Which is one characteristic of producers?
They recycle nutrients.
They do not eat other organisms.
They break down waste for energy.
They use other organisms for energy.

Answer:

Explanation:

RX + AgNO₃ = R⁺ ( carbocation ) + AgX + NO₃⁻

C₂H₅OH ( a nucleophile ) + R⁺ = ROC₂H₅

C₅H₁₁X + AgNO₃ = C₅H₁₁⁺ + AgX + NO₃⁻

In the first case carbocation produced is CH₃CH₂CH₂CH₂CH₂⁺

CH₃CH₂CH₂CH₂CH₂⁺ ⇒  CH₃CH₂CH₂C⁺HCH₃ ( secondary carbocation more stable )

CH₃CH₂CH₂C⁺HCH₃ + C₂H₅OH ⇒ CH₃CH₂CH₂CH(OC₂H₅)CH₃

Hence option D is correct .

b )

In the second case carbocation produced is

CH₃CH₂CH₂CH⁺CH₃

CH₃CH₂CH₂C⁺HCH₃ + C₂H₅OH ⇒ CH₃CH₂CH₂CH(OC₂H₅)CH₃

The product formed is same as in case of first

Option B is correct

Answer:

ΔH = -338.8kJ

Explanation:

it is possible to sum the enthalpy changes of some reactions to obtain the enthalpy change of the whole reaction (Hess's law).

Using the reactions:

R₁ O₂(g) → 2O(g) ΔH₁°= 449.2 kJ

R₂ O₃(g) + Cl(g) → O₂(g) + ClO(g) ΔH₂° = -126 kJ

R₃ ClO (g) + O (g) → O₂ (g) + Cl (g) ΔH₃°= -268 kJ

By the sum 2R₂ + 2R₃:

(2R₂ + 2R₃) = 2O(g) + 2O₃(g) → 4O₂(g)

ΔH = 2ₓ(-126kJ) + (2ₓ-268kJ) = -788kJ

Now, this reaction + R₁

2O₃(g) → 3O₂(g)

ΔH = -768kJ + 449.2kJ

Answer:

The concentration of one or more of the products is small.

The reaction will not proceed very far to the right.

The reaction will generally form more reactants than products  

Explanation:

We often write

K =[Products]/[Reactants]

Thus, if K is small

A. is wrong. Usually, if K < 1, the concentration of reactants is greater than that of the products.

Answer:

The enthalpy of formation of CaF₂ is -1224.4 kJ.

Explanation:

The enthalpy of formation of CaF₂ can be calculated as follows:

[tex] \Delta H_{f} = \frac{q}{n_{CaF_{2}}} [/tex]

Where:

q: is the heat liberated in the reaction = -251 kJ

The number of moles of CaF₂ is:

[tex] n_{CaF_{2}} = \frac{m}{M} [/tex]

Where:

m: is the mass of CaF₂ = 16 g

M: is the molar mass of CaF₂ = 78.07 g/mol

[tex] n_{CaF_{2}} = \frac{m}{M} = \frac{16 g}{78.07 g/mol} = 0.205 moles [/tex]

Now, the enthalpy of formation of CaF₂ is:

[tex]\Delta H_{f} = \frac{q}{n_{CaF_{2}}} = \frac{-251 \cdot 10^{3} J}{0.205 moles} = -1224.4 kJ/mol[/tex]

Therefore, the enthalpy of formation of CaF₂ is -1224.4 kJ.

I hope it helps you!      

Answer:

C

Explanation:

I had this question and C is the right answer

One idea that Dalton taught about atoms was that all atoms of one type were identical in mass and properties.

An atom is defined as the smallest unit of matter which forms an element. Every form of matter whether solid,liquid , gas consists of atoms . Each atom has a nucleus which is composed of protons and neutrons and shells in which the electrons revolve.

The protons are positively charged and neutrons are neutral and hence the nucleus is positively charged. The electrons which revolve around the nucleus are negatively charged and hence the atom as a whole is neutral and stable due to presence of oppositely charged particles.

