Answer:
The transition with the greatest distance is 5p → 1s, which is n = 5 going to n = 1. This means this transition also has the largest energy and frequency. Therefore, the electron transition that produces light of the highest frequency in the hydrogen atom is a. 5p → 1s.
Explanation:
The energy requirement order for excitation for different transitions is as follows. n→∏* transition requires lowest energy while σ→σ* requires highest amount of energy
Answer:
The Two reasons are :-
1. Combustion efficiency
2. Standardization of test conditions as non-luminous flame.
Explanation:
A luminous flame isn't suitable for heating as it gives out soot (A black powdery or flaky substance consisting largely of amorphous carbon, produced by the incomplete burning of organic matter)
Answer:
The mass of silver chloride produced = 2.202 g
Explanation:
Equation of the reaction is given below
3AgNO₃(aq) + FeCl₃(aq) ----> 3AgCl(s) + Fe(NO₃)₃(aq)
molar mass of AgNO₃ = 170 g/mol
molar mass of FeCl₃ = 233.5 g/mol
molar mass of AgCl = 143.5 g/mol
3 moles of silver nitrate reacts with 1 mole of iron (iii) chloride to give 3 moles of silver nitrate
4.15 grams of AgNO₃ = 4.15/170 = 0.0244 moles of AgNO₃
1.11 grams of FeCl₃ = 1.11/233.5 = 0.0047 moles of FeCl₃
mole ratio of AgNO₃ to FeCl₃ = 0.0244/0.0047 = 5 : 1
therefore, FeCl₃ is the limiting reactant
0.0047 moles of FeCl₃ reacting will produce 0.0047 * 3 moles of AgCl = 0.0141 moles of AgCl
0.0141 moles of AgCl = 0.0141 * 143.5 g of AgCl = 2.02 g of AgCl =
Therefore mass of silver chloride produced = 2.202 g
Answer: 1) Maximum mass of ammonia 198.57g
2) The element that would be completely consumed is the N2
3) Mass that would keep unremained, is the one of the excess Reactant, that means the H2 with 3,44g
Explanation:
- In order to calculate the Mass of ammonia , we first check the Equation is actually Balance:
N2(g) + 3H2(g) ⟶2NH3(g)
Both equal amount of atoms side to side.
- Now we verify which reagent is the limiting one by comparing the amount of product formed with each reactant, and the one with the lowest number is the limiting reactant. ( Keep in mind that we use the molecular weight of 28.01 g/mol N2; 2.02 g/mol H2; 17.03g/mol NH3)
Moles of ammonia produced with 163.3g N2(g) ⟶ 163.3g N2(g) x (1mol N2(g)/ 28.01 g N2(g) )x (2 mol NH3(g) /1 mol N2(g)) = 11.66 mol NH3
Moles of ammonia produced with 38.77 g H2⟶ 38.77 g H2 x ( 1mol H2/ 2.02 g H2 ) x (2 mol NH3 /3 mol H2 ) = 12.79 mol NH3
- As we can see the amount of NH3 formed with the N2 is the lowest one , therefore the limiting reactant is the N2 that means, N2 is the element that would be completey consumed, and the maximum mass of ammonia will be produced from it.
- We proceed calculating the maximum mass of NH3 from the 163.3g of N2.
11.66 mol NH3 x (17.03 g NH3 /1mol NH3) = 198.57 g NH3
- In order to estimate the mass of excess reagent, we start by calculating how much H2 reacts with the giving N2:
163.3g N2 x (1mol N2/28.01 g N2) x ( 3 mol H2 / 1 mol N2)x (2.02 g H2/ 1 mol H2) = 35.33 g H2
That means that only 35.33 g H2 will react with 163.3g N2 however we were giving 38.77g of H2, thus, 38.77g - 35.33 g = 3.44g H2 is left
