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:
The molecules in a solid are close together and they vibrate in place. The molecules in a liquid move quickly and farther apart from eachother. The particles in a gas move freely at high speeds and slide past eachother.
We need the dissociation constant of benzoic acid which is 6.3x10^'5. Then using the dissociation formula, ka = x2 / (Mo - x) where Mo is the initial concentration. x is determined then. percent ionization is computed as (x/Mo)* 100%. This is then the final answer.
Speed = distance/time
=> speed = 817/19 m/sec = 43 m/sec
hope it helps!
Answer:
43.2 g of table salt occupies 20.0 cm2of space- Appropriate for calculating density
150 g of iron density of iron 79.0 g/cm3- Appropriate for calculating volume
21.0 mL of methanol density of methanol 0.79g/mL - Appropriate for calculating mass
3.00 g of white pine, density of white pine 0.40 g/cm3- Appropriate for calculating volume
5.00 cm,3 of table salt weighs 10.8g-Appropriate for calculating density
Explanation:
Density is defined as mass per unit volume. Hence in science;
Density= mass/volume
When mass and volume are both given, the information can be used to determine the density of a given substance.
When volume and density are both given, the information can be used to determine the mass of a given substance.
When the mass and density of a substance are both given, the information can be used to determine the volume of the substance.
Hence the answers given above.