Atoms of the same element are similar as they have number of sub- atomic particles which on combination do not alter the chemical properties of the substances.

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Answer:

[tex][H^+]=0.000285[/tex]

[tex]pH=3.55[/tex]

Explanation:

In this, we can with the ionization equation for the hydrazoic acid ([tex]HN_3[/tex]). So:

[tex]HN_3~<->~H^+~+~N_3^-[/tex]

Now, due to the Ka constant value, we have to use the whole equilibrium because this is not a strong acid. So, we have to write the Ka expression:

[tex]Ka=\frac{[H^+][N_3^-]}{[HN_3]}[/tex]

For each mol of [tex]H^+[/tex] produced we will have 1 mol of [tex]N_3^-[/tex]. So, we can use "X" for the unknown values and replace in the Ka equation:

[tex]Ka=\frac{X*X}{[HN_3]}[/tex]

Additionally, we have to keep in mind that [tex]HN_3[/tex] is a reagent, this means that we will be consumed. We dont know how much acid would be consumed but we can express a subtraction from the initial value, so:

[tex]Ka=\frac{X*X}{0.004-X}[/tex]

Finally, we can put the ka value and solve for "X":

[tex]2.2X10^-^5=\frac{X*X}{0.004-X}[/tex]

[tex]2.2X10^-^5=\frac{X^2}{0.004-X}[/tex]

[tex]X= 0.000285[/tex]

So, we have a concentration of 0.000285 for [tex]H^+[/tex]. With this in mind, we can calculate the pH value:

[tex]pH=-Log[H^+]=-Log[0.000285]=3.55[/tex]

I hope it helps!

The [H+] and pH of a 0.0040 M hydrazoic acid solution is 0.000296648 and  3.527759

What information do we have?

Hydrazoic acid solution = 0.0040 M

Ka of hydrazoic acid = 2.20 × 10⁻⁵

We know that weak acids

[H+] = √( Ka × C)

[H+] = √( 2.2 × 10⁻⁵ × 0.0040)

[H+] = 0.000296648

So,

pH = -log [H+]

pH = -log [0.000296648]

Using log calculator

pH = 3.527759

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Answer:

[tex]m_{Ti}=13.0g[/tex]

Explanation:

Hello,

In this case, based on the given, we can infer that as titanium is hot and water cold, it cools down whereas the water is heated up, therefore, in terms of heat, we have that the heat lost by the titanium is gained by the water:

[tex]-Q_{Ti}=Q_{H_2O}[/tex]

That in terms of mass, specific heat and temperatures is:

[tex]-m_{Ti}Cp_{Ti}(T_2-T_{Ti})=m_{H_2O}Cp_{H_2O}(T_2-T_{H_2O})[/tex]

In such a way, for computing the mass of titanium, considering the heat capacity of water 4.18 J/g°C, we have:

[tex]m_{Ti}=\frac{m_{H_2O}Cp_{H_2O}(T_2-T_{H_2O})}{-Cp_{Ti}(T_2-T_{Ti})} \\\\m_{Ti}=\frac{50.0g*4.18\frac{J}{g\°C}(22.6-20.0)\°C}{-0.54\frac{J}{g\°C}*(22.6-100.0)\°C} \\\\m_{Ti}=13.0g[/tex]

Regards.

A process that is not a reversible reaction is frying an egg.

Reversible reactions are those reactions in which product will again change into the reactant.

Hence option (D) is correct.

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Answer:

A tetraquark in physics is an exotic meson composed of four valence quarks.

Explanation:

It has been suspected to be allowed by quantum chromodynamics, the modern story of strong interactions.

Hope it helps.

Answer:

filtration, drying, and weighing

Explanation:

The procedures that would facilitate the measurement of the actual yield of the solid would be filtration of the precipitate, drying of the precipitate, and weighing of the precipitate respectively.

The liquid/solid mixture resulting from the reaction can be separated by the process of filtration using a filter paper. The residue in the filter paper would be the solid while the filtrate would be the liquid portion of the reaction's product.

