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Licemer1 [7]
2 years ago
9

Consider the reaction 2CO * O2 —> 2 CO2 what is the percent yield of carbon dioxide (MW= 44g/mol) of the reaction of 10g of c

arbon monoxide (MW=28g/mol) with the excess O2 produces 10g of carbon dioxide? Answer in units of %
Chemistry
1 answer:
Arturiano [62]2 years ago
8 0

Answer:

Y = 62.5%

Explanation:

Hello there!

In this case, for the given chemical reaction whereby carbon dioxide is produced in excess oxygen, it is firstly necessary to calculate the theoretical yield of the former throughout the reacted 10 grams of carbon monoxide:

m_{CO_2}^{theoretical}=10gCO*\frac{1molCO}{28gCO}*\frac{2molCO_2}{2molCO}  *\frac{44gCO_2}{1molCO_2}\\\\ m_{CO_2}^{theoretical}=16gCO_2

Finally, given the actual yield of the CO2-product, we can calculate the percent yield as shown below:

Y=\frac{10g}{16g} *100\%\\\\Y=62.5\%

Best regards!

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<u>Answer:</u> This therapy is available for 15 days.

<u>Explanation:</u>

We are given:

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So, the volume required for 300 mg of solution will be = \frac{5mL}{75mg}\times 300mg=20mL

The total volume of the ranitidine bottle = 300 mL

To calculate the number of days, we divide the total volume of the bottle by the volume of dose taken per night, we get:

\text{Number of days}=\frac{\text{Total volume}}{\text{Volume of dose taken per night}}=\frac{300mL}{20mL}=15

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3 years ago
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Answer:

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Now that Snape and Dumbledore has taught you the finer points of hydration calculations they have a slightly more challenging pr
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Answer:

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