Please help thaank youu

1 answer:

6 0

First, let's start off by finding the mass of this whole hydrate.
(Note: the unit of measurement for mass will be amu)

Let's find the molecular mass of each element.
$Co=58.933$
$Cl=35.45$
$H=1.008$
$O=15.999$

Now, let's find the mass of each compound.

$CoCl_2=58.933+2(35.45)=129.833$
$H_2O=2(1.008)+15.999=18.015$

We have 6 molecules of H2O, so multiply 18.015 by 6 then add that with the weight of CoCl2.

$6(18.015)=108.09$
$129.833+108.09=237.923$

Now divide 108.09 (mass of all the H2O in the hydrate) by 237.923 (total mass of hydrate).

$\dfrac{108.09}{237.923}$

$\approx0.45431$

Turn that into a percentage and you get 45.431%.
Hope this helps! :)

You might be interested in

what is the limiting reactant in the reaction of 12.0 g of SO2 with 8.0 g of H2S and their reaction? what mass of sulfur will be

larisa86 [58]

The balanced equation for the reaction is as follows;
2H₂S + SO₂ —> 2H₂O + 3S
Stoichiometry of H₂S to SO₂ is 2:1
Limiting reactant is fully used up in the reaction and amount of product formed depends on amount of limiting reactant present.
Number of H₂S moles - 8.0 g / 34 g/mol = 0.24 mol of H₂S
Number of SO₂ moles = 12.0 g / 64 g/mol = 0.188 mol of SO₂
According to molar ratio of 2:1
If we assume H₂S to be the limiting reactant
2 mol of H₂S reacts with 1 mol of SO₂
Therefore 0.24 mol of H₂S requires - 1/2 x 0.24 = 0.12 mol of SO₂
But 0.188 mol of SO₂ is present therefore SO₂ is in excess and H₂S is the limiting reactant.
H₂S is the limiting reactant
Amount of S produced depends on amount of H₂S present
Stoichiometry of H₂S to S is 2:3
2 mol of H₂S forms 3 mol of S
Therefore 0.24 mol of H₂S forms - 3/2 x 0.24 mol = 0.36 mol of S
Mass of S produced = 0.36 mol x 32 g/mol = 11.5 g of S is produced

7 0

3 years ago

Aluminum carbide powder decomposes in water:

omeli [17]

Answer:

Explanation:

I think option d is correct

4 0

3 years ago

What is the viscosity and flammability of refinery gas?

Liono4ka [1.6K]

Hydrocarbons

Hope this helps :)