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3 years ago

5

C6h12o6 + 6o2 → 6h2o+ 6co2 when you respire, your cells are undergoing a chemical reaction. glucose, one of the reactants, is be

ing broken down or decomposed. what is one of the products of cellular respiration?
a.c6h12o6
b.co2
c.h2
d.o2

2 answers:

Arte-miy333 [17]3 years ago

7 0

I really thinks its b

11Alexandr11 [23.1K]3 years ago

4 0

The correct answer is co2

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The three reactions involved in the Ostwald process for the conversion of NH3 to HNO3 are:

4NH3(g) + 5O2(g) ==> 4NO(g) + 6H2O(l) ------(1)

2NO(g) + O2(g) ==> 2NO2(g) ------------------------(2)

3NO2(g) + H2O(l) ==> 2HNO3(aq) + NO(g) ---(3)

Mass of HNO3 produced = 1.81 tons

In grams, the mass of HNO3 = 1.81*2000*453.6 = 1642032 g

Molar mass of HNO3 = 63 g/mol

Thus, # moles of HNO3 produced = 1642032/63 = 26064 moles

Based on the stoichiometry of reaction (3):

Theoretical moles of NO2 = 1.5(Moles HNO2 ) = 1.5(26064) = 39096

Assuming 80% yield:

Actual moles of NO2 = 0.8*39096 = 31277 moles

Based on the stoichiometry in reaction (2):

Theoretical moles of NO = 31277 moles

Assuming 80% yield:

Actual moles of NO = 0.8*31277 = 25022 moles

Bases on stoichiometry in reaction(3):

Moles of NH3 = 25022 moles

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A sample of CaCO3 (molar mass 100. g) was reported as being 30. percent Ca. Assuming no calcium was present in any impurities, c

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Answer:

Approximately 75%.

Explanation:

Look up the relative atomic mass of Ca on a modern periodic table:

Ca: 40.078.

There are one mole of Ca atoms in each mole of CaCO₃ formula unit.

The mass of one mole of CaCO₃ is the same as the molar mass of this compound: $\rm 100\; g$ .
The mass of one mole of Ca atoms is (numerically) the same as the relative atomic mass of this element: $\rm 40.078\; g$ .

Calculate the mass ratio of Ca in a pure sample of CaCO₃:

$\displaystyle \frac{m(\mathrm{Ca})}{m\left(\mathrm{CaCO_3}\right)} = \frac{40.078}{100} \approx \frac{2}{5}$ .

Let the mass of the sample be 100 g. This sample of CaCO₃ contains 30% Ca by mass. In that 100 grams of this sample, there would be $\rm 30 \% \times 100\; g = 30\; g$ of Ca atoms. Assuming that the impurity does not contain any Ca. In other words, all these Ca atoms belong to CaCO₃. Apply the ratio $\displaystyle \frac{m(\mathrm{Ca})}{m\left(\mathrm{CaCO_3}\right)} \approx \frac{2}{5}$ :

$\begin{aligned} m\left(\mathrm{CaCO_3}\right) &= m(\mathrm{Ca})\left/\frac{m(\mathrm{Ca})}{m\left(\mathrm{CaCO_3}\right)}\right. \cr &\approx 30\; \rm g \left/ \frac{2}{5}\right. \cr &= 75\; \rm g \end{aligned}$ .

In other words, by these assumptions, 100 grams of this sample would contain 75 grams of CaCO₃. The percentage mass of CaCO₃ in this sample would thus be equal to:

$\displaystyle 100\%\times \frac{m\left(\mathrm{CaCO_3}\right)}{m(\text{sample})} = \frac{75}{100} = 75\%$ .

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