Claudio S.

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Chemistry

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Question:

In the reaction shown below, if 58.58 g of Na$$_{2}$$S are reacted with 169.95 g of AgNO$$_{3}$$, which of the two reagents is the limiting reagent and how much (in g) of the excess reagent will be left over once the reaction has gone to completion? Na$$_{2}$$S + 2 AgNO$$_{3}$$ $$\rightarrow$$ Ag$$_{2}$$S + 2 NaNO$$_{3}$$

Claudio S.

Answer:

We are are given the masses of both reactants used in the question and must first determine the limiting reagent. The formula weight of AgNO$$_{3}$$ is 169.9 grams per formula unit; the formula weight of Na$$_{2}$$S is 78.1 grams per formula unit. We can now determine that we are given 0.75 mol Na$$_{2}$$S and 1 mol AgNO$$_{3}$$ simply by converting the grams of each reactant used to mols with its respective formula weight. (If you are confused about the conversion, please see example calculation below) Conversion from grams to mol 58.58 g Na$$_{2}$$S $$*\frac{1 mol Na_{2}S}{78.1 g Na_{2}S} = 0.75 mol$$ Na$$_{2}$$S The same logic can be applied for AgNO$$_{3}$$ According to the stoichiometry of the reaction, we need 2 mol of AgNO$$_{3}$$ for every mol of Na$$_{2}$$S. This tells us that if we use all 0.75 mol of Na$$_{2}$$S, we would need 1.5 mol of AgNO$$_{3}$$ to satisfy the stoichiometry of the reaction. Since we do not have that much AgNO$$_{3}$$ at our disposal, we have to use the entire 1 mol AgNO$$_{3}$$ that we do have and only use 0.5 mol Na$$_{2}$$S; thus making AgNO_{3} our limiting reagent. If we only use 0.5 mol Na$$_{2}$$S, we have 0.25 mol excess Na$$_{2}$$S left over once the reaction has gone to completion. 0.25 mol Na$$_{2}$$S equates to ~ 19.5 g Na$$_{2}$$S. Therefore, we have 19.5 g of the excess reagent Na$$_{2}$$S left over once this reaction has gone to completion with the amount of reactants given in the beginning.

Biology

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Question:

In a nonevolving large population, there are two alleles, $$T$$ and $$t$$, which code for the same trait. The frequency of $$T$$ is 40%. What are the frequencies of all the possible genotypes?

Claudio S.

Answer:

We must first analyze the question and recognize that this is in fact a Hardy-Weinberg population because evolution is not occurring, the gene frequencies are not changing and we are dealing with a large population with a stable gene pool. Thus, the population is in Hardy-Weinberg equilibrium and we can apply the following equation: $$ p^{2} + 2pq + q^{2}=1$$ where $$p^{2}$$ is the frequency of the homozygous dominant genotype $$(TT)$$, $$2pq$$ is the frequency of the heterozygous dominant genotype $$(Tt)$$ and $$q^{2}$$ is the frequency of the homozygous recessive $$(tt)$$ genotype. Lets use the information provided in the question to set up our equations. We are given the frequency of $$T$$ being 40%, and as such, $$ p = 0.40$$. The frequency of the recessive gene $$t$$ is 100% - 40% = 60%; therefore $$ q = 0.60 $$ The frequency of the genotypes, according to Hardy-Weinberg equlibirum are, $$p^{2} = TT $$, $$2pq = Tt$$, and $$q^{2} = tt$$. Therefore, the frequency of the genotypes are: $$p^{2} =(0.4)^{2} = 0.16 $$ = 16% TT $$ 2pq = 2 * 0.4 * 0.6 = 0.48$$ = 48% Tt $$q^{2} = (0.6)^{2} = 0.36 $$ = 36% tt To double check our work, if we add these together, we do in fact get 100%. 16% + 48% + 36% = 100%

Algebra

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Question:

Timmy studies the change in elephant population of the Serengeti Plains over time. The relationship between the elapsed time $$ t $$, in decades, since Timmy started studying the population, and the number of elephants $$ N(t) $$, is modeled by the following function: $$N(t) = 710 * (\frac {8}{125} )^t$$ The elephant population loses $$\frac{3}{5}$$ of its size every _ decades.

Claudio S.

Answer:

On initial analysis of this function: $$N(t) = 710 * (\frac {8}{125} )^t$$ we can deduce that the initial population (710) is changing at a rate of $$\frac{8}{125}$$ per unit of time, $$t$$. We also must recognize our units. The time $$ t $$ is given in decades and Timmy is studying the elephant population $$N(t)$$ as a function of time in decades. The answer is asked for in decades so we can think of one decade as being equivalent to one unit of time, $$t$$. With this in mind, lets tackle the problem. If the population is losing $$\frac{3}{5}$$ of its population in $$t$$ decades we can assume that we have $$\frac{2}{5}$$ of the initial population remaining in $$t$$ decades. Lets plug this into the initial function and solve for $$t$$ to arrive at our answer. $$710 * \frac {2}{5} = 710 * (\frac {8}{125} )^t$$ When we see this, our instinct should be to cancel the 710 from each side by dividing by 710. This leaves us with $$ \frac {2}{5} = (\frac {8}{125} )^t$$ Thinking back on our Rules of Exponents, if we have a common base, we can equate our exponents to each other. If we take the cube root of 8 and 125 we are left with $$ \frac {2}{5}$$. This allows us to simplify the equation above to the following: $$ \frac {2}{5}^{1} = (\frac {2}{5})^{3t} $$ Which can be further simplified to $$1=3t$$ If we solve for $$t$$ now, we arrive at the answer $$\frac{1}{3}$$ Therefore, keeping our units in mind: The elephant population loses $$\frac{3}{5}$$ of its size every $$\frac{1}{3}$$ or $$ 0.33$$ decades. This is one of several ways to go about solving this problem. But if you know your Rules of Exponents it is a simple way that doesn't require the use of a calculator. : )

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