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Ber [7]
3 years ago
9

Fill in the blanks: 87 pg = [? ]x10^[?]g

Chemistry
1 answer:
jeyben [28]3 years ago
5 0

87 picograms are equivalent to 8.7\times 10^{-11} grams.

This is a unit problem, in which a <em>derived</em> unit must translated into a <em>fundamental</em> unit. A picogram is a equivalent to <em>a trillionth</em> of a gram, which is the fundamental SI <em>mass</em> unit. Mathematically speaking, we have the following equivalence:

1\,pg = 1\times 10^{-12}\,g

Now we proceed to find the equivalent value for this case:

87\,pg

87\times 10^{-12}\,g

8.7\times 10^{-11}\,g

87 picograms are equivalent to 8.7\times 10^{-11} grams.

To learn more on unit conversions, we kindly invite to check this verified question: brainly.com/question/13016491

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The combustion of caffeine with the molecular masses is given below. If you have 0.150 grams of caffeine, how much NO2 in grams
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2. 0.27 g of CO2.

Explanation:

The balanced equation for the reaction is given below:

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Next, we shall determine the mass of caffeine, C8H10N4O2 that reacted and the masses of nitrogen (iv) oxide, NO2 and carbon (iv) oxide, CO2 produced from the balanced equation. This can be obtained as follow:

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Summary:

From the balanced equation above,

388.38 g of caffeine, C8H10N4O2 reacted to produce 704.16 g of CO2 and 368.08 g of NO2.

1. Determination of the mass of NO2 produced from the reaction.

This can be obtained as follow:

From the balanced equation above,

388.38 g of caffeine, C8H10N4O2 reacted to produce 368.08 g of NO2.

Therefore, 0.15 g of caffeine, C8H10N4O2, will react to produce = (0.15 × 368.08) / 388.38 = 0.14 g of NO2.

Therefore, 0.14 g of NO2 was obtained from the reaction.

2. Determination of the mass of CO2 produced from the reaction.

This can be obtained as follow:

From the balanced equation above,

388.38 g of caffeine, C8H10N4O2 reacted to produce 704.16 g of CO2.

Therefore, 0.15 g of caffeine, C8H10N4O2, will react to produce = (0.15 × 704.16) / 388.38 = 0.27 g of CO2.

Therefore, 0.27 g of CO2 was obtained from the reaction.

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