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3 years ago

What is the mass percent of water in copper (II) sulfate?

Chemistry

1 answer:

Veseljchak [2.6K]3 years ago

8 0

The answer is sulfate pentahydrate

Hope this helps you :D

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How many moles of aluminum are needed to react completely with 1.2 mol of feo

Amanda [17]

Answer:

0.8 mol.

Explanation:

The balanced equation for the reaction between Al and FeO is represented as:

2Al + 3FeO → 3Fe + Al₂O₃,

It is clear that 2 mol of Al react with 3 mol of FeO to produce 3 mol of Fe and 1 mol of Al₂O₃.

Using cross multiplication:

2 mol of Al needs → 3 mol of FeO, from stichiometry.

??? mol of Al needs → 1.2 mol of FeO.

∴ The no. of moles of Al are needed to react completely with 1.2 mol of FeO = (2 mol)(1.2 mol)/(3 mol) = 0.8 mol.

3 0

3 years ago

59 Q.No. 28 b Hydrogen peroxide can act as oxidizing as well as reducing agent with example.

professor190 [17]

Answer:

Hydrogen peroxide can function as an oxidizing agent as well as reducing agent.

H2O2 act as oxidizing agent in acidic medium.

Explanation:

Example : 2FeSO4 +H2SO4 +H2O2 —>

(ferrous sulphate)

Fe2(SO4)3 +2H2O

(ferric sulphate)

3 0

2 years ago

How many atoms are in 34.2 grams of carbon?

Oksana_A [137]

Answer:

6.02*1022

Explanation:

I don't know why

8 0

3 years ago

Help ASAP If you know what to do comment if u don’t I’ll report you ! Points added

mylen [45]

Proton:

Positive

Found in Nucleus

Mass of 1 AMU

Neutron:

Neutral

Found in Nucleus

Mass of 1 AMU

Electron:

Negative

Found in orbitals

Mass of 0 AMU

5 0

2 years ago

I have 2 samples of solid chalk (aka calcium carbonate). Sample A has a total mass of 4.12 g and Sample B has a total mass of 19

IRINA_888 [86]

Answer:

A) Sample B has more calcium carbonate molecules

Explanation:

M = Molar mass of calcium carbonate = 100.0869 g/mol

$N_A$ = Avogadro's number = $6.022\times 10^{23}\ \text{mol}^{-1}$

For the 4.12 g sample

Moles of a substance is given by

$n=\dfrac{m}{M}\\\Rightarrow n=\dfrac{4.12}{100.0869}\\\Rightarrow n=0.0411\ \text{mol}$

Number of molecules is given by

$nN_A=0.0411\times 6.022\times 10^{23}=2.48\times 10^{22}\ \text{molecules}$

For the 19.37 g sample

$n=\dfrac{19.37}{100.0869}\\\Rightarrow n=0.193\ \text{mol}$

Number of molecules is given by

$nN_A=0.193\times 6.022\times 10^{23}=1.16\times 10^{23}\ \text{molecules}$

$1.16\times 10^{23}\ \text{molecules}>2.48\times 10^{22}\ \text{molecules}$

So, sample B has more calcium carbonate molecules.

The ratio of the elements of carbon, oxygen, calcium atoms, ions, has to be same in both the samples otherwise the samples cannot be considered as calcium carbonate. Same is applicable for impurities. If there are impurites then the sample cannot be considered as calcium carbonate.

7 0

3 years ago