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FrozenT [24]
3 years ago
5

If a 750 mL of a gas at a pressure of 100.7 kPa has a decrease of pressure to 99.8 kPa, what is the new volume? Show work

Chemistry
1 answer:
SVEN [57.7K]3 years ago
4 0

Explanation:

P1V1 = P2V2

(100.7 kPa)(0.75 L) = (99.8 kPa)V2

V2 = (100.7 kPa)(0.75 L)/(99.8 kPa)

= 0.757 L

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Given an initial cyclopropane concentration of 0.00560 m, calculate the concentration of cyclopropane that remains after 1.50 ho
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<span>We can solve this problem by assuming that the decay of cyclopropane follows a 1st order rate of reaction. So that the equation for decay follows the expression:</span>

A = Ao e^(- k t) 

Where,

A = amount remaining at time t = unknown (what to solve for) <span>
Ao = amount at time zero = 0.00560 M </span><span>
<span>k = rate constant
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The rate constant should be given in the problem which I think you forgot to include. For the sake of calculation, I will assume a rate constant which I found in other sources:

k = 5.29× 10^–4 s–1                     (plug in the correct k value)

<span>Plugging in the values in the 1st equation:</span>

A = 0.00560 M * e^(-5.29 × 10^–4 s–1 * 5400 s )

A = 3.218 <span>× 10^–4 M           (simplify as necessary)</span>

8 0
3 years ago
Which surface has the most friction
Nesterboy [21]

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3 0
4 years ago
Read 2 more answers
You have a 100 ml stock solution of 100 mg/ml ampicillin in deionized water. You want to make 30 ml of 25 mg/ml ampicillin in de
Georgia [21]

Explanation:

Volume of the stock solution is V_{1} = ?

Initial concentration of ampicillin is M_{1} = 100 mg/ml

Final volume (V_{2}) = 30 ml

Final concentration of  ampicillin (M_{2}) = 25 mg/ml

Therefore, calculate the volume of given stock as follows.

              M_{1} \times V_{1} = M_{2} \times V_{2}

Now, putting the given values into the above formula as follows.

            M_{1} \times V_{1} = M_{2} \times V_{2}

            100 mg/ml \times V_{1} = 25 g/ml \times 30 ml    

                   V_{1} = 7.5 ml

Now, we will calculate the volume of water added into it as follows.

            Volume of water added = V_{2} - V_{1}

                                                  = 30 ml - 7.5 ml

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Thus, we can conclude that required solution is 22.5 ml of deionized water.

3 0
3 years ago
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