Answer:
Explanation:
At , the heat of vaporization of water is given by:
The water here condenses and gives off heat given by the product between its mass and the heat of vaporization:
The block of aluminum absorbs heat given by the product of its specific heat capacity, mass and the change in temperature:
According to the law of energy conservation, the heat lost is equal to the heat gained:
or:
Rearrange for the final temperature:
We obtain:
Then:
Liquids have more kinetic energy in their particles compared to solids. this allows the particles to move more freely, hence why they are fluids
Liquids diffuse from a region of high concentration to a region of low concentration, until equilibrium is reached
When heat is applied the particles gain more kinetic energy so they now have enough energy to overcome the bonds holding them in the liquid. this means they can evaporate off
Find the mass of C in the 2.657 g CO2:
(2.657 g CO2) / (44.01 g/mol) = 0.06037 mol CO2
Since each mole of CO2 also has 1 mole of C, this is equivalent to 0.06037 mol C.
Find the mass of H in the 1.089 g H2O:
(1.089 g H2O) / (18.02 g/mol) = 0.06043 mol H2O
Since 1 mol H2O has 2 mol H, this is equivalent to (0.06043)*2 = 0.1209 mol H.
Taking the ratio of H to C: 0.1209 / 0.06037 = 2.002 ~ 2
Therefore, the empirical formula of isobutylene is CH2.
Answer:
2.5 × 10⁻⁵ M H₃O⁺ and 4.0 × 10⁻¹⁰ M OH⁻.
Explanation:
<em>∵ pH = - log[H₃O⁺]</em>
∴ 4.6 = - log[H₃O⁺].
∴ log[H₃O⁺] = - 4.6.
∴ [H₃O⁺] = 2.51 x 10⁻⁵.
∵ [H₃O⁺][OH⁻] = 10⁻¹⁴.
[H₃O⁺] = 2.51 x 10⁻⁵ M.
∴ [OH⁻] = 10⁻¹⁴/[H₃O⁺] = 10⁻¹⁴/(2.51 x 10⁻⁵ M) = 3.98 × 10⁻¹⁰ M ≅ 4.0 × 10⁻¹⁰ M.
<em>So, the right choice is: 2.5 × 10⁻⁵ M H₃O⁺ and 4.0 × 10⁻¹⁰ M OH⁻.</em>
Answer:
1. pH = 1.23.
2.
Explanation:
Hello!
1. In this case, for the ionization of H2C2O4, we can write:
It means, that if it is forming a buffer solution with its conjugate base in the form of KHC2O4, we can compute the pH based on the Henderson-Hasselbach equation:
Whereas the pKa is:
The concentration of the base is 0.347 M and the concentration of the acid is 0.347 M as well, as seen on the statement; thus, the pH is:
2. Now, since the addition of KOH directly consumes 0.070 moles of acid, we can compute the remaining moles as follows:
It means that the acid remains in excess yet more base is yielded due to the effect of the OH ions provided by the KOH; therefore, the undergone chemical reaction is:
Which is also shown in net ionic notation.
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