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Savatey [412]
3 years ago
6

Calculate the amount of heat needed to raise 190.0 g of water vapor at 118 C to 150. C

Chemistry
1 answer:
Mama L [17]3 years ago
5 0
<h3>Answer:</h3>

12,281.6 Joules

<h3>Explanation:</h3>
  • The amount of heat energy is calculated by multiplying the mass of a substance by its specific heat capacity and then by the change in temperature.
  • That is, Q = m × c ×Δt

In our case;

Mass of water vapor, m = 190.0 g

Change in temperature, Δt = 150°C - 118°C = 32°C

Taking the specific heat capacity of water vapor = 2.02 J/g°C

Therefore;

Amount of heat = 190.0 g × 2.02 J/g°C × 32°C

                          = 12,281.6 Joules

Hence, the amount of heat needed is 12,281.6 Joules

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Radioactive radium has a half-life of approximately 1,599 years. The initial quantity is 13 grams. How much (in grams) remains a
Luda [366]

The quantity of substance remains after 850 years is 8.98g if the half life of radioactive radium is 1,599 years.

<h3>What is half life period? </h3>

The time taken by substance to reduce to its half of its initial concentration is called half life period.

We will use the half- life equation N(t)

N e^{(-0.693t) /t½}

Where,

N is the initial sample

t½ is the half life time period of the substance

t2 is the time in years.

N(t) is the reminder quantity after t years .

Given

N = 13g

t = 350 years

t½ = 1599 years

By substituting all the value, we get

N(t) = 13e^(0.693 × 50) / (1599)

= 13e^(- 0.368386)

= 13 × 0.691

= 8.98

Thus, we calculated that the quantity of substance remains after 850 years is 8.98g if the half life of radioactive radium is 1,599 years.

learn more about half life period:

brainly.com/question/20309144

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4 0
2 years ago
A sample of benzene was vaporized at 25◦C. When 37.5 kJ of heat was supplied, 95.0 g of the liquid benzene vaporized. What is th
grandymaker [24]

Answer:

Enthalpy of vaporization = 30.8 kj/mol

Explanation:

Given data:

Mass of benzene = 95.0 g

Heat evolved = 37.5 KJ

Enthalpy of vaporization = ?

Solution:

Molar mass of benzene = 78 g/mol

Number of moles = mass/ molar mass

Number of moles = 95 g/ 78 g/mol

Number of moles = 1.218 mol

Enthalpy of vaporization =  37.5 KJ/1.218 mol

Enthalpy of vaporization = 30.8 kj/mol

8 0
3 years ago
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