Pyruvate is the end product of glycolysis. Its further metabolism depends on the organism and on the presence or absence of oxygen. Draw the structure of the product from each reaction as it would exist at pH 7. Include the appropriate hydrogen atoms. Reaction A: aerobic conditions in humans or yeast
Answers
The given question is incomplete. The image present in the question for Reaction A is attached below along with the answer.
Explanation:
Pyruvate molecule reacts with Coenzyme A in the presence of oxygen and it results in the formation of acetyl Coenzyme A and carbon dioxide.
The enzyme pyruvae dehydrogenase helps in catalyzing this reaction. As in this biochemical reaction [tex]NAD^{+}[/tex] gets converted into NADH.
This reaction is shown in the image attached below.
Related Questions
what is the balanced equation for calcium sulfate?
Answers
Answer:
CaSO4
Explanation:
Calcium sulfate (or calcium sulphate) is the inorganic compound with the formula CaSO4 and related hydrates.
Consider the reaction in a commercial heat pack: 4 Fe (s) + 3 O2(g) ® 2 Fe2O3 (s) DH = -1652 kJ a) How much heat is released when 1.00 g iron is reacted with excess O2? b) What mass of O2 must react with iron in order to generate 2150 kJ of heat?
Answers
Answer:
a) -7.395kJ of energy are released.
b) 125g of O₂ must react.
Explanation:
Based on the reaction:
4 Fe (s) + 3 O₂(g) → 2 Fe₂O₃ (s) ΔH = -1652 kJ
4 moles of iron with an excess of oxygen release -1652kJ of energy
a) The heat released is:
1.00g Fe (molar mass: 55.845g/mol)
1.00g × (1mol / 55.845g) = 0.0179 moles de Fe.
As 4 moles release -1652kJ, 0.0179 moles release:
0.0179 mol Fe × (-1652kJ / 4mol Fe) = -7.395kJ of energy are released.
b) As 3 moles of oxygen produce -1652kJ, 2150kJ are released when react:
2150kJ × (3 mol O₂ / 1652kJ) = 3.9 moles of O₂
As molar mass of O₂ is 32g/mol, mass of 3.9 moles of O₂ is:
3.9 mol O₂ × (32g / mol) = 125g of O₂ must react.
Which diagram represents this molecule?
Answers
Answer:
C
Explanation:
The molecule has 8 carbon atoms joined by 7 C-C bonds.
The first two diagrams show 6 carbon atoms, not 8.
The last two diagrams show line segments representing C-C bonds. Only choice C shows 7 such segments.
The appropriate choice is C.
Answer:
C.
Explanation:
Consider this reaction:
2Cl2O5 —> 2Cl2 + 5O2
At a certain temperature it obeys this rate law.
rate = (2.7.M^-1•s^-1) [Cl2O5]^2
Suppose a vessel contains Cl2O5 at a concentration of 0.600M. calculate how long it takes for the concentration of Cl2O5 to decrease by 94%. you may assume no other reaction is important. round your answer to two digits
Answers
Answer:
[tex]t=9.7s[/tex]
Explanation:
Hello,
In this case, we have a second order kinetics given the second power of the concentration of chlorine (V) oxide in the rate expression, thus, the integrated equation for the concentration decay is:
[tex]\frac{1}{[Cl_2O_5]}=kt+\frac{1}{[Cl_2O_5]_0}[/tex]
Thus, the final concentration for a 94% decrease is:
[tex][Cl_2O_5]=0.600M-0.600M*0.94=0.036M[/tex]
Therefore, we compute the time for such decrease:
[tex]kt=\frac{1}{[Cl_2O_5]}-\frac{1}{[Cl_2O_5]_0}=\frac{1}{0.036M}-\frac{1}{0.60M} =26.1M^{-1}[/tex]
[tex]t=\frac{26.1M^{-1}}{k}= \frac{26.1M^{-1}}{2.7M^{-1}*s^{-1}}\\\\t=9.7s[/tex]
Regards.
