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azamat
1 year ago
15

3 Cells use oxygen to release energy during

Chemistry
2 answers:
Tamiku [17]1 year ago
4 0

Answer:

Explanation: Cellular respiration is the process that occurs in the mitochondria of organisms (animals and plants) to break down sugar in the presence of oxygen to release energy in the form of ATP. This process releases carbon dioxide and water as waste products.

Elis [28]1 year ago
3 0
Cell division.
synthesis of proteins from amino acids.
active transport.
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Help me what the landform
mariarad [96]
A.) Delta . Delta is a landform that forms from Deposition.
8 0
3 years ago
Read 2 more answers
32 gm of O2 to mole of O2
lana [24]

Molar mass of O2:-

\\ \rm\longmapsto 2(16u)=32g/mol

Now

\boxed{\sf No\:of\;moles=\dfrac{Given\:Mass}{Molar\:Mass}}

\\ \rm\longmapsto No\:of\;moles=\dfrac{32}{32}

\\ \rm\longmapsto No\:of\;moles=1mol

5 0
3 years ago
A solution with a hydrogen ion concentration of 3.25 × 10-2 m is ________ and has a hydroxide concentration of _______
Romashka-Z-Leto [24]

To know the acidity of a solution, we calculate the pH value. The formula for pH is given as:

<span>pH = - log [H+]            where H+ must be in Molar</span>

We are given that H+ = 3.25 × 10-2 M

Therefore the pH is:

pH = - log [3.25 × 10-2] 

pH = 1.488

Since pH is way below 7, therefore the solution is acidic.

 

To find for the OH- concentration, we must remember that the product of H+ and OH- is equivalent to 10^-14. Therefore,

[H+]*[OH-] = 10^-14 <span>
</span>[OH-] = 10^-14 / [H+]

[OH-] = 10^-14 / 3.25 × 10-2

[OH-] = 3.08 × 10-13 M

 

Answers:

Acidic

[OH-] = 3.08 <span>× 10-13 M</span>

6 0
2 years ago
Which statements about the operation of an automated micropipettor are true? The volume dial has already been set, do not alter
yanalaym [24]

Answer:

The correct statement is that the point of initial resistance is the level of depression that will fill the pipette with the desired volume of solution.

Explanation:

The pipette can be otherwise stored vertically or horizontally, this has nothing to do with the operation, and before the operation, the dial needs to be set. Also, the plastic pipette tip should not be ejected at all, thus, no new disposable plastic tip is required for each sample.

The correct statement is that the point of initial resistance is the level of depression that will fill the pipette with the desired volume of solution.

4 0
2 years ago
When 5.00 g of Al2S3 and 2.50 g of H2O are reacted according to the following reaction: Al2S3(s) + 6 H2O(l) → 2 Al(OH)3(s) + 3 H
Debora [2.8K]

Answer:

Y=58.15\%

Explanation:

Hello,

For the given chemical reaction:

Al_2S_3(s) + 6 H_2O(l) \rightarrow 2 Al(OH)_3(s) + 3 H_2S(g)

We first must identify the limiting reactant by computing the reacting moles of Al2S3:

n_{Al_2S_3}=5.00gAl_2S_3*\frac{1molAl_2S_3}{150.158 gAl_2S_3} =0.0333molAl_2S_3

Next, we compute the moles of Al2S3 that are consumed by 2.50 of H2O via the 1:6 mole ratio between them:

n_{Al_2S_3}^{consumed}=2.50gH_2O*\frac{1molH_2O}{18gH_2O}*\frac{1molAl_2S_3}{6molH_2O}=0.0231mol  Al_2S_3

Thus, we notice that there are more available Al2S3 than consumed, for that reason it is in excess and water is the limiting, therefore, we can compute the theoretical yield of Al(OH)3 via the 2:1 molar ratio between it and Al2S3 with the limiting amount:

m_{Al(OH)_3}=0.0231molAl_2S_3*\frac{2molAl(OH)_3}{1molAl_2S_3}*\frac{78gAl(OH)_3}{1molAl(OH)_3} =3.61gAl(OH)_3

Finally, we compute the percent yield with the obtained 2.10 g:

Y=\frac{2.10g}{3.61g} *100\%\\\\Y=58.15\%

Best regards.

7 0
2 years ago
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