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2 years ago

Cocl2 is often used in hygrometers. search the internet to determine why? how does this relate to this experiment?

1 answer:

6 0

The answer to this question would be: Because CoCl2 will change color when humidity changes.

Cobalt chloride or CoCl2 can change its color depends on the humidity. The dry cobalt chloride will have pink color but it will changes into blue if you mix it with water. If the humidity is high, the water in the air might condense into the cobalt chloride and change its color to blue. So high humidity will cause more cobalt chloride turn into blue.

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Octane (C8H18) is a component of gasoline. Complete combustion of octane yields H2O and CO2. Incomplete combustion produces H2O

mestny [16]

$86.5\; \%$ of octane had been converted to carbon dioxide CO₂.

<h3>Explanation</h3>

Octane has a molar mass of

$12.01 \times 8 + 1.008 \times 18 = 114.22 \; \text{g} \cdot \text{mol}^{-1}$

1.000 gallon of this fuel would have a mass of 2.650 kilograms or $2.65 \times 10^{3} \; \text{g}$ , which corresponds to $2.65 \times 10^{3} / 114.22 = 23.2\; \text{mol}$ of octane.

Octane undergoes complete combustion to produce carbon dioxide and water by the following equation:

$2\; \text{C}_8\text{H}_{18} + 25 \; \text{O}_2 \to 16 \; \text{CO}_2 + 18 \; \text{H}_2\text{O}$

An incomplete combustion of octane that gives rise to carbon monoxide and water but no carbon dioxide would consume not as much oxygen:

$2\; \text{C}_8\text{H}_{18} + 17 \; \text{O}_2 \to 16 \; \text{CO} + 18 \; \text{H}_2\text{O}$

The mass of the product mixture is $11.53 - 2.65 = 8.88 \; \text{kg}$ heavier than that of the octane supplied. Thus $8.88 \; \text{kg} = 8.88 \times 10^{3} \; \text{g}$ of oxygen were consumed in the combustion. There are $277.5 \; \text{mol}$ of oxygen molecules in $8.88 \times 10^{3} \; \text{g}$ of oxygen.

Let the number of moles of octane that had undergone complete combustion as seen in the first equation be $x$ ( $0 \le x \le 23.2$ ). The number of moles of octane that had undergone incomplete combustion through the second equation would thus equal $23.2 - x$ .

25 moles of oxygen gas is consumed for every two moles of octane that had undergone complete combustion and 17 moles if the combustion is incomplete.

$n(\text{O}_2, \; \text{Complete Combustion}) + n(\text{O}_2, \; \text{Incomplete Combustion} ) = n(\text{O}_2, \; \text{Consumed})\\$

$\frac{25}{2} \; x + \frac{17}{2} \; (23.2 - x) = 277.5\\4 \; x = 277.5 - \frac{17}{2} \times 23.2\\x = 20.1$

Therefore $20.1 \; \text{mol}$ out of the 23.2 moles of octane had undergone complete combustion to produce carbon dioxide.

$\%n(\text{Complete Combustion}) = 20.1 / 23.2 \times 100 \; \%= 86.5\; \%$

7 0

2 years ago

What is the absolute structural necessity for an alcohol to be oxidized with chromium trioxide?

4vir4ik [10]

The alcohol being oxidized must not be a tertiary alcohol.

A tertiary alcohol is one in which the -OH group is attached to a carbon atom which is attached to three other carbon atoms. This "closes off" the alcohol group and prevents the formation of oxidation products. This is the reason why tertiary alcohols do not undergo oxidation in mild conditions.

8 0

2 years ago

Suppose you measure the absorbance of a yellow dye solution in a 1.00 cm cuvette. The absorbance of the solution at 427 nm is 0.

Dimas [21]

Answer:

The concentration of the solution, $C=7.2992\times 10^{-6} M$

Explanation:

The absorbance of a solution can be calculated by Beer-Lambert's law as:

$A=\varepsilon Cl$

Where,

A is the absorbance of the solution

ɛ is the molar absorption coefficient ( $L.mol^{-1}.cm^{-1}$ )

C is the concentration ( $mol^{-1}.L^{-1}$ )

l is the path length of the cell in which sample is taken (cm)

Given,

A = 0.20

ɛ = 27400 $M^{-1}.cm^{-1}$

l = 1 cm

Applying in the above formula for the calculation of concentration as:

$A=\varepsilon Cl$

$0.20= 27400\times C\times 1$

$C = \frac{0.20}{27400\times 1} M$

So , concentration is:

$C=7.2992\times 10^{-6} M$

4 0

1 year ago

495 cm3 of oxygen gas and 877 cm3 of nitrogen gas, both at 25.0 C and 114.7 kpa, are injected into an evacuated 536 cm3 flask. F

I am Lyosha [343]

Answer:

Total pressure of the flask is 2.8999 atm.

Explanation:

Given data:

Volume of oxygen (O2) gas= 495 cm3

= 0.495 L (1 cm³ = 1 mL = 0.001 L)

Volume of nitrogen (N2) gas = 877 cm3

= 0.877 L (1 cm³ = 1 mL = 0.001 L)

volume of falsk = 536 cm3

= 0.536 L (1 cm³ = 1 mL = 0.001 L)

Temperature = 25 °C

T = (25°C + 273.15) K

= 298.15 K

Pressure = 114.7 kPa

= 114.700 Pa

Pressure (torr) = 114,700 / 101325

= 1.132 atm

Formula:

PV=nRT (ideal gas equation)

P = pressure

V = volume

R (gas constnt)= 0.0821 L.atm/K.mol

T = temperature

n = number of moles for both gases

Solution:

Firstly we will find the number of moles for oxygen and nitrogen gas.

For Oxygen:

n = PV / RT

n = 1.132 atm × 0.495 L / 0.0821 L.atm/K.mol × 298.15 K

= 0.560 / 24.47

= 0.0229 moles

For Nitrogen:

n = PV / RT

n = 1.132 atm × 0.877 / 0.0821 L.atm/K.mol × 298.15 K

n = 0.992 / 24.47

= 0.0406

Total moles = moles for oxygen gas + moles for nitrogen gas

= 0.0229 moles + 0.0406 moles

n = 0.0635 moles

Now put the values in formula

PV=nRT

P = nRT / V

P = 0.0635 × 0.0821 L.atm/K.mol × 298.15 K / 0.536 L

P = 1.554 / 0.536

P = 2.8999 atm

Total pressure in the flask is 2.8999 atm, while assuming the temperature constant.

7 0

2 years ago

Read 2 more answers

Calculate the number of grams of carbon dioxide produced from complete combustion of one liter of octane by placing the conversi

lesya [120]

Answer:

15.71g

Explanation:

The general combustion equation for all hydrocarbons is

CxHy + (x+y/4) O2 = xCO2 + (y/2) H2O

For octane, C8H18 :

C8H18 + ( 8 + 18/4 ) O2 = 8CO2 + 9H2O

C8H18 + 50/4 O2 = 8CO2 + 9H2O

C8H18 + 25/2 O2 = 8CO2 + 9H2O

2C8H18 + 25 O2 = 16 CO2 + 18H2O (balanced)

From the balanced equation,

2 x 22.4 L of octane produced 16 [ 12 + (16 x 2)] of carbon dioxide

That is,

44.8 L of octane produced 704g of carbon dioxide

So, 1L of octane will produce 1 L x 704g/44.8 L = 15.71g of carbon dioxide

Therefore, 15.71g of carbon dioxide will be produced by the complete combustion of 1 L of octane.

7 0

2 years ago