Pls help really need it pls

Consider the reaction. X ( g ) + Y ( g ) − ⇀ ↽ − Z ( g ) K p = 1.00 at 300 K In which direction will the net reaction proceed fo

marta [7]

Answer:

Explanation:

We have in this question the equilibrium

X ( g ) + Y ( g ) ⇆ Z ( g )

With the equilibrium contant Kp = pZ/(pX x pY)

The moment we change the concentration of Y, we are changing effectively the partial pressure of Y since pressure and concentration are directly proportional

pV = nRT ⇒ p = nRT/V and n/V is molarity.

Therefore we can calculate the reaction quotient Q

Qp = pZ/(pX x pY) = 1/ 1 x 0.5 atm = 2

Since Qp is greater than Kp the system proceeds from right to left.

We could also arrive to the same conclusion by applying LeChatelier´s principle which states that any disturbance in the equilibrium, the system will react in such a way to counteract the change to restore the equilibrium. Therefore, by having reduced the pressure of Y the system will react favoring the reactants side increasing some of the y pressure until restoring the equilibrium Kp = 1.

4 0

3 years ago

Complete the reaction

GarryVolchara [31]

1) (C2H5)2CBrCH2CH3 is the answer

explaiation:-

so when HBr is added to an alkene , according to the Markonicoff's rule ...H atoms are bonded to the C containing the most amount of H and Br is added to the other C.

2) Just add alkoholic KOH∆

6 0

3 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

The standard unit for concentration, Molarity (M) can also be written as?

emmasim [6.3K]

mole is the standardized form of molarity

8 0

2 years ago

Which metal atoms can form ionic bonds by losing electrons from both the outermost and next to outermost principal energy levels

RoseWind [281]

Full question options;

(Fe, Pb, Mg, or Ca)

Answer:

Iron - Fe

Explanation:

We understand tht metals pretty much form bonds by losing their valence (outermost electrons). But this question specifically asks for metals that lose beyond their outermost electrons; next to outermost principal energy levels.

Pb, Mg, and Ca only lose their outermost electrons to form the following ions;

Pb2+, Mg2+, and Ca2+.

This is because their ions have achieved a stable octet configuration - the dreamland of atoms where they are satisfied and don't need to go into reactions again.

Iron on the other hand has the following electronic configurations;

Fe: [Ar]4s2 3d6

Fe2+: [Ar]4s0 3d6

Fe3+: [Ar]4s0 3d5

This means ion can lose both the ooutermost electrons (4s) and next to outermost principal energy levels (3d). So correct option is Iron.

5 0

3 years ago