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Trava [24]
3 years ago
14

What are the products of the combustion of a hydrocarbon?

Chemistry
2 answers:
tiny-mole [99]3 years ago
8 0

Answer:

carbon dioxide and water

Keith_Richards [23]3 years ago
8 0

Answer:

The answer is C I just took the test and got a 100.

Explanation:

Have an amazing day and do a great job on your school work : )

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At constant temperature and pressure, 5.0 L of SO2 is
Ne4ueva [31]
I've done this a few times and keep coming up with 5.0 L . I used the mole ratio, and SO 2 as the LR. That would make 2 moles of SO3 5.0 L but its not one of your choices. the temp and pressure are constant, so according to n/v = p/rt the volume is the same as the moles. hope this helps
4 0
3 years ago
¿Cuántos moles de C3H8 contienen 9.25 x 10^24 moléculas?
Igoryamba

Answer:

\large \boxed{\text{e. 15.4 mol C$_{3}$H}_{8}}

Explanation:\text{Moles of propane} = 9.25 \times 10^{24}\text{ molecules} \times \dfrac{\text{1 mol }}{6.022 \times\ 10^{23} \text{molecules}} = \textbf{15.4 mol}

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3 years ago
Can some one help I'm lost iam being timed:( Show using two conversion factors how you would convert from 0.020 kg into mg
dezoksy [38]

Answer:

20000 Mg

Explanation:

5 0
3 years ago
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HACTEHA [7]

Answer:

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Explanation:

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6 0
3 years ago
Read 2 more answers
The SN1 reaction yields Entry field with incorrect answer two (number) products. This is because ______________________. Entry f
vagabundo [1.1K]

Answer:

Four substitution products are obtained. The carbocation that forms can react with either nucleophile (H2O or CH3OH) from either the top or bottom side of the molecule

Explanation:

An SN1 reaction usually involves the formation of a carbocation in the slow rate determining step. This carbocation is now attacked by a nucleophile in a subsequent fast step to give the desired product.

However, the product is obtained as a racemic mixture because the nucleophile may attack from the top or bottom of the carbocation hence both attacks are equally probable.

The attacking nucleophile in this case may be water or CH3OH

4 0
4 years ago
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