A vegetable soup recipe requires one teaspoonful of salt. A chef accidentally puts in one tablespoonful. Now the soup is much too salty.
a) What can the chef do to reduce the salty taste of the soup?
b) What effects would your suggestion in a) have on the soup?
Answer:
a. Put a piece of fresh sliced yam with a bore into it into the soup.
Explanation:
b. Osmosis may occur
The chef can put a slice of yam in the soup with a hole in it as it will absorb excess of salt by process of diffusion.
What is diffusion?
Diffusion is defined as the process of movement of molecules which takes place under concentration gradient. It helps in movement of substances in and out from the cell.The molecules move from lower concentration region to a higher concentration region till the concentration becomes equal.
There are 2 main types of diffusion:
1) simple diffusion-process in which substances move through a semi-permeable membrane without the aid of transport proteins.
2) facilitated diffusion- It is a passive movement of molecules across cell membrane from higher concentration region to lower concentration.
There are 2 types of facilitated diffusion one is osmosis and dialysis.
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Methods in electrochemistry that can be used for the separation of proteins and enzymes?
Answer:
Some of the methods in Methods in electrochemistry that can be used for the separation of proteins and enzymes are as follows:
Redox transformations: In this method enzymes or proteins would be adsorbed on the electrode surface and facilitates direct electron transfer that causes denaturation and loss of their electrochemical activities and bioactivities. It is widely used in biosensors and biofuel cells.
Protein electrophoresis: In this process proteins are seperated by placing them in a gel matrix in the presence of an electrical field. In this method a negative charge is applied so that proteins move towards a positive charge.
Consider the following reaction where Kc = 1.29×10-2 at 600 K: COCl2 (g) CO (g) + Cl2 (g) A reaction mixture was found to contain 0.104 moles of COCl2 (g), 4.66×10-2 moles of CO (g), and 3.76×10-2 moles of Cl2 (g), in a 1.00 liter container. Indicate True (T) or False (F) for each of the following:
1. In order to reach equilibrium COCl2(g) must be consumed.
A. True B. False
2. In order to reach equilibrium Kc must increase.
A. True B. False
3. In order to reach equilibrium CO must be consumed.
A. True B. False
4. Qc is greater than Kc.
A. True B. False
5. The reaction is at equilibrium. No further reaction will occur.
A. True B. False
Answer:
1. In order to reach equilibrium COCl₂(g) must be consumed.
B. False
2. In order to reach equilibrium Kc must increase.
B. False .
3. In order to reach equilibrium CO must be consumed.
A. True.
4. Qc is greater than Kc.
A. True
5. The reaction is at equilibrium. No further reaction will occur.
B. False.
Explanation:
Based on the reaction:
COCl₂(g) → CO (g) + Cl₂(g)
And Kc is defined as:
Kc = 1.29x10⁻² = [CO] [Cl₂] / [COCl₂]
Molar concentrations of each species are:
[COCl₂] = 0.104 moles of COCl₂ / 1L = 0.104M
[CO] = 4.66×10⁻² moles of CO / 1L = 4.66×10⁻²M
[Cl₂] = 3.76×10⁻² moles of Cl₂ / 1L = 3.76×10⁻²M
Replacing in Kc formula:
4.66×10⁻²M × 3.76×10⁻²M / 0.104M = 1.68x10⁻²
As the concentrations are not in equilibrium, 1.68x10⁻² is defined as the reaction quotient, Qc.
As Qc > Kc, the reaction will shift to the left producing more COCl₂ and consuming CO and Cl₂. Thus
1. In order to reach equilibrium COCl₂(g) must be consumed.
B. False
2. In order to reach equilibrium Kc must increase.
B. False . Kc is a constant that never change.
3. In order to reach equilibrium CO must be consumed.
A. True.
4. Qc is greater than Kc.
A. True
5. The reaction is at equilibrium. No further reaction will occur.
B. False. The reaction is in equilibrium when Qc = Kc
A gaseous system undergoes a change in temperature and volume. What is the entropy change for a particle in this system if the final number of microstates is 0.842 times that of the initial number of microstates
Answer: -2.373 x 10^-24J/K(particles
Explanation: Entropy is defined as the degree of randomness of a system which is a function of the state of a system and depends on the number of the random microstates present.
The entropy change for a particle in a system depends on the initial and final states of a system and is given by Boltzmann equation as
S = k ln(W) .
where S =Entropy
K IS Boltzmann constant ==1.38 x 10 ^-23J/K
W is the number of microstates available to the system.
