Plz will give brainliest, 50 ! use your knowledge of the law of conversation of mass to answer the following question: from a laboratory process, a student collects 28.0g of hydrogen and 224.0g of oxygen. how much water was originally involved in the process?

According to law of conservation of mass, in a chemical reaction mass can neither be destroyed nor created thus, the total mass created here will be 224 g+28 g=252 g thus, mass of involved should be 252 g.

This can be experimentally proved as follows:

The balanced chemical reaction for the above process will be as follows:

Considering any one of the reactant.

2 mol of are produced from 2 mol of thus, 1 mol of produced from 1 mol of .

Now, mass of hydrogen gas is 28.0 g, molar mass is 2 g/mol, converting mass into number of moles:

thus, 14 mol of produced from 1×14=14 mol of .

Molar mass of is 18 g/mol. Converting number of moles into mass as follows:

m=n×M=14 mol×18 g/mol=252 g

Thus, mass of involved in the reaction will be 252 g.

answer:

the answer is 6 valence electrons.

explanation:

valence is the # of electrons on the outer most shell of an atom.

oxygen has 6.

although, the valency of oxygen is 2, because it only needs to make 2 covalent bonds, or gain 2 electrons, to achieve the stable octet configuration of a noble gas.

that's it!

if you need further assistance, let me know: )

~ trueboss72

consider the reaction below.

 no rx

at 500 k, the reaction is at equilibrium with the following concentrations.

[pci5]= 0.0095 m

[pci3] = 0.020

[ci2] = 0.020 m

what is the equilibrium constant for the given reaction?

0.042

0.42

2.4

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