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# Tutor profile: Cole C.

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Cole C.
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## Questions

### Subject:Chemistry

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Question:

How much $$CH_4$$ (methane) can I react with 50g $$O_2$$ in a combustion reaction? How much water is created? Keep 4 decimal places for calculations, and round to 4 significant figures for final answer.

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Cole C.

Great question! First, let's define what concepts we're talking about. Put simply, we're going to: - Define combustion - Balance the equation - Convert from grams to moles and back to find the answers === Combustion === Combustion is when a hydrocarbon (made of C, H and O) reacts with oxygen to form $$CO_2$$ and $$H_2O$$, and heat. In our case, it would look like this: $$CH_4 + O_2 ==> CO_2 + H_2O$$ === Balancing === The question asks how much $$CH_4$$ can react with 50g of $$O_2$$. Now how do we find how much $$CH_4$$ and $$O_2$$ to use, or how much water and carbon dioxide are created? We need to balance the equation! Why? Atoms aren't created or destroyed very often, so we need to get out what we put in. If we look at the equation: $$CH_4 + O_2 ==> CO_2 + H_2O$$ We have on the left side (reactants) 1 C 4 H 2 O And on the right side (products) 1 C 3 O 2 H That doesn't add up! We need to balance the equation so what goes in, comes out. We start with the least common atom - C in this case. It's already balanced, so let's move on to H. To get more H on the right side, we double the $$coefficient$$ of the $$H_2O$$ $$CH_4 + O_2 ==> CO_2 + 2H_2O$$ Now on the left we have: 1 C 4 H 2 O And on the right side we have: 1 C 4 O 4 H We can double the amount of $$O_2$$ on the left side to complete the equation. $$CH_4 + 2O_2 ==> CO_2 + 2H_2O$$ Now on the left we have: 1 C 4 H 4 O And on the right side we have: 1 C 4 O 4 H Amazing! === Molar Mass === The ratios in the equation above are in terms of number of atoms, so how do we find the number of atoms from the mass? We use the molar mass! The molar mass is how many grams make up one mole. A mole is a number of atoms - $$6.02*10^{23}$$ atoms. It's just a number though, it could be $$6.02*10^{23}$$ of anything - even puppies! (sidenote, that many puppies would be 1/3 the weight of the earth! A mole is a massive number of atoms.) Why is this number important? It's a ratio between the weight of a single atom, in atomic mass units (amu) and the weight of a "reasonable" quantity of substance, measured in grams - where the weight in grams is the same number as the weight in amu! In other words, if one carbon atom is 12.0107 amu, Then $$6.02*10^{23}$$ carbon atoms weighs 12.0107 grams It's a weird looking number, but it's important for converting "reasonable" (easy to work with) weights of substances and finding the number of atoms involved, so we can figure out how many atoms react with how many other atoms. Then we can measure products and reactants in easy-to-use grams instead of counting out each individual atom in a reaction - that might take a while! Each $$molecule$$ of $$O_2$$ has two oxygen atoms, each weighing 15.9994 amu, or 31.9988 amu total. So, each $$mol$$ of $$O_2$$ weighs 31.9988g! If we take 50g of O_2, $$50g * \frac{1 mol O_2} {31.9988 g/mol} = 1.5626$$ mol $$O_2$$ Now we have our $$O_2$$ in terms of number of atoms, so it works in our equation. Amazing! $$CH_4 + 2O_2 ==> CO_2 + 2H_2O$$ How much $$CH_4$$ do we need? Mass $$O_2$$ -> moles $$O_2$$ -> moles $$CH_4$$ -> mass $$CH_4$$ The molar mass of $$CH_4$$ is $$12.0107g + 4(1.0079g) = 16.0423$$ g/mol $$CH_4$$ And its ratio with oxygen means we need: $$1.5626 mol O_2 * \frac{1 CH_4}{2 O_2} = 0.7813$$ mol $$CH_4$$ To react, so, $$0.7813 mol CH_4 * \frac{16.0423 g CH_4}{1 mol CH_4} = 12.5338 g CH_4$$ combusted. How much water is created? $$CH_4 + 2O_2 ==> CO_2 + 2H_2O$$ We need 2 $$H_2O$$ for every 2 $$O_2$$, And the molar mass of water is $$1.0079g * 2 + 15.9994g) = 18.0152$$ g/mol $$H_2O$$ So $$(1.5626 mol O_2)(\frac{2H_2O}{2O_2})(\frac{18.0152g H_2O}{1 mol}) = 28.1506 g H_2O$$ created. For our final, rounded answers: 12.53 g $$CH_4$$ combusted. 28.15 g $$H_2O$$ created.

### Subject:Calculus

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Question:

Find the derivative of: $$f(x) = (e^{3x}+13x)^4$$

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Cole C.

Let's look at what we have! Our primary derivative rule is the chain rule. If we have any function, let's say $$f(x) = ax^n$$ then $$f'(x) = nax^{n-1}$$ Same rule applies to big functions made up of multiple parts, except we get this: $$f(x) = a[g(x)]^n$$ $$f'(x) = na[g(x)]^{n-1}[g'(x)]$$ Or in other words, we treat the 'inside function' [g(x)] as one big chunk, then multiply the 'outer equation' by the derivative, g'(x) when it's all said and done. Now, down to business. If we have: $$f(x) = (e^{3x}+13x)^4$$ Then $$g(x) = e^{3x}+13x$$ Since now we have two separate functions added together, we find the derivative of each individually If $$y = e^{ax}$$ Then $$y' = ae^{ax}$$ And if $$y = 13x$$ Then $$y' = 13$$ So we get $$g'(x) = 3e^{3x} + 13$$ Take a breath, we've solved the inside part. Now remember, $$f'(x) = na[g(x)]^{n-1}[g'(x)]$$ So we can just fill in the blanks in the equation we wrote earlier. $$f(x) = (e^{3x}+13x)^4$$ $$f'(x) = 4[(e^{3x}+13x)^3][3e^{3x} + 13]$$ And this is our derivative! Congrats!

