Explanation:
When carbon atom tends to form single bonds then its hybridization is 
, when carbon atom tends to form double bond then its hybridization is 
and when a carbon atom is attached to a triple bond or with two double bonds then its hydridization is sp.
For example, in HCN molecule there is a triple existing between the carbon and nitrogen atom.
So, hybridization of carbon in this molecules is sp. Moreover, nitrogen atom is also attached via triple bond and it also has a lone pair of electrons. Hence, the hybridization of nitrogen atom is also sp.
Thus, we can conclude that s and p type of orbitals overlap to form the sigma bond between C and N in H−C≡N:
Chemical reaction: Sr(OH)₂ + 2HNO₃ → Sr(NO₃)₂ + 2H₂O.
part A) m(Sr(OH)₂) = 21,78 g.
n(Sr(OH)₂) = m(Sr(OH)₂) ÷ M(Sr(OH)₂).
n(Sr(OH)₂) = 21,78 g ÷ 87,62 g/mol.
n(Sr(OH)₂) = 0,248 mol.
c(Sr(OH)₂) = n(Sr(OH)₂) ÷ V(Sr(OH)₂).
c(Sr(OH)₂) = 0,248 mol ÷ 0,075 L.
c(Sr(OH)₂) = 3,3 mol/L.
part B) V(Sr(OH)₂) = 34,21 mL = 0,03421 L.
n(Sr(OH)₂) = 0,03421 L · 3,3 mol/L = 0,112 mol.
From chemical reaction: n(Sr(OH)₂) : n(HNO₃) = 1 : 2.
n(HNO₃) = 0,225 mol.
V(HNO₃) = 0,225 mol ÷ 0,25 mol/L.
V(HNO₃) = 0,9 L = 900 ml.
Answer:
Temperature does not affect the radioactive decay
Explanation:
because Temperature has no factor in radioactivity
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