This is right
Reduce the concentration of NICO4
Answer:
no picture you need add it
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 standred units of measurements are important to scientists because they are the same measurements that all scientist take to make sure that they all have the correct results.
Answer:
16 mg of water can be produced by 7.1×10⁻³ g of CH₄
Explanation:
This is the reaction:
CH₄ + 2O₂ → CO₂ + 2H₂O
In a combustion, oxgen is a reactant with another compound, and the products are always water and carbon dioxide
1 mol of methane can produce 2 moles of water. Ratio is 1:2
If we convert the mass to moles → 7.1×10⁻³ g . 1 mol/ 16g = 4.43×10⁻⁴ mol
In this reaction I would produce the double of moles I have from methane, so If I have 4.43×10⁻⁴ moles of methane I would produce 8.87×10⁻⁴ moles of water.
What mass of water, corresponds to 8.87×10⁻⁴ moles?
8.87×10⁻⁴ mol . 18g / 1mol = 0.016 g which is actually the same as 16 mg