The residue can then be allowed to dry, and then weighed using a laboratory-grade weighing balance. The weight of the solid represents the actual yield of the solid.

Answer:

21.2 mL

Explanation:

Step 1: Write the balanced equation.

NaOH + HBr ⇒ NaBr + H₂O

Step 2: Calculate the reacting moles of HBr

25.0 mL of 0.0720 M hydrobromic acid react.

[tex]0.0250 L \times \frac{0.0720mol}{L} = 1.80 \times 10^{-3} mol[/tex]

Step 3: Calculate the reacting moles of NaOH

The molar ratio of NaOH to HBr is 1:1. The reacting moles of NaOH are 1/1 × 1.80 × 10⁻³ mol = 1.80 × 10⁻³ mol.

Step 4: Calculate the required volume of NaOH

[tex]1.80 \times 10^{-3} mol \times\frac{1,000mL}{0.0850mol} = 21.2 mL[/tex]

Explanation:

system at equilibrium, will the reaction shift towards reactants ~

--?'

2. (2 Pts) Consider the reaction N2(g) + 3H2(g) =; 2NH3(g). The production of ammonia is an

exothermic reaction. Will heating the equilibrium system increase o~e amount of

ammonia produced? . .co:(

3. (2 Pts) Consider the reaction N2(g) + 3H2(g) =; 2NH3(g). Ifwe use a catalyst, which way will

the reaction shift? ':'\

.1.+- w~t s~,H (o')l r'eo.c. e~ ei~i"liht-,·u.fn\ P~~,

4. (3 Pts) ff 1ven th e o £ 11 owmg d t a a £ or th ere action: A(g) + 2B(s) =; AB2(g)

Temperature (K) Kc

300 1.5x104

600 55 k ' pr, cl l<..J~

e- ~ r fee, ct o. ~ 1<

900 3.4 X 10-3

Is the reaction endothermic or exothermic (explain your answer)?

t d- IS o.,;r-. \4\a..i~1f't~ °the te.Y'il(lf1,:J'u.r-a a•~S. j lrvdu..c,,.) +~H~to{' '\

exothe-rnh't.-- ,.. ..,. (/.., ,~.

5. (4 Pts) Consider the reaction, N2(g) + 3H2(g) =; 2NH3(g). Kc= 4.2 at 600 K.

What is the value of Kc for 4 NH3(g) =; 2N2(g) + 6H2(g)

N ... ~l + 3 H~(ri ~ ~Nli3~) kl,= ~:s;H,J3 # 4. J..

~ ;)N~~) ~ ~ H ~) ~\-_ == [A!;J:t D~~Jb

J. [,v 1+3] ~

I

4,:i.~ = 0,05

5.00 mol of ammonia are introduced into a 5.00 L reactor vessel and when the equilibrium is reached, 1.00 mole remains. The concentration equilibrium constant is 17.3.

Initially, there are 5.00 mol of ammonia in a 5.00 L reactor vessel. The initial concentration of ammonia is:

[tex][NH_3]_i = \frac{5.00mol}{5.00L} = 1.00 M[/tex]

At equilibrium, there is 1.00 mole of ammonia in the 5.00 L vessel. The concentration of ammonia at equilibrium is:

[tex][NH_3]eq = \frac{1.00mol}{5.00L} = 0.200 M[/tex]

We can calculate the concentrations of all the species at equilibrium using an ICE chart.

       2 NH₃(g) ⇄ 3 H₂(g) + N₂(g)

I          1.00              0          0

C         -2x              +3x        +x

E       1.00-2x          3x          x

Since the concentration of ammonia at equilibrium is 0.200 M,

[tex]1.00-2x = 0.200\\\\x = 0.400 M[/tex]

The concentrations of all the species at equilibrium are:

[tex][NH_3] = 0.200 M\\[H_2] = 3x = 1.20 M\\[N_2] = x = 0.400 M[/tex]

The concentration equilibrium constant (Kc) is:

[tex]Kc = \frac{[H_2]^{2} [N_2]}{[NH_3]^{2} } = \frac{(1.20^{3})(0.400) }{0.200^{2} } = 17.3[/tex]

5.00 mol of ammonia are introduced into a 5.00 L reactor vessel and when the equilibrium is reached, 1.00 mole remains. The concentration equilibrium constant is 17.3.

