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kirza4 [7]
3 years ago
12

when carbon is burned in the air, it reacts with oxygen to form carbon dioxide. when 22.8 g of carbon were burned in the presenc

e of 79.0 of oxygen, 18.2g of oxygen remained unreacted. what mass if carbon dioxide was produced?​
Chemistry
1 answer:
OverLord2011 [107]3 years ago
4 0

Answer:

So there is83.6g CO2 produced

Explanation:

Burning carbon with air has the following equation

C + O2 → CO2

For 1 mol Carbon, we have 1 mol O2 and 1 mol CO2

Step 2: Calculating moles

mole C = 22.8g / 12g/mole

Mole C = 1.9 mole

1.9 mole C will completely react

Since for each mole C there is 1 mole O2 and 1 mole CO2

This means there will also react 1.9 mole of 02, to be formed 1.9 mole of CO2

mole CO2 = mass CO2 / Molar mass CO2

mass CO2 = 1.9 mole CO2 * 44g/mole =<u>83.6g CO2</u>

In this reaction 18.2 g of O2 remained unreacted

we can control this: 79g - 18.2 g = 60.8g

1.9 mole * 32g/mol = 60.8g

So there is83.6g CO2 produced

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Read 2 more answers
When adjusted for any changes in ΔHΔH and ΔSΔS with temperature, the standard free energy change ΔG∘TΔGT∘Delta G_{T}^{\circ} at
STALIN [3.7K]

The equilibrium constant is 0.0022.

Explanation:

The values given in the problem is

ΔG° = 1.22 ×10⁵ J/mol

T = 2400 K.

R = 8.314 J mol⁻¹ K⁻¹

The Gibbs free energy should be minimum for a spontaneous reaction and equilibrium state of any reaction is spontaneous reaction. So on simplification, the thermodynamic properties of the equilibrium constant can be obtained as related to Gibbs free energy change at constant temperature.

The relation between Gibbs free energy change with equilibrium constant is ΔG° = -RT ln K

So, here K is the equilibrium constant. Now, substitute all the given values in the corresponding parameters of the above equation.

We get,

1.22 * 10^{5} = - 8.314* 2400 * ln K

\\ 1.22 * 10^{5} = -19953.6 * ln K

ln K = \frac{-1.22*10^{5} }{19953.6} =-6.114\\\\k =e^{-6.114}=0.0022

So, the equilibrium constant is 0.0022.

4 0
3 years ago
A cubic piece of platinum metal (specific heat capacity = 0.1256 J/°C・g) at 200.0°C is dropped into 1.00 L of deuterium oxide ('
polet [3.4K]

Answer:

a=5.65cm

Explanation:

Hello,

In this case, for this heat transfer process in which the heat lost by the hot platinum is gained by the cold deuterium oxide based on the equation:

Q_{Pt}=-Q_{Deu}

We can represent the heats in terms of mass, heat capacities and temperatures:

m_{Pt}Cp_{Pt}(T_f-T_{Pt})=-m_{Deu}Cp_{Deu}(T_f-T_{Deu})

Thus, we solve for the mass of platinum:

m_{Pt}=\frac{-m_{Deu}Cp_{Deu}(T_f-T_{Deu})}{Cp_{Pt}(T_f-T_{Pt})} \\\\m_{Pt}=\frac{-1.00L*1110g/L*4.211J/(g\°C)*(41.9-25.5)\°C}{0.1256J/(g\°C)*(41.9-200.0)\°C} \\\\m_{Pt}=3860.4g

Next, by using the density of platinum we compute the volume:

V_{Pt}=\frac{3860.4g}{21.45g/cm^3}\\ \\V_{Pt}=180cm^3

Which computed in terms of the edge length is:

V=a^3

Therefore, the edge length turns out:

a=\sqrt[3]{180cm^3}\\ \\a=5.65cm

Best regards.

6 0
3 years ago
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