The change in entropy is given as
S2 -S1 = kln W2 - klnW1
dS = k ln (W2/W1)
where w1 and w2 are initial and final microstates
from the question, W2(final) = 0.842 x W1(initial), so:
= 1.38*10-23 ln (0.842)
=1.38*10-23 x -0.1719
= -2.373 x 10^-24J/K(particles)
What would cause a balloon to expand if taken to the top of a mountain?
O A. Increased molecular collision
O B. Increased amount of molecules
O C. Lowered temperature
D. Lowered pressure
Answer:
D. Lowered pressure
Explanation:
As you go to more altitude or height, the atmospheric pressure significantly lowers so the gas molecules are free to expand and take up as room as possible.
This is best explained by Boyle's law where pressure and volume are inversely related, where if one thing goes up another goes down. Here the pressure goes down, so volume increases and ballon expands.
In this lab, you are to carry out the formation of a Grignard reagent from 1-bromo-benzene andits subsequent reaction with solid carbon dioxide (CO2) followed by acidic workup (using HCl asthe acid). Write the overall reaction and product(s) that you expect.
Answer:
(A). C6H5Br + Mg(in ether) -----------> C6H5MgBr.
(B). C6H5MgBr + O = C = O -----------> C6H5-COO^- Mg^+ Br.
(C). C6H5-COO^- Mg^+ Br + HCl --------> C6H5-COOH + Mg^+Br(OH).
PRODUCTS=> C6H5-COOH and Mg^+Br(OH).
Explanation:
A Grignard reagent is a reagent that/which is an organometallic compound that is R -Mg- X. The R = alkyl, vinyl or allyl and the X = halogens.
It must be noted that an important reaction of Grignard reagent is its reaction with compounds containing the Carbonyl that is -CO functional group and this kind of Reaction is known as a Grignard Reaction.
So, in this question we are told that;
=> "1-bromo-benzene andits subsequent reaction with solid carbon dioxide (CO2) followed by acidic workup (using HCl asthe acid). "
Thus;
(A). C6H5Br + Mg(in ether) -----------> C6H5MgBr.
(B). C6H5MgBr + O = C = O -----------> C6H5-COO^- Mg^+ Br.
(C). C6H5-COO^- Mg^+ Br + HCl --------> C6H5-COOH + Mg^+Br(OH).
In Chapter 4, we will learn that single bonds experience free rotation at room temperature, while double bonds do not. Consider the two C-N bonds in the structure. One of these bonds exhibits free rotation, as expected for a single bond, but the other C-N bond exhibits restricted rotation. Identify the C-N bond with restricted rotation, and justify your answer by drawing resonance structures.
Answer:
Explanation:
The main objective here is to draw a diagram of an heterocyclic compound containing two C-N bonds in the structure. One with free rotation, as expected for a single bond, but the other C-N bond exhibits restricted rotation. After that ; we will identify the C-N bond with restricted rotation, and also justify our answer by drawing resonance structures.
So; the first image below shows the structure of the heterocyclic compound containing two C-N bonds in the structure with One with free rotation, as expected for a single bond, but the other C-N bond exhibits restricted rotation. From the first diagram. the squared area indicates the C-N bond that exhibits restricted rotation.
The amide bonds in the C-N bonds offers the resonance characteristics and thus exhibits restricted rotation. The resonance is shown in the second image below
Which statement describes a chemical property of an object? A:The object is white in color.B:The object has a powdery texture.C:The object’s density is 2.11 g/cm3.D:The object reacts with acid to form water.
Answer:
D
Explanation:
Color, texture, and density are all physical properties but reactivity is a chemical property so the answer is D.
Consider the following reaction where Kc = 1.80×10-2 at 698 K:
2HI(g) → H2(g) + I2(g)
A reaction mixture was found to contain 0.280 moles of HI (g), 2.09×10^-2 moles of H2 (g), and 4.14×10^-2 moles of I2 (g), in a 1.00 liter container.
Required:
a. Is the reaction at equilibrium?
b. What direction must it run in order to reach equilibrium?
c. The reaction
1. must run in the forward direction to reach equilibrium.
2. must run in the reverse direction to reach equilibrium.
3. is at equilibrium.
Answer:
The system is not in equilibrium and the reaction must run in the forward direction to reach equilibrium.
Explanation:
The reaction quotient Qc is a measure of the relative amount of products and reagents present in a reaction at any given time, which is calculated in a reaction that may not yet have reached equilibrium.
