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asambeis [7]
3 years ago
7

Can someone help me please and please show work

Chemistry
1 answer:
FinnZ [79.3K]3 years ago
4 0

Answer:

Average of the trial is: 288.50 C

Percent Error: 3.83%

Explanation:

(291 + 287 + 295 + 281) : 4 = 288.50 C

Average: 288.50 C

Percent Error: {(300 - 288.50) : 300} x 100%  = 3.83 %

                   

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A patient is given 200 mg of a drug with the formula: C32H34N6O10. How many mg of the drug is carbon?
Mice21 [21]

200 mg of C₃₂H₃₄N₆O₁₀ contains 116 mg of carbon.

To obtain the answer to the question, we'll begin by calculating the molar mass of C₃₂H₃₄N₆O₁₀. This can be obtained as follow:

Molar mass of C₃₂H₃₄N₆O₁₀ = (12×32) + (1×34) + (14×6) + (16×10)

= 384 + 34 + 84 + 160

<h3>= 662 g/mol </h3>

From the molar mass of C₃₂H₃₄N₆O₁₀, we can see that:

<h3>662 g of C₃₂H₃₄N₆O₁₀ contains 384 g of carbon. </h3>

Converting 662 g of C₃₂H₃₄N₆O₁₀ to mg, we have

1 g = 1000 mg

Therefore,

662 g = 662 × 1000

<h3>662 g of C₃₂H₃₄N₆O₁₀ = 662000 mg</h3>

Converting 384 g of carbon to mg, we have,

1 g = 1000 mg

Therefore,

384 g = 384 × 1000

<h3>384 g of carbon = 384000 mg</h3>

Thus, we can say that:

662000 mg of C₃₂H₃₄N₆O₁₀ contains 384000 mg of carbon.

Finally, we shall determine the mass (in mg) of carbon in 200 mg of C₃₂H₃₄N₆O₁₀. This can be obtained as follow:

662000 mg of C₃₂H₃₄N₆O₁₀ contains 384000 mg of carbon.

Therefore, 200 mg of C₃₂H₃₄N₆O₁₀ will contain = \frac{200 * 384000}{662000} = 116 mg of carbon.

Thus, we can conclude that 200 mg of C₃₂H₃₄N₆O₁₀ contains 116 mg of carbon.

Learn more: brainly.com/question/24572517

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