The height of the sun in the sky at noontime is called
Answers
Answer:
It is called 'solar noon'
Explanation:
It's when the sun is at its highest
And also the moment the sun crosses the meridian.
A 25.0 mLsample of an acetic acid solution is titrated with a 0.175 M NaOH solution. The equivalence point is reached when 37.5 mL of the base is added. The concentration of acetic acid in the sample was:_______,
A) 0.263
B) 0.365
C) 0.175
D) 1.83×10−4
E) 0.119
Answers
Answer:
0.263M of CH₃COOH is the concentration of the solution.
Explanation:
The reaction of acetic acid (CH₃COOH) with NaOH is:
CH₃COOH + NaOH → CH₃COO⁻Na⁺ + H₂O
1 mole of acetic acid reacts per mole of NaOH to produce sodium acetate and water.
In the equivalence point, moles of acetic acid are equal to moles of NaOH and moles of NaOH are:
0.0375L × (0.175 moles / L) = 6.56x10⁻³ moles of NaOH = moles of CH₃COOH.
As the sample of acetic acid had a volume of 25.0mL = 0.025L:
6.56x10⁻³ moles of CH₃COOH / 0.0250L =
0.263M of CH₃COOH is the concentration of the solution
how many moles of helium gas occupy 22.4 L at 0 degreeC at 1 atm pressure
Answers
Answer:
1 mole of the gas occupies 22. 4L at O0C at 1atm pressure. Hence, the correct option is C.
Explanation:
What is the Arrhenius definition of an acid? A substance that increases H3O+ concentration when it is dissolved in water. A substance that increases OH– concentration when it is dissolved in water. A compound that donates protons. A compound that accepts protons.
Answers
Answer:
A substance that increases H3O+ concentration when it is dissolved in water.
Explanation:
Note that H3O+ and H+ are used quite interchangeably in chemistry.
An acid makes the H+ content higher, thereby decreasing the pH.
Answer:
a
A substance that increases H3O+ concentration when it is dissolved in water.
Explanation:
What is the limiting reactant if 8 g of Ba reacts with 2.8 g of Al2(SO4)3?
Answers
Answer:
Al2(SO4)3 is the limiting reactant
Explanation:
We'll begin by writing the balanced equation for the reaction. This is illustrated below:
3Ba + Al2(SO4)3 → 2Al + 3BaSO4
Next, we shall determine the mass of Ba and the mass of Al2(SO4)3 that reacted from the balanced equation. This is illustrated below:
Molar mass of Ba = 137g/mol
Mass of Ba from the balanced equation = 3 x 137 = 411g
Molar mass of Al2(SO4)3 = 2x27 + 3[32 + (16x4)]
= 54 + 3[32 + 64]
= 54 + 3[96]
= 54 + 288 = 342g/mol
Mass of Al2(SO4)3 from the balanced equation = 1 x 342 = 342g
Summary:
From the balanced equation above,
411g of Ba reacted with 342g of Al2(SO4)3.
Finally, we shall determine the limiting reactant as follow:
From the balanced equation above,
411g of Ba reacted with 342g of Al2(SO4)3.
Therefore, 8g of Ba will react with
= (8 x 342/411 = 6.66g of Al2(SO4)3.
From the calculations made above, we can see that it will take a higher mass of Al2(SO4)3 i.e 6.66g than what was given i.e 2.8g to react completely with 8g of Ba.
Therefore, Al2(SO4)3 is the limiting reactant and Ba is the excess reactant.
A student sets up the following equation to convert a measurement. (The ? stands for a number the student is going to calculate.) Fill in the missing part of this equation. Note: your answer should be in the form of one or more fractions multiplied together. (23. Pa cm^3)____?kPa . m^3
Answers
Answer:
The correct answer will be "-6.7 × 10¹⁰ kg.m/s".