For the reversible reaction aA + bB⇔ cC + dD, where a, b, c and d are the stoichiometric coefficients of the balanced equation, Qc is calculated by:
[tex]Qc=\frac{[C]^{c}*[D]^{d} } {[A]^{a}*[B]^{b}}[/tex]
In this case:
[tex]Qc=\frac{[H_{2} ]*[I_{2} ] } {[HI]^{2}}[/tex]
Since molarity is the concentration of a solution expressed in the number of moles dissolved per liter of solution, you have:
[tex][H_{2} ]=\frac{2.09*10^{-2} moles}{1 Liter}[/tex]=2.09*10⁻² [tex]\frac{moles}{liter}[/tex][tex][I_{2} ]=\frac{4.14*10^{-2} moles}{1 Liter}[/tex]=4.14*10⁻² [tex]\frac{moles}{liter}[/tex][tex][I_{2} ]=\frac{0.280 moles}{1 Liter}[/tex]= 0.280 [tex]\frac{moles}{liter}[/tex]So,
[tex]Qc=\frac{2.09*10^{-2} *4.14*10^{-2} } {0.280^{2} }[/tex]
Qc= 0.011
Comparing Qc with Kc allows to find out the status and evolution of the system:
If the reaction quotient is equal to the equilibrium constant, Qc = Kc, the system has reached chemical equilibrium.
If the reaction quotient is greater than the equilibrium constant, Qc> Kc, the system is not in equilibrium. In this case the direct reaction predominates and there will be more product present than what is obtained at equilibrium. Therefore, this product is used to promote the reverse reaction and reach equilibrium. The system will then evolve to the left to increase the reagent concentration.
If the reaction quotient is less than the equilibrium constant, Qc <Kc, the system is not in equilibrium. The concentration of the reagents is higher than it would be at equilibrium, so the direct reaction predominates. Thus, the system will evolve to the right to increase the concentration of products.
Being Qc=0.011 and Kc=1.80⁻²=0.018, then Qc<Kc. The system is not in equilibrium and the reaction must run in the forward direction to reach equilibrium.
A student states that the graduated cylinder contains 150 mL of water his statement is
A. A prediction
B. An observation
C. A theory
D. A hypothesis
The correct answer is B. An observation
Explanation:
An observation is defined as a statement or conclusion you made after observing or measuring a phenomenon, this includes statements based on precise instruments. For example, if you conclude a plant grows 2 inches every month by measuring the plant during this time, this is classified as an observation. The conclusion of the student is also an observation because he concludes this after analyzing the volume of the water in the cylinder through the lines in the graduated cylinder, considering the water is just in the middle of 100 mL and 200 mL which indicates there are 150 mL of water.
Answer:
B. An observation
Explanation:
Hello,
Given the illustration, such statements is considered as an observation, since it came up from something the student realized with his/her own eyes, as in the volumetric cylinder the level of the liquid reached 150 mL of water. Predictions are not observed but assumed, theories are stated when experimentation is already deeply studied and hypothesis are assumptions before experimenting.
Regards.
Calculate the mass of CaCl2•2H2O required to make 100.0 mL of a 0.100 M solution. Each of the calculations below will take you through the necessary steps. You will be asked to show your answer and calculations for each. Calculate the moles of CaCl2•2H2O in 100.0 mL of a 0.100 M solution Enter your answer:
Answer:
The mass is 1.4701 grams and the moles is 0.01.
Explanation:
Based on the given question, the volume of the solution is 100 ml or 0.1 L and the molarity of the solution is 0.100 M. The moles of the solute (in the given case calcium chloride dihydride (CaCl2. H2O) can be determined by using the formula,
Molarity = moles of solute/volume of solution in liters
Now putting the values we get,
0.100 = moles of solute/0.1000
Moles of solute = 0.100 * 0.1000
= 0.01 moles
The mass of CaCl2.2H2O can be determined by using the formula,
Moles = mass/molar mass
The molar mass of CaCl2.2H2O is 147.01 gram per mole. Now putting the values we get,
0.01 = mass / 147.01
Mass = 147.01 * 0.01
= 1.4701 grams.
The mass should be considered as the 1.4701 grams and the moles should be 0.01.
Calculation of the mass and moles:Since we know that
Molarity = moles of solute/volume of solution in liters
So,
0.100 = moles of solute/0.1000
Moles of solute = 0.100 * 0.1000
= 0.01 moles
Now The mass should be
Moles = mass/molar mass
0.01 = mass / 147.01
Mass = 147.01 * 0.01
= 1.4701 grams.
hence, The mass should be considered as the 1.4701 grams and the moles should be 0.01.
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What allows two or more atoms to be held together?
Answer:
Atoms come together to form molecules because of their electrons. Electrons can join (or bond) atoms together in two main ways. When two atoms share electrons between them, they are locked together (bonded) by that sharing. These are called covalent bonds.
Explanation:
Answer:
Electrons
Explanation:
took the test got 100%
Which of the following are not created by an arrangement of electric charges
or a current (the flow of electric charges)?
