It’s fluorine it has a big jump from the 6th to 7th ionisation energy.
It’s electronic configuration is 1s2 2s2 2p5 as it has 9 electrons and 7 electrons in outermost shell.
Therefore it’s in group 7, as after the 7 electrons are removed, the 8th electron is removed from a quantum shell closer to the nucleus.
This results in greater ionisation energy due to stronger electrostatic forces of attraction between positively charge nuclei and electrons ( + shorter distance)
Answer:
1) The limiting reactant is N₂ because it is present with the lower no. of moles than H₂.
2) The amount (in grams) of excess reactant H₂ = 4.39 g.
Explanation:
- Firstly, we should write the balanced equation of the reaction:
<em>N₂ + 3H₂ → 2NH₃.</em>
<em>1) To determine the limiting reactant of the reaction:</em>
- From the stichiometry of the balanced equation, 1.0 mole of N₂ reacts with 3.0 moles of H₂ to produce 2.0 moles of NH₃.
- This means that <em>N₂ reacts with H₂ with a ratio of (1:3).</em>
- We need to calculate the no. of moles (n) of N₂ (5.23 g) and H₂ (5.52 g) using the relation:<em> n = mass / molar mass.</em>
The no. of moles of N₂ in (5.23 g) = mass / molar mass = (5.23 g) / (28.00 g/mol) = 0.1868 mol.
The no. of moles of H₂ (5.52 g) = mass / molar mass = (5.52 g) / (2.015 g/mol) = 2.74 mol.
- From the stichiometry, N₂ reacts with H₂ with a ratio of (1:3).
The ratio of the reactants of N₂ (5.23 g, 0.1868 mol) to H₂ (5.52 g, 2.74 mol) is (1:14.67).
∴ The limiting reactant is N₂ because it is present with the lower no. of moles than H₂.
0.1868 mol of N₂ react completely with 0.5604 mol of H₂ and the remaining of H₂ is in excess.
<em>2) To determine the amount (in grams) of excess reactant of the reaction:</em>
- As showed in the part 1, The limiting reactant is N₂ because it is present with the lower no. of moles than H₂.
- Also, 0.1868 mol of N₂ react completely with 0.5604 mol of H₂ and the remaining of H₂ is in excess.
- The no. of moles are in excess of H₂ = 2.74 mol - 0.5604 mol (reacted with N₂) = 2.1796 mol.
- ∴ The amount (in grams) of excess reactant H₂ = n (excess moles) x molar mass = (2.1796 mol)((2.015 g/mol) = 4.39 g.
Answer:
17 kJ
Explanation:
Calculation for the Calculate the energy required to heat 0.60kg of ethanol from 2.2°C to 13.7°C.
Using this formula
q = mC∆T
Where,
q represent Energy
m represent Mass of substance=0.60kg=600g
C represent Specific heat capacity=2.44J·g−1K−1.
∆T represent change in Temperature=2.2°C to 13.7°C.
Let plug in the formula
q=(0.60 kg x 1000 g/kg)(2.44 J/gº)(13.7°C-2.2°C)
q = (600g)(2.44 J/gº)(11.5º)
q=16.836 kJ
q= 17 kJ (Approximately)
Therefore the energy required to heat 0.60kg of ethanol from 2.2°C to 13.7°C will be 17 kJ
The charge that 3 protons, 3 neutrons and 2 electrons have is that it will be a cation of Li with a charge of +1.
Li^+1.
Answer:
A) 4P + 5O₂ → 2P₂O₅
Explanation:
A) 4P + 5O₂ → 2P₂O₅
This equation is balanced. There are four phosphorus and ten oxygen atoms are on both side of equation.
B) 5P + 4O₂ → 2P₄O₅
This equation is not balanced. There are five phosphorus and eight oxygen atoms on left, eight phosphorus ten oxygen on right side of equation.
C) 2P + O₂ → P₂O₅
This equation is not balanced. There are two phosphorus, two oxygen atoms on left and two phosphorus five oxygen on right side of equation.
D) 4P + 2O₂ → 2P₄O₅
This equation is not balanced. There are four phosphorus, four oxygen atoms on left and eight phosphorus ten oxygen on right side of equation.