Explanation:
The required conversions are:
⇒ [tex]1 \ kg=1000 \ g[/tex]
⇒ [tex]1 \ m=100 \ cm[/tex]
Now,
The complete conversion will be:
= [tex][-6.7\times 10^5 \ \frac{kg \ m}{s} ]\times [\frac{10^3 \ g}{kg}\times \frac{10^2 \ cm}{1 \ m}][/tex]
On cancelling the terms, we get
= [tex]-6.7\times 10^{10} \frac{kg \ m}{s}[/tex]
So that the missing terms will be [tex][\frac{10^3 \ g}{kg}\times \frac{10^2 \ cm}{1 \ m}][/tex] and [tex][-6.7\times 10^{10}\frac{kg \ m}{s}][/tex]
Which compound has the lowest melting point? KCl CaCl2 Na2O C6H12O6
Answers
Alkyl derivatives of mercury are highly toxic and can cause mercury poisoning in humans. Dimethylmercury is one of the strongest known neurotoxins. Although it is said to have a slightly sweet smell, inhaling enough to discern this would be hazardous.
Give the empirical formula of dimethylmercury.
Answers
Answer:
The empirical formula of dimethylmercury is C2H6Hg
Explanation:
Dimethylmercury, as it says in the name, presents not only the mercury metal in its structure (Hg) but also two radical groups called methyl, which is why its name begins with the prefix DI, referring to the fact that there are two methyl.
A compound has an empirical formula of CHN. What is the molecular formula, if it’s molar mass is 135.13 g/mol? (C=12.01 amu, H=1.008 amu, N= 14.01)
Answers
Answer:
well the MF is 224.78 g/mol
Explanation:
just times them all by the molor mass and divide it by 3
If a pork roast must absorb 1500 kJkJ to fully cook, and if only 14% of the heat produced by the barbeque is actually absorbed by the roast, what mass of CO2CO2 is emitted into the atmosphere during the grilling of the pork roast?
Answers
You need to know the amount of heat generated by the combustion reaction.
Assuming propane as fuel, you can use thiis data:
C3H8(g)+5O2(g)---3CO2(g)+4H2O(g) ΔH= -2217 KJ
So when 3 moles of CO2 is emmitted 2217 kJ of heat is produced.
The molar wegiht of CO2 is 12 g/mol + 2 * 16 g/mol = 44 g/mol.
Then 3 mol * 44 g / mol = 132 g of CO2 are produced with 2217 kJ of heat.
Now you have to calculate how much energy you need to produce if only 12% is abosrbed by the pork
Energy absorbed by the pork = 12% * total energy =>
total energy = energy absorbed by the pork / 0.12 = 1700 kJ / 0.12 = 14,166.67 kJ.
Now, state the proportion:
132 g CO2 / 2217 kJ = x / 14,166.7 kJ =>
x = 14,166.67 * 132 / 2217 = 843.48 g CO2.
Answer: 843 g of CO2
When 106 g of water at a temperature of 21.4 °C is mixed with 64.3 g of water at an unknown temperature, the final temperature of the resulting mixture is 46.8 °C. What was the initial temperature of the second sample of water? (The specific heat capacity of liquid water is 4.184 J/g ⋅ K.)
Answers
Answer:
THE INITIAL TEMPERATURE OF THE SECOND SAMPLE IS 4.93 C OR 277.93 K
Explanation:
Mass of first sample of water = 106 g
Initial temp of first sample = 21.4 °C = 21.4 + 273 K = 294.4 K
Mass of second sample = 64.3 g
Final temp of theresulting mixture = 46.8 °C = 46.8 + 273 K = 319.8 K
Specific heat capacity of water = 4.184 J/g K
It is worthy to note that;
Heat gained by the first sample = Heat lost by the second sample
Since heat = mass * specific heat capacity * change in temperature, we have
Mass * specific heat * change in temp of the first sample = Mass * specific heat * change in temp. of the second sample
MC (T2 - T1) = MC (T2-T1)
106 * 4.184 * ( 319.8 - 294.4) = 64.3 * 4.184 * ( 319.8 - T1)
106 * 4.184 * 25.4 = 269.0312 ( 319.8 - T1)
11 265.0016 = 269.0312 (319.8 - T1)
Since the change in temperature = 319.8 -T1
Change in temperature =11265.0016 / 269.0312
Change in temperature = 41.87
Change in temperature = 319.8 -T1
41.87 = 319.8 - T1
T1 = 319.8 - 41.87
T1 = 277.93 K
T1 = 4.93 °C
So therefore, the initial temperature of the sacond sample is 4.73 °C or 277.93 K
What will be the final temperature when a 25.0 g block of aluminum (initially at 25 °C) absorbs 10.0 kJ of heat? The specific heat of Al is 0.900 J/g·C.
