Dalton's Law of Partial Pressures, commonly applied to ideal gases, explains that the partial pressures of individual, non-reacting gases are equal to the total pressure exerted by the gas mixture. The given gas mixture composed of 90% argon and 10% carbon dioxide has the following partial pressures: 3.6 atm for argon and 0.4 atm for carbon dioxide (answer).
On temperature 25°C (298,15K) and pressure of 1 atm each gas has same amount of substance:
n(gas) = p·V ÷ R·T = 1 atm · 20L ÷ <span>0,082 L</span>·<span>atm/K</span>·<span>mol </span>· 298,15 K
n(gas) = 0,82 mol.
1) m(He) = 0,82 mol · 4 g/mol = 3,28 g.
d(He) = 10 g + 3,28 g ÷ 20 L = 0,664 g/L.
2) m(Ne) = 0,82 mol · 20,17 g/mol = 16,53 g.
d(Ne) = 26,53 g ÷ 20 L = 1,27 g/L.
3) m(CO) = 0,82 mol ·28 g/mol = 22,96 g.
d(CO) = 32,96 g ÷ 20L = 1,648 g/L.
4) m(NO) = 0,82 mol ·30 g/mol = 24,6 g.
d(NO) = 34,6 g ÷ 20 L = 1,73 g/L.
Let's look at the molar weight of the answers:
NO is 30 g/mol
NO2 is 46
N2O is 44
N2O4 is 124
<span>We have the grams of the product, so we need the moles in order to calculate the molar weight. We us PV=nRT for this, assuming standard temperature and pressure. </span>
You were given the liters (.120L)
Std pressure is 1 atmosphere
You're looking for n, the number of moles
<span>Temp is 293.15 kelvin, thats standard </span>
And r is the gas constant in liters-atm per mol kelvin
(.120 liters)(1atm)=n(293.15K)(.08206)
Solving for n is .0049883835 mol
<span>.23g divided by .0049883 mol is about 46g/mol. You're answer is B I think, NO2
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Answer:
sublime, melt, condense, deposit
Explanation:
1. When ice is warmed at a steady pressure 0.00512 atm, it will be sublime.
2. It will be melt when ice is warmed at a consistent pressure of 1 atm.
3. If water vapour pressure is continued to increase at a temperature of 100 C, it will be condense.
4. If water vapour pressure is continued to increase at a temperature of -50 C, it will be deposited.
Answer:
Explanation:
Fe⁺² (aq) + 2e⁻ = Fe (s) ; E⁰ = - .44 V
Fe⁺³ (aq) + e⁻ = ® Fe²⁺ (aq) ; E⁰ = + .77 V
Reduction potential of second reaction is more , so it will take place , ie Fe⁺³ will be reduced and Fe will be oxidised .
So reaction in the combined cell will be
2Fe⁺³ + Fe = 3Fe⁺²
cell potential = .77 - ( - .44 )
= 1.21 V .