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Answer:

[tex]m_{Si}=37.2gSi[/tex]

Explanation:

Hello,

In this case, for the undergoing balanced chemical reaction:

[tex]SiCl_4 + 2Mg \rightarrow Si + 2MgCl_2[/tex]

We must first identify the limiting reactant given the 225 g of SiCl4 and 101 g of Mg. Thus, we compute the available moles of SiCl4:

[tex]n_{SiCl_4}=225gSiCl_4*\frac{1molSiCl_4}{169.8963gSiCl_4}=1.324molSiCl_4[/tex]

Next, by using the 1:2 mole ratio between SiCl4 and Mg, we compute the moles of SiCl4 consumed by 101 g of Mg:

[tex]n_{SiCl_4}^{consumed}=101gMg*\frac{1molMg}{24.3050gMg} *\frac{1molSiCl_4}{2molMg} =2.08molSiCl_4[/tex]

Thus, since less moles of SiCl4 are available, we can infer it is the limiting reactant whereas the Mg is in excess. In such a way, the produced grams of Si are computed considering the 1:1 molar ratio between SiCl4 and Si:

[tex]m_{Si}=1.324molSiCl_4*\frac{1molSi}{1molSiCl_4} *\frac{28.0855gSi}{1molSi} \\\\m_{Si}=37.2gSi[/tex]

Best regards.

Answer:

See figure 1

Explanation:

In this question, we have to start with the protonation of the double bond. In carvone we have two double bonds, so, we have to decide first which one would be protonated.

The problem states that the terminal alkene is the one that would is protonated. Therefore, we have to do the protonation in the double bond at the bottom to produce the carbocation number 1. Then, a hydride shift takes place to produce the carbocation number 2. A continuation, an elimination reaction takes place to produce the conjugated diene. Then the diene is protonated at the carbonyl group and with an elimination reaction of an hydrogen in the alpha carbon we can obtain carvacol.

Answer:

Explanation:

KHT is a salt which ionises in water as follows

KHT ⇄ K⁺ + HT⁻

Solubility product Kw= [ K⁺ ] [ HT⁻ ]

product of concentration of K⁺ and HT⁻ in water

In KCl solution , the solubility product of KHT will be decreased .

In KCl solution , there is already presence of K⁺  ion in the solution . So

in the equation  

[ K⁺ ] [ HT⁻ ]  = constant

when K⁺ increases [ HT⁻ ] decreases . Hence less of KHT dissociates due to which its  solubility decreases . It is called common ion effect . It is so because here the presence of common ion that is K⁺ in both salt to be dissolved and in solvent , results in decrease of solubility of the salt .

Answer:

OH⁻ is the Bronsted-Lowry base.

Explanation:

A Bronsted-Lowry base is a substance that accepts protons. In the chemical equation, OH⁻ accepts a proton from H₂PO₄⁻ to become H₂O. H₂PO₄⁻ would be a Bronsted-Lowry acid because it donates a proton to OH⁻ and becomes HPO₄²⁻.

Hope that helps.

Answer:

The pressure of the gas sample will be  0.954 atm.​

Explanation:

Boyle's law states that the pressure of a gas in a closed container is inversely proportional to the volume of the container, when the temperature is constant. That is, if the pressure increases, the volume decreases; conversely if the pressure decreases, the volume increases.

Boyle's law is expressed mathematically as:

Pressure * Volume = constant

o P * V = k

To determine the change in pressure or volume during a transformation at constant temperature, the following is true:

P1 · V1 = P2 · V2

That is, the product between the initial pressure and the initial volume is equal to the product of the final pressure times the final volume.