Answers
Q
=
725 J
m
=
55.0 g
c
=
0.900 J/(°C⋅g)
Δ
T
=
final temperature - initial temperature
Δ
T
=
(
x
−
27.5
)
°C
We solve for
Δ
T
.
725 J
=
55.0 g
⋅
0.900 J/(°C⋅g)
(
x
−
27.5
)
°C
NOW IT'S JUST BASIC ALGREBRA
725
=
49.5
x
−
1361
2086
=
49.5
x
42.1
=
x
The final temperature is 42.1 °C.
42.1°C is the final temperature when a 25.0 g block of aluminum (initially at 25 °C) absorbs 10.0 kJ of heat.
What is temperature?The physical concept of temperature indicates in numerical form how hot or cold something is. A thermometer is used to determine temperature. Thermometers are calibrated using a variety of temperature scales, which historically defined distinct reference points or thermometric substances.
The most popular scales include the Celsius scale, sometimes known as centigrade, with the unit symbol °C, the scale of degrees Fahrenheit (°F), or the Kelvin scale (K), with the latter being mostly used for scientific purposes. One of the United Nations System of Units' (SI) seven base units is the kelvin.
Q=725 J
m=55.0 g
c=0.900 J/(°C⋅g)
ΔT=final temperature - initial temperature
ΔT=(x−27.5)°C
725 J=55.0 g⋅0.900 J/(°C⋅g)(x−27.5)
X=42.1°C
Therefore, 42.1°C is the final temperature when a 25.0 g block of aluminum (initially at 25 °C) absorbs 10.0 kJ of heat.
To know more about temperature, here:
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At 25 °C, what is the hydroxide ion concentration, [OH−] , in an aqueous solution with a hydrogen ion concentration of [H+]=1.9×10−9 M?
Answers
Answer:
The correct answer will be "5.26 × 10⁻⁶".
Explanation:
The given values is:
[tex][H^{+}]=1.9\times 10^{-9} M[/tex]
As we know,
⇒ [tex]pH+pOH=14[/tex]
On taking log, we get
⇒ [tex]-log[H^{+}] + -log[OH^{-}] = 14[/tex]
Now,
Taking "log" as common, we get
⇒ [tex]log[H^{+}][OH^{-}]= -14[/tex]
⇒ [tex][H^{+}][OH^{-}]= 10^{-14}[/tex]
⇒ [tex][OH^{-}]=\frac{10^{-14}}{[H^{+}]}[/tex]
On putting the estimated value of "[tex][H^{+}][/tex]", we get
⇒ [tex]=\frac{10^{-14}}{1.9\times 10^{-9}}[/tex]
⇒ [tex]=5.26\times 10^{-6}[/tex]
AHP for the formation of rust (Fe2O3) is -826 kJ/mol. How much energy is
involved in the formation of 5.00 grams of rust?
A 25.9 kJ
B 25.9 J
C 66.0 kJ
D 66.0)
Answers
Answer:
A- 25.9 kJ
Explanation:
ΔH of formation is defined as the amount of energy that is involved in the formation of 1 mole of substance.
ΔH of rust is -826kJ/mol, that means when 1 mole of rust is formed, there are released -826kJ.
