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3 years ago

Which of the following explains the conservation of mass during cellular respiration?

Chemistry

1 answer:

lord [1]3 years ago

8 0

Answer:

The total number of atoms when glucose and oxygen reacts stays the same when carbondioxide and water are produced.

Explanation:

Chemical reaction:

C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O

We can see that the number of atoms of each element remain same on both side of reaction so law of conservation of mass is followed by this reaction. Six number of carbon atoms twelve number of hydrogen atoms and eighteen number of oxygen atoms are present on both side.

There are two types of respiration:

1. Aerobic respiration

2. Anaerobic respiration

Aerobic respiration

It is the breakdown of glucose molecule in the presence of oxygen to yield large amount of energy. Water and carbon dioxide are also produced as a byproduct.

Glucose + oxygen → carbon dioxide + water + 38ATP

Anaerobic Respiration

It is the breakdown of glucose molecule in the absence of oxygen and produce small amount of energy. Alcohol or lactic acid and carbon dioxide are also produced as byproducts.

Glucose→ lactic acid/alcohol + 2ATP + carbon dioxide

This process use respiratory electron transport chain as electron acceptor instead of oxygen. It is mostly occur in prokaryotes. Its main advantage is that it produce energy (ATP) very quickly as compared to aerobic respiration.

You might be interested in

For the reaction o(g) + o2(g) → o3(g) δh o = −107.2 kj/mol given that the bond enthalpy in o2(g) is 498.7 kj/mol, calculate the

Aneli [31]

The reaction;
O(g) +O2(g)→O3(g), ΔH = sum of bond enthalpy of reactants-sum of food enthalpy of products.
ΔH = ( bond enthalpy of O(g)+bond enthalpy of O2 (g) - bond enthalpy of O3(g)
-107.2 kJ/mol = O+487.7kJ/mol =O+487.7 kJ/mol +487.7kJ/mol =594.9 kJ/mol
Bond enthalpy (BE) of O3(g) is equals to 2× bond enthalpy of O3(g) because, O3(g) has two types of bonds from its lewis structure (0-0=0).
∴2BE of O3(g) = 594.9kJ/mol
Average bond enthalpy = 594.9kJ/mol/2
=297.45kJ/mol
∴ Averange bond enthalpy of O3(g) is 297.45kJ/mol.

8 0

3 years ago

Ok so I have 3 beakers of clear liquid each holding a different substance, Water, Magnesium Sulphate and Sodium Carbonate how ca

nataly862011 [7]

For water you could add oil..ex: cooking oil separates form water because water is heavier than oil.

For Magnesium Sulfate you could add Sodium Carbonate..ex: Sodium Carb reacts to Mg Sulfate adding a darker hue to the liquid and adding a lot of bubbles.

For Sodium Carbonate you could add Sulfuric Acid..ex: Sulfuric Acid would add a reaction to the Sodium Carb that would resembling water boiling

H0P3 It H3LPS :)

7 0

3 years ago

At what Celsius temperature does 0.750 mol of an ideal gas occupy a volume of 35.9 L at 114 kPa?

shepuryov [24]

382.85 Celsius is the temperature does 0.750 moles of an ideal gas occupy a volume of 35.9 L at 114 kPa.

Explanation:

Given data:

number of moles of the gas = 0.75 moles

volume of the gas = 35.9 liters

pressure of the gas = 114 KPa or 1.125 atm

R = 0.0821 latm/moleK

temperature of the gas T = ?

The equation used to calculate temperature from above data is ideal gas law equation.

the equation is :

PV = nRT

T = $\frac{PV}{nR}$

Putting the values in the above rewritten equation:

T = $\frac{1.125 X 35.9}{0.75 X 0.0821}$

T = 655.9 K

To convert kelvin into celsius, formula used is

K = 273.15+ C

putting the values in the equation

C = 656 - 273.15

= 382.85 Celsius

8 0

3 years ago

You are given a solid that is a mixture of na2so4 and k2so4.

Murljashka [212]

Here we have to calculate the amount of $SO_{4}^{2-}$ ion present in the sample.

In the sample solution 0.122g of $SO_{4}^{2-}$ ion is present.

The reaction happens on addition of excess BaCl₂ in a sample solution of potassium sulfate (K₂SO₄) and sodium sulfate [(Na)₂SO₄] can be written as-

K₂SO₄ = 2K⁺ + $SO_{4}^{2-}$

(Na)₂SO₄=2Na⁺ + $SO_{4}^{2-}$

Thus, BaCl₂+ $SO_{4}^{2-}$ = BaSO₄↓ + 2Cl⁻ .

(Na)₂SO₄ and K₂SO₄ is highly soluble in water and the precipitation or the filtrate is due to the BaSO₄ only. As a precipitation appears due to addition of excess BaCl₂ thus the total amount of $SO_{4}^{2-}$ ion is precipitated in this reaction.

The precipitate i.e. barium sulfate (BaSO₄)is formed in the reaction which have the mass 0.298g.

Now the molecular weight of BaSO₄ is 233.3 g/mol.

We know the molecular weight of sulfate ion ( $SO_{4}^{2-}$ ) is 96.06 g/mol. Thus in 1 mole of BaSO₄ 96.06 g of $SO_{4}^{2-}$ ion is present.

Or. we may write in 233.3 g of BaSO₄ 96.06 g of $SO_{4}^{2-}$ ion is present. So in 1 g of BaSO₄ $\frac{96.06}{233.3}=0.411$ g of $SO_{4}^{2-}$ ion is present.

Or, in 0.298 g of the filtered mass (0.298×0.411)=0.122g of $SO_{4}^{2-}$ ion is present.

5 0

3 years ago

How does heat transfer through conduction

Alexeev081 [22]

Answer: Conduction is the process by which heat energy is transmitted through collisions between neighboring atoms or molecules. ... The fire's heat causes molecules in the pan to vibrate faster, making it hotter. These vibrating molecules collide with their neighboring molecules, making them also vibrate faster.

HOPE THIS HELPS

8 0

3 years ago