Answer:
27.4 gram is the solution it's simple dude...
Explanation:
don't be afraid of huge question they confuse you you need not to be confused
now see simple solution
molality is denoted by m
so
m= moles of solute / mass of solvent in kg.
i hope your know the meaning of solute and solvent....
so moles are given 0.467
and molar mass is given 58.44
so just take out the gram means
by applying formula
58.44×0.467
it will give 27.4 grams simple.....
We need to increase the concentration of common ion first, in order to promote the common ion effect
<h3>What is the Common ion effect?</h3>
It is an effect that suppresses the dissociation of salt due to the addition of another salt having common ions.
For example, a saturated solution of silver chloride in equilibrium has Ag⁺ and Cl⁻ . Sodium Chloride is added to the solution and has a common ion Cl⁻. As a result, the equilibrium shifts to the left to form more silver chloride. Thus, solubility of AgCl decreases.
The Equilibrium law states that if a process is in equilibrium and is subjected to a change
- in temperature,
- pressure,
- the concentration of reactant or product,
then the equilibrium shifts in a particular direction, according to the condition.
Thus, an increase in the concentration of common ion promotes the common ion effect.
Learn more about common ion effect:
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1. Nickel (II) Bromide
2. Iron (II) Oxide
3. Iron (III) Oxide
4. Tin (IV) Chloride
5. Lead (IV) tetrachloride
6. Tin (II) Bromide
7. Chromium (III) Phosphide
8. Iron (II) Fluoride
9. Gold (III) Chloride
I hope this helps. I'm more than 100% sure that all the answers except for number 7 are correct. I knew all of them off the top of my head except for this one. I hope the other answer has the correct answer for that one. Good luck and have a great day.
The formula of Iron(III) oxide is Fe2O3
In order to calculate the mass of iron in a given sample of iron(III) oxide, we must first know the mass percentage of iron in iron(III) oxide. This is calculated by:
[mass of iron in one mole of iron(III) oxide/ mass of one mole of iron(III) oxide] * 100
= [(moles of iron * Mr of iron) / (moles of Iron * Mr of Iron + moles of Oxygen * Mr of Oxygen)] * 100
= [(2 * 56) / (2 * 56 + 3 * 16)] * 100
= (112 / 160) * 100
= 70%
Thus, in a 100g sample, the weight of iron will be:
100 * 70%
= 70 grams