Moles of 5.00g of Fe₂O₃ (Molar mass: 159.69g/mol) are:
5.00g ₓ (1 mole / 159.69g) = 0.0313 moles of Fe₂O₃.
If 1 mole release -826kJ, 0.0313 moles release:
0.0313 moles ₓ (-826kJ / 1 mole) = -25.9kJ
Thus, heat involved is:
A- 25.9 kJHow many moles of CO2 are produced when 84 0 mol O2 completely react?
Answers
Answer:
Explanation:boom
Which process is a physical change
Answers
Answer:
a physical change is something that has not been modified chemically and can possibly be changed back to the state it was once before. A physical change keeps all the same atoms and none of them is modified.
Example:
When a block of clay is morphed into a giraffe statue, it can be morphed back to its original state. If someone burnt the block of clay, the atoms would be modified and it would be unable to go back to its previous state.
-----------------------------------------------------------------------------------------------------------------
(If you're referring to a question with these answers)
A. iron rusting
B. milk turning to curd
C. water boiling
D. paper burning
E. hard water staining pipes
-----------------------------------------------------------------------------------------------------------------
Answer:
C. Water Boiling
(If you are referring to a question with these answers I think this is the correct answer if not I do apologize)
-----------------------------------------------------------------------------------------------------------------
When you look at an ant up close, using a convex lens, what do you see?
Answers
Answer:
You would be able to see the ants clearly with the unique body parts.
Explanation:
Convex lens is also known as Converging lens. It helps in converging rays of light to become a principal focus which is usually very clear and visible to the eyes. The converging lens when used to view the ant makes the ant appear very visible and the individual is able to see all the unique body parts of the ant. This type of lens is used in individuals who are longsighted.
what is the color of benzene and bromine
Answers
Benzene is clear or colorless
Explanation:
Benzene is colorless, with a sweet odour.
Color of Bromine is reddish brown .
Hope this helps.
A gas has a volume of 6.6 L at a temperature of 40 C. What is the volume of
the gas if the temperature changes to 15 C?
Answers
Answer:
6.07 L
Explanation:
It appears that the reading has been made at constant pressure .
At constant pressure , the gas law formula is
V/T = constant V is volume and T is temperature of the gas.
V₁ / T₁ = V₂ / T₂
V₁ = 6.6 L ,
T₁ = 40°C
= 273 + 40
= 313 K
T₂ = 15+ 273
= 288K
V₂ = ?
Putting the values in the formula above
6.6 / 313 = V₂ / 288
V₂ = 6.07 L.
how many grams of H2 are needed to produce 14.34 g of NH3?
please help, this hw is due in a few hours
Answers
Answer:
im pretty sure its 2.54g H2
Explanation:
14.34gNH3 / 17.03gNH3 <-- molar mass
.842g x 3 mol <-- mols of H2
2.52 / 2 mol <-- mols of NH3
1.26 x 2.016gH2= 2.54gH2
Interpret the following equation for a chemical reaction using the coefficients given: CO(g) Cl2(g) COCl2(g) On the particulate level: _________ of CO(g) reacts with _________ of Cl2(g) to form _________ of COCl2(g). On the molar level: _________ of CO(g) reacts with _________ of Cl2(g) to form _________ of COCl2(g).
Answers
Answer:
On the particulate level: 6.02 * 10²³ particles of CO(g) reacts with 6.02 * 10²³ particles of Cl₂(g) to form 6.02 * 10²³ particles of COCl2(g).
On the molar level: 1 mole of CO(g) reacts with 1 mole of Cl2(g) to form 1 mole of COCl₂(g).
Explanation:
The particulate level refers to the microscopic or atomic level of substances. It also involves the ions, protons, neutrons and molecules present in substances.
The molar level refers to the quantitative measure of substances in terms of the mole, where a mole represents the amount of substances containing the Avogadro number of particles which is equal to 6.02 * 10³ particles.
Equation of the reaction: CO(g) + Cl₂(g) ----> COCl₂(g)
From the equation above, I mole of CO gas reacts with 1 mole of Cl₂ gas to produce 1 mole of COCl₂ gas.
