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3 years ago

12

By the reaction of carbon & oxygen , a mixture of CO &CO2 is obtained. What is the composition by mass of the mixture ob

taind when 20 grams of O2 reacts with 12 grams of carbon ?

1 answer:

Hatshy [7]3 years ago

4 0

M(O₂)=20g
M(O₂)=32.0 g/mol
n(O₂)=20/32.0=0.625 mol

m(C)=12 g
M(C)=12.0 g/mol
n(C)=12/12.0=1.0 mol

2C + O₂ → 2CO
1 mol 0.625 mol 1 mol
0.625-0.5=0.125 mol

2CO + O₂ → 2CO₂
0.250 mol 0.125 mol 0.250 mol

n(CO)=1 mol - 0.250 mol = 0.750 mol
M(CO)=28.0 g/mol
m(CO)=0.750*28.0=21.0 g

n(CO₂)=0.250 mol
M(CO₂)=44.0 g/mol
m(CO₂)=0.250*44.0=11.0 g

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3. At 35 C, a sample of gas has a volume of 256 ml and a pressure of 720.torr. What would the volume

STatiana [176]

Answer:

The volume will be 185.83 mL.

Explanation:

Gay-Lussac's law indicates that when there is a constant volume, as the temperature increases, the pressure of the gas increases. And when the temperature is decreased, the pressure of the gas decreases. Gay-Lussac's law can be expressed mathematically as follows:

$\frac{P}{T} =k$

Where P = pressure, T = temperature, k = Constant

Boyle's law says that the volume occupied by a given gaseous mass at constant temperature is inversely proportional to pressure. Boyle's law is expressed mathematically as:

P*V=k

Where P = pressure, V = volume, k = Constant

Finally, Charles's Law consists of the relationship that exists between the volume and the temperature of a certain quantity of ideal gas, which is kept at a constant pressure. For a given sum of gas at a constant pressure, as the temperature increases, the volume of the gas increases and as the temperature decreases, the volume of the gas decreases because the temperature is directly related to the energy of the movement of the gas molecules. .

In summary, Charles's law is a law that says that when the amount of gas and pressure are kept constant, the quotient that exists between the volume and the temperature will always have the same value:

$\frac{V}{T} =k$

Combined law equation is the combination of three gas laws called Boyle's, Charlie's and Gay-Lusac's law:

$\frac{P*V}{T} =k$

Studying two different states, an initial state 1 and a final state 2, it is satisfied:

$\frac{P1*V1}{T1} =\frac{P2*V2}{T2}$

In this case:

P1= 720 torr
V1= 256 mL
T1= 35 C= 308 K (being 0 C= 273 K)
P2= 1.25 atm= 950 torr (being 1 atm= 760 torr)
V2= ?
T2= 22 C= 295 K

Replacing:

$\frac{720 torr*256 mL}{308 K} =\frac{950 torr*V2}{295 K}$

Solving:

$V2= \frac{295K}{950 torr} *\frac{720 torr*256 mL}{308 K}$

V2= 185.83 mL

<u><em>The volume will be 185.83 mL.</em></u>

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3 years ago

Suppose that you add 24.3 g of an unknown molecular compound to 0.250 kg of benzene, which has a K f of 5.12 oC/m. With the adde

pochemuha

Solution :

We know that :

$\Delta T_f = k_f.m$ and $m=\frac{w_2}{m_2 \times w_1}$

Then, $\Delta T_f = k_f.\frac{w_2}{m_2.w_1}$ ..................(1)

Where,

$w_1$ = amount of solvent (in kg)

$w_2$ = amount of solute (in kg)

$m_2$ = molar mass of solute (g/mole)

$m$ = molality of solution (mole/kg)

Given :

$\Delta T_f$ = $3.14\ ^\circ C$ , $k_f= 5.12\ ^\circ C/m$

$=5.12 \ ^\circ C/mole/kg$

$=5.12 \ ^\circ C \ kg/mole$

$w_1$ = 0.250 kg, $w_2$ = 24.3 g

Then putting this values in the equation is (1),

$3.14 = \frac{5.12 \times 24.3}{m_2 \times 0.250}$

$m_2 = \frac{5.12 \times 24.3}{3.14 \times 0.250}$

$m_2= 158.49$ g/mole

So, the molar mass of the unknown compound is 158.49 g/mole.

7 0

3 years ago