Elements of Group 1 and group 2 in the periodic
table contain elements so reactive that they are never found in the free state
<u>Explanation</u>:
The metals in group 1 of periodic table consisting of 'alkali metals' which include lithium, potassium, sodium, rubidium, Francium and caesium. They are highly reactive because they have low ionisation energy and larger radius. The group 2 metals consist of 'alkaline earth metals' which include calcium, strontium, barium, beryllium, radium and magnesium. These alkaline earth metal have +2 oxidation number, hence are highly reactive.
These both group metals are mostly reactive and so are never found in a free state. When they are exposed to air they would immediately react with oxygen. Hence, are stored in oils to avoid oxidation.
Answer:
Explanation:
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In this case, since the study of the bond energy allows us to compute the enthalpies of some reactions, for this combination reaction by which ammonia is yielded, we understand the enthalpy of reaction equals the enthalpy of formation of ammonia, and, in terms of the bonds energy we can write:
Whereas the bonds enthalpy of those bonds that get broken cover the N≡N and the three H-H bonds at the reactants side and the enthalpy of those bonds that are formed cover the six N-H bonds at the products; which means we obtain:
Which differs from the theoretical value that is -46 kJ/mol.
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Answer:
The answer to your question is V2 = 5.09 l
Explanation:
Data
Volume 1 = V1 = 5.0 L
Temperature 1 = T1 = 5°C
Volume 2 = V2 = ?
Temperature 2 = T2 = 10°C
Formula (Charles law)
V1/T1 = V2/T2
-Solve for V2
V2 = V1T2 / T1
-Convert temperature to °K
T1 = 5 + 273 = 278°K
T2 = 10 + 273 = 283°K
-Substitution
V2 = (5)(283) / (278)
-Simplification
V2 = 1415 / 278
-Result
V2 = 5.09 l
