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ANTONII [103]
3 years ago
9

How much heat is absorbed by 15.5 g of water when its temperature is increased from 20.0°C to 50.0°C? The specific heat of water

is 4.184 J/(g°C).,
Chemistry
2 answers:
Zigmanuir [339]3 years ago
5 0
Q = m . C . Δ T
q : Heat absorbed
m : mass of the sample
C : The specific heat of the substance.
Δ T : The change in temperature (Final T - initial T)
So: 
q = 15.5 x 4.18 x (50 - 25) = 1619.75 J 
postnew [5]3 years ago
4 0

Answer : The heat absorbed by the water is 19.4 kJ

Explanation :

Formula used :

Q=m\times c\times \Delta T

or,

Q=m\times c\times (T_2-T_1)

where,

Q = heat absorbed = ?

m = mass of water = 15.5 g

c = specific heat of water = 4.184J/g^oC

T_1 = initial temperature  = 20.0^oC

T_2 = final temperature  = 50.0^oC

Now put all the given value in the above formula, we get:

Q=15.5g\times 4.184J/g^oC\times (50.0-20.0)^oC

Q=1945.56J=1.94\times 10^3J=1.94kJ

Therefore, the heat absorbed by the water is 19.4 kJ

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7 0
3 years ago
Suppose a gas mixture used for anesthesia contains 4.60 mol oxygen (O₂) and 6.00 mol nitrous oxide (N₂O). The total pressure of
julia-pushkina [17]

Considering the definition of mole fraction, the mole fraction of O₂ in the mixture is 0.434.

<h3>Definition of mole fraction</h3>

The molar fraction is a way of measuring the concentration that expresses the proportion in which a substance is found with respect to the total moles of the solution.

In other words, the mole fraction expresses the concentration of solute in a solution as the ratio of moles of substance to total moles of solution:

mole fraction=\frac{moles of substance}{moles of solution}

<h3>Mole fraction of O₂ in this mixture</h3>

In this case, you know a gas mixture used for anesthesia contains 4.60 mol oxygen (O₂) and 6.00 mol nitrous oxide (N₂O).

So, the total moles of the solution can be calculated as:

Total moles = moles of oxygen (O₂) + moles of nitrous oxide (N₂O)

Then:

Total moles= 4.60 moles + 6 moles

Total moles= 10.60 moles

Finally, the more fraction of O₂ can be calculated as follow:

Mole fraction of O_{2} =\frac{moles of O_{2}}{total moles}

Mole fraction of O_{2} =\frac{4.60 moles}{10.6o moles}

Solving:

<u><em>Mole fraction O₂ = 0.434</em></u>

Finally, the mole fraction of O₂ in the mixture is 0.434.

Learn more about mole fraction:

brainly.com/question/14434096

brainly.com/question/10095502

#SPJ1

3 0
2 years ago
5. What are the relative rates of diffusion for methane, CH, and oxygen, O2? If O2 la travels 1.00 m in a certain amount of time
11Alexandr11 [23.1K]

Answer:

The relative rates of diffusion for methane and oxygen is 1.4142.

Methane gas will be able to travel 1.4142 meter in the same conditions.

Explanation:

To calculate the rate of diffusion of gas, we use Graham's Law.

This law states that the rate of effusion or diffusion of gas is inversely proportional to the square root of the molar mass of the gas. Mathematically written as:

\text{Rate of diffusion}\propto \frac{1}{\sqrt{\text{Molar mass of the gas}}}

We are given:

Molar mass of methane gas, m = 16 g/mol

Molar mass of oxygen gas,m' = 32 g/mol

By taking their ratio, we get:

\frac{d_{CH_4}}{d_{O_2}}=\sqrt{\frac{m'}{m}}

\frac{d_{CH_4}}{d_{O_2}}=\sqrt{\frac{32}{16}}=1.4142

The relative rates of diffusion for methane and oxygen is 1.4142.

If oxygen gas travels 1 meters in time t.

Rate of diffusion of oxygen =d_{O_2}=\frac{1 m}{t}

If methane gas travels travels in y meters in time t.

Rate of diffusion of methane=d_{CH_4}=\frac{y }{t}

\frac{d_{CH_4}}{d_{O_2}}=\frac{\frac{y }{t}}{\frac{1 m}{t}}=1.4142

y = 1.4142 m

Methane gas will be able to travel 1.4142 meter in the same conditions.

8 0
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