<h3>
Answer:</h3>
0.144 moles
<h3>
Explanation:</h3>
- The relationship between mass of a compound, number of moles and molar mass of the compound is given by;
- Number of moles = Mass ÷ Molar mass
- Molar mass is equivalent to the relative formula mass of the compound that is calculated the atomic masses of the elements making the compound.
In this case;
Our compound, KClO3 will have a molar mass of;
= 39 + 35.5 + 4(16)
= 138.5 g/mol
Mass of KClO3 is 20 g
Therefore;
Number of moles = 20 g ÷ 138.5 g/mol
= 0.144 moles
Thus, the number of moles in 20 g of KClO3 is 0.144 moles
Answer:Hence, the bond length in HCl is 125 pm.
Explanation:
Bond length : It is an average distance between the nuclei of two bonded atoms in a molecule.
Also given that bond length is the distance between the centers of two bonded atoms. on the potential energy curve, the bond length is the inter-nuclear distance between the two atoms when the potential energy of the system reaches its lowest value. Beyond this if atoms come closer to each other then their will be repulsion between them.
So, the bond length between the Hydrogen and Chlorine atom in HCl molecule is :
Hence, the bond length in HCl is 125 pm.
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
I think it would be C.100.5cm or D.100.5ml hope that helps
