Answer: (a) cesium and bromine :
(b) sulfur and barium: 
(c) calcium and fluorine :
Explanation:
For formation of a neutral ionic compound, the charges on cation and anion must be balanced. The cation is formed by loss of electrons by metals and anions are formed by gain of electrons by non metals.
The nomenclature of ionic compounds is given by:
1. Positive is written first.
2. The negative ion is written next and a suffix is added at the end of the negative ion. The suffix written is '-ide'.
(a) cesium and bromine: Here cesium is having an oxidation state of +1 called as 
cation and bromine 
is an anion with oxidation state of -1. Thus they combine and their oxidation states are exchanged and written in simplest whole number ratios to give neutral 
called as cesium bromide.
(b) sulfur and barium : Here barium is having an oxidation state of +2 called as 
cation and sulphur 
is an anion with oxidation state of -2. Thus they combine and their oxidation states are exchanged and written in simplest whole number ratios to give neutral . called as barium sulfide.
(c) calcium and fluorine: Here calcium is having an oxidation state of +2 called as 
cation and fluorine 
is an anion with oxidation state of -1. Thus they combine and their oxidation states are exchanged and written in simplest whole number ratios to give neutral . called as calcium fluoride.
The answer is 3. The releasing of energy means exothermic reaction. So the ΔH should be negative. And the greatest quantity of energy released means that the greatest number. So according to the table I, the answer is 3.
An electron in the 3s orbital. The order of electron orbital energy levels starting from lowest to highest is as follows: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p.
Answer:
Explanation:
Combustion reaction is given below,
C₂H₅OH(l) + 3O₂(g) ⇒ 2CO₂(g) + 3H₂O(g)
Provided that such a combustion has a normal enthalpy,
ΔH°rxn = -1270 kJ/mol
That would be 1 mol reacting to release of ethanol,
⇒ -1270 kJ of heat
Now,
0.383 Ethanol mol responds to release or unlock,
(c) Determine the final temperature of the air in the room after the combustion.
Given that :
specific heat c = 1.005 J/(g. °C)
m = 5.56 ×10⁴ g
Using the relation:
q = mcΔT
- 486.34 = 5.56 ×10⁴ × 1.005 × ΔT
ΔT= (486.34 × 1000 )/5.56×10⁴ × 1.005
ΔT= 836.88 °C
ΔT= T₂ - T₁
T₂ = ΔT + T₁
T₂ = 836.88 °C + 21.7°C
T₂ = 858.58 °C
Therefore, the final temperature of the air in the room after combustion is 858.58 °C
Answer:
2.77g of 
can be produced.
Explanation:
Use stoichiometry.
Balanced equation: 
⇒ 
3 : 1 **
0.0623** : 
= 0.0623
m= (0.020767)(133.34)
m= 2.77g
