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insens350 [35]
3 years ago
8

Cells take in small, energy-rich molecules from food. These molecules include sugar molecules. Cells use the molecules in a long

series of chemical reactions. These reactions are known together as cellular respiration. Cellular respiration breaks down energy-rich molecules to form different molecules. This process also releases energy. Which of the following statements are true about cellular respiration? Select all that apply.
1.Cellular respiration occurs inside cells. 2.Cellular respiration releases energy. 3.Cellular respiration involves many chemical reactions.
4.Cellular respiration produces sugar molecules. ​
Chemistry
1 answer:
Solnce55 [7]3 years ago
4 0

Answer:

1,2, and 3 are correct.

Explanation:

1 is true because cellular respiration begins in the cytoplasm of a cell. 2 is true because cellular respiration releases energy aerobically which uses glucose and oxygen or anaerobically which just uses glucose. Either way, energy is being released. 3 is true, I don't know how to explain why. Just keep the cellular respiration equation in mind, (C6H12O6 + 6O2 → 6CO2 + 6H2O + Chemical Energy (in ATP)), the chemical reactions are basically summed up into an equation. Hope that helped.

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Soil covered, saturated, submerged, flooded w water, standing water
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How many moles in Co are there in 6 billion Co atoms
nasty-shy [4]
Hello there!

<span> Use Avogadro's number, 1 mol = 6.02 x 10^23 atoms. 
</span>Convert atoms to moles and you get:

<span>9.97x10^(-15) moles Co
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Hope This Helps You!
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3 0
3 years ago
How many moles are in 2.98x10^23 particles?
Novosadov [1.4K]

Answer:

\boxed {\boxed {\sf 0.495 \ mol}}

Explanation:

We are given a number of particles and asked to convert to moles.

<h3>1. Convert Particles to Moles </h3>

1 mole of any substance contains the same number of particles (atoms, molecules, formula units) : 6.022 *10²³ or Avogadro's Number. For this question, the particles are not specified.

So, we know that 1 mole of this substance contains 6.022 *10²³ particles. Let's set up a ratio.

\frac { 1 \ mol }{6.022*10^{23 } \ particles}}

We are converting 2.98*10²³ particles to moles, so we multiply the ratio by that value.

2.98*10^{23} \ particles *\frac { 1 \ mol }{6.022*10^{23 } \ particles}}

The units of particles cancel.

2.98*10^{23}  *\frac { 1 \ mol }{6.022*10^{23 } }}

\frac { 2.98*10^{23}}{6.022*10^{23 } }}  \ mol

0.4948522086 \ mol

<h3>2. Round</h3>

The original measurement of particles (2.98*10²³) has 3 significant figures, so our answer must have the same.

For the number we found, 3 sig figs is the thousandth place.

The 8 in the ten-thousandth place (0.4948522086) tells us to round the 4 up to a 5 in the thousandth place.

0.495 \ mol

2.98*10²³ particles are equal to approximately <u>0.495 moles.</u>

3 0
3 years ago
The oxidation of Cu₂O(s) to CuO(s) is an exothermic process:
Alex73 [517]

Answer:

w = 7376.6 J

Explanation:

To calculate the work done in the system, we need first to calculate the number of moles of all compounds.

The following expression, that comes from the 1° law of thermodynamic, will help you to get the work:

w = ΔnRT

Where:

Δn: difference in the number of moles between products and reactants

R: constant universal of gases (8.314 J / mol K)

T: Temperature in Kelvin.

As we are doing this reaction in STP, then the Pressure is 1 atm, and temperature is 0 °C or 273 K.

Now, we already have the moles of Cu₂O, let's see, according to the balanced reaction, how many moles we should have of O₂ and CuO.

If 1 mole of O₂ reacts with 2 moles of Cu₂O then 6.5 moles of Cu₂O will be:

moles O₂ = 6.5 moles Cu₂O * (1 mole O₂/2 moles Cu₂O) = 3.25 moles O₂

If 2 moles of Cu₂O produces 4 moles of CuO, then 6.5 moles will be:

moles CuO = 6.5 moles Cu₂O*(4 moles CuO/2 moles Cu₂O) = 13 moles CuO

Now that we have the moles, let's calculate the value of Δn.

Δn = moles product - moles reactants

Δn = 13 - (6.5 + 3.25) = 3.25 moles

Now, we can calculate the work done in the system:

w = 3.25 * (8.314 * 273)

<h2>w = 7376.6 J or 7.3766 kJ</h2>

Hope this helps

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