Answer:
The answer to your question is: 6.55 x 10 ²³ atoms of Br
Explanation:
CH2Br2 = 37.9 g
MW CH2Br2 = (12 x 1) + (2 x 1) + (80 x 2) = 174 g
174 g of CH2Br2 ------------------ 160 g of Br2
37.9 g of CH2Br2 --------------- x
x = 37.9 x 160/174 = 34.85 g of Br
1 mol of Br ----------------- 160 g Br2
x ---------------- 174 g Be2
x = 174 x 1 /160 = 1.088 mol of Br2
1 mol of Br ----------------- 6.023 x 10 ²³ atoms
1.088 mol of Br ------------- x
x = 1.088 x 6.023 x 10 ²³ / 1 = 6.55 x 10 ²³ atoms
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
To find the ratio of the the combination for the ion, write the charge of the cation as the subscript for the anion, and the charge of the anion as the subscript of the cation. This will make the charges effectively cancel and you will be left with a neutral ionic compound. Remember, that an ionic compound is made up of a metal and a nonmetal.
For Ca2+ and Cl-, you will get the neutral compound to be CaCl₂.
