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
Answer:
True
Explanation:
The given statement is true. The elements on the upper right side of periodic table accept the electrons because of higher electronegativity values and form anions.
For example:
Consider the halogens. All halogens accept the one electron per atom to complete the octet and form anion. when they combine with metals metals loses their electrons to halogens atoms and form cations while halogens accept the electrons and form anion.
Consider the sodium chloride salt. The sodium metal loses its one valance electron which is accepted by chlorine.
Na⁺Cl⁻
Magnesium bromide:
To complete the octet bromine required one electron. By gaining the one electron it showed -1 oxidation state.
When bromine combine with magnesium both formed MgBr₂.
The one atom of Mg have +2 oxidation state while bromine have -1 that's why to make the overall compound neutral two bromine atoms combine with one atom of magnesium.
Answer:
MgC15
Explanation:
because of the weight it carries
Electrochemical cell representation for above reaction is,
Br-/Br2//I2/I-
Reaction at Anode: Br2 + 2e- → 2Br- (1)
Reaction at Cathode: 2I- → I2 + 2e- (2)
Standard reduction potential for Reaction 1 = Ered(anode) = 1.066 v
Standard reduction potential for Reaction 2 = Ered(cathode) = 0.535 v
Eo cell = Ered(cathode) - Ered(anode)
= 0.535 - 1.066
= -0.531v
Now, we know that ΔGo = -nF (Eo cell) ..............(3)
Also, ΔGo = RTln(K) ..........(4)
Equation 3 and 4 we get,
ln (K) = nF (Eo cell) / RT
= 2 X 96500 X (-0.531)/ (8.314 X 298)
∴ K = 1.085 X 10^-18.