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yulyashka [42]
3 years ago
15

Why do atoms of elements take part in chemical reaction?​

Chemistry
1 answer:
Marina CMI [18]3 years ago
4 0

Explanation:

During a chemical reaction, the atoms of the original substances gain, lose or share their electrons with those of the substances with which they are reacting. The reaction creates new substances made up of a new combination of atoms and a different configuration of electrons.

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Why does one Oxygen atom bond with two Hydrogen atoms to form water? Why don't they bond in a different ratio?
Vladimir79 [104]

This is because oxygen (2.8.6) requires two electrons on its valence shell to attain stable configuration (2.8.8). Hydrogen (1) on the other hand requires one electron on its valence shell to attain stable configuration (2). Therefore in a covalent bond, it requires two hydrogen and one oxygen to share electrons and achieve stable configuration.  

4 0
3 years ago
Hi there can i plz have help :)
Kobotan [32]

Answer:

Help on  what exactly??

Explanation:

5 0
3 years ago
Please help<br> It’s super quick
UkoKoshka [18]

Answer:

1. a

Explanation:

2. c

7 0
3 years ago
Read 2 more answers
In the following reaction, what element is gaining mass?
vagabundo [1.1K]

Mg gained mass because it went from being a single element (on the reactant side) to being a molecule (on the product side).

8 0
3 years ago
When 0.5141 g of biphenyl (C12H10) undergoes combustion in a bomb calorimeter, the temperature rises from 25.823 °C to 29.419 °C
natka813 [3]

Answer:

\Delta_{r}U of the reaction is -6313 kJ/mol

\Delta_{r}H of the reaction is -6312 kJ/mol

Explanation:

Temperature\,\,change= \Delta U = 29.419-25.823 =3.506^{o}C

q_{cal}= C \times \Delta T

=5.861 \times 3.596 = 21.076\,kJ

q_{rxn}= -q_{cal}= -21.076\,kJ

\Delta_{r}U= -21.076 \times \frac{154}{0.5141}= -6313\, kJ/mol

Therefore, \Delta_{r}U of the reaction is -6313 kJ/mol.

The chemical reaction in bomb calorimeter  is as follows.

C_{12}H_{10}(s)+\frac{27}{2}O_{2}(g)\rightarrow 12CO_{2}(g)+5H_{2}O(g)

Number\,of\,moles\Delta n=(12+5)-\frac{27}{2}=3.5

\Delta_{r}H=\Delta E+ \Delta n. RT

=-6313+3.5\times 8.314\times 10^{-3} \times 3.596=-6312\,kJ/mol

Therefore, \Delta_{r}H of the reaction is -6312 kJ/mol.

3 0
3 years ago
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