Answer: the exciation of molecules is brount by absorption of energy in spectroscpy
Explanation:
ultraviolet photon (λ = 58.4nm) from a helium gas discharge tube is absorbed by a hydrogen molecule which is at rest. Since momentum is conserved, what is the velocity of the hydrogen molecule after absorbing the photon? What is the translational energy of the hydrogen molecule in Jmol-1.
[h = 6.626 x 10-34 Js; NA = 6.022 x 1023 mol-1]
Answer:
Translation energy of 1 mole of H2 molecules = KE x Avogadros number
[tex]= 1.923 * 10^{-26} * 6.022 * 10^{23}\\\\= 0.0116 J \\\\= 1.16 * 10^{-2} \ J[/tex]
Explanation:
Photon wavelength [tex]= 58.4 nm = 58.4 * 10^{-9} m[/tex]
Photon momentum = h/wavelength
[tex]= (6.626 * 10^{-34})/(58.4 * 10^{-9})\\\\ = 1.1346 * 10^{-26} \ kg.m/s[/tex]
Mass of H2 molecule m = molar mass/Avogadros number
[tex]= (2.016)/(6.022 * 10^{23})\\\\= 3.3477 * 10^{-24} \ g = 3.3477 * 10^{-27} \ kg[/tex]
Since momentum is conserved:
Photon momentum = H2 molecule momentum = mass x velocity of H2
[tex]1.1346 * 10^{-26} = 3.3477 * 10^{-27} * v[/tex]
velocity [tex]v = 3.389 m/s = 3.39 m/s[/tex]
Translation energy of 1 H2 molecule = kinectic energy (KE) = (1/2)mv^2
[tex]= 1/2 * 3.3477 * 10^{-27} * 3.389^2\\\\= 1.923 * 10^{-26} J[/tex]
Translation energy of 1 mole of H2 molecules = KE x Avogadros number
[tex]= 1.923 * 10^{-26} * 6.022 * 10^{23}\\\\= 0.0116 J \\\\= 1.16 * 10^{-2} \ J[/tex]
Ni
Express your answer in condensed form in the order of orbital filling as a string without blank space between orbitals. For example, [He]2s22p2 should be entered as [He]2s^22p^2.
Answer:
[Ar]3d^84s^2
Explanation:
From the question given, we are asked to write the condensed form of electronic configuration of nickel, Ni.
To do this, we simply write the symbol of the noble gas element before Ni in a squared bracket followed by the remaining electrons to make up the atomic number of Ni.
This is illustrated below:
The atomic number of Ni is 28.
The noble gas before Ni is Argon, Ar.
Therefore, the condensed electronic configuration of Ni is written as:
Ni(28) => [Ar]3d^84s^2
Answer:
[Ar] 4s^23d^8
Explanation:
Identify the person who made the correct statement.
Mike said petrified fossils are hard and heavy like rock.
Bobby said that petrified fossils have the same appearance as when they were alive.
Neither Mike nor Bobby is correct.
Mike is correct.
Bobby is correct.
Both Mike and Bobby are correct.
Answer: Both Mike and Bobby are correct.
Explanation:
Petrifcation can be defined as the process in which the organic material of the dead living being becomes fossil by the replacement of mineral deposition in the bony, hard material.
Thus although the body components gets decomposed wiped out due to this process. The body shape of the dead organism remains the same as that was in living.
Thus the statements made by Mike and Bobby both are correct. The fossils are hard and have the same appearance as when they were alive.
Photochromic lenses contain Group of answer choices both AgCl and CuCl embedded in the glass. only AgCl embedded in the glass. neither AgCl nor CuCl embedded in the glass. only CuCl embedded in the glass.
Answer:
both AgCl and CuCl embedded in the glass
Explanation:
Photochromic lenses contain both AgCl and CuCl embedded in the glass.
They are light-sensitive lenses that adapt to environmental changes. They appear clear when in an apartment or a building and automatically darken when outside as a result of exposure to sunlight. The darkening is activated by the UV component of the sunlight.
Photochromic lenses are otherwise known as light-adaptive or intelligent lenses and they are formed by coating lenses with silver chloride compounds whose concentration ranges from 0.01 to 0.001 %. Copper (I) chloride is also included in addition to the silver halide.