A. An electric field
B. A magnetic field
C. A quantum field
D. A gravitational field
Answer:
gravitational and quantum ARE NOT, but electric and magnetic ARE. there is a similar question to this but it's the exact opposite, so don't get confused
. The pI is called ________________. The rule of calculating pI of an amino acid is that first, write the dissociation equation from fully protonated form to fully deprotonated form, label the charge of each form; second, identify the zwitterionic form (zero charge) and find the closest pKs (left and right side in the dissociation equation); third, average these two pKs. Write the dissociation equations for amino acids, glutamate, histidine, and calculate their pIs.
Answer:
The isoelectric point is that the pH at which the compound is in an electronically neutral form.
For diss equations, please find them in the enclosed file.
The pIs of 2 amino acids:
Glutamate: pI = 3,2Histidine: pI = 7,6Explanation:
Formula for the pI calculation: pI = (pKa1 + pKa2)/2
Given 3 pKa :
Acid glutamic with an acid sidechain:Use the lower 2 pKas (corresponding with 2 -COOH groups)
pKa1 = 2,19; pKa2 = 4,25; so pI = 3,2
Histidine with 2 amino groups:Use the higher 2 pKas ( -COOH group and -NH= group)
pKa1 = 6; pKa2 = 9,17; so pI = 7,6
A pure sample of the R enantiomer of a compound has a specific rotation, [ α], of +20 °. A solution containing 0.2 g/mL of a mixture of enantiomers rotates plane polarized light by −2 ° in a 1 dm polarimeter. What is the enantiomeric excess (%ee) of the mixture?
Answer:
Explanation:
The specific rotation of the sample is -2 degrees/0.2 g/mL of mixture
This equals -10 degrees/g/mL of sample.
let the proportion of the R (+) enantiomer be x. The proportion of the S (-) enantiomer in the mixture will be given by (1-x).
specific rotation of the mixture = proportion of R enantiomer* its specific rotation + proportion of S enantiome * its specific rotation
i.e.
-10 = x *(+20) + (1-x)*(-20)
-10 = 20x-20 + 20x
-10+20 = 40x
+10 = 40 x
x=10/40 = 25%
Since the proportion of the other enantiomer is 1-x, it is 0.75 or 75%
So the mixture contains 25% R, 75% S, giving you an excess of 50%.
Answer:
10%
Explanation:
Enantiomeric excess is a way of describing how optically pure a mixture is by calculating the purity of the major enantiomer. It can range from 0%-100%. Enantiomeric excess ( ee ) can also be defined as the absolute difference between the mole fractions of two enantiomers.
Enantiomeric excess is also called optical purity. This is because chiral molecules cause the rotation of plane-polarized light and are said to be optically active. An enantiomerically pure sample has an enantiomeric excess of 100 percent
Enantiomeric excess = observed specific rotation/specific rotation of the pure enantiomer x 100
From the data given in the question;
observed specific rotation= -2°
specific rotation of the pure enantiomer = +20°
Therefore;
ee= 2/20 ×100
ee= 10%
Quantum number of boron
Answer:
The answer is 5.
An aqueous KNO3 solution is made using 72.5 g of KNO3 diluted to a total solution volume of 2.00 L. Calculate the molarity, molality, and mass percent of the solution. (Assume the density of 1.05 g/mL for the solution.)
Answer:
The molarity is 0.359[tex]\frac{moles}{L}[/tex]
The molality is 0.354 [tex]\frac{moles}{kg}[/tex]
The mass percent of the solution es 3.45%
Explanation:
Molarity is a unit of concentration that indicates the amount of moles of solute that appear dissolved in each liter of the mixture. It is determined by:
[tex]Molarity (M)=\frac{number of moles of solute}{volume}[/tex]
Being:
K: 39 g/moleN: 14 g/moleO: 16 g/moleThe molar mass of KNO₃ is:
KNO₃= 39 g/mole + 14 g/mole + 3*16 g/mole= 101 g/mole
You can apply the following rule of three: if 101 grams of KNO₃ are present in 1 mole, 72.5 grams in how many moles are present?
[tex]moles of KNO_{3}=\frac{72.5 grams*1 mole}{101 grams}[/tex]
moles of KNO₃= 0.718
So you have:
moles of KNO₃= 0.718volume= 2 LApplying this quantity in the definition of molarity:
[tex]molarity=\frac{0.718 moles}{2 L}[/tex]
Molarity= 0.359[tex]\frac{moles}{L}[/tex]
The molarity is 0.359[tex]\frac{moles}{L}[/tex]
Molality is a way of measuring the concentration of solute in solvent and indicates the amount of moles of solute in each kilogram of solvent.
