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natita [175]
3 years ago
12

How many grams of CO2 are used when 6.50 g of O2 are produced?

Chemistry
1 answer:
RoseWind [281]3 years ago
7 0

Answer:

Approximately 5.05 g of CO2 are used when 6.50 g of O2 are produced.

Explanation:

The answer to this question is the implication that the production of 6.5g of O2 requires 5g-6g CO2, since it is 63% oxygen by mass and 37% carbon dioxide by mass (not including water vapor). This would mean that 4g-5g CO2 productions could be expected to yield 2.25 or so grams oxygen for each gram CO2 produced, depending on how low the temperature was at which the reaction took place (a product like dry ice produces only cold enough to use 1/16th its volume in volume rather than 1/200th).

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An astronomer observes an asteroid in the solar system. He notes that the asteroid is three times farther from the Sun than Eart
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Answer:

3 AU

Explanation:

The distance from the Earth to the Sun is known as 1 AU, or 1 Astronomical Unit. If an asteroid is three times this distance, it is 3 AU away.

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3 years ago
Which spectroscopic tool would be best for distinguishing a sample of 1,2,2-tribromopropane from 1,1,2-tribromopropane?
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1-H NMR spectroscopy tool will be used for distinguishing a sample of 1,2,2-tribromopropane from 1,1,2-tribromopropane.

The preferred method for determining or validating the structure of organic molecules or those containing protons is H NMR. When compared to other nuclei, a solution-state proton spectrum may be obtained relatively quickly, and it contains a wealth of knowledge regarding a compound's structure.

It can be calculated by simply counting the number of unique hydrogens on one side of the symmetry plane will give you the count of signals individual molecules emit in a 1H NMR spectrum.

Therefore, 1-H NMR spectroscopy tool will be used for distinguishing a sample of 1,2,2-tribromopropane from 1,1,2-tribromopropane.

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5 0
2 years ago
A cylinder with a moving piston expands from an initial volume of 0.250 L against an external pressure of 2.00 atm. The expansio
kondor19780726 [428]

Answer:

The final volume of the cylinder is 1.67 L

Explanation:

Step 1: Data given

Initial volume = 0.250 L

external pressure = 2.00 atm

Expansion does 288 J of work on the surroundings

Step 2: Definition of reversible work:

Wrev = -P(V2-V1) = -288 J

The gas did work, so V2>V1  (volume expands) and the work has a negative sign.(Wrev<0)

V2 = (-Wrev/P)  + V1

⇒ with Wrev = reverse work (in J)

⇒ with P = the external pressure (in atm)

⇒ with V1 = the initial volume

We can see that your pressure is in  atm  and energy in J

To convert from J to L * atm we should use a convenient conversion unit using the universal gas constants :

R = 8.314472 J/mol *K and R= 0.08206 L*atm/K*mol

V2 =- (-288 J * (0.08206 L*atm/K*mol  /8.314 J/mol *K))/2.00 atm  + 0.250L

V2 = 1.67 L

The final volume of the cylinder is 1.67 L

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