In summary, photochromic lenses contain both AgCl and CuCl.
Mass is:
measured in kilograms
measured using a scale
affected by gravity
all of the above
8) What is the molarity (M) of an aqueous solution containing 22.5 g of sucrose (C12H22011) in 35.5 mL of solution?
A) 3.52 M
B) 1.85 x 10-2M
C) 0.104 M
D) 0.0657 M
E) 1.85 M
Answer:
E) 1.85 M
Explanation:
M(C12H22O11) = 342.3 g/mol
22.5 g * 1mol/342.3 g = 0.0657 mol
35.5 mL = 0.0355 L
Molarity = mol solute/L solution = 0.0657 mol/0.0355L =1.85 mol/L = 1.85 M
The molarity of the aqueous solution is 1.85 M. The correct option is E) 1.85 M
From the question,
We are to determine the molarity (that is, concentration) of the given sucrose solution
First, we will determine the number of moles present in the given mass of sucrose
Mass of sucrose = 22.5 g
Using the formula
[tex]Number\ of\ moles = \frac{Mass}{Molar\ mass}[/tex]
Molar mass of sucrose = 342.2965 g/mol
∴ Number of moles of sucrose present = [tex]\frac{22.5}{342.2965}[/tex]
Number of moles of sucrose present = 0.0657325 moles
Now, for the molarity (concentration) of the sucrose solution
From the formula
Number of moles = Concentration × Volume
Then,
[tex]Concentration = \frac{Number\ of\ moles}{Volume}[/tex]
From the question,
Volume = 35.5 mL = 0.0355 L
∴ [tex]Concentration = \frac{0.0657325}{0.0355}[/tex]
Concentration = 1.85 M
Hence, the molarity of the aqueous solution is 1.85 M. The correct option is E) 1.85 M
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One compound in the list below is a gas at room conditions, while all of the others are liquids or solids. For each compound, indicate whether it is the gas or the force most responsible for it being a liquid or solid.
First list:
A. CH3OH
1. a gas2. dispersion forces 3. dipolar forces4. hydrogen bonds5. ionic bonds6. covalent bondsB. CH4
1. a gas2. dispersion forces 3. dipolar forces4. hydrogen bonds5. ionic bonds6. covalent bondsC. CaCO3
1. a gas2. dispersion forces 3. dipolar forces4. hydrogen bonds5. ionic bonds6. covalent bondsD. C6H14
1. a gas2. dispersion forces 3. dipolar forces4. hydrogen bonds5. ionic bonds6. covalent bondsSecond list:
A. H2O
B. C2H2
C. CCl4
D. KCl
Answer:
First list
A. CH3OH----hydrogen bonds
CH4----dispersion forces
CaCO3---ionic bonds
C6H14----dispersion forces
Second list
H2O------ liquid----hydrogen bonds
C2H2----gas---dispersion forces
CCl4---liquid---dispersion forces
KCl----solid---ionic bonds
Explanation:
For every compound, the intermolecular forces decide whether the substance will be solid liquid or gas. Molecules are known to associate with each other in any particular state of matter. These molecules are held together by different intermolecular interactions with varying degrees of strength. The strength of the intermolecular interaction between the molecules of a substance will decide if the substance will be a solid, liquid or gas.
When the intermolecular forces are very strong such as in ionic solids and covalent network solids, the substance exists as a solid. When the intermolecular forces are not so strong such as dispersion forces and hydrogen bonds, the substance exists as a liquid. However, very weak intermolecular dispersion forces are found in gases hence the molecules are relatively free when compared to molecules of liquids and solids.
iron oxide + oxygen equals to ?
Answer:
It's ferric oxide Fe2O3
Explanation:
I don't say u must have to mark my ans as brainliest but if it has really helped u plz don't forget to thank me plz...
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%
If you have 101 g of hydrogen gas (H2) and excess amount of nitrogen gas (N2), how many grams of ammonia gas (NH3) can you make?
