Answer:
357 g of the transition metal are present in 630 grams of the compound of the transition metal and iodine
Explanation:
In any sample of the compound, the percentage by mass of the transition metal is 56.7%. This means that for a 100 g sample of the compound, 56.7 g is the metal while the remaining mass, 43.3 g is iodine.
Given mass of sample compound = 630 g
Calculating the mass of iodine present involves multiplying the percentage by mass composition of the metal by the mass of the given sample;
56.7 % = 56.7/100 = 0.567
Mass of transition metal = 0.567 * 630 = 357.21 g
Therefore, the mass of the transition metal present in 630 g of the compound is approximately 357 g
The pressure of the gas is expected to increase in accordance to Boyle's law.
<h3>What is Boyle's law?</h3>
Boyle's law states that, the volume of a given mass of gas is inversely proportional to its pressure at constant temperature.
By implication, when the piston is lowered and the volume of the gas is decreased, the pressure of the gas is expected to increase in accordance to Boyle's law.
Learn more about Boyle's law: brainly.com/question/1437490
Answer:
Se detailed explanation.
Explanation:
Hello,
In this case, since both magnesium and calcium ions are in group IIA, we can review the following similar properties:
- Since both calcium and magnesium are in group IIA they have two valence electrons, it means that the both of them have two electrons at their outer shells.
- They are highly soluble in water when forming ionic bonds with nonmetals such as those belonging to halogens and oxygen's family.
- Calcium has 18 electrons and magnesium 10 which are two less than the total protons (20 and 12 respectively) since the both of them have lost two electrons due their ionized form.
- Their electron configurations are:
It means that the both of them are at the region since it is the last subshell at which their electrons are.
Best regards.
<span>Answer is: Van't Hoff factor
(i) for this solution is 1.051 .
Change in boiling point from pure solvent to solution: ΔT
=i · Kb · b.
Kb - </span><span>molal boiling point elevation constant</span><span> is 0.512°C/m.
b - molality, moles of solute per kilogram of solvent.
b = 1.26 m.
ΔT = 101.63°C - 100</span>°C = 1.63°C.
i = 1.63°C ÷ (0.512°C/m · 1.26 m).
i = 1.051.