Then the molality is calculated by:
[tex]Molality=\frac{moles of solute}{mass of solvent in kilograms}[/tex]
Density is defined as the property that matter, whether solid, liquid or gas, has to compress in a given space. That is, it is the amount of mass per unit volume. So, if the density of 1.05 g / mL for the solution indicates that in 1 mL of solution there are 1.05 grams of solution, in 2000 mL (where 2L = 2000 mL, because 1 L = 1000mL) how much mass is there?
[tex]mass=\frac{2000 mL*1.05 grams}{1 mL}[/tex]
mass= 2100 grams
Since mass solution = mass water + mass KNO₃
then mass water = mass solution - mass KNO₃
Being mass solution 2100 grams and mass KNO₃ 72.5 grams, and replacing you get: mass water= 2100 grams - 72.5 grams
mass water= 2,027.5 grams
Then, being:
moles of KNO₃= 0.718mass of solvent in kilograms= 2.0275 kg (being 2,027.5 grams= 2.0275 kilograms because 1,000 grams= 1 kilogram)Replacing in the definition of molality:
[tex]molality=\frac{0.718 moles}{2.0275 kg}[/tex]
molality= 0.354 [tex]\frac{moles}{kg}[/tex]
The molality is 0.354 [tex]\frac{moles}{kg}[/tex]
The mass percent of a solution is the number of grams of solute per 100 grams of solution. Then the mass percent is the mass of the element or solute divided by the mass of the compound or solute and the result of which is multiplied by 100 to give a percentage.
[tex]mass percent=\frac{mass of solute}{mass of solution} *100[/tex]
So, in this case:
[tex]mass percent=\frac{72.5 grams}{2100 grams} *100[/tex]
mass percent= 3.45 % KNO₃ by mass
The mass percent of the solution es 3.45%
The molarity of the solution is 0.36 mol/L. The molality of the solution is 0.34 m. The mass percent of the solution is 3.33%.
Number of moles of KNO3 = mass/molar mass = 72.5 g/101 g/mol = 0.72 moles
Molarity = Number of moles / volume = 0.72 moles/ 2.00 L = 0.36 mol/L
The molality = Number of moles of solute/Mass of solution in kilograms
mass of solution = 1.05 g/mL × 2000 mL = 21000 g or 2.1Kg
Molality of solution = 0.72 moles/2.1 Kg = 0.34 m
Mass percent of solution = mass of solute/mass of solution × 100/1
Mass percent of solution = 72.5 g/ (72.5 g + 2100 g) × 100/1
= 3.33%
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Which of these scientists diagnosed smallpox and measles?
A. Nicolaus Copernicus
B. Al-Razi
C. Archimedes
D. Rosalind Franklin
Answer:
B
Explanation:
Draw every stereoisomer for 1-bromo-2-chloro-1,2-difluorocyclopentane. Use wedge-and-dash bonds for the substituent groups, and be sure that they are drawn on the outside of the ring, adjacent to each other. The skeletal structure of one molecule is included to indicate the proper format.
Answer:
Explanation:
The objective here is mainly drawing the diagrams of every stereoisomer for 1-bromo-2-chloro-1,2-difluorocyclopentane.
Stereoisomerism is the difference of the spatial arrangement of atoms in a molecule or a compound with the same molecular formula.
For 1-bromo-2-chloro-1,2-difluorocyclopentane.
We have the stereoisomers as follows:
(1R,2S)-1-bromo-2-chloro-1,2-difluorocyclopentane.
(1S,2R)-1-bromo-2-chloro-1,2-difluorocyclopentane.
(1S,1S)-1-bromo-2-chloro-1,2-difluorocyclopentane.
(1R,1R)-1-bromo-2-chloro-1,2-difluorocyclopentane.
Their diagrams are drawn and shown in the attached file below in the order with which they are listed above.
When a saturated solution of NH4Br dissolved in 100 grams of water is cooled from 60°C to 30°C, how much NH4Br will precipitate?
Answer:
[tex]m_{precipitated}=24.8g[/tex]
Explanation:
Hello,
In this case, since at 60 °C, 108 grams of ammonium bromide are completely dissolved in 100 grams of water for a saturated solution, once it is cooled to 30 °C, wherein only 83.2 grams are completely dissolved in 100 grams of water, the following mass will precipitate:
[tex]m_{precipitated}=108g-83.2g\\\\m_{precipitated}=24.8g[/tex]
Best regards.
An excess of AgNO3 reacts with 185.5 mL of an AlCl3 solution to give 0.325 g of AgCl. What is the concentration, in moles per liter, of the AlCl3 solution? Must show your work on scratch paper to receive credit. AlCl3(aq) + 3 AgNO3(aq) → 3 AgCl(s) + Al(NO3)3(aq)
Answer:
4.07x10⁻³M AlCl₃.