In this case:

Replacing:

0.609 atm* 19.9 L= P2* 12.7 L

Solving:

[tex]P2=\frac{0.609 atm* 19.9 L}{12.7 L}[/tex]

P2= 0.954 atm

The pressure of the gas sample will be  0.954 atm.​

Answer:

The initial temperature was  [tex]36.4^\circ \:C[/tex]

Explanation:

[tex]\Delta t=\frac{q}{m\cdot C_s}=\frac{5.83\times10^5}{2190\times 4.184}\\\\=63.6^\circ\:C[/tex]

The temperature difference [tex]=100-63.6=36.4^\circ\:C[/tex]

Best Regards!

Answer:

41L

Explanation:

Of a 4.8x10⁻⁵M mercury (II) iodide that contains 900mg of mercury (II) iodide. 2 significant digits

Molarity, M, is an unit of concentration in chemistry defined as the ratio between moles of solute (Mercury (II) iodide in this case) per Liter of solution.

A 4.8x10⁻⁵M solution contains 4.8x10⁻⁵ moles of solute per liter.

Now, 900mg = 0.900g of mercury (II) iodide (Molar mass: 454.4g/mol) are:

0.900g × (1mol / 454.4g) = 1.98x10⁻³moles of HgI₂

If in 1L there are 4.8x10⁻⁵ moles of HgI₂, There are 1.98x10⁻³moles of HgI₂ in:

1.98x10⁻³moles of HgI₂ ₓ (1L / 4.8x10⁻⁵moles) =

Answer:

Explanation:

CH₃CHOHCOOH    ⇄    CH₃CHOHCOO⁻   +    H⁺

ionisation constant = 1.36 x 10⁻⁴ .

molecular weight of lactic acid = 90 g

moles of acid used = 20 / 90

= .2222

it is dissolved in one litre so molar concentration of lactic acid formed

C = .2222M

Let n be the fraction of moles ionised  

CH₃CHOHCOOH    ⇄    CH₃CHOHCOO⁻   +    H⁺

C  - nC                                          nC                  nC

By definition of ionisation constant Ka

Ka = nC x nC / C - nC

= n²C ( neglecting n in the denominator )

n² x .2222 = 1.36 x 10⁻⁴

n = 2.47  x 10⁻²

nC = 2.47  x 10⁻² x .2222

= 5.5 x 10⁻³

So concentration of hydrogen or hydronium ion = 5.5  x 10⁻³ g ion per litre .

The concentration of hydrogen or hydronium ion = 5.5  x 10⁻³ g ion per liter .

Ionization of lactic acid can be represented as:

CH₃CHOHCOOH⇄ CH₃CHOHCOO⁻  + H⁺

Given:

ionization constant = 1.36 x 10⁻⁴

mass= 20.0 g

Now, Molecular weight of lactic acid = 90 g

[tex]\text{Number of moles}=\frac{20}{90} =0.22mol[/tex]

It is dissolved in 1.00L so molar concentration of lactic acid formed will be

C = 0.22M

Consider "n" to be the fraction of moles ionized  

CH₃CHOHCOOH    ⇄    CH₃CHOHCOO⁻   +    H⁺

C  - nC                                          nC                  nC

By definition of ionization constant Ka

[tex]K_a =\frac{nC*nC}{C-nC}[/tex]

[tex]K_a= n^2C[/tex] ( neglecting n in the denominator )

On substituting the values we will get:

[tex]n^2 *0.22 = 1.36 *10^{-4}\\\\n = 2.47 * 10^{-2}[/tex]

To find the concentration of hydronium ion in the solution,

[tex]nC = 2.47 *10^{-2} *0.22\\\\nC= 5.5 * 10^{-3}[/tex]

So, concentration of hydrogen or hydronium ion = 5.5  x 10⁻³ g ion per liter.  