### Subject:Physics

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Question:

A 200 kg cannon is placed inside a 2000 kg, 20m long train boxcar. The cannon is placed 1m from the left end of the car with 40 cannonballs. It fires all of the 10 kg cannonballs at the other end of the traincar, where they come to rest against the wall. Once the cannon has finished firing: 1. How fast is the traincar moving? 2. How far has the traincar been displaced? 3. Where is the center of mass? Assume friction between the traincar and the ground is negligible, as is air resistance.

Inactive
Cole C.

First, let's look at the concepts we can possibly uncover in the question. Center of mass - we have a system with several different masses in relation to each other Relative coordinates - we have several masses moving in relation to each other Maybe kinetics and kinematics - we have moving objects Maybe dynamics - the cannonballs and the cannon exert forces We may also want to draw a picture of the scenario BEFORE =============== *** = cannonballs | === | === = cannon | ^ *** | ^ ------------------------- Cm = center of mass 00 00 | x | 0m Cm 20m And we know: $$m_c = 200 kg$$ $$m_b = 10 kg$$ $$m_t = 2000 kg$$ $$c_{c,x,i} = 1m$$ $$c_{b,x,i} = 1m$$ $$c_{t,x,i} = \frac{20m}{2} = 10m$$ 1. Now, how fast will the train car be moving after the collision? Since there's no friction, and we know the mass - we can assume an outside force will cause acceleration and increase the speed. Only problem is - there's no outside force. The cannon releases energy in shooting the cannonballs, but the kickback force from each shot is canceled by the equal and opposite impact when the balls hit the inside of the traincar. No energy in or out, and no force applied means there is no speed increase. We can assume the traincar is at rest once all the cannonballs have been shot. 2. Now, how far has the train car moved? Well, we know the center of mass of a group of objects must always stay the same in relation to an outside object. Or, put simply - if one mass moves, the rest have to shift the opposite way to keep the center of mass aligned where it originally was. If the balls shoot to the other side of the traincar, the traincar will have to roll back in the opposite direction to compensate. Equal and opposite reaction. What we can do is calculate the center of mass of the group before the shots are fired, then the center of mass when the cannonballs come to rest at the other end of the cart - then find the difference between the two centers of mass. So, working from what we know: $$c_{x,f} - c_{x,i}$$ gives us the distance the car has moved, if we use our handy equation $$c_x = \frac {\sum_{i=1}^{n} c_{xi}m_i} {\sum_{i=1}^{n} m_i }$$ Or, the sum of all the masses times their centers of mass, divided by the total mass. This gives us a weighted average (ha!) of where the center of mass is - think of it as the balance point where you could lift the traincar up and not have it tip to either side. $$c_{x,i} = \frac {(c_{c,x,i} m_c + c_{b,x,i} m_b + c_{t,x,i} m_t)}{m_c + m_b + m_t}$$ We can see we have all these values, so: $$c_{x,i} = \frac {(1m)(200 kg) + (1m)(40)(10 kg) + (10m)(2000 kg)} {200 kg + (40)(10 kg) + 2000 kg}$$ $$c_{x,i} = \frac {(1m)(200 kg) + (1m)(40)(10 kg) + (10m)(2000 kg)} {200 kg + (400 kg) + 2000 kg}$$ $$c_{x,i} = \frac {20600 kg*m} {2600 kg}$$ $$c_{x,i} = 7.92m$$ Now, for the center after the balls have been fired. In reference to the train, we know the balls have traveled to the righthand wall of the traincar at x = 20m. $$c_{b,x,f} = 20m$$ We know the cannon stays in the same spot, so $$c_{c,x,f} = c_{c,x,i} = 1m$$ Same for the train car, the center is still in the middle at $$c_{t,x,i} = c_{t,x,f} = 10m$$ So the equation for the center of mass will look like: $$c_{x,f} = \frac {(c_{c,x,f} m_c + c_{b,x,f} m_b + c_{t,x,f} m_t)}{m_c + m_b + m_t}$$ $$c_{x,f} = \frac {(1m)(200 kg) + (20m)(40)(10 kg) + (10m)(2000 kg)} {200 kg + (40)(10 kg) + 2000 kg}$$ $$c_{x,f} = \frac {28200 kg*m} {2600 kg}$$ $$c_{x,f} = 10.85m$$ Now, 10.85m - 7.92m = 2.93m, so the train car would have to shift 2.93m to the left when the cannonballs shift the center of mass 2.93m to the right. So the traincar rolls 2.93 meters as the balls are firing, then stops! 3. In order to keep the center of mass in the same place (since no outside forces are acting to move the whole system to a new place), the center of mass will be at the original location, x = 7.92m in our drawing. AFTER =============== | === | | ^ ***| ------------------------- 00 00 | x | 0m Cm 20m Here if you have any questions!

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