Answer:
572. 3 g of NH3
Explanation:
Equation of the reaction: 3H2 + N2 ----> 2NH3
From the equation of reaction, 3 moles of H2 reacts with 1 mole of N2 to produce 2 moles of NH3.
Since N2 is in excess in the given reaction, H2 is the limiting reactant.
Molar mass of H2 = 2 g/mol
Molar mass of NH3 = 17 g/mol
Therefore 3 * 2 g of H2 reacts to produce 2 * 17 g of NH3
6 g of H2 produces 34 g of NH3
101 g of H2 will produce (34 * 101)/6 g of NH3 = 572.3 g of NH3
Therefore, 572.3 g of NH3 are produced
Answer:
572.33g of NH3.
Explanation:
We'll begin by writing the balanced equation for the reaction. This is given below:
N2 + 3H2 —> 2NH3
Next, we shall determine the mass of the H2 that reacted and the mass of NH3 produced from the balanced equation. This is illustrated below:
Molar Mass of H2 = 2x1 = 2g/mol
Mass of H2 from the balanced equation = 3 x 2 = 6g
Molar Mass of NH3 = 14 + (3x1) = 17g/mol
Mass of NH3 from the balanced equation = 2 x 17 = 34g.
From the balanced equation above,
6g of H2 reacted to produce 34g of NH3.
Finally, we can determine the mass of ammonia (NH3) produced by reacting 101g of H2 as follow:
From the balanced equation above,
6g of H2 reacted to produce 34g of NH3.
Therefore, 101g of H2 will react to produce = ( 101 x 34) / 6 = 572.33g of NH3.
Therefore, 572.33g of NH3 is produced from the reaction.
How many grams of 02 are needed to react with 7.50g of ethanol
Determine the amount of heat (in kJ) associated with the production of 5.71 × 104 g of ammonia according to the following equation. N2(g) + 3H2(g) 2NH3ΔH°rxn = −92.6 kJ Assume that the reaction takes place under standard-state conditions at 25°C.
Answer:
[tex]Q=-3.11x10^5kJ[/tex]
Explanation:
Hello,
In this case, for the given reaction, we are given the standard enthalpy of reaction per mole of ammonia that is -92.6 kJ, it means, that forming one mole of ammonia will release 92.6 kJ of energy. In such a way, for the formation of 5.71x10⁴ g of ammonia, the following amount of heat will be released:
[tex]Q=5.71x10^4gNH_3*\frac{1molNH_3}{17gNH_3}*-92.6\frac{kJ}{molNH_3}\\ \\Q=-3.11x10^5kJ[/tex]
Best regards.
The amount of the heat associated with the production of 5.71×10⁴ g of ammonia, NH₃ is –311026.732 KJ
We'll begin by calculating the number of mole in 5.71×10⁴ g of NH₃
Mass of NH₃ = 5.71×10⁴ g
Molar mass of NH₃ = 14 + (3×1) = 17 g/mol
Mole of NH₃ =?Mole = mass / molar mass
Mole of NH₃ = 5.71×10⁴ / 17
Mole of NH₃ = 3358.82 moles Finally, we shall determine the heat required to produce 3358.82 moles (i.e 5.71×10⁴ g) of NH₃. This can be obtained as follow:N₂(g) + 3H₂(g) —> 2NH₃(g) ΔH°rxn = −92.6 kJ
Since reaction took place at standard conditions, it means:
1 moles of NH₃ required −92.6 kJ
Therefore,
3358.82 moles of NH₃ will require = 3358.82 × –92.6 = –311026.732 KJ
Thus, the amount of the heat associated with the production of 5.71×10⁴ g of ammonia, NH₃ is –311026.732 KJ
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Suppose the amount of a certain radioactive substance in a sample decays from to over a period of days. Calculate the half life of the substance. Round your answer to significant digit.
The given question is incomplete, the complete question is:
Suppose the amount of a certain radioactive substance in a sample decays from 1.30 mg to 100. ug over a period of 29.5 days. Calculate the half life of the substance Round your answer to 2 significant digits.
Answer:
The correct answer is 7.974 days.
Explanation:
Based on the given question, the concentration of a radioactive substance present in a sample get decays to 100 micro grams from 1.30 milligrams in 29.5 days. There is a need to find the half-life of the substance.