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Answer:

See explanation

Explanation:

In molecules with the same number of electron groups but different molecular geometries, the bond angles differ markedly owing to the presence of lone pairs on the central atom. Recall that lone pairs of electrons take up more space around the central atom and causes more repulsion thus squeezing the bond angle and making it less than the value expected on the basis of the Valence Shell Electron Pair Repulsion Theory.

As the number of bonding groups increases, the bond angle increases since the repulsion due to lone pairs of electrons is being progressively removed by increase in the number of bonding groups.

For 5 electron group molecules, the axial groups are oriented at a bond angle of 90° while the equatorial groups are oriented at a bond angle of 120°. In the presence of lone pairs, the equatorial bonds are removed because the equatorial bonds often have a greater bond length than the axial bonds.

In the tetrahedral geometry, four groups are bonded to the central atom while in a bent molecular geometry, only two groups are bonded to the central atom with two lone pairs present in the molecule.

Answer:

272.33 J/Kg°C

Explanation:

Data obtained from the question include the following:

Weight of metal = 30 N

Heat used (Q) = 1.25×10⁴ J

Change in temperature (ΔT) = 15.0 °C.

Specific heat capacity (C) =..?

Next, we shall determine the mass of the metal.

The mass of the metal can be obtained as follow:

Weight (W) = mass (m) x acceleration due to gravity (g)

W = mg

Weight of metal = 30 N

Acceleration due to gravity = 9.8 m/s²

Mass (m) =..?

W = mg

30 = m x 9.8

Divide both side by 9.8

m = 30/9.8

m = 3.06 Kg

Finally, we shall determine the specific heat capacity of the metal as show below:

Heat used (Q) = 1.25×10⁴ J

Change in temperature (ΔT) = 15.0 °C.

Mass (m) = 3.06 Kg

Specific heat capacity (C) =..?

Q = mCΔT

1.25×10⁴ = 3.06 x C x 15

Divide both side by 3.06 x 15

C = (1.25×10⁴) / (3.06 x 15)

C = 272.33 J/Kg°C

Therefore, the specific heat capacity of metal is 272.33 J/Kg°C.

Answer:

[tex]\large \boxed{\text{D. 23.34 min}}[/tex]

Explanation:

1. Find the order of reaction

Use information from the graph to find the order.

If a plot of ln[A] vs time is linear, the reaction is first order and the slope = -k.

2. Find the  half-life

[tex]k = \dfrac{\ln2}{ t_{\frac{1}{2}}}\\\\k = \text{-slope} = -(-2.97 \times 10^{-2} \text{ min}^{-1}) =2.97 \times 10^{-2} \text{ min}^{-1} \\ t_{\frac{1}{2}} =\dfrac{\ln 2}{k} = \dfrac{\ln 2}{2.97 \times 10^{-2}\text{ min}^{-1}} =\textbf{23.34 min}\\\\\text{The half-life is $\large \boxed{\textbf{23.34 min}}$}[/tex]

The half life of the reaction is 23.33 minutes.

We know that for a first order reaction;

ln[A]t = ln[A]o  - kt

A plot of ln[A]t  against time (t) will yield a straight line graph with a slope of -k.

From the question, the slope is -2.97 x 10-2 min-1.

So, -2.97 x 10-2 min-1 = - k

k = 2.97 x 10-2 min-1

The half life of a first order reaction is obtained from;

t1/2 = 0.693/k

t1/2 = 0.693/2.97 x 10-2 min-1

t1/2 = 23.33 minutes

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Answer:

THE FRACTION OF THE SAMPLE REMAINING AFTER THREE HALF LIVES IS 0.125 OR 125/1000

Explanation:

A radioactive isotope of mercury decay to gold with a disintegration constant of 0.0108 h^-1

To calculate the fraction of sample remaining after three half life, we first calculate the half life of the decay.

Half life = ln 2 / Y

Y = disintegration constant

So therefore,

half life = ln 2 / 0.0108

half life = 0.693 / 0.0108

half life = 64.18 hours.

So a decay occurs after 64.18 hours.

