Answer:- Formula of the hydrate is 
and it's name is Iron(III)sulfate pentahydrate.
Solution:- As per the given information, there is 18.4% water in the hydrate. If we assume the mass of the hydrate as 100 grams then there would be 18.4 grams of water and 81.6 grams of Iron(III)sulfate present in the hydrate.
Molar mass for Iron(III)sulfate is 399.88 gram per mol and the molar mass for water is 18.02 gram per mol.
We will calculate the moles of Iron(III)sulfate and water present in the compound on dividing their grams by their molar masses as:
= 
= 
Now, the next step is to calculate the mol ratio and for this we divide the moles of each by the least one of them means whose moles are less. Here, the moles of Iron(III)sulfate are less than moles of water. So, we divide the moles of each by 0.204.
= 1
= 5
There is 1:5 mol ratio between Iron(III)sulfate and water. So, the formula of the hydrate is and the name of the hydrate is Iron(III)sulfate pentahydrate.
Answer: 120g/mol
Explanation:
The first step we are to take is to calculate the freezing point depression of the solution.
ΔT(f) = freezing point of pure solvent - freezing point of solution
ΔT(f) = 5.48 - 3.77
ΔT(f) = 1.71°C
Next we are to calculate the molal concentration of the solution using freezing point depression
ΔT(f) = K(f) * m
m = ΔT(f)/K(f)
m = 1.71/5.12
m = 0.333 molal
Now, we calculate the molecular weight of the unknown...
m = 0.333 mol = 0.333 mol X per kg of benzene
moles of X = 0.333 mol of X per kg of benzene * 0.5kg of benzene
moles of X = 0.1665
molecular weight of X = 20g of X/0.1665
molecular weight of X = 120/mol
Every electron carries one elementary negative charge. Concerning mass,
it takes roughly 1,840 electrons to make enough mass for 1 proton or 1 neutron.
Electrons don't necessarily have to stay connected to an atom, but when they do,
they circle the nucleus.
So you should select (C): ==> Negative, ==> light, ==> circling the nucleus.
Answer:
Mass of Sodium = 574.75 g
Mass of Chlorine = 886.25 g
Explanation:
The balance chemical equation for the synthesis of NaCl is,
2 Na + Cl₂ → 2 NaCl
Step 1: <u>Find out moles of each reactant required,</u>
According to balance chemical equation,
2 moles of NaCl is produced by = 2 moles of Na
So,
25 moles of NaCl will be produced by = X moles of Na
Solving for X,
X = 25 mol × 2 mol / 2 mol
X = 25 moles of Na
Similarly for Cl₂,
According to balance chemical equation,
2 moles of NaCl is produced by = 1 mole of Cl₂
So,
25 moles of NaCl will be produced by = X moles of Cl₂
Solving for X,
X = 25 mol × 1 mol / 2 mol
X = 12.5 moles of Cl₂
Step 2: <u>Convert each moles to mass as;</u>
Mass = Moles × Atomic Mass
For Na,
Mass = 25 mol × 22.99 g/mol
Mass = 574.75 g
For Cl₂,
Mass = 12.5 mol × 70.90 g/mol
Mass = 886.25 g