Radioactive decay is an illustration of first order reaction.
K = (2.303 / t) log [a/(a-x)]
Here a is 1.30 mg and (a-x) is 100 micrograms = 100 * 10^-3 mg or 0.1 mg, and t is 29.5 days. Now putting the values we get,
K = (2.303 /29.5)log (1.30/0.1)
= 2.303/29.5 log13
= 2.303/29.5 * 1.1139
K = 0.0869
The half-life or t1/2 is calculated by using the formula, 0.693 / K
= 0.693 / 0.0869
= 7.974 days.
What is the molar mass of CH2O2 ? ( C= 12.01 g/mol, H=1.008 g/mol, O=16.00)
Answer:
Molar Mass of CH2O2 is 46.026
Explanation:
What is the molar mass of CH2O2 ? ( C= 12.01 g/mol, H=1.008 g/mol, O=16.00)
C = 12.01g/mol
H = 1.008g/mol
O = 16g/mol
CH2O2 = 12.01+1.008x2+16x2 = 46.026g/mole
When researchers need to prepare many reactions for polymerase chain reaction (PCR) amplification, they often create a "master mix" solution. A master mix contains the reagents common to all the planned PCR amplifications, regardless of the target DNA. Making a master mix is a way to minimize the number of pipetting steps.Suppose a researcher needs to PCR amplify seven different genes of interest from different organisms. The researcher prepares a master mix and dispenses it to seven different PCR tubes, one for each gene of interest.Select the PCR components the researcher must add to each of the seven tubes of master mix to selectively amplify each gene of interest.dNTPsMg2+-Mg2+-based bufferprimersDNA polymeraseDNA template
Answer:
The master mix contains the following reagents: dNTPs, DNA Polymerase, PCR buffer and MgCl2.
Explanation:
The DNA templates are the gene fragments to amplify by PCR, thereby they have to be added separately in each tube. Moreover, the primer pairs are specific for each gene, thereby they have to be added separately in each tube.
Deoxynucleotide triphosphates (dNTPs) are the building blocks of the DNA molecules: dGTP, dATP, TTP, and dCTP.
The PCR buffer provides a suitable medium for the activity of the DNA polymerase, often it contains Tris-Hcl and KCl.
MgCl2 is a cofactor for the activity of the DNA Polymerase.
The DNA Polymerase is an enzyme that amplifies DNA by adding nucleotides to the 3' end.
What did John Dalton publish?
Answer:
An early theory describing properties of atoms.
Explanation:
Apex
What would form a solution?
O A. Mixing two insoluble substances
O B. Mixing a solute and a solvent
O C. Mixing a solute and a precipitate
O D. Mixing two solutes together
Answer:
B. Mixing a solute and a solvent
Explanation:
Hello,
In this case, solutions are defined as liquid homogeneous mixtures formed when two substances having affinity are mixed. It is important to notice that the two substances are known as solute, which is added to other substance that is the solvent. Therefore, answer is B. Mixing a solute and a solvent.
Notice that when two insoluble substances are mixed no solution is formed. Furthermore, if two solutes together or a solute and a precipitate are mixed, no liquid homogeneous solution is formed, as commonly solutes are solid, nevertheless, when liquid, one should have to act as the solvent.
Best regards.
Answer:
B. Mixing a solute and a solvent
Explanation:
ap3x
The equilibrium constant for the reaction NO2(g)+NO3(g)→N2O5(g) is 2.1x10-20 , therefore: a. At equilibrium, the concentration of products and reactants is about the same. b. At equilibrium, the concentration of products is greater than the reactants. c. At equilibrium, the concentration of reactants is greater than the products
Answer: c. At equilibrium, the concentration of reactants is greater than the products
Explanation:
Equilibrium constant for a reaction is the ratio of concentration of products to the concentration of reactants each raised to the power its stoichiometric coefficients.