Since 1 mole of a substance contains 6.02 * 10²³ particles, on a particulate level, 6.02 * 10²³ particles of CO gas reacts with 6.02 * 10²³ particles of Cl₂ gas to produce 6.02 * 10²³ particles of COCl₂ gas.
Now construct a different electrochemical cell. You put a zinc metal anode in contact with a 0.052 M solution of zinc nitrate and a silver cathode in contact with a 0.0042 M solution of silver(I) nitrate. What is the value of the electric potential at the moment the reaction begins
Answers
Answer:
[tex]1.66~V[/tex]
Explanation:
We have to start with the half-reactions for both ions:
[tex]Zn^+^2~+2e^-~->Zn[/tex] V= -0.76
[tex]Ag^+~e^-~->~Ag[/tex] V= +0.80
If we want a spontaneous reaction (galvanic cell) we have to flip the first reaction, so:
[tex]Zn~->~Zn^+^2~+2e^-~[/tex] V= +0.76
[tex]Ag^+~+~e^-~->~Ag[/tex] V= +0.80
If we want to calculate ºE we have to add the two values, so:
ºE=0.76+0.80 = 1.56 V
Now, we have different concentrations. So, if we want to calculate E we have to use the nerts equation:
[tex]E=ºE~+~\frac{0.059}{n}LogQ[/tex]
On this case, Q is equal to:
[tex]Q=\frac{[Zn^+^2]}{[Ag^+]^2}[/tex]
Because the total reaction is:
[tex]Zn~+~2Ag^+~->~Zn^+^2~+~2Ag[/tex]
So, the value of "Q" is:
[tex]Q=\frac{[0.052 M]}{[0.0042]^2}=2947.84[/tex]
Now, we can plug all the values in the equation (n=2, because the amount of electrons transferred is 2). So:
[tex]E=1.56~V~+~\frac{0.059}{2}Log(2947.84)=1.66~V[/tex]
I hope it helps!
If a jet’s cruising altitude is 32,200ft(to three significant figures),the distance in km is :(1 mile=1.61km;1 mile=5280 ft)
Answers
Answer:
9.82 km.
Explanation:
Hello,
In this case, given the conversion factors from miles to kilometres and from miles to feet, we can directly compute the jet’s cruising altitude in kilometres as shown below:
[tex]32,200ft\times \frac{1mile}{5280ft}\times \frac{1.61km}{1mile} \\\\=9.82km[/tex]
Best regards.
WILL GIVE BRAINLIEST!!!!! will give brainliest!!!! ******According to the reaction in question I above, how many grams of solid copper will theoretically be
produced when 14.4 g of aluminum are reacted with 14.4 g of copper (II) sulfate? Which reactant is the
limiting reactant? Show your work. Be sure to include units!
Answers
Answer:
5.73 g Cu
Explanation:
M(CuSO4) = 159.6 g/mol
M(Al) = 27.0 g/mol
M(Cu) = 63.5 g/mol
14.4 g Al * 1 mol/27.0 g = 0.5333 mol Al
14.4 g CuSO4 * 1 mol/159.6 g = 0.0902 mol CuSO4
2 Al + 3CuSO4 -------> 3Cu + Al2(SO4)3
from reaction 2 mol 3 mol
given (0.5333 mol )x 0.0902 mol
needed 0.0601 mol
x= 2*0.0902/3 = 0.0601 mol Al
Al is excess, CuSO4 is a limiting reactant.
