Answer:
M(AgNO3) = 169,8731 g mol
Explanation:
M(Ag) = 107,8682 g mol
M(N) = 14.0067 g mol
M(O) = 15.9994 g mol
M(AgNO3) = M(Ag) + M(N) + 3 x M(O) = 107.8682 + 14.0067 + 3 x 15.9994 =
= 169,8731 g mol
Just sum up all the respective elements' molar mass.
Metals are on the left side of the table and nonmetals are on the left with metalloids between them. And the noble gases are all in group 18 of the periodic table.
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
mass MgCl₂ = mol x MM MgCl₂ = 0.05 x 95.211 g/mol = 4.76 g
mass Cl in MgCl₂ :
= (2 x AM Cl)/MM MgCl₂ x mass MgCl₂
= (2 x 35.5 g/mol)/95.211 g/mol x 4.76
= 3.55 g
% mass Cl in the mixture :
= (mass Cl / mass mixture) x 100%
= 3.55 / 9.8 x 100%
= 36.22%
