Answer:
Reactants: Ethanol (C₅H₅OH), Oxygen (O₂)
Products: Carbon dioxide (CO₂) and Water (H₂O)
Explanation:
Let’s rewrite each part of the described situation in chemical equation terms.
Ethanol - C₂H₅OH
is burned in the prescence of - +
oxygen - O₂
producing - →
carbon dioxide - CO₂
and - +
water - H₂O
Chemical equation:
C₂H₅OH + O₂ → CO₂ + H₂O
The terms on the left side of a chemical equation are reactants, and the terms on the right side of a chemical equation are products.
Therefore, since we know this, ethanol (C₂H₅OH) & oxygen (O₂) are the reactants, and carbon dioxide (CO₂) & water (H₂O) are the products.
Answer:
0.238 M
Explanation:
A 17.00 mL sample of the dilute solution was found to contain 0.220 M ClO₃⁻(aq). The concentration is an intensive property, so the concentration in the 52.00 mL is also 0.220 M ClO₃⁻(aq). We can find the initial concentration of ClO₃⁻ using the dilution rule.
C₁.V₁ = C₂.V₂
C₁ × 24.00 mL = 0.220 M × 52.00 mL
C₁ = 0.477 M
The concentration of Pb(ClO₃)₂ is:

Answer:
ΔH = - 2020.57 kJ/mol
Explanation:
Given that :
mass of propanol = 1.685 g
the molar molar mass = 60 g/mol
Thus; the number of moles = mass/molar mass
= 1.685 g/60 g/mol
= 0.028 g/mol
However ;
ΔH = heat capacity C × Δ T
Given that:
The temperature increases from 298.00 K to 302.16 K.
Then ;
Δ T = 302.16 K - 298.00 K
Δ T = 4.16 K
heat capacity C = 13.60 kJ/K
∴
ΔH = 13.60 kJ/K × 4.16 K
ΔH = 56.576 kJ
The equation of the given reaction can be represented as :

Thus for 0.028 mol of heat liberated; ΔH = 56.576 kJ
For 1 mole of heat liberated now:
ΔH = 56.576 kJ/0.028 mol
ΔH = 2020.57 kJ/mol
SInce , Heat is liberated, the reaction undergoes an exothermic reaction thus;
ΔH = - 2020.57 kJ/mol
Answer : The pH of the solution is, 2.67
Explanation :
The equilibrium chemical reaction is:

Initial conc. 0.450 0 0
At eqm. (0.450-x) x x
As we are given:

The expression for equilibrium constant is:

Now put all the given values in this expression, we get:


The concentration of
= x = 0.00212 M
Now we have to calculate the pH of solution.
![pH=-\log [H^+]](https://tex.z-dn.net/?f=pH%3D-%5Clog%20%5BH%5E%2B%5D)


Therefore, the pH of the solution is, 2.67