When parallel rays exit a concave lens, the light rays are divergent.
The rays diverge or bend away from the axis it has been traveling upon entering the lens when it reaches the other side of the lens. These rays appear to have come from the same focal point before entering the concave lens. When these parallel rays are extended, it will be traced back to a single point of origin.
Answer:
The IUPAC name of given compound is 3−5−ethyl−5−−3−methylheptane. Explanation: The parental chain is of 7 carbons with single bonds hence it is heptane. Two substituents ethyl and methyl group are attached from an equal distance. Hence according to the alphabetical order preference, counting starts from carbon which is close to an ethyl group.
Answer is: A. 1.1 3 1023 NiCl2 formula units.
m(NiCl₂) = 24.6 g; mass of nickel(II) chloride.
M(NiCl₂) = 129.6 g/mol; molar mass of nickel(II) chloride.
n(NiCl₂) = m(NiCl₂) ÷ M(NiCl₂).
n(NiCl₂) = 24.6 g ÷ 129.6 g/mol.
n(NiCl₂) = 0.19 mol; amount of nickel(II) chloride.
Na = 6.022·10²³ 1/mol; Avogadro constant.
N(NiCl₂) = n(NiCl₂) · Na.
N(NiCl₂) = 0.19 mol · 6.022·10²³ 1/mol.
N(NiCl₂) = 1.13·10²³; number of formula units.
RbOH is a strong base that dissociates completely and HCl is a strong acid that too dissociates completely. the complete reaction between the acid and base is;
RbOH + HCl ---> RbCl + H₂O
stoichiometry of acid to base is 1:1
At neutralisation point
H⁺ mol = OH⁻ mol
mol = molarity x volume
if Ma - molarity of acid and Va - volume of acid reacted
Mb - molarity of base and Vb - volume of base reacted
Ma x Va = Mb x Vb
0.5 M x 52.8 mL = Mb x 60.0 mL
Mb = 0.44 M
molarity of base - 0.44 M
