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Tutor profile: Andrew K.

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Andrew K.
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Questions

Subject: Chemistry

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Question:

Assuming ideal behavior, calculate how many moles of nitrogen must be present in an $$\textrm{80.0 L}$$ container at $$25.5^{\circ} C$$ at $$\text{2.00 atm}$$, given $$R=0.082057\ \text{L} \cdot \text{atm} \cdot \text{mol}^{-1} \cdot \text{K}^{-1}$$?

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Andrew K.
Answer:

For this problem, we must use the ideal gas law. The ideal gas law states: $(PV=nRT$)where $$P$$ = pressure, $$V$$= volume, $$n$$= number of moles, $$R$$= the ideal gas constant, and $$T$$= temperature. In this case, we want to solve for the number of moles, so we should algebraically manipulate the ideal gas law equation to also equal this value: $(n=\frac{PV}{RT}$)Next, we must make sure that all of our units match. We have been given $$R$$ in terms of $$\text{L} \cdot \text{atm} \cdot \text{mol}^{-1} \cdot \text{K}^{-1}$$. The only unit that doesn't match is temperature. To convert temperature from $$^{\circ}\text{C}$$ to $$\text{K}$$, we use the following equation: $(T_{\text{Kelvin}}=T_{\text{Celsius}}+273.15$) $(T_{\text{Kelvin}}=25.5 + 273.15=298.65\ \text{K}$)We will round this number to 298.7 K to preserve significant figures. Now, we can enter all of our given values into our modified ideal gas law equation to give our final answer: $(n=\frac{2.00\ \text{atm} \cdot 80.0\ \text{L}}{0.082057\ \text{L} \cdot \text{atm} \cdot \text{mol}^{-1} \cdot \text{K}^{-1} \cdot 298.7\ \text{K}}$) $(n=6.53\ \text{mol}$)

Subject: Biology

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Question:

Explain the functional difference between an ion pump and an ion channel, and describe what roles they play in maintaining or dissipating concentration gradients across a lipid bilayer.

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Andrew K.
Answer:

When a solution of high concentration sits across a semi-permeable membrane from a solution of low concentration, it is energetically favorable for the concentrations of these solutions to equalize. Only small, uncharged, non-polar molecules can easily pass through a lipid bilayer. Charged particles, or ions, cannot pass through a lipid bilayer by themselves. In order to take advantage of this energy gradient, cells have ion pumps and ion channels embedded in their cell membranes to help facilitate the flow of ions through the lipid bilayer that would otherwise be impermeable to ions. Ion pumps utilize active transport to move ions against a gradient, or from an area of low concentration to an area of high concentration. This might be used to drive a certain cellular process that relies on the maintenance of a concentration gradient for usable energy. Ion channels use passive transport to allow the ions to naturally flow down their gradient, or from an area of high concentration to an area of low concentration. This might be used to relieve the potential energy of a concentration gradient across a membrane.

Subject: Algebra

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Question:

Alfred and Beth each bought cars worth $$$37,000$$. Alfred's car's value depreciates by half every three years. Beth's car's value depreciates by half every five years. How much more will Beth's car be worth than Alfred's car after seven years, to the nearest dollar?

Inactive
Andrew K.
Answer:

To start, we must derive an equation that is appropriate to solve this sort of problem. We have the principle value of the cars, which is $$$37,000$$. Next, we have the amount that their value depreciates by, which is $$1/2$$, since the problem states that their value depreciates by half. Next, we must figure out how to properly scale the depreciation amount, since each car's value depreciates by a different amount over time. To do this, we must take the depreciation amount, $$1/2$$, and raise it to the power of how many times the car has depreciated by this amount. If we say $$t$$ is the amount of time that passes, and $$n$$ is the amount of years that must pass for the value to depreciate by half, then we get the following equation: $(V=($37,000)(0.5)^{t/n}$) In this case, Alfred's car depreciates by half every three years, so the value of Alfred's car is: $(V_{A}=($37,000)(0.5)^{7/3}$) $(V_{A}=$7,342$) Similarly, Beth's car depreciates by half every five years, so the value of Beth's car is: $(V_B=($37,000)(0.5)^{7/5}$) $(V_B=$14,020$) To find the difference of value and therefore our final answer, we must simply calculate $$V_B - V_A$$: $($14,020-$7,342=$6,678$)

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