To calculate the fraction remaining after 3 half life:

N(t) = N(o) e ^-Yt

where t = 3 half life

So, N / No = e^-Y ( 3 t1/2)

Since t 1/2 = ln 2 / Y, so we can re-write the formula as:

Nt / No = e^-Y ( 3 ln 2/ Y)

Nt / No = e^-3 ln2

Nt / No = e^-3 * 0.693

Nt / No = e^-2.079

Nt / No = 0.125

So the fraction of the sample remaining after 3 half lives is 125/ 1000 or 0.125

The given question is incomplete, the complete question is:

A chemist adds 200.0 ml of a 0.52M barium chlorate (Ba(CIO3)2) solution to a reaction flask. Calculate the mass in grams of barium chlorate the chemist has added to the flask. Round your answer to significant digits.

Answer:

The correct answer is 32 grams.

Explanation:

Based on the given solution, the molarity of barium chlorate solution given is 0.52 M, this shows that the solution will comprise 0.52 moles in 1 L or 1000 ml of the solution.  

Therefore, in 200 ml, it will comprise 0.52/1000 × 200 moles of Ba(ClO₃)₂,  

= 0.52/1000 × 200 = 0.104 moles

The molecular mass of Ba(ClO₃)₂ is 304.23 gram per mole

So, the mass of Ba(ClO₃)₂ in 0.104 moles will be,  

= 304.23 g/mol × 0.104

= 31.639 grams or 32 grams.  

Answer:

The hydrogen can be gotten from the added Acid or water during "workup".

Explanation:

Basically we can say that the reaction describe in this question is a Reduction reaction because of the chemical compound called sodium borohydride. In the reaction described above we can see that there is a Reduction of ketone to alcohol by the compound; sodium borohydride.

For the reduction Reaction to occur the C-O bond must break so as to enable the formation of O-H bond and C-H bond.

So, "the reaction mixture is treated with water and H2SO4 to produce the desired alcohol", thus, the oxygen will definitely pick up the hydrogen from H2SO4 or H2O.

Answer:

D

Explanation:

Answer:

See figure 1

Explanation:

For this question, we have to remember that a hydrogen bond is an interaction in which we have a partial attraction between a positive dipole and a negative dipole, and in this attraction, we have in the middle a hydrogen atom.

In this interaction, we can have a donor (positive dipole) or a receptor (negative dipole). The receptor is a heteroatom (an atom different to carbon or hydrogen) with high electronegativity. The donor is usually hydrogen atom bonded to the heteroatom.  

I hope it helps!

Answer:

Acidic proteins are proteins that move faster than serum albumin on zone electrophoresis (starch or acrylamide gel) and which bind most strongly to the basic ion exchangers used in protein chromatography.

Basic protein is a late gene product associated with the viral DNA within the nucleocapsid. The harnessing of this promoter allows the expression of foreign genes at earlier times than those using the very late phase promoters of the polyhedron and p10 genes.

Answer:

Acidic proteins are proteins that move faster than serum albumin on zone electrophoresis (starch or acrylamide gel) and which bind most strongly to the basic ion exchangers used in protein chromatography.

Basic protein is a late gene product associated with the viral DNA within the nucleocapsid. The harnessing of this promoter allows the expression of foreign genes at earlier times than those using the very late phase promoters of the polyhedron and p10

Explanation:

Answer:

Explanation:

Work function of potassium = 2.29 eV = 3.67 X 10⁻¹⁹ J

So the minimum energy of photon must be equal to 3.67 X 10⁻¹⁹ J .

energy of photon of wavelength λ = hc / λ

where h = 6.67  x 10⁻³⁴

c = 3 x 10⁸

Putting the values in the equation above

6.67  x 10⁻³⁴  x  3 x 10⁸ / λ =  3.67 X 10⁻¹⁹

λ  = 6.67  x 10⁻³⁴  x  3 x 10⁸ /  3.67 X 10⁻¹⁹

= 5.452 x 10⁻⁷

= 5452 x 10⁻¹⁰ m

= 5452 A .