1) At tne same temperature and with the same volume, initially the chamber 1 has the dobule of moles of gas than the chamber 2, so the pressure in the chamber 1 ( call it p1) is the double of the pressure of chamber 2 (p2)
=> p1 = 2 p2
Which is easy to demonstrate using ideal gas equation:
p1 = nRT/V = 2.0 mol * RT / 1 liter
p2 = nRT/V = 1.0 mol * RT / 1 liter
=> p1 / p2 = 2.0 / 1.0 = 2 => p1 = 2 * p2
2) Assuming that when the valve is opened there is not change in temperature, there will be 1.00 + 2.00 moles of gas in a volumen of 2 liters.
So, the pressure in both chambers (which form one same vessel) is:
p = nRT/V = 3.0 mol * RT / 2liter
which compared to the initial pressure in chamber 1, p1, is:
p / p1 = (3/2) / 2 = 3/4 => p = (3/4)p1
So, the answer is that the pressure in the chamber 1 decreases to 3/4 its original pressure.
You can also see how the pressure in chamber 2 changes:
p / p2 = (3/2) / 1 = 3/2, which means that the pressure in the chamber 2 decreases to 3/2 of its original pressure.
Answer:
Explanation:
Group one elements are alkali metals. All alkali metal have one valance electron. They loses their one valance electron and from cation with charge of +1.
Charges on group one.
Hydrogen = +1
Lithium = +1
Sodium = +1
Potassium = +1
Rubidium = +1
Cesium = +1
Francium = +1
Group two elements are alkaline earth metals. All alkaline earth metal have two valance electron. They loses their two valance electron and from cation with charge of +2.
Charges on group two.
Beryllium = +2
Magnesium = +2
Calcium = +2
Strontium = +2
Barium= +2
Radium = +2
Group 13 elements are boron family. All elements have three valance electrons. They loses their three valance electron and from cation with charge of +3.
Charges on group 13.
Boron = +3
Aluminium = +3
Gallium = +3
Indium = +3
Thallium= +3
Group 13 elements are also shows +1 charge by losing one valance electron.
Answer:
The answer to your question is
1.-Fe₂O₃
2.- 280 g
3.- 330 g
Explanation:
Data
mass of CO = 224 g
mass of Fe₂O₃ = 400 g
mass of Fe = ?
mass of CO₂
Balanced chemical reaction
Fe₂O₃ + 3CO ⇒ 2Fe + 3CO₂
1.- Calculate the molar mass of Fe₂O₃ and CO
Fe₂O₃ = (56 x 2) + (16 x 3) = 160 g
CO = 12 + 16 = 28 g
2.- Calculate the proportions
theoretical proportion Fe₂O₃ /3CO = 160/84 = 1.90
experimental proportion Fe₂O₃ / CO = 400/224 = 1.78
As the experimental proportion is lower than the theoretical, we conclude that the Fe₂O₃ is the limiting reactant.
3.- 160 g of Fe₂O₃ --------------- 2(56) g of Fe
400 g of Fe₂O₃ --------------- x
x = (400 x 112) / 160
x = 280 g of Fe
4.- 160 g of Fe₂O₃ --------------- 3(44) g of CO₂
400 g of Fe₂O₃ -------------- x
x = (400 x 132)/160
x = 330 gr
Answer:
1) Length - Meter
2) Mass - Pound
3) Time - Minute
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