Mostly located on the bottom of the leaf, open close the stomata

What is the fate of glucose 6‑phosphate, glycolytic intermediates, and pentose phosphate pathway intermediates in this cell? Gly

madreJ [45]

Answer:

The Phosphorylated glucose(glucose +inorganic phosphate), with the energy supplied from ATP hydrolysis formed glucose 6- phosphate, which is later converted to 2 molecules of fructose 6-phosphate- this is phosphorylation.And represented the fate of glucose -6-phosphate.

The fructose 6-phosphate are converted to triose phosphate- which is a 2-molecules of 3C compound. The latter is oxidized by NAD→ NADH+ to form intermediates in the glycolytic pathways .

These intermediates are converted to ribose 5-phosphates in the presence of transketolase and transaldolase enzymes.And they are finally converted to pyruvate in the glycolytic pathway with the production of 2ATPs per molecule of glucose.

Basically the phosphate pathway reaction is very slow due to enzyme catalysis.

3 0

3 years ago

For many purposes we can treat methane as an ideal gas at temperatures above its boiling point of . Suppose the temperature of a

Elza [17]

Answer:

The volume decreases 5.5%

Explanation:

First, the question is incomplete, you are not giving the values of the temperatures and the pressure. However, I managed to find one similar question, and the given data is the temperature is lowered from 21 °C to -8°C, and the pressure decreased by 5%. If your data is different, you should only replace your data in the procedure, and you'll get an accurate result.

Now, with this data, let's see what we can do.

If this is an ideal gas, the equation to use is:

PV = nRT

Now, we know that this gas is suffering a decrease in temperature and pressure, but the moles stay the same so:

n₁ = n₂ = n

The constant R, is the same for both conditions. The only thing that differs here is the volume, temperature, and pressure. Therefore:

P₁V₁ = nRT₁ -----> n = P₁V₁ / RT₁

Doing the same with the pressure and volume 2 we have:

n = P₂V₂ / RT₂

Equalling both expressions and solving for V₂:

P₁V₁ / RT₁ = P₂V₂ / RT₂

V₂ = P₁T₂V₁ / P₂T₁

Now, as we know that P2 is 5% decreased from P1, so P2 = 0.95P1:

V₂ = P₁T₂V₁ / 0.95P₁T₁

The values of temperature in K:

T1 = 21+273 = 294 K

T2 = -8 + 273 = 265 K

Finally, let's calculate the volume:

V₂ = 264*P₁*V₁ / 294*0.95*P₁ ----> P cancels out

V₂ = 264V₁ / 294*0.95

V₁ = 0.945V₂

With this, we can day that Volume 2 decreases.

Now the percentage change would be using the following expression:

%V = (V₁ - V₂ / V₁) * 100

Replacing the data we have:

%V = V1 - 0.945V₁ / V₁

%V = 0.055V₁ / V₁ * 100

%V = 5.5%

7 0

3 years ago

Do sound waves travel best in dense or less dense areas?

velikii [3]

Thus, sound will travel at a slower rate in the denser object. If sound waves of the same energy were passed through a block of wood and a block of steel, which is more dense than the wood, the molecules of the steel would vibrate at a slower rate. Thus, sound passes more quickly through the wood, which is less dense.

3 0

2 years ago

Read 2 more answers

Be sure to answer all parts.The equilibrium constant, Kc, for the formation of nitrosyl chloride from nitric oxide and chlorine,

djverab [1.8K]

<u>Answer:</u> The reaction proceeds in the forward direction

<u>Explanation:</u>

For the given chemical equation:

$2NO(g)+Cl_2(g)\rightleftharpoons 2NOCl(g)$

Relation of $K_p\text{ with }K_c$ is given by the formula:

$K_p=K_c(RT)^{\Delta n_g}$

where,

$K_p$ = equilibrium constant in terms of partial pressure = ?

$K_c$ = equilibrium constant in terms of concentration = $6.5\times 10^4$

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature = $35^oC=[35+273]K=308K$

$\Delta n_g$ = change in number of moles of gas particles = $n_{products}-n_{reactants}=2-3=-1$

Putting values in above equation, we get:

$K_p=6.5\times 10^4\times (0.0821\times 500)^{-1}\\\\K_p=1583.43$

$K_p$ is the constant of a certain reaction at equilibrium while $Q_p$ is the quotient of activities of products and reactants at any stage other than equilibrium of a reaction.