Answer:
The percent by mass of nitric acid in the mixture is 5.48 %
Explanation:
Givn that
Mass of HNO3 = 9.03 grams
Volume of KOH = 12.0 mL = 0. 012 L
Molarity of KOH = 0.655 M
The balanced equation
Ba(OH)2 + HNO3 → Ba(NO3)2 + H2O
Calculate the moles of KOH
Moles of Ba(OH)2 = molarity KOH * volume
Moles Ba(OH)2 = 0.655 M * 0.012 L
Moles Ba(OH)2 = 0.00786 moles
Calculate moles of HNO3
For 1 mol of Ba(OH)2 we need 1 mol of HNO3
For 0.00786 moles of Ba(OH)2 we need 0.00786 moles of HNO3
Calculate mass of HNO3
Mass HNO3 = moles HNO3 * molar mass HNO3
Mass HNO3 = 0.00786 moles * 63.01 g/mol
Mass HNO3 = 0.495 grams
Calculate mass % HNO3 in sample
mass % = (0.495 grams / 9.03 grams)*100%
mass % = 5.48 %
The percent by mass of nitric acid in the mixture is 5.48 %
When using a Punnett square, you first need to identify the genotype of each parent. We use capital and lower case letters to express dominant and recessive traits where capital letters are dominant and lower case letters are recessive.
In the case of your first question Black fur is dominant over white. Let Black (B) and white (b). The male is heterozygous so its genotype is both Black and white thus, it is expressed as such: Bb. The female is homozygous dominant meaning its genotype is both the dominant black color BB.
With this we can put it into our punnett square:
B b
B BB Bb Because the dominant trait is always expressed, whether or
not it is paired with a recessive trait, the chances of a white
B BB Bb off-spring is 0%
For numbers 2 to 5, we will apply the same principles:
2. Let: F - free earlobes and f - attached earlobes
Parent genotype: Both heterozygous
Question: Probability of a child with free earlobes
F f
F FF Ff The dominant trait is expressed in 3 out of 4 squares, thus the
probability of the child having free earlobes is 75%
f Ff ff
3. Using the same given, the question asks for the probability of having a homozygous dominant. This means the offspring should have an FF genotype. Using the same Punnett square, you can see that only 1 out of the 4 squares have a genotype of FF, so the probability is 25%.
4. Let: S - wrinkled seeds and s - smooth seeds
Parent genotypes: Homozygous recessive = ss
Heterozygous = Ss
Question: Probability of yielding off spring with smooth seeds: ss
s s
S Ss Ss Because 2 out of 4 squares have the genotype ss, we
can say that there is a 50% probability.
s ss ss
5. This question uses the same Punnet square because it assigns the same genotype for the parents and the dominant and recessive traits are the same. It asks the probability for a homozygous dominant offspring, which means its genotype has to be SS. Looking at the Punnet square, there is no square that has a genotype of SS, so the probability is 0%.
6. There are 3 possible genetic mutations that can occur:
Substitution, Insertion or Deletion.
You can easily determine the type just by looking at the resulting DNA sequence. If it is longer than the original, insertion occurred. If it is shorter then deletion occurred. If it did not get longer nor shorter, substitution occurred. So in the case of your problem, it is SUBSTITUTION, specifically missense substitution because only 1 amino acid changed.
7. The genetic variation that occurred in the scenario given where a black rabbit and a white rabbit have an off spring that is both white and black is called Codominance. This occurs when both traits of the parents are expressed in the offspring.
8. When an insertion mutation occurs, one of the noticeable changes is that there is an extra amino acid in the chain. If you look at the first option, <span>T T C G A A T C T G T A T G A C, there is one amino acid inserted in the original sequence. In the other selections, you will see that the sequence is in complete disarray where some were substituted and at the same time, there were insertions in between.
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9. Environmental influences can affect genetic expression. Light and temperature for example can affect the genetic expression of certain organisms. Chemical exposure can also have a direct impact on development.
10. The process of breeding certain livestock to yield certain traits in their off spring is called selective breeding. The process is also sometimes called artificial selection.
11. Karyotypes are done by matching up homologous chromosomes. Homologous chromosomes should be of the same size and the same structure.
Refrigerant may be of four type such as R-22, R 410A, R- 290, R-417 A
The refrigerator R410A having standing pressure of about 260 psi whereas saction pressure of about 120 to 130 psi and discharge pressure of about 450 to 500 psi.
Refrigerant R-410 is a cooling machine consists of equipments and chillers for air conditioning purpose along with coolant gas. This refrigerant have high cooling capacity which consumes less power. The suction pressure is 118 psi and discharge pressure is 400 psi.
It has a decrease in oxidation number
The process of reduction is when the oxidation number of a chemical specie is lowered, or when the specie gains electrons. A chemical specie undergoing reduction will act as an oxidizing agent, because it will oxidize the other reactant by itself getting reduced.
Answer:
C
Explanation:
Since they all are 2 mols worth, all you have to do is count the number of atoms in each molecule
H2SO4 contains 2 hydrogen 1 sulfur 4 oxygen = 7 atoms
H2S contains 2 hydrogen and 1 sulfur = 3 atoms
H3PO4 contains 3 hydrogens 1 phosphorus 4 oxygens = 8 atoms
HNO3 contains 1 hydrogen 1 nitrogen and 3 oxygens = 5 atoms
You have the right answer.