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cricket20 [7]
3 years ago
5

How many moles of O2 are needed to produce 50.0 moles of CO2?

Chemistry
1 answer:
-Dominant- [34]3 years ago
6 0

Answer:

4 moles

Explanation:

Please see the step-by-step solution in the picture attached below.

Hope this answer can help you. Have a nice day!

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A baseball player hits baseball weighing 0.2 kg (200n g = .25
Ilya [14]

Answer:

its study harder

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3 years ago
What happens when a condition of metal oxide is tested with (i)blue litmus (ii)red litmus
Artemon [7]

it has no effect on the blue litmus.

When it's tested on the red litmus paper it turns blue

7 0
4 years ago
Draw the major product of the reaction between 1-butanol and na2cr2o7, h2so4, h2o
Tju [1.3M]

The major product of reaction between 1-butanol and Na2Cr2O7 is butanoic acid.

When a primary alcohol like 1-butanol (OH is bonded to a primary carbon) is begin to oxidize in the presence of strong oxidizing reagent such as sodium dichromate (Na2Cr2O7) and H2SO4, sulfuric acid, the stepwise oxidation take place as above firstly to the corresponding aldehyde which undergoes further oxidation to the corresponding carboxylic acid.

You can find that the formed aldehyde after first oxidation is butanal and the only organic product, due to the strong oxidizing reagent is butanoic acid.

Thus, the major product formed is butanoic acid.

learn more about oxidation:

brainly.com/question/13059271

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5 0
2 years ago
What evidence of a chemical reaction might you see dropping an alka seltzer
aleksandrvk [35]
Bubbling if you put it in a liquid

3 0
3 years ago
A certain watch’s luminous glow is due to zinc sulfide paint that is energized by beta particles given off by tritium, the radio
sp2606 [1]

Answer:

The watch is 40.9 years old.

Explanation:

To know how many years old is the watch we need to use the following equation:

I_{(t)} = I_{0}e^{-\lambda t}   (1)

Where:

I_{(t)}: is the brightness in a time t = (1/10)I₀

I_{0}: is the initial brightness

λ: is the decay constant of tritium

The decay constant is given by:

\lambda = \frac{ln(2)}{t_{1/2}}   (2)

Where:

t_{1/2}: is the half-life of tritium = 12.3 years

By entering equation (2) into (1)  we have:

I_{(t)} = I_{0}e^{-\lambda t} = I_{0}e^{-\frac{ln(2)}{t_{1/2}}t}

\frac{I_{(t)}}{I_{0}} = e^{-\frac{ln(2)}{t_{1/2}}t}

By solving the above equation for "t" we have:

ln(\frac{I_{(t)}}{I_{0}}) = -\frac{ln(2)}{t_{1/2}}t

t = -\frac{ln(\frac{I_{(t)}}{I_{0}})}{\frac{ln(2)}{t_{1/2}}} = -\frac{ln(\frac{1}{10})}{\frac{ln(2)}{12.3}} = 40.9 y

Therefore, the watch is 40.9 years old.

 

I hope it helps you!

7 0
3 years ago
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