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Nata [24]
3 years ago
15

Which of the following is released during cellular respiration? A.Carbon dioxide B.Glucose C.Mitochondria D.Oxygen

Chemistry
2 answers:
slamgirl [31]3 years ago
8 0
A. Carbon dioxide is released during cellular respiration
Phantasy [73]3 years ago
4 0

Answer: A.Carbon dioxide

Explanation: Cellular respiration is the process by which  glucose molecule breakdown into energy form (ATP) in presence of oxygen. As a result of this reaction ATP is formed and carbon dioxide is released.

Process of cellular respiration completed in following 4 steps :

Glycolysis, Pyruvate Oxidation, Citric acid (Krebs cycle), and Oxidative Phosphorylation.

Reaction of cellular respiration is as follows -

C6H12O6 + 6O2 ------------ 6CO2 +6H2O  +38 ATP

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If 12.5 grams of strontium hydroxide is reacted with 150 mL of 3.5 M carbonic acid, identify the limiting reactant.
vesna_86 [32]

Answer:

Sr(OH)2

Explanation:

We'll begin by calculating the number of mole of carbonic acid in 150mL of 3.5 M carbonic acid solution. This is illustrated below:

Molarity = 3.5M

Volume = 150mL = 150/1000 = 0.15L

Mole of carbonic acid, H2CO3 =..?

Mole = Molarity x Volume

Mole of carbonic acid, H2CO3 = 3.5 x 0.15 = 0.525 mole.

Next, we shall convert 0.525 mole of carbonic acid, H2CO3 to grams.

Mole of H2CO3 = 0.525 mole

Molar mass of H2CO3 = (2x1) + 12 + (16x3) = 62g/mol.

Mass of H2CO3 =..?

Mass = mole x molar mass

Mass of H2CO3 = 0.525 x 62 = 32.55g

Next, we shall write the balanced equation for the reaction. This is given below:

Sr(OH)2 + H2CO3 → SrCO3 + 2H2O

Next, we shall determine the mass of Sr(OH)2 and H2CO3 that reacted from the balanced equation. This is illustrated below:

Molar mass of Sr(OH)2 = 88 + 2(16 + 1) = 88 + 2(17) = 122g/mol

Mass of Sr(OH)2 from the balanced equation = 1 x 122 = 122g

Molar mass of H2CO3 = (2x1) + 12 + (16x3) = 62g/mol.

Mass of H2CO3 from the balanced equation = 1 x 62 = 62g.

From the balanced equation above, 122g of Sr(OH)2 reacted with 62g of H2CO3.

Finally, we shall determine the limiting reactant as follow:

From the balanced equation above, 122g of Sr(OH)2 reacted with 62g of H2CO3.

Therefore, 12.5g of Sr(OH)2 will react with = (12.5 x 62)/122 = 6.35g.

We can see evidently from the calculations made above that it will take 6.35g out 32.55g of H2CO3 to react with 12.5g of Sr(OH)2. Therefore, Sr(OH)2 is the limiting reactant and H2CO3 is the excess reactant

5 0
3 years ago
Read 2 more answers
Draw the molecules listed below and circle any functional groups present. Provide the following information for each molecule:Mo
aksik [14]

The question requires us to draw the structural formula, provide the name and highlight any functional groups for the compound: diethyl ether.

The molecule diethyl ether can be represented as it follows, with two ethyl groups (-CH2CH3) bonded to a oxygen atom:

Note that the functional group ether (R-O-R) is present in the structre and highlighted in blue in the image. The official name of diethyl ether is ethoxyethane.

3 0
1 year ago
What expression approximates the volume of O2 consumed, measure at STP, when 55 g of Al reacts completely with excess O2?2 Al(s)
nikdorinn [45]

Answer : The volume of O_2 consumed are, 0.75\times 2\times 22.4 L.

Explanation :

The balanced chemical reaction will be:

4Al(s)+3O_2(g)\rightarrow 2Al_2O_3(s)

First we have to calculate the moles of Al.

\text{Moles of }Al=\frac{\text{Mass of }Al}{\text{Molar mass of }Al}

Molar mass of Al = 27 g/mole

\text{Moles of }Al=\frac{55g}{27g/mol}=2mole

Now we have to calculate the moles of O_2.

From the reaction we conclude that,

As, 4 mole of Al react with 3 moles of O_2

So, 2 mole of Al react with \frac{3}{4}\times 2=0.75\times 2 moles of O_2

Now we have to calculate the volume of O_2 consumed.

As we know that, 1 mole of substance occupies 22.4 liter volume of gas.

As, 1 mole of O_2 occupies 22.4 liter volume of O_2 gas

So, 0.75\times 2 mole of O_2 occupies 0.75\times 2\times 22.4 liter volume of O_2 gas

Therefore, the volume of O_2 consumed are, 0.75\times 2\times 22.4 L.

3 0
3 years ago
How many moles of chlorine gas would occupy a volume of 35.5 L at a pressure of 100.0 kPa and a
Nikolay [14]

Answer:

this was the last of .the last 555543 years of 55my in

Explanation:

t

4 0
3 years ago
Each element has a unique number of ___
Zigmanuir [339]
the answer is protons
4 0
3 years ago
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