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2 years ago

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If an individual is eating a diet in which 25% of their ATP production is a result of utilizing carbohydrates, 50% of their ATP

production is a result of metabolizing fats, 25% of their ATP production is a result of metabolizing proteins, their cardiac output is 4.7 L/min, and their arterial-venous oxygen difference is 6.2 mL/dL, how much carbon dioxide are they producing

1 answer:

NISA [10]2 years ago

5 0

Answer:

Oxidative phosphorylation proceeds with the formation of energy laden molecules i.e; carbondioxide and water.

Therefore, Total CO₂ production is directly related to VCO₂ = R x VO₂

where, R is the respiratory quotient varing among 0.7 to 1.0 according to the energy intake (ATP) ie 0.25 of the total diet consumed .

VO₂ is, as mentioned above arterial venous oxygen difference = 6.2ml/dl

therefore, VCO₂ = 0.25 x 6.2

= 1.55 ml/dl

ie; VO₂ : VCO₂ = 6.2 : 1.55.

Explanation:

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If a gas is cooled from 315K to 231K and the volume is kept constant, what would the final pressure be if the original pressure

goblinko [34]

Answer:

455.4 mmHg

Explanation:

To solve this problem we can use <em>Gay Lussac's Law</em>, which describe the temperature and pressure of a gas, at constant volume and composition:

P₁T₂ = P₂T₁

In this case:

P₁ = 621 mmHg

P₂ = ?

T₁ = 315 K

T₂ = 231 K

We <u>input the data</u>:

621 mmHg * 231 K = P₂ * 315 K

And <u>solve for P₂</u>:

P₂ = 455.4 mmHg

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2 years ago

If you had 100 ml of juice, how many milliliter would de fruit juice?

Julli [10]

Since the measurement is not changing, the answer is 100 mL. Hope this helps.

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3 years ago

What is the concentration of silver ions where silver iodide, Agl, is in a solution of hydroiodic

jonny [76]

Answer:

$[Ag^+]=2.82x10^{-4}M$

Explanation:

Hello there!

In this case, for the ionization of silver iodide we have:

$AgI(s)\rightleftharpoons Ag^+(aq)+I^-(aq)\\\\Ksp=[Ag^+][I^-]$

Now, since we have the effect of iodide ions from the HI, it is possible to compute that concentration as that of the hydrogen ions equals that of the iodide ones:

$[I^-]=[H^+]=10^{-3.55}=2.82x10^{-4}M$

Now, we can set up the equilibrium expression as shown below:

$Ksp=8.51x10^{-17}=(x)(x+2.82x10^{-4})$

Thus, by solving for x which stands for the concentration of both silver and iodide ions at equilibrium, we have:

$x=[Ag^+]=2.82x10^{-4}M$

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3 years ago

What has to happen to join two monosaccharide units into a disaccharide?

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A condensation reaction forming a glycosidic bond. so in other words a monosaccharide joining together to form a disaccharide.

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3 years ago

Consider the following reaction: A(g)⇌2B(g). Find the equilibrium partial pressures of A and B for each of the following differe

Illusion [34]

Answer:

a. Kp=1.4

$P_{A}=0.2215 atm$

$P_{B}=0.556 atm$

b.Kp=2.0 * 10^-4

$P_{A}=0.495atm$

$P_{B}=0.00995 atm$

c.Kp=2.0 * 10^5

$P_{A}=5*10^{-6}atm$

$P_{B}=0.9999 atm$

Explanation:

For the reaction

A(g)⇌2B(g)

Kp is defined as:

$Kp=\frac{(P_{B})^{2}}{P_{A}}$

The conditions in the system are:

A B

initial 0 1 atm

equilibrium x 1atm-2x

At the beginning, we don’t have any A in the system, so B starts to react to produce A until the system reaches the equilibrium producing x amount of A. From the stoichiometric relationship in the reaction we get that to produce x amount of A we need to 2x amount of B so in the equilibrium we will have 1 atm – 2x of B, as it is showed in the table.

Replacing these values in the expression for Kp we get:

$Kp=\frac{(1-2x)^{2}}{x}$

Working with this equation:

$x*Kp=(1-2x)^{2} - -> x*Kp=4x^{2}-4x+1- - >4x^{2}-(4+Kp)*x+1=0$

This last expression is quadratic expression with a=4, b=-(4+Kp) and c=1

The general expression to solve these kinds of equations is:

$x=\frac{-b(+-)*\sqrt{(b^{2}-4ac)}}{2a}$ (equation 1)

We just take the positive values from the solution since negative partial pressures don´t make physical sense.

Kp = 1.4

$x_{1}=\frac{(1.4+4)+\sqrt{(-(1.4+4)^{2}-4*4*1)}}{2*4}=1.128$

$x_{1}=\frac{(1.4+4)-\sqrt{(-(1.4+4)^{2}-4*4*1)}}{2*4}=0.2215$

With x1 we get a partial pressure of:

$P_{A}=1.128 atm$

$P_{B}=1-2*1.128 = -1.256 atm$

Since negative partial pressure don´t make physical sense x1 is not the solution for the system.

With x2 we get:

$P_{A}=0.2215 atm$

$P_{B}=1-2*0.2215 = 0.556 atm$

These partial pressures make sense so x2 is the solution for the equation.

We follow the same analysis for the other values of Kp.

Kp=2*10^-4

X1=0.505

X2=0.495

With x1

$P_{A}=0.505atm$

$P_{B}=1-2*0.505 = -0.01005 atm$

Not sense.

With x2

$P_{A}=0.495atm$

$P_{B}=1-2*0.495 = 0.00995 atm$

X2 is the solution for this equation.

Kp=2*10^5

X1=50001

$X2=5*10^{-6}$

With x1

$P_{A}=50001atm$

$P_{B}=1-2*50001=-100001atm$

Not sense.

With x2

$P_{A}= 5*10^{-6}atm$

$P_{B}=1-2*5*10^{-6}= 0.9999 atm$

X2 is the solution for this equation.

8 0

3 years ago