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# Tutor profile: Rachel D.

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Rachel D.
Undergraduate Student at Johns Hopkins University
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## Questions

### Subject:Chemistry

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Question:

Calculate what temperature the following reaction becomes spontaneous: \$\$ Cl_{2} \$\$ (aq) ---> \$\$ Cl_{2} \$\$ (g) \$\$ \Delta \$\$ S = 102 J/K*mol \$\$ \Delta \$\$ H = 23 KJ/mol

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Rachel D.

Step 1: Set up inequality In order for a reaction to be spontaneous \$\$ \Delta \$\$ G must be less than 0 \$\$ \Delta \$\$ G = \$\$ \Delta \$\$ H - T\$\$ \Delta \$\$ S 0 > \$\$ \Delta \$\$ H - T\$\$ \Delta \$\$ S Step 2: Plug in values for H and S (Make sure the have the same units). 0 > 23 KJ/mol - T x 0.102 KJ/K*mol (Get rid of units to make it look simpler) 0 > 23 - 0.102 T Solve for T (Remember that when you divide by a negative number, you must flip the inequality sign). 225 < T or T > 225 K

### Subject:Basic Chemistry

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Question:

Complete and balance the following reaction: \$\$ C_{5} H_{12} \$\$ + \$\$ O_{2} \$\$ ->

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Rachel D.

Step 1: Recognize this is a combustion reaction. Combustion reactions always produce Carbon Dioxide ( \$\$ CO_{2} \$\$ ) and Water ( \$\$ H_{2}O \$\$ ) \$\$ C_{5} H_{12} \$\$ + \$\$ O_{2} \$\$ -> \$\$ CO_{2} \$\$ + \$\$ H_{2}O \$\$ Step 2: Balance. All Combustion reactions are balanced in this order: Carbon, Hydrogen, then Oxygen. 5 Carbons on the left so place a 5 in front of \$\$ CO_{2} \$\$ \$\$ C_{5} H_{12} \$\$ + \$\$ O_{2} \$\$ -> 5 \$\$ CO_{2} \$\$ + \$\$ H_{2}O \$\$ 12 Hydrogen on the left so place a 6 in front of \$\$ H_{2}O \$\$ \$\$ C_{5} H_{12} \$\$ + \$\$ O_{2} \$\$ -> 5 \$\$ CO_{2} \$\$ + 6 \$\$ H_{2}O \$\$ 16 Oxygens on the right so place an 8 in front of \$\$ O_{2} \$\$ \$\$ C_{5} H_{12} \$\$ + 8 \$\$ O_{2} \$\$ -> 5 \$\$ CO_{2} \$\$ + 6 \$\$ H_{2}O \$\$

### Subject:Calculus

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Question:

Find two positive numbers whose sum is 200 and whose product is a maximum.

Inactive
Rachel D.

Step 1: Set up equations that model the question (1) x + y = 200 (2) Z= xy Step 2: Solve for Y using equation (1) and plug it into equation (2) so that equation (2) is only in terms of one variable. Make sure to simplify. y = 200 - x ----> Z = x (200-x) ----> Z = 200x - \$\$ x^{2} \$\$ equation (3) Step 3: Find critical points for equation (3) by taking its derivative, setting it equal to zero, and solving for x. Z = 200x - \$\$ x^{2} \$\$ ----> Z' = 200 - 2x ----> 200 - 2x = 0 ---> x = 100 Step 4: Check your critical point to make sure it is a maximum using a number line test or second derivative test. I will be using second derivative test. Second derivative test: Take a second derivative, then plug in the critical point, if second derivative is negative, that means it is concave down and therefore a maximum. Z" = -2 therefore our critical point is a maximum. Step 5: Now that we know our value for x is 100, we simply plug that into equation (1) to solve for y. x = 100 y = 100

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