We have to calculate the percentage (by mass) of Ba in the mixture of BaBr₂ and inert material.
Given, bromide ion is precipitated as AgBr by the addition of an excess of silver nitrate as per following reaction:
BaBr₂(aq) + 2AgNO₃(aq)→2AgBr(s)+Ba(NO₃)₂(aq). From this balanced chemical reaction, it is clear that one mole BaBr₂ reacts with two moles of AgNO₃ and gives two moles of AgBr. We have to calculate 0.6226 g AgBr contains how many moles of AgBr. Using that information we can get how many moles of BaBr₂ reacted to give 0.6226g AgBr and one mole BaBr₂ contains one mole of Ba. By multiplying number of moles of Ba with atomic mass of Ba we can get amount of Ba present in the mixture. Accordingly we can calculate mass percentage of Ba in the mixture.
Atomic mass of Ba= 137.327 g.mol, Molecular mass of BaBr₂=297.1 g/mol and molecular mass of AgBr=187.7 g/mol. Mass of AgBr is 0.6226 g which contains 0.6226/187.7 mole= 3.31X10⁻³ moles of AgBr. So, moles of BaBr₂ reacts= (3.31X10⁻³)/2 moles= 1.65X10⁻³ moles. One mole BaBr₂ contains one mole of Ba. So, 1.65X10⁻³ moles of BaBr₂ contains 1.65X10⁻³ moles of Ba atoms whose mass is=1.65X10⁻³X137.327g=0.2265 g. Out of 0.7207 g sample amount of Ba present is 0.2265 g. So, mass percentage is =31.42 %
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Answer:
Genotypic ratio = 1 Bb: 1 bb
Phenotypic ratio = 1 black rabbit : 1 brown rabbit
Explanation:
This question involves a gene coding for hair color in rabbits. The allele for Black hair (B) is dominant to the allele for brown hair (b).
According to this question, a black rabbit whose mother is brown haired (bb) is crossed with a brown hair rabbit. This means that the black rabbit will have a genotype 'Bb' while the brown rabbit will have a genotype 'bb'. Each parent will produce the following gametes:
Bb - B and b
bb - b and b
Using these gametes in a punnet square (see attached image), the following proportion of offsprings will be produced:
Bb, Bb, bb and bb
This means that the;
Genotypic ratio = 1 Bb: 1 bb
Phenotypic ratio = 1 black rabbit : 1 brown rabbit
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