Explanation:
Based on the reaction:
AlCl₃(aq) + 3 AgNO₃(aq) → 3 AgCl(s) + Al(NO₃)₃(aq)
That means 1 mole of AlCl₃ reacts with 3 moles of AgNO₃ to produce 3 moles of AgCl.
As 0.325g of AgCl are produced. Moles of AgCl are (Molar mass AgCl: 143.32g/mol):
0.325g AgCl ₓ ( 1 mol / 143.32g) = 2.27x10⁻³ moles of AgCl
As 3 moles of AgCl are produced from 1 mole of AlCl₃, moles of AlCl₃ that produce 2.27x10⁻³ moles of AgCl are:
2.27x10⁻³ moles of AgCl ₓ (1 mole AlCl₃ / 3 moles AgCl) =
7.56x10⁻⁴ moles AlCl₃
As volume of the AlCl₃ solution that reacts is 185.5mL = 0.1855L, molar concentration of the solution is:
7.56x10⁻⁴ moles AlCl₃ / 0.1855L =
4.07x10⁻³M AlCl₃At that volume is measured to be 755 mm of Hg. If the lungs are compressed to a newA healthy male adult has a lung capacity around 6.00 liters. The pressure in the lungs volume of 3.81 liters, what would be the new pressure in the lungs? What would happen to the air in the lungs?
Answer:
1188.976 mmHg
Explanation:
Initial pressure P1= 755 mmHg
Initial volume V1 = 6.00 litres
Final volume V2 = 3.81 litres
Final pressure P2= the unknown
Now applying Boyle's Law,we have;
P1V1 = P2V2
Since P2 is the unknown then it has to be made the subject of the formula.
P2=P1V1/ V2
P2= 755 × 6.00/ 3.81
P2= 1188.976 mmHg
Therefore, the new pressure is; 1188.976 mmHg
Which correctly lists the three processes that are affected by freeze and thaw cycles?
creep, landslide, and deposition
deposition, creep, and weathering
landslide, slump, and deposition
O slump. weathering, and creep
Answer:
slump. weathering, and creep
Explanation:
Freezing and thawing cycle in geology is the process in which water gets in between soil space or rock cracks, freeze in a cold season, and then melt in a warmer season, exerting a force on the soil or rock around it. This force is due to the expansion and contraction of water when it changes from ice to liquid water.
The three geological processes slump, weathering and creep all depend on thawing and freezing cycle among other factors.
Slump: Slump is a type of geological process that occurs when coherent mass of loosely consolidated materials or a rock layer moves a short distance down a slope. The movement of a slump is characterized by sliding along a concave-upward or planar surface. Causes includes earthquake shocks, thorough wetting, freezing and thawing, undercutting, and loading of a slope.Weathering: This is a geological process that results in the gradual disintegration of rocks into smaller sizes. It is one of the most important soil formation process, and is different from erosion by the degree of movement of the soil formed. Weathering does not move the soil from its origin. Thawing and freezing cycle plays a major role in weathering by helping crack up the rocks and by also tearing the rock apart. plays a major role.Creep: This is the slow, often imperceptible downslope movement of soil or other debris. The effects of creep is often seen in the presence of physical characteristics like bent trees, tilted fences, and cracked walls. Creep is caused by multiple factors, of which heaving is likely the most important process. Heaving involves the expansion and contraction of rock fragments, and occurs during cycles of wetting and drying, as well as freezing and thawing.Answer:
It is slump, weathering and creep
Explanation:
Took the test on edg
Phosphofructokinase is a four‑subunit protein with four active sites. Phosphofructokinase catalyzes step 3 of glycolysis, converting fructose‑6‑phosphate to fructose‑1,6‑bisphosphate. Phosphoenolpyruvate (PEP) is the product of step 9 of glycolysis. The PEP concentration in the cell affects phosphofructokinase activity.Select the true statements about PEP regulation of phosphofructokinase.
1. PEP is a feedback inhibitor of phosphofructokinase.
2. The apparent affinity of phosphofructokinase for its substrate increases when PEP binds.
3. PEP is a positive effector of phosphofructokinase.
4. PEP inhibition of phosphofructokinase yields a sigmoidal velocity versus substrate curve.
5. PEP competes with fructose-6-phosphate for the active site of phosphofructokinase.
6. The binding of PEP to one phosphofructokinase subunit causes a conformation change that affects the ability of the substrate to bind to the other subunits.
