Answer: The predicted change in the boiling point of water is Δt = 0.0148 °C
Solution:
We will use the equation for boiling point elevation Δt
Δt = i Kb m
where the van't Hoff Factor i is equal to 3 since one molecule of barium chloride in aqueous solution will produce one Ba2+ ion and two Cl- ions. The molality m of the solution of 4.00 g of barium chloride dissolved in 2.00 kg of water can be calculated using the molar mass of barium chloride:
m = [4.00g BaCl2 * (1 mol BaCl2 / 208.233g BaCl2)] / 2.00kg H2O
= 0.009605 mol/kg
Therefore, the amount Δt the boiling point increases is
Δt = i Kb m
= (3) (0.512 °C·kg/mol) (0.009605 mol/kg)
= 0.0148 °C
We can also find the new boiling point T for the solution since we know that pure water boils at 100 °C:
Δt = T - 100°C T = Δt + 100°C = 0.0148 °C + 100°C = 100.0148°C
4HCl + MnO₂ → MnCl₂ + 2H₂O + Cl₂
2Cl⁻ → Cl₂ + 2e⁻
MnO₂ + 4H⁺ + 2e⁻ → Mn²⁺ + 2H₂O
2Cl⁻ + MnO₂ + 4H⁺ → Cl₂ + Mn²⁺ + 2H₂O
Elements in the same group<span> in the </span>periodic table<span> have </span>similar<span> chemical properties. This is because their atoms have the </span>same<span> number of electrons in the highest occupied energy level. </span>Group<span> 1 </span>elements<span> are reactive metals called the alkali metals.</span>Group<span> 0 </span>elements<span> are unreactive non-metals called the noble gases.</span>
I believe the answer is "ions" because the definition of that word is "an atom or molecule with a net electric charge due to the loss or gain of one or more electrons."
Explanation:
We know that relation between and is as follows.
= 14
As it is given that is 8.18. Therefore, calculate the value of as follows.
= 14
= 14
= 14 - 8.18
= 5.82
Similarly, as value of pH is given as 7.18. Therefore, value of pOH will be as follows.
pH + pOH = 14
7.18 + pOH = 14
pOH = 6.82
Let us take that B represents the enzyme. Hence, its reaction with proton will be as follows.
(protonated active enzyme)
Hence, pOH =
6.82 = 5.82 +
= 10
Therefore, percentage of active enzyme = % =
% = 90.9%
Thus, we can conclude that 90.9% is the percentage of the enzyme which is active in a buffer at pH 7.18.