2 Al + 3CuSO4 -------> 3Cu + Al2(SO4)3
from reaction 3 mol 3 mol
given 0.0902 mol x mol
x = 0.0902 mol Cu
0.0902 mol Cu * 1 mol/63.5 g Cu = 5.73 g Cu
Identify the Lewis acids and Lewis bases in the following reactions:
1. H+ + OH- <-> H2O Lewis acid: Lewis base:
2. Cl- + BCl3 <-> BCl4- Lewis acid: Lewis base:
3. K+ + 6H2O <-> K(H2O)6+ Lewis acid: Lewis base:
Answers
Answer: 1. [tex]H^++OH^-\rightarrow H_2O[/tex] Lewis acid : [tex]H^+[/tex], Lewis base : [tex]OH^-[/tex]
2. [tex]Cl^-+BCl_3\rightarrow BCl_4^-[/tex] Lewis acid : [tex]BCl_3[/tex], Lewis base : [tex]Cl^-[/tex]
3. [tex]K^++6H_2O\rightarrow K(H_2O)_6[/tex] Lewis acid : [tex]K^+[/tex], Lewis base : [tex]H_2O[/tex]
Explanation:
According to the Lewis concept, an acid is defined as a substance that accepts electron pairs and base is defined as a substance which donates electron pairs.
1. [tex]H^++OH^-\rightarrow H_2O[/tex]
As [tex]H^+[/tex] gained electrons to complete its octet. Thus it acts as lewis acid.[tex]OH^-[/tex] acts as lewis base as it donates lone pair of electrons to electron deficient specie [tex]H^+[/tex].
2. [tex]Cl^-+BCl_3\rightarrow BCl_4^-[/tex]
As [tex]BCl_3[/tex] is short of two electrons to complete its octet. Thus it acts as lewis acid. [tex]Cl^-[/tex] acts as lewis base as it donates lone pair of electrons to electron deficient specie [tex]BCl_3[/tex].
3. [tex]K^++6H_2O\rightarrow K(H_2O)_6[/tex]
As [tex]K^+[/tex] is short of electrons to complete its octet. Thus it acts as lewis acid. [tex]H_2O[/tex] acts as lewis base as it donates lone pair of electrons to electron deficient specie [tex]K^+[/tex].
A semipermeable sac containing 4% NaCl, 9% glucose, and 10% albumin is suspended in a solution with the following composition: 10% NaCl, 10% glucose, and 40% albumin. Assume that the sac is permeable to all substances except albumin. State whether each of the following will (a) move into the sac, (b) move out of the sac, or (c) not move.
glucose: a. moves into sac
water: b. moves out of sac
albumin: c; does not move
NaCl: a; moves into sac
Answers
Answer:
Glucose: (a) moves into sac
Water: (b) moves out of sac
Albumin: (c) does not move
NaCl: (a) moves into sac
Explanation:
A semi-permeable membrane is a membrane that allow certain molecules or ions to pass through by diffusion.
Diffusion is the movement of molecules from a region of higher concentration to a region of lower concentration until equilibrium is attained. A special form of diffusion known as osmosis transports water molecules across a semi-permeable membrane.
Osmosis is the movement of water molecules from a region lower solute concentration (high water molecules concentration) to a region of higher solute concentration (low water molecules concentration) until equilibrium is attained.
From the above definitions and the given data;
Glucose concentration is higher in solution outside the sac, thus, glucose molecules will move into the sac.
Water molecules are higher in the sac as a result of the lower concentration of solutes, therefore, water molecules will move out of the sac into the solution outside.
Since the sac is impermeable to albumin, it does not move.
NaCl concentration is lower in the sac, therefore, it will move ro the solution outside into the sac.
Glucose will move into the sac, water will move out of the sac, albumin will neither move in nor out, and NaCl will move into the sac.
The molecules of each of the substances will move by diffusion from the region of higher concentration to the region of lower concentration of each substance as long as there is a permeable membrane. Water, on the other hand, will move by osmosis from the region of high to low water potential through a permeable membrane. Regions with higher concentrations of substances usually have low water potentials and vice versa.Thus, both glucose and NaCl molecules will diffuse from the solution into the sac, and water molecules will move from the sac into the surrounding solution. Since the sac is not permeable to albumin, then the movement in or out is inhibited.
More on osmosis and diffusion can be found here: https://brainly.com/question/19867503?referrer=searchResults