Answer:
1. PEP is a feedback inhibitor of phosphofructokinase.
4. PEP inhibition of phosphofructokinase yields a sigmoidal velocity versus substrate curve.
6. The binding of PEP to one phosphofructokinase subunit causes a conformation change that affects the ability of the substrate to bind to the other subunits.
Explanation:
Phosphofructokinase-1, PFK-1, is an allosteric enzymes composed of four protein subunits.
Allosteric enzymes are enzymes that function through non-covalent binding of allosteric modulators which may be activators or inhibitors. They produce a characteristic velocity versus substrate sigmoidal curve. PFK-1 has a separate binding site for its substrate, fructose-6-phosphate and it's allosteric modulators: ATP, ADP or phosphoenolpyruvate, PEP.
The enzyme can exist in two conformations, the T-state (tense) or the R-state (resting). Binding of substrate causes a conformational change from T-state to R-state, whereas binding of allosteric inhibitors returns it to the T-state.
PEP, the product of step 9 in glycolysis, is an allosteric inhibitor of PFK-1. When it binds to the the allosteric site, it leads to conformational changes in PFK-1 from the R-state to the T-state which reduces the enzymes ability to bind the substrate. These changes are responsible for the sigmoidal velocity/substrate curve in allosteric enzymes.
Therefore, the true statements from the options above are 1, 4, 6.
Options 2,3 and 5 are wrong because PEP is a negative effector of PFK-1, thus its binding reduces the affinity of PFK-1 for its substrate. Also, PFK-1 being an allosteric enzyme has separate binding sites for its substrate and its modulators. Thus, there is no competition for active site binding by substrate and modulators.
A 8.00g of a certain Compound X, known to be made of carbon, hydrogen and perhaps oxygen, and to have a molecular molar mass of 44./gmol, is burned completely in excess oxygen, and the mass of the products carefully measured: product mass carbon dioxide 24.01g water 13.10g Use this information to find the molecular formula of X.
Answer:
C3H6.
Explanation:
Data obtained from the question:
Mass of the compound = 8g
Mass of CO2 = 24.01g
Mass of H2O = 13.10g
Next, we shall determine the mass of C, H and O present in the compound. This is illustrated below:
Molar Mass of CO2 = 12 + (2x16) = 44g/mol
Molar Mass of H2O = (2x1) + 16 = 18g/mol
Mass of C in compound = Mass of C/Molar Mass of CO2 x 24.01
=> 12/44 x 24.01 = 6.5g
Mass of H in the compound = Mass of H/Molar Mass of H2O x 13.1
=> 2x1/18 x 13.1 = 1.5g
Mass of O in the compound = Mass of compound – (mass of C + Mass of H)
=> 8 – (6.5 + 1.5) = 0
Next, we shall determine the empirical formula of the compound. This is illustrated below:
C = 6.5g
H = 1.
Divide by their molar mass
C = 6.5/12 = 0.54
H = 1.4/1 = 1.
Divide by the smallest
C = 0.54/0.54 = 1
H = 1/0.54 = 2
Therefore, the empirical formula is CH2
Finally, we shall determine the molecular formula as follow:
The molecular formula of a compound is a multiple of the empirical formula.
Molecular formula = [CH2]n
[CH2]n = 44
[12 + (2x1)]n = 44
14n = 44
Divide both side by 14
n = 44/14
n = 3
Molecular formula = [CH2]n = [CH2]3 = C3H6
Therefore, the molecular formula of the compound is C3H6
what is a mitochondrion
Explanation:
Mitochondria (sing. mitochondria) are organelles, or parts of the eukaryote cell. They are in the cytoplasm, not the nucleus. They make the most cell supply of adenosine triphosphate (ATP), a molecule that cells use as an energy source. ... This means that mitochondria are known as '' the powerhouse of the cell'' or ''cell strength".
Good Luck, and have a great day..
A gas company in Massachusetts charges $2.80 for 15.0 ft3 of natural gas (CH4) measured at 20.0°C and 1.00 atm. Calculate the cost of heating 2.00 × 102 mL of water (enough to make a cup of coffee or tea) from 20.0°C to 100.0°C. Assume that only 50.0% of the heat generated by the combustion is used to heat the water; the rest of the heat is lost to the surroundings. Assume that the products of the combustion of methane are CO2(g) and H2O(l).
