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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.
Answer:

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.
Answer:

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.

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Rachel D.
Answer:

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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