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# Tutor profile: Pat B.

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Pat B.
Biomedical Engineering Student
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## Questions

### Subject:Inorganic Chemistry

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Question:

What volume of a 0.15 M KCl solution will completely react with 0.150 L of a 0.175 M Pb(NO3)2 solution?

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Pat B.

The first step is to write out the balanced chemical equation for the reaction between KCl and Pb(NO3)2. 2KCl(aq) + Pb(NO3)2 yields PbCl2(s) + 2KNO3(aq). Our first step is to find the number of Pb(NO2)2 molecules resent in in the solution. To do this we multiply 0.15L by 0.175M to get 0.02625 moles of Pb(NO3)2 present in solution. Since the balanced chemical equation has a 2 coefficient next to the KCl there must be twice as many KCl molecules as Pb(NO3)2 molecules to completely react. So we multiply the 0.02625 moles by two to get 0.0525 moles of KCl. The problem asks for volume so we must divide 0.0525 moles KCl by the concentration 0.15M to get 0.35 L KCl solution needed

### Subject:Biology

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Question:

Explain how ATP in synthesized in the electron transport chain.

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Pat B.

The electron transport chain exists along a phospholipid bilayer membrane that prevents the passage of water and ions across the membrane. NADH adds an electron to the beginning of the chain. This electron is attracted to the Oxygen at the end of the electron transport chain. This is because oxygen is a highly electronegative element and attracts electrons. As the electron passes along the chain, its energy is used to pump protons through the proteins in the membrane into the cytosol outside the membrane. Eventually a gradient or imbalance in the number of protons is formed, with more protons concentrated outside the membrane than within. The energy from this gradient is used as the protons flow back into the membrane through a protein called ATP synthase. The energy from this proton is used to add a free phosphate group to adenosine diphosphate (ADP) to adenosine triphosphate (ATP). ATP is then used to provide energy for almost all reactions within living things.

### Subject:Algebra

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Question:

Train A leaves Oklahoma City heading north at 4 miles per hour. Train B leaves the same station at the same time heading west at 3 miles per hour. After how long will the two trains be 5 miles apart?

Inactive
Pat B.

In order, to simplify this problem we first need to draw a diagram. Your diagram should show a point for the station with a line going straight up labeled 4mph and a line pointing left labeled 5mph. then draw a line connecting the ends of the lines labeled 5 miles. You should recognize the figure as a right triangle and think of the pythagorean theorem, a^2 + b^2 = c^2. Use this equation to set up an equation with your known information. C, the length of the hypotenuse, is 50 miles. We do not know the distance of the other two sides of the triangle but we know they have been going 4 and 3 miles per hour for the same amount of time. So we assign the a and b values to 4 x t and 3 x t (Speed times time equals distance). Plug our values into the pythagorean theorem, (4t)^2 + (3t)^2 = 5^2. Now we solve for t. Apply the squares in the equation, 16t^2 + 9t^2 = 25. Combine the terms on the left, 25t^2 = 25. Divide by 25, t^2 = 1. square root of one is equal to one so the trains were 5 miles apart after 1 hour.

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