Answer:
-2.3 ºC
Explanation:
Kf (benzene) = 5.12 ° C kg mol – 1
1st - We calculate the moles of condensed gas using the ideal gas equation:
n = PV / (RT)
P = 748/760 = 0.984 atm
T = 270 + 273.15 = 543.15 K
V = 4 L
R = 0.082 atm.L / mol.K
n = (0.984atm * 4L) / (0.082atm.L / K.mol * 543.15K) = 0.088 mol
Then, you calculate the molality of the solution:
m = n / kg solvent
m = 0.088 mol / 0.058 kg = 1.52mol / kg
Then you calculate the decrease in freezing point (DT)
DT = m * Kf
DT = 1.52 * 5.12 = 7.8 ° C
Knowing that the freezing point of pure benzene is 5.5 ºC, we calculate the freezing point of the solution:
T = 5.5 - 7.8 = -2.3 ºC
First let's convert the minutes to hours (multiply minutes by 60 to get hours):
30 × 60 = 1,800 drops/hour
Now that you know how much drops there are per hour, you can multiply this answer by 3 to work out how many drops there are in 3 hours:
1,800 × 3 = 5,400 drops
We know that 5 drops is equal to 1 ml, so we can divide 5,400 by 5 to work out the amount of ml:
5,400 ÷ 5 = 1,080 ml
Therefore, your final answer is 1,080 millilitres (ml)
Answer:
On the Moh's scale of hardness, aluminum oxide is positioned just below to diamond due to which it is considered as one of the hardest known compounds. This also shows that the compound exhibit an enormous amount of lattice energy, as to transform the oxide into its constituent ions, the energy is required to overcome.
Based on the chemical formula of the compound, that is, Al2O3, it is shown that the ions of Al3+ and O2- are kept close due to the activity of the strong electrostatic ionic bonds. The electrostatic forces and the ionic bonding between the ions are extremely robust due to the presence of the ions high charge density. Therefore, to dissociate the bonds, an enormous amount of energy is needed, and at the same time, a high amount of lattice energy is present.
Answer: 3
Explanation:
An oxide-reduction reaction or, simply, redox reaction, is a <u>chemical reaction in which one or more electrons are transferred between the reactants</u>, causing a change in their oxidation states, which is the hypothetical electric charge that the atom would have if all its links with different elements were 100% ionic.
For there to be a reduction-oxidation reaction, in the system there must be an element that yields electrons and another that accepts them:
-
The oxidizing agent picks up electrons and remains with a state of oxidation inferior to that which it had, that is, it is reduced.
- The reducing agent supplies electrons from its chemical structure to the medium, increasing its oxidation state, ie, being oxidized.
To balance a redox equation you must <u>identify the elements that are oxidized and reduced and the amount of electrons that they release or capture, respectively.
</u>
In the reaction that arises in the question the silver (Ag) is reduced <u>because it decreases its oxidation state from +1 to 0</u> and the aluminum (Al) is oxidized because <u>its oxidation state increases from 0 to +3</u>, releasing 3 electrons (e⁻). Then we can raise two half-reactions:
Ag⁺ + e⁻ → Ag⁰
Al⁰ → Al⁺³ + 3e⁻
In order to obtain the balanced equation, we must multiply the first half-reaction by 3 so that, when both half-reactions are added, the electrons are canceled. In this way:
(Ag⁺ + e⁻ → Ag⁰ ) x3
Al⁰ → Al⁺³ + 3e⁻ +
-------------------------------------
3Ag⁺ + Al⁰ → 3Ag⁰ + Al⁺³
So, the coefficient of silver in the final balanced equation is 3.
