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3 years ago

Please answer both questions.

Chemistry

1 answer:

Deffense [45]3 years ago

6 0

Answer:

1)Krypton

2)11H

Explanation:

electrons=protons

protons=atomic number

mass number=protons+neutrons

mass number is the superscript

atomic number is the subscript.

1)The answer is Krypton because its atomic number= number of protons=number of electrons is 36.

mass number is 46+36=82.

2)subscript=atomic number=number of protons=number of electrons

i. H = electrons=1

=neutrons=0

ii. Cl=electrons=17

=neutrons=35-17=18

iii. Na=electrons=11

=neutrons=23-11= 12

so the answer is Hydrogen because it has 1 electron and 0 neutron.

I hope this helps.

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How does the reaction, A + BC ⇌ AB + C, differ from the reaction, N2 + 3H2 ⇌ 2NH3 + energy, by which ammonia is synthesized?

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Answer:

N₂+3H₂ ⇄2NH₃ is a thermochemical reaction whereas A+BC⇄AB is not.

A+BC⇄AB is a reaction of pure a element with a compound while N₂+3H₂ ⇄2NH₃ is a reaction between two pure elements.

Explanation:

Let A+BC⇄AB be equation i and N₂+3H₂ ⇄2NH₃ be equation ii.

The two reactions differ in that ii is a thermo-chemical reaction whereas i is not. This is because energy is included in reaction ii but not included in reaction i.

Also i is a reaction of pure a element with a compound while ii is a reaction between two pure elements. The compound is BC while the pure element is A.

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4 years ago

Complete and balance the following acid-base equations:

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Answer:

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b) $H_2SO_4(aq) + 2NaOH(aq) \rightarrow Na_2SO_4 (aq) + 2H_2O (l)$

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Explanation:

when $HClO_4$ is added to LiOH, lithium chlorate and water is formed.

$HClO_4 (aq) + LiOH \rightarrow LiClO_4 + H_2O$

Balancing of above reaction,

It can be seen that all the atoms in both the sides are balanced. so it is a balanced reaction.

(b) When aqueous $H_2SO_4$ is added to NaOH, $Na_2SO_4$ and $H_2O$ is formed. It is a neutrilization reaction.

$H_2SO_4(aq) + NaOH(aq) \rightarrow Na_2SO_4 (aq) + H_2O (l)$

Balancing of above reaction,

First balance all the atoms except O and H

S atom is already balanced in either side

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No. of Na atom in right hand side = 2

So multiply NaOH by 2, now the reaction becomes:

$H_2SO_4(aq) + 2NaOH(aq) \rightarrow Na_2SO_4 (aq) + H_2O (l)$

No. of O atoms in left hand side = 6

No. of O atoms in right hand side = 5

So multiply H2O by 2, now the reaction becomes:

$H_2SO_4(aq) + 2NaOH(aq) \rightarrow Na_2SO_4 (aq) + 2H_2O (l)$

Now, it can been seen that all the atoms are balanced.

c) When $Ba(OH)2$ reacts with HF, baroum fluoride and water is formed.

$Ba(OH)2 + HF\rightarrow BaF_2 + H_2O$

Balancing of above reaction,

Barium atoms are already balanced on either side.

No. of F atoms on left hand side = 1

No. of F atoms on right hand side = 2

So, multiply HF by 2, now reaction becomes

$Ba(OH)2 + 2HF\rightarrow BaF_2 + H_2O$

In order to balance H and O on either side, multiply H2O by 2.

$Ba(OH)2 + 2HF\rightarrow BaF_2 + 2H_2O$

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