<span>ANSWER:
Electrical energy accelerates the electrons in the neon gas. The gas ionizes and becomes plasma, containing both positive and negative ions.</span>
Answer:
Mole fraction H₂ = 0.29
Partial pressure of H₂ → 88.5 kPa
Explanation:
You need to know this relation to solve this:
Moles of a gas / Total moles = Partial pressure of the gas / Total pressure
Total moles = 3 mol + 7.3 mol → 10.3 moles
Mole fraction H₂ → 3 moles / 10.3 moles = 0.29
Mole fraction = Partial pressure of the gas / Total pressure
0.29 . 304 kPa = Partial pressure of H₂ → 88.5 kPa
Answer:
H₂ is excess reactant and O₂ the limiting reactant
Explanation:
Based on the chemical reaction:
2H₂(g) + O₂(g) → 2H₂O
<em>2 moles of H₂ react per mole of O₂</em>
<em />
To find limiting reactant we need to convert the mass of each reactant to moles:
<em>Moles H₂ -Molar mass: 2.016g/mol-:</em>
10g H₂ * (1mol / 2.016g) = 4.96 moles
<em>Moles O₂ -Molar mass: 32g/mol-:</em>
22g O₂ * (1mol / 32g) = 0.69 moles
For a complete reaction of 0.69 moles of O₂ are needed:
0.69mol O₂ * (2mol H₂ / 1mol O₂) = 1.38 moles of H₂
As there are 4.96 moles,
<h3>H₂ is excess reactant and O₂ the limiting reactant</h3>
Answer:
b. muscles is the answer
Explanation:
muscles is a meat, animals eat meat
Answer:
0.78 M
Explanation:
First, we need to know which is the value of Kc of this reaction. In order to know this, we should take the innitial values of N2, O2 and NO and write the equilibrium constant expression according to the reaction. Doing this we have the following:
N2(g) + O2(g) <------> 2NO(g) Kc = ?
Writting Kc:
Kc = [NO]² / [N2] * [O2]
Replacing the given values we have then:
Kc = (0.6)² / (0.2)*(0.2)
Kc = 9
Now that we have the Kc, let's see what happens next.
We add more NO, until it's concentration is 0.9 M, this means that we are actually altering the reaction to get more reactants than product, which means that the equilibrium is being affected. If this is true, in the reaction when is re established the equilibrium, we'll see a loss in the concentration of NO and a gaining in concentrations of the reactants. This can be easily watched by doing an ICE chart:
N2(g) + O2(g) <------> 2NO(g)
I: 0.2 0.2 0.9
C: +x +x -2x
E: 0.2+x 0.2+x 0.9-2x
Replacing in the Kc expression we have:
Kc = [NO]² / [N2] * [O2]
9 = (0.9-2x)² / (0.2+x)*(0.2+x) ----> (this can be expressed as 0.2+x)²
Here, we solve for x:
9 = (0.9-2x)² / (0.2+x)²
√9 = (0.9-2x) / (0.2+x)
3(0.2+x) = 0.9-2x
0.6 + 3x = 0.9 - 2x
3x + 2x = 0.9 - 0.6
5x = 0.3
x = 0.06 M
This means that the final concentration of NO will be:
[NO] = 0.9 - (2*0.06)
[NO] = 0.78 M
