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2 years ago

leaves change carbon dioxide and water into A.sugar and oxygen B.soil and sugar C.oxygen and sunlight D. none of the above

1 answer:

3 0

Hey there! :D

During the process of photosynthesis, leaves turn carbon dioxide and water into glucose and oxygen. (glucose is just a fancy word for sugar)

So "A" is correct.

We need trees to make oxygen as a waste, or product, of the reaction. Similarly, we take in oxygen and breathe out carbon dioxide. As a cycle.

I hope this helps!
~kaikers

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A 6.175 gram sample of an organic compound containing only C, H, and O is analyzed by combustion analysis and 13.30 g CO2 and 5.

Bingel [31]

Answer:

Empirical and molecular formulas are the same, C₅H₁₀O₂.

Explanation:

Hello!

In this case, when determining the empirical and molecular formulas of organic compounds via combustion analysis, we first need to compute the moles of carbon and hydrogen via the yielded mass of carbon dioxide and water:

$n_C=13.30gCO_2*\frac{1molCO_2}{44.01gCO_2}*\frac{1molC}{1molCO_2}=0.30molC\\\\n_H=5.447gH_2O*\frac{1molH_2O}{18.02gH_2O}*\frac{2molH}{1molH_2O}=0.60molH$

Next, we need to compute the mass of oxygen by subtracting the mass of carbon and hydrogen to the mass of the sample of the compound:

$m_O =6.175g-0.3molC*12.01gC/molC-0.6molH*1.01gH/molH =1.966gO$

And consequently the moles:

$n_O=0.12molO$

Now, we need to divide the moles of each atom by the fewest moles, it in this case, those of oxygen to obtain the subscripts in the empirical formula:

$C=\frac{0.30}{0.12} =2.5\\\\H=\frac{0.60}{0.12} =5\\\\O=\frac{0.12}{0.12} =1$

Thus, the empirical formula, taken the nearest whole number is:

$C_5H_{10}O_2$

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Best regards!

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2 years ago

Consider the following reaction, which is spontaneous at room temperature. C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g) Is ΔS positive o

Ugo [173]

Answer:

∆H = negative and ∆S = positive.

Explanation:

The reaction given in the question is spontaneous at room temperature ,

hence ,

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According to the formula ,

∆G = ∆H -T∆S

The value of ∆G can be negative , if ∆H has a negative value and ∆S has a positive value , because , T∆S , has a negative sign .

Hence , the answer will be , ∆H = negative and ∆S = positive.

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