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givi [52]
3 years ago
15

Lous

Chemistry
1 answer:
ratelena [41]3 years ago
6 0

Answer:

In the last chapter we looked at the particle model of matter and specifically at changes of state. Do you remember heating and cooling candle wax to observe it melt and then solidify. The wax first changed from a solid into a liquid and then back to a solid again. These are physical changes. The chemical properties of the substance does not change.

We are now going to look at what happens when we get chemical changes in substances. These take place during chemical reactions.

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The SI unit for distance is metres - m.

The SI unit for speed or velocity is metres per second - m/s.

The SI unit for acceleration is metres per second squared. - m/s^2.
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4 years ago
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Calculate the enthalpy of combustion of 1 mol decane, C10H22, (l), to form CO2 and H2O. ∆Hf0 for decane is —300.9 kJ/mol using t
worty [1.4K]

Answer:

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Explanation:

5 0
3 years ago
The pressure exerted by 1.5 mol of gas in a 13 L flask at 22 °C is ____ kPa
sukhopar [10]

Answer:

282.7KPa

Explanation:

Step 1:

Data obtained from the question.

Number of mole of (n) = 1.5 mole

Volume (V) = 13L

Temperature (T) = 22°C = 22 + 273°C = 295K

Pressure (P) =..?

Gas constant (R) = 0.082atm.L/Kmol

Step 2:

Determination of the pressure exerted by the gas.

This can be obtained by using the ideal gas equation as follow:

PV = nRT

P = nRT /V

P = 1.5 x 0.082 x 295 / 13

P = 2.79atm.

Step 3:

Conversion of 2.79atm to KPa.

This is illustrated below:

1 atm = 101.325KPa

Therefore, 2.79atm = 2.79 x 101.325 = 282.7KPa

Therefore, the pressure exerted by the gas in KPa is 282.7KPa

8 0
3 years ago
Which arrow or arrows represent reactions that demonstrate a conservation of mass and energy? Explain your answer.
liberstina [14]

Answer:

A

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A mixture of 15.0 g of the anesthetic halothane (C2HBrClF3 197.4 g/mol) and 22.6 g of oxygen gas has a total pressure of 862 tor
AlexFokin [52]

Answer : The partial pressure of C_2HBrClF_3 and O_2 are, 84 torr and 778 torr respectively.

Explanation : Given,

Mass of C_2HBrClF_3 = 15.0 g

Mass of O_2 = 22.6 g

Molar mass of C_2HBrClF_3 = 197.4 g/mole

Molar mass of O_2 = 32 g/mole

First we have to calculate the moles of C_2HBrClF_3 and O_2.

\text{Moles of }C_2HBrClF_3=\frac{\text{Mass of }C_2HBrClF_3}{\text{Molar mass of }C_2HBrClF_3}=\frac{15.0g}{197.4g/mole}=0.0759mole

and,

\text{Moles of }O_2=\frac{\text{Mass of }O_2}{\text{Molar mass of }O_2}=\frac{22.6g}{32g/mole}=0.706mole

Now we have to calculate the mole fraction of C_2HBrClF_3 and O_2.

\text{Mole fraction of }C_2HBrClF_3=\frac{\text{Moles of }C_2HBrClF_3}{\text{Moles of }C_2HBrClF_3+\text{Moles of }O_2}=\frac{0.0759}{0.0759+0.706}=0.0971

and,

\text{Mole fraction of }O_2=\frac{\text{Moles of }O_2}{\text{Moles of }C_2HBrClF_3+\text{Moles of }O_2}=\frac{0.706}{0.0759+0.706}=0.903

Now we have to partial pressure of C_2HBrClF_3 and O_2.

According to the Raoult's law,

p^o=X\times p_T

where,

p^o = partial pressure of gas

p_T = total pressure of gas

X = mole fraction of gas

p_{C_2HBrClF_3}=X_{C_2HBrClF_3}\times p_T

p_{C_2HBrClF_3}=0.0971\times 862torr=84torr

and,

p_{O_2}=X_{O_2}\times p_T

p_{O_2}=0.903\times 862torr=778torr

Therefore, the partial pressure of C_2HBrClF_3 and O_2 are, 84 torr and 778 torr respectively.

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3 years ago
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