Answer:
$0.0238
Explanation:
The energy you need to increase the temperature of water from 20°C to 100°C is obtained from:
Q = C×m×ΔT
Where Q is the energy, C is specific heat of water (4.184J/g°C), m is mass of water (2.00x10²g - Density of water 1g/mL), ΔT is change in temperature (100.0°C - 20.0°C)
Replacing:
Q = 4.184J/g°C × 2.00x10²g × 80.0°C
Q = 66944J = 66.944kJ
As you are assuming the energy of combustion will be just 50.0% to heat the water the energy you need is 66.944kJ × 2 = 133.888kJ
The combustion of methane is:
CH4(g) + 2O2(g) ⟶ CO2(g) + 2H2O(l) ΔH = −890.8kJ
That means 1 mole of methane produce 890.8kJ. As you need 133.888kJ, moles of methane are:
133.888kJ × (1 mol CH₄ / 890.8kJ) = 0.150 moles of CH₄.
Using PV = nRT, moles of 15.0ft³ (424.8L) at 20.0°C (293.15K) and 1.00atm:
1.00atmₓ424.8L = moles CH₄ₓ0.082atmL/molKₓ293.15K
17.67 = moles CH₄
As 17.67 moles of CH₄ cost $2.80, the cost of 0.150 moles of CH₄ is:
0.150 moles CH₄ ₓ ($2.80 / 17.67 moles) =
$0.0238During chemical reaction 7.55gKI and 9.06g were allowed to react. How many grams of excess reagent are left over after the reaction is complete. Reaction: Pb(NO3)2(s) + 2KCI(s) > 2KNO3(s) + PbI(s)
Answer: 7.45 g of [tex]Pb(NO_3)_2[/tex] excess reagent are left over after the reaction is complete.
Explanation:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
a) [tex]{\text{Number of moles of} KI}=\frac{7.55g}{166g/mol}=0.045moles[/tex]
b) [tex]{\text{Number of moles of} Pb(NO_3)_2}=\frac{9.06g}{331.2g/mol}=0.027moles[/tex]
The balanced chemical reaction is :
[tex]Pb(NO_3)_2(s)+2KI(s)\rightarrow 2KNO_3(s)+PbI(s)[/tex]
According to stoichiometry :
2 moles of [tex]KI[/tex] require = 1 mole of [tex]Pb(NO_3)_2[/tex]
Thus 0.045 moles of [tex]KI[/tex] will require=[tex]\frac{1}{2}\times 0.045=0.0225moles[/tex] of [tex]Pb(NO_3)_2[/tex]
Thus [tex]KI[/tex] is the limiting reagent as it limits the formation of product and [tex]Pb(NO_3)_2[/tex] is the excess reagent as (0.045-0.0225) = 0.0225 moles are left
Mass of [tex]Pb(NO_3)_2=moles\times {\text {Molar mass}}=0.0225moles\times 331.2g/mol=7.45g[/tex]
Thus 7.45 g of [tex]Pb(NO_3)_2[/tex] of excess reagent are left over after the reaction is complete.
Propane (C3H8) is widely used in liquid form as a fuel for barbecue grills and camp stoves. For 67.7 g of propane, determine the following.(a) Calculate the moles of compound.mol(b) Calculate the grams of carbon.g
Answer:
A. 1.54 mole.
B. 55.39g of carbon
Explanation:
A. Determination of the number of mole in 67.7g of C3H8.
Mass of C3H8 = 67.7g
Molar mass of C3H8 = (3x12) + (8x1) = 44g/mol
Number of mole of C3H8 =..?
Number of mole = Mass/Molar Mass
Number of mole of C3H8 = 67.7/44
Number of mole of C3H8 = 1.54 mole
B. Determination of the mass of carbon in the compound.
This is illustrated below:
The mass of C in compound can be obtained as follow:
=> 3C/C3H8 x 67. 7
=> 3x12 / 44 x 67.7
=> 36/44 x 67.7
=> 55.39g
Therefore, 55.39g of carbon is present in the compound.
Arrange the following compounds in order of increasing solubility in
water and explain your sequence.
C7H15OH C6H13OH C6H6 C2H5OH
Answer:
C6H6<C7H15OH<C6H13OH<C2H5OH
Explanation:
Organic substances are ordinarily nonpolar. This means that they do not dissolve in water. However, certain homologous series of organic compounds actually dissolve in water because they possess certain functional groups that effectively interact with water via hydrogen bonding.
A typical example of this is alcohol family. All members of this homologous series contain the -OH functional group. This group can effectively interact with water via hydrogen bonding, leading to the dissolution of low molecular weight alcohols in water.
Low molecular weight alcohols are miscible with water in all proportions. This implies that they are highly soluble in water. However, as the size of the alkyl moiety in the alcohol increases, the solubility of the alcohol in water decreases due to less effective interaction of the -OH group with water via hydrogen bonding. This explains the fact that C2H5OH is the most soluble alcohol in the list.
C6H6 is insoluble in water since it is purely a hydrocarbon with no -OH group capable of interaction with water via hydrogen bonding.