Answer: D. Mutation in coding sequences are more likely to be deleterious to the organism than mutations in noncoding sequences.
Explanation: It was not likely to be that the coding sequences are replicated more often. The only possible explanation is that the mutations in coding is more likely to be deleterious to the organism than mutations because it is in a non coding sequence.
There are several ways of expressing concentration of solution. Few of them are listed below
1) mass percentage
2) volume percentage
3) Molarity
4) Normality
5) Molality
In most of the drugs, concentration is expressed either in terms of mass percentage or volume percentage. For, solid in liquid type systems, mass percentage is convenient way of expressing concentration, while for liquid in liquid type solutions, expressing concentration in terms of volume percentage is preferred. Present system is an example of liquid in liquid type solution
Here, concentration of H2O2 is given antiseptic = 3.0 % v/v
It implies that, 3ml H2O2 is present in 100 ml of solution
Thus, 400 ml of solution would contain 4 X 3 = 12 ml H2O2
Answer:
8.20 % → Percent yield reaction
Explanation:
To find the percent yield of reaction we apply this:
(Produced yield / Theoretical yield) . 100 = %
Produced yield = 112.9 g
Theoretical yield = 1375.5 g
We replace → (112.9g / 1375.5 g) . 100
8.20 % → Percent yield reaction
Answer:
The new concentration will be 0.01 M.
Explanation:
To determine the new concentration we use the following formula.
concentration (1) × volume (1) = concentration (2) × volume (2)
concentration (1) = 0.1 M
volume (1) = 100 mL
concentration (2) = unknown
volume (2) = 100 mL + 900 mL = 1000 mL
concentration (2) = [concentration (1) × volume (1)] / volume (2)
concentration (2) = (0.1 × 100) / 1000 = 0.01 M
Answer:
Explanation:
Hello.
In this case, since the undergoing chemical reaction is only between the sodium bicarbonate and the acid HA:
For 0.561 g of yielded carbon dioxide (molar mass 44 g/mol), the following mass of sodium bicarbonate (molar mass 84 g/mol) that reacted was:
Considering the 1:1 mole ratio between CO2 and NaHCO3. Finally, the percent by mass of NaHCO3 is computed by dividing the mass of reacted NaHCO3 and t the mixture:
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