For the reaction:
[tex]NO_2(g)+NO_3(g)\rightleftharpoons N_2O_5(g)[/tex]
Equilibrium constant is given as:
[tex]K_{eq}=\frac{[N_2O_5]}{[NO_2]\times [NO_3]}[/tex]
[tex]2.1\times 10^{-20}=\frac{[N_2O_5]}{[NO_2]\times [NO_3]}[/tex]
When
a) K > 1, the concentration of products is greater than the concentration of reactants
b) K < 1, the concentration of reactants is greater than the concentration of products
c) K= 1, the reaction is at equilibrium, the concentration of reactants is equal to the concentration of products
Thus as [tex]K_{eq}[/tex] is [tex]2.1\times 10^{-20}[/tex] which is less than 1,
the concentration of reactants is greater than the concentration of products
Indicate whether each of the following indicates that a physical or chemical change has taken place when a piece of magnesium metal is studied: (a) Can be cut into tiny pieces (b) Fizzling occurs when placed water (c) Light is emitted when burned (d) Turns to ash
Answer:
a) Can be cut into tiny pieces - Physical Change
b) Fizzling occurs when placed water -Chemical Change
c) Light is emitted when burned -Chemical Change
d) Turns to ash -Chemical Change
Explanation:
How many protons are in an ion with 36 electrons and a -1 charge
Answer:
Explanation:
There are 35 protons.
The number of electrons = 36 electrons gives a -1 charge.
Where did all the other minus charges go?
They must be balanced by 35 protons.
List three ways the rate of solvation of sodium chloride in water may be
increased
Answer:
1) Increasing temperature
2) Stirring
3) Increasing surface area of salt by grinding it
Precision can be defined as the?
Answer:Precision can be defined as the. reproducibility of a measured value. Precision is how close the measured values are to each others. In contrast with accuracy, accuracy is the agreement between a measured value and an accepted value.
Explanation:
What is the molar mass of P2O5?
Answer:
142 grams
Explanation:
To find the molar mass of a molecule or compound, you simply need to add together the molar masses of all of the atoms that comprise it. Phosphorus has a molar mass of about 31, while oxygen has one of about 16, meaning that the molar mass of this molecule is:
2(31)+5(16)=62+80=142
Hope this helps!
Consider the following reaction. I– 2 H2O2 (l) 2 H2O (l) + O2 (g) A solution contains 15 mL 0.1 M KI, 15 mL of DI water and 5 mL of 3% H2O2. After the decomposition of H2O2 is complete, you titrate the solution with 0.1 M AgNO3. If the catalyst, I–, is not consumed in the reaction and is completely recovered, what volume of the 0.1 M AgNO3 is required to reach the end point?
Answer:
Explanation:
The given chemical reaction is:
[tex]2H_2O_{(l)} \to^{I^-}} 2H_2O_{(l)}+O_2_{(g)}[/tex]
From above equation [tex]I^-[/tex] serves as catalyst which is not consumed by the reaction and also it is completely recovered; as a result to that , the full volume of KI will definitely react with AgNO₃.
Given that :
the volume of potassium iodide [tex]V_{KI} = 15 \ ml[/tex]
the molarity of potassium [tex]M_{KI} = 0.1 \ M[/tex]
the volume of distilled water [tex]V_W = 15 \ mL[/tex]
The volume of 3% [tex]H_2O_2 \ \ V_{H_2O_2} = 5 \ mL[/tex]
Molarity of AgNO₃ [tex]M_{AgNO_3} = 0.1 \ M[/tex]
Let take an integral look with the reaction between KI and AgNO₃; we have
[tex]KI + AgNO_3 \to KNO_3 + AgI[/tex]
At the end point; the moles of KI will definitely be equal to the moles of AgNO₃
So;
[tex]M_{KI}V_{KI}= M_{AgNO_3}V_{AgNO_3} \\ \\ V_{AgNO_3} = \dfrac{M_{KI}V_{KI}}{M_{AgNO_3}} \\ \\ \\ V_{AgNO_3} = \dfrac{ 0.1*15}{0.1}[/tex]
[tex]V_{AgNO_3} = 15 \ ml[/tex]
Thus; the volume of 0.1 M AgNO₃ needed to reach the end point is 15 mL
The type of nuclear decay an unstable nucleus will undergo depends on its ratio of neutrons to protons. The radioisotope cobalt-65 has a ratio of neutrons to protons of 1.41, which is too high for a nucleus of this size. What nuclear changes could reduce this ratio
Answer:
Explanation:
In cobalt - 65 ,
no of protons is 27 ( p )
no of neutron = 65 - 27 ( n )
= 38
n / p ratio
= 38 / 27
= 1.41
If case of emission of alpha particle
no of proton p = 27 - 2 = 25
no of neutrons = 38 - 2 = 36
n / p ratio = 36 / 25
= 1.44
So it increases
In case of emission of beta particle
No of neutron n = 38 - 1 = 37
No of proton = 27 + 1 = 28
n / p ratio = 37 / 28
= 1.32
Hence ratio decreases.
