4NH3(0) + 302(9) = 2N2(9) + 6H2O(g) If a chemist mixed 2.930 moles of ammonia (NH3) gas with 1.978 moles of oxygen (02) gas in a 7.95 Liter air tight metal flask at 127°C and after the pressures stabilizes at equilibrium and then draws an air sample and found 0.683 moles of nitrogen (N2) gas. Now that you have calculated the Ke value which way will the chemical equilibrium shift?

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4NH39) + 302(g) = 2N2(9) + 6H2O(9)
Q8
If a chemist mixed 2.930 moles of ammonia (NH3) gas with 1.978 moles of oxygen
(02) gas in a 7.95 Liter air tight metal flask at 127°C and after the pressures stabilizes
at equilibrium and then draws an air sample and found 0.683 moles of nitrogen (N2)
gas.
Now that you have calculated the Ke value which way will the chemical equilibrium
shift?
For the chemical reaction shown below select the correct the equilibrium constant expression for
Q1 K
4NH3(9) + 302(9)
2 2N2(9)
6H20(g)
+
O (PN2)? x (PH20)6 / (PNH3)ª x (Po2)3
Transcribed Image Text:4NH39) + 302(g) = 2N2(9) + 6H2O(9) Q8 If a chemist mixed 2.930 moles of ammonia (NH3) gas with 1.978 moles of oxygen (02) gas in a 7.95 Liter air tight metal flask at 127°C and after the pressures stabilizes at equilibrium and then draws an air sample and found 0.683 moles of nitrogen (N2) gas. Now that you have calculated the Ke value which way will the chemical equilibrium shift? For the chemical reaction shown below select the correct the equilibrium constant expression for Q1 K 4NH3(9) + 302(9) 2 2N2(9) 6H20(g) + O (PN2)? x (PH20)6 / (PNH3)ª x (Po2)3
Q2
For the chemical reaction shown below select the correct the equilibrium constant expression for
4NH3(g) + 302(g) = 2N2(9) + 6H2O(g)
Q3
4NH3(9) + 302(9) = 2N2(9) + 6H2O(9)
If a chemist mixed 2.930 moles of ammonia (NH3) gas with 1.978 moles of oxygen
(02) gas in a 7.95 Liter air tight metal flask at 127°C and after the pressures stabilizes
at equilibrium and then draws an air sample and found 0.683 moles of nitrogen (N2)
gas then how many moles of ammonia (NH3) is present at equilibrium?
4NH3(9) + 302(9) 2N2(9) + 6H2O(g)
Q4
If a chemist mixed 2.930 moles of ammonia (NH3) gas with 1.978 moles of oxygen
(02) gas in a 7.95 Liter air tight metal flask at 127°C and after the pressures stabilizes
at equilibrium and then draws an air sample and found 0.683 moles of nitrogen (N2)
gas then how many moles of water (H,O) is present at equilibrium?
4NH3(9) + 302(g) 2N2(9) + 6H2O(9)
Q5
If a chemist mixed 2.930 moles of ammonia (NH3) gas with 1.978 moles of oxygen
(02) gas in a 7.95 Liter air tight metal flask at 127°C and after the pressures stabilizes
at equilibrium and then draws an air sample and found 0.683 moles of nitrogen (N2)
gas then what is the Molarity (M) of oxygen (02) gas present at equilibrium?
4NH3(9) + 302(g) = 2N2(9) + 6H20(g)
Q6
If a chemist mixed 2.930 moles of ammonia (NH3) gas with 1.978 moles of oxygen
(02) gas in a 7.95 Liter air tight metal flask at 127°C and after the pressures stabilizes
at equilibrium and then draws an air sample and found 0.683 moles of nitrogen (N2)
gas.
Now that you have the moles and Molarity of each reactant and product calculate the
Kc value for this chemical equilibrium reaction.
4NH3(9) + 302(9) = 2N2(9) + 6H2O(9)
Q7
If a chemist mixed 2.930 moles of ammonia (NH3) gas with 1.978 moles of oxygen
(02) gas in a 7.95 Liter air tight metal flask at 127°C and after the pressures stabilizes
at equilibrium and then draws an air sample and found 0.683 moles of nitrogen (N2)
gas.
Now that you have calculated the Kc value use the equation K, = Ke (RT)A" to solve
for the Kp value.
Transcribed Image Text:Q2 For the chemical reaction shown below select the correct the equilibrium constant expression for 4NH3(g) + 302(g) = 2N2(9) + 6H2O(g) Q3 4NH3(9) + 302(9) = 2N2(9) + 6H2O(9) If a chemist mixed 2.930 moles of ammonia (NH3) gas with 1.978 moles of oxygen (02) gas in a 7.95 Liter air tight metal flask at 127°C and after the pressures stabilizes at equilibrium and then draws an air sample and found 0.683 moles of nitrogen (N2) gas then how many moles of ammonia (NH3) is present at equilibrium? 4NH3(9) + 302(9) 2N2(9) + 6H2O(g) Q4 If a chemist mixed 2.930 moles of ammonia (NH3) gas with 1.978 moles of oxygen (02) gas in a 7.95 Liter air tight metal flask at 127°C and after the pressures stabilizes at equilibrium and then draws an air sample and found 0.683 moles of nitrogen (N2) gas then how many moles of water (H,O) is present at equilibrium? 4NH3(9) + 302(g) 2N2(9) + 6H2O(9) Q5 If a chemist mixed 2.930 moles of ammonia (NH3) gas with 1.978 moles of oxygen (02) gas in a 7.95 Liter air tight metal flask at 127°C and after the pressures stabilizes at equilibrium and then draws an air sample and found 0.683 moles of nitrogen (N2) gas then what is the Molarity (M) of oxygen (02) gas present at equilibrium? 4NH3(9) + 302(g) = 2N2(9) + 6H20(g) Q6 If a chemist mixed 2.930 moles of ammonia (NH3) gas with 1.978 moles of oxygen (02) gas in a 7.95 Liter air tight metal flask at 127°C and after the pressures stabilizes at equilibrium and then draws an air sample and found 0.683 moles of nitrogen (N2) gas. Now that you have the moles and Molarity of each reactant and product calculate the Kc value for this chemical equilibrium reaction. 4NH3(9) + 302(9) = 2N2(9) + 6H2O(9) Q7 If a chemist mixed 2.930 moles of ammonia (NH3) gas with 1.978 moles of oxygen (02) gas in a 7.95 Liter air tight metal flask at 127°C and after the pressures stabilizes at equilibrium and then draws an air sample and found 0.683 moles of nitrogen (N2) gas. Now that you have calculated the Kc value use the equation K, = Ke (RT)A" to solve for the Kp value.
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