When a glucose molecule loses a hydrogen atom as the result of an oxidation-reduction reaction in glycolysis, the glucose is?

How well do you think your system will digest food?

yaroslaw [1]

While it depends on what you eat so say you ate sushi or duck or what ever and you have never ate it before then your body will take a while to digest what its taking in!

7 0

4 years ago

. Using the pH scale below what would be the pH value of an Acid? *

earnstyle [38]

Answer : The pH value of an acid is below 7.

Explanation :

pH : It is defined as the negative logarithm of hydrogen ion or hydronium ion concentration.

Mathematically,

$pH=-\log [H^+]$

When the pH less than 7 then the solution is acidic and the concentration of hydrogen ion is greater than hydroxide ion.

When the pH more than 7 then the solution is basic and the concentration of hydrogen ion is less than hydroxide ion.

When the pH is equal to 7 then the solution is neutral and the concentration of hydrogen ion is equal to the hydroxide ion.

Hence, the pH value of an acid is below 7.

6 0

4 years ago

The equilibrium constant for the reaction

Hitman42 [59]

The question is incomplete, here is the complete question:

The equilibrium constant for the reaction

N₂O₄(g)⇌2NO₂ at 2°C is Kc = 2.0

If each yellow sphere represents 1 mol of N₂O₄(g) and each gray sphere 1 mol of NO₂ which of the following 1.0 L containers represents the equilibrium mixture at 2°C?

The image is attached below.

Answer: The system which represents the equilibrium having value of $K_c=2.0$ is system (b)

Explanation:

Equilibrium constant in terms of concentration is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as $K_c$

For a general chemical reaction:

$aA+bB\rightarrow cC+dD$

The expression for $K_{c}$ is written as:

$K_{c}=\frac{[C]^c[D]^d}{[A]^a[B]^b}$

For the given chemical equation:

$N_2O_4(g)\rightleftharpoons 2NO_2$

The expression of $K_c$ for above equation follows:

$K_c=\frac{[NO_2]^2}{[N_2O_4]}$ .......(1)

We are given:

Volume of the container = 1.0 L

Value of $K_c$ = 2.0

Molarity of the substance is calculated by using the equation:

$\text{Molarity}=\frac{\text{Number of moles}}{\text{Volume}}$

For the given images:

For a:

Number of Gray spheres = 8 moles

Number of yellow spheres = 4 moles

Putting values in expression 1, we get:

$K_c=\frac{(8/1)^2}{(4/1)}\\\\K_c=16$

For b:

Number of Gray spheres = 4 moles

Number of yellow spheres = 8 moles

Putting values in expression 1, we get:

$K_c=\frac{(4/1)^2}{(8/1)}\\\\K_c=2$

For c:

Number of Gray spheres = 6 moles

Number of yellow spheres = 6 moles

Putting values in expression 1, we get:

$K_c=\frac{(6/1)^2}{(6/1)}\\\\K_c=6$

For d:

Number of Gray spheres = 2 moles

Number of yellow spheres = 8 moles

Putting values in expression 1, we get:

$K_c=\frac{(2/1)^2}{(8/1)}\\\\K_c=\frac{1}{2}$

Hence, the system which represents the equilibrium having value of $K_c=2.0$ is system (b)

3 0

4 years ago

According to the oxygen-hemoglobin dissociation curve, PO2 in the lungs of 100 mm Hg results in Hb being 98% saturated. At high

Ad libitum [116K]

Answer:

Hb would be 78.4% saturated.

Explanation:

This problem can be solved by using simple unitary method.

At 100 mm Hg pressure of oxygen, Hb is saturated by 98%

So, at 1 mm Hg pressure of oxygen, Hb is saturated by $\frac{98}{100}$ %

Hence, at 80 mm Hg pressure of oxygen, Hb is saturated by $\frac{98\times 80}{100}$ % or 78.4%

Therefore, at 80 mm Hg pressure of oxygen in the lungs, Hb would be 78.4% saturated.

4 0

3 years ago

16.25 g of water at 54 C relaeases 402.7 J. What will be its final temp?

leonid [27]

Data:
Q = 402.7 J → releases → Q = - 402.7 J
m = 16.25 g
T initial = 54 ºC
adopting: c = 4.184J/g/°C
ΔT (T final - T initial) = ?

Solving:

Q = m*c*ΔT
-402.7 = 16.25*4.184*ΔT
-402.7 = 67.99*ΔT
$\Delta\:T = \frac{-402.7}{67.99}$
$\boxed{\Delta\:T \approx -5.92\:^0C}$

If: ΔT (T final - T initial) = ?
$-5.92^0 = T_{final} - 54^0$
$T_{final} = 54^0 - 5.92^0$
$\boxed{\boxed{T_{final} = 48.08\:^0C}}\end{array}}\qquad\quad\checkmark$

8 0

4 years ago