Shifting the layers of an ionic crystal causes the crystal to

Sunny_sXe [5.5K]

In 3.8 moles of li, there are about 0.144071459 grams. I assume that you are talking about lithium.

7 0

4 years ago

If each NADHNADH generates 3 ATPATP molecules and each FADH2FADH2 generates 2 ATPATP molecules, calculate the number of ATPATP m

Nadusha1986 [10]

Answer:

$128~ATP$

Explanation:

The metabolic pathway by which energy can be obtained from a fatty acid is called "beta-oxidation". In this route, acetyl-Coa is produced by removing 2 carbons from the fatty acid for each acetyl-Coa produced. In other words, for each round, 1 acetyl Coa is produced and for each round 2 carbons are removed from the initial fatty acid. Therefore, the first step is to calculate the number of rounds that will take place for an 18-carbon fatty acid using the following equation:

$Number~of~Rounds=\frac{n}{2}-1$

Where "n" is the number of carbons, in this case "18", so:

$Number~of~Rounds=\frac{18}{2}-1~=~8$

We also have to calculate the amount of Acetyl-Coa produced:

$Number~of~Acetyl-Coa=\frac{18}{2}~=~9$

Now, we have to keep in mind that in each round in the beta-oxidation we will have the production of 1 $FADH_2$ and 1 $NADH$ . So, if we have 8 rounds we will have 8 $FADH_2$ and 8 $NADH$ .

Finally, for the total calculation of ATP. We have to remember the yield for each compound:

-) $1~FADH_2~=~2~ATP$

-) $1~NADH~=~3~ATP$

-) $Acetyl~CoA~=~10~ATP$

Now we can do the total calculation:

$(8*2)~+~(8*3)~+~(9*10)=130~ATP$

We have to subtract "2 ATP" molecules that correspond to the activation of the fatty acid, so:

$130-2=128~ATP$

In total, we will have 128 ATP.

I hope it helps!

6 0

3 years ago

As part of a soil analysis on a plot of land, a scientist wants to determine the ammonium content using gravimetric analysis wit

jeka94

Answer:

Mass percentage of NH₄Cl = 3.54%

Mass percentage of K₂CO₃ = 1.01%

Explanation:

If a 200.0 mL aliquot produced 0.105 g of KB(C₆H₅)₄, then a 100.0 mL aliquot would produce 1/2 * 0.105 g = 0.0525 g of KB(C₆H₅)₄.

Therefore, mass of NH₄B(C₆H₅)₄ in the 100.0 ml aliquot = (0.277 - 0.0525)g = 0.2245 g

Number of moles of NH₄B(C₆H₅)₄ in 0.2245 g = 0.2245 g/ 337.27 g/mol = 0.0006656 moles

In 500 ml solution, number of moles present = 0.0006656 * 500/100 = 0.003328 moles.

From equation of the reaction; mole ratio of NH₄⁺ and NH₄B(C₆H₅)₄ = 1:1

Similarly, mole ratio of NH₄⁺ and NH₄Cl = 1:1

Therefore, moles of NH₄Cl in 500 ml sample = 0.003328 moles

Mass of NH₄Cl = 0.003328 mol * 53.492 g/mol = 0.178 g

Mass percentage of NH₄Cl = (0.178/5.025) * 100% = 3.54%

Number of moles of KB(C₆H₅)₄ in 0.105 g (precipitated from 200.0 ml aliquot) = 0.105 g/ 358.33 g/mol = 0.000293 moles

In 500 ml solution, number of moles present = 0.000293 * 500/200 = 0.0007326 moles.

From equation of the reaction; mole ratio of K⁺ and KB(C₆H₅)₄ = 1:1

Similarly, mole ratio of K⁺ and K₂CO₃ = 2:1

Therefore, moles of K₂CO₃ in 500 ml sample = 0.0007326/2 moles = 0.0003663 moles

Mass of K₂CO₃ = 0.0003663 mol * 138.21 g/mol = 0.05063 g

Mass percentage of K₂CO₃ = (0.05063/5.025) * 100% = 1.01%

7 0

4 years ago

A solution has a [OH−] of 1 x 10−12. What is the pOH of this solution? 2 7 10 12

strojnjashka [21]

-log (1×10^-12) is how you calculate the pOH which in this case is 12

7 0

3 years ago

Read 2 more answers

Help! I can't find this answer anywhere!

Gnom [1K]

What is the question that needs an answer?

8 0

3 years ago