

Vb = 25.00mL

Solving for Cb

Cb = Ca Va / Vb

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Using the dilution equation;

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Initial Volume, V1 = 25.00 mL

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3 0
3 years ago
Calculate the missing variables in each experiment below using Avogadro’s law.
blagie [28]

Answer:

The answer to your question is: letter c

Explanation:

Data

V1 = 612 ml    n1 = 9.11 mol

V2 = 123 ml    n2 = ?

Formula

                               \frac{V1}{n1}  =  \frac{V2}{n2}

                                         n2 = \frac{n1V2}{V1}

                                         n2 = \frac{(9.11)((123)}{(612)}

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5 0
3 years ago
If 8.00 g NH4NO3 is dissolved in 1000 g of water, the water decreases in temperature from 21.00 degrees Celsius to 20.39 degrees
telo118 [61]

Answer:

25.7 kJ/mol

Explanation:

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heat of solution of NH₄NO₃ + heat from water = 0

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∴ n =   0.100 mol NH₄NO₃

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ΔHsoln  =  +2570.19 J  /0.100 mol  =  +25702 J/mol  =  +25.7 kJ/mol

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5 0
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