Hence beta ray decay will result in decrease in n / p ratio.
Which accurately labels the lysosome?
Answer:
One of the organelles in eukaryotic cells that carry out digestion and waste removal.
Answer:
It's X
Explanation:
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.
The cryosphere is part of which sphere of the Earth system?
atmosphere
biosphere
geosphere
hydrosphere
Answer:
Ice (frozen water) is part of the hydrosphere, but it's given its own name, the cryosphere.
The cryosphere is part of the hydrosphere of the Earth system. The correct option is D.
What is the cryosphere?The cryosphere contains all the frozen parts of the earth. The term is made up of the Greek word “krios” which means cold. All the frozen water of the oceans and snow comes under the cryosphere.
The atmosphere contains all spheres, it is an envelope of gases. The geosphere is the land part of the earth, and the biosphere is the part where the living part is present.
The cryosphere is h habitat of many living creatures, and the climate of the earth is highly dependent on this sphere. The warmth of the earth is increasing and the cryosphere part is decreasing day by day, which is having problems for many animals.
Thus, the correct option is D, hydrosphere.
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Question 8
1 pts
A closed flask contains a 0.25 moles of O2 which exerts a pressure of
0.50 atm. If 0.75 moles of CO, is added to the container what is the
total pressure in the flask?
Answer:
\large \boxed{\text{2.0 atm}}
Explanation:
We can use Dalton's Law of Partial Pressures:
Each gas in a mixture of gases exerts its pressure separately from the other gases.
0.25 mol of O₂ exerts 0.50 atm.
If you add 0.75 mol of CO, the total amount of gas is
0.25 mol + 0.75 mol = 1.00 mol
[tex]p_{\text{total}} = \text{1.00 mol} \times \dfrac{\text{0.50 atm}}{\text{0.25 mol}}= \textbf{2.0 atm}\\\\\text{The total pressure in the flask is $\large \boxed{\textbf{2.0 atm}}$}[/tex]
The pressure of the closed flask after the addition of 0.75 moles of CO has been 2 atm.
Partial pressure can be defined as the pressure exerted by each gas in a given solution.
The total moles of gas in the container by the addition of CO has been:
Total moles = moles of oxygen + moles of CO
Total moles = 0.25 + 0.75
Total moles = 1 mol.
By using Dalton's law of partial pressure:
Total pressure = total moles [tex]\rm \times\;\dfrac{pressure\;of\;oxygen}{moles\;of\;oxygen}[/tex]
Total pressure = 1 [tex]\rm \times\;\dfrac{0.50}{0.25}[/tex]
Total pressure = 2 atm.
The pressure of the closed flask after the addition of 0.75 moles of CO has been 2 atm.
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g Reduction involves the A) loss of neutrons, gain of electrons, and an increase in oxidation state. B) loss of neutrons. C) increase in oxidation state. D) gain of electrons and an increase in oxidation state. E) gain of electrons.
Answer:
E. Gain of electrons
Explanation:
A reduction reaction is one part of the two concurrent reactions that take place in a redox (reduction-oxidation) reaction.
During reduction, an atom gains electrons from a donor atom, and it's oxidation number becomes smaller.
Option A is wrong because reduction does not increase oxidation state nor are neutrons involved
Option B is wrong because reduction is not a nuclear reaction (does not involve the nucleons)
Option C is wrong because reduction leads to reduction in oxidation state
Option D is wrong leads to a reduction in oxidation state when electrons are gained
Option E is correct because reduction involves gain of electrons
