The standard redox potentials of isolated components of an electron transport chain in a cyanobacterium are found to be as follows:
Complex A: standard redox potential: -100 mV
Complex B: standard redox potential: -780 mV
Complex C: standard redox potential: +510 mV
Complex D: standard redox potential: +310 mV
Plastocyanin: standard redox potential: +360 mV
Which complex will likely have a binding site with high affinity for reduced plastocyanin?
A. Complex A.
B. Complex B.
C. Complex C.
D. Complex D.

Answers

Answer 1

Answer:

B. Complex B.

Explanation:

Complex B will have binding site with high affinity for reduced plastocyanin due to greater redox potential. The high number of redox potential will will transport electron chain in cyanobacteria.

Answer 2

Redox potential is the measure of the electron gain or loss to the electrode. For reduced plastocyanin complex B will have the highest affinity.

What is electron affinity?

Electron affinity is the energy released when the atom gets attached to the atom or other molecule. The high number of redox potential increases the electron transport in the cell.

The greater the redox potential more will be its tendency to show electron affinity. To bond with reduced species, the oxidized species must have greater redox potential.

Therefore, option B. complex b will have the highest affinity.

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Related Questions

How many electrons are shown in the following electron
configuration: 1s22s22p63s 23p64s23d104p65s24d105p66s2 ?
Express your answer numerically as an integer.

Answers

Answer:

1s22s22p6

Explanation:

Neon is an element in the periodic table and has an atomic number of 10, which means it has 10 protons in its nucleus and thus since the number of protons and electrons is the same then it has 10 electrons.

Therefore, it has 2 electrons in the first energy shell and 8 electrons in the second energy shell. To elaborate further, the first shell has a single s-sub shell that contains a single s-orbital that can hold two electrons. The second energy shell has a single s-sub-shell whose s-orbital will occupy 2 electrons, and also has a p-orbital which can hold 6 electrons, making the second shell to have 8 electrons.

Which of the following are examples of physical properties of ethanol? Select all that apply.

The boiling point is 78.37°C

It is a clear, colorless liquid

It is flammable

It is a liquid at room temperature

Answers

Ethanol is: flammable, liquid at room temperature, the boiling point is 78.37 ° C.

An endothermic reaction will start when the required
energy is received from the environment or solution.
AH
activation
thermal
kinetic

Answers

Answer:

A: ΔH

Explanation:

Endothermic reactions are this that occur as a result of absorption of heat energy from the surroundings by the reactants to form new products.

Thus, we can say it is one with an increase in enthalpy (ΔH) of the system.

Thus, option A is correct.

2.50 L of 0.700 M phosphoric acid reacts with 5.25 moles of sodium hydroxide. How many moles of hydrogen ions will completely neutralize the moles of hydroxide ions present in this amount of sodium hydroxide? a) 0.583 b) 1.75 c) 3.00 d) 15.75 e) 5.25

Answers

Answer:

5.25 moles of protons. Option e

Explanation:

Reaction between phosphoric acid and sodium hydroxide is neutralization.

We can also say, we have an acid base equilibrium right here:

H₃PO₄  +  3NaOH →  Na₃PO₄  +  3H₂O

Initially we have 5.25 moles of base.

We have data from the acid, to state its moles:

M = mol/L, so mol = M . L

mol = 1.75 moles of acid

If we think in the acid we know:

H₃PO₄  →  3H⁺  +  PO₄⁻³

We know that 1 mol of acid can give 3 moles of protons (hydrogen ions)

If we have 1.75 moles of acid, we may have

(1.75 . 3) /1 = 5.25 moles of protons

These moles will be neutralized by the 5.25 moles of base

H₃O⁺  +  OH⁻  ⇄  2H₂O     Kw

In a titration of a weak acid and a strong base, we have a basic pH

DATA SHEET p 45. TRIAL 1 TRIAL 2 1. Mass of the ground pretzel 1.00 gram 1.03 g 2. Initial volume of the AgNO3 solution 0.00 mL 9.10 mL 3. Final volume of the AgNO3 solution 9.10 mL 17.25 mL 4. Volume of AgNO3 solution used 9.10 mL 8.15 mL Line 3 – Line 2 5. Volume of AgNO3 solution in liters _____ L _____ L 6. Molarity of AgNO3 solution 0.01 M 0.01 M (given) 7. Number of moles of AgNO3 ______ mol _____ mol (Line 5 × Line 6) 8. Number of mol of NaCl present in pretzel ______ mol _____ mol (Line 7) number of mol NaCl = number of mol AgNO3 9. Mass of NaCl present in the titrated sample ______ gram _____ gram (Line 8) × 58.5 g/mol

Answers

Answer:

1. 1.00 gm

2. 50 ml

3. 38.93 ml

4. 11.07 ml

5. 0.01107 L

6. 0.010 moles / L

7. 0.0001107 moles

8. 0.0001107 moles

9. 0.00647042 grams

Explanation:

Silver nitrate can react with various compounds to form different products. The weight of products may be different from the original solution introduced due to combustion reaction, as heat energy is released during the chemical process.

0.50 g of hydrogen chloride (HCl) is dissolved in water to make 4.0 L of solution. What is the pH of the resulting hydrochloric acid solution

Answers

Explanation:

Given the mass of HCl is ---- 0.50 g

The volume of solution is --- 4.0 L

To determine the pH of the resulting solution, follow the below-shown procedure:

1. Calculate the number of moles of HCl given by using the formula:

[tex]number of moles of a substance=\frac{given mass of the substance}{its molecular mass}[/tex]

2. Calculate the molarity of HCl.

3. Calculate pH of the solution using the formula:

[tex]pH=-log[H^+][/tex]

Since HCl is a strong acid, it undergoes complete ionization when dissolved in water.

[tex]HCl(aq)->H^+(aq)+Cl^-(aq)[/tex]

Thus, [tex][HCl]=[H^+][/tex]

Calculation:

1. Number of moles of HCl given:

[tex]number of moles of a substance=\frac{given mass of the substance}{its molecular mass}\\=0.50g/36.5g/mol\\=0.0137mol[/tex]

2. Concentration of HCl:

[tex]Molarity of HCl=\frac{number of moles of HCl}{its molar mass}\\=\frac{0.0137 mol}{4.0 L} \\= 0.003425 M[/tex]

3. pH of the solution:

[tex]pH=-log[H^+]\\=-log(0.003425)\\=2.47[/tex]

Hence, pH of the given solution is 2.47.

For a gas sample containing equimolar amounts of carbon monoxideand heliumat 300 K, heliumhas _____________average speed and _____________ average kinetic energy compared tocarbon monoxidegas.a.a lower; the same b. the same; the same c. a higher; the same d. a higher; higher

Answers

Answer:

Option C (a higher; the same) is the appropriate response.

Explanation:

Given:

Temperature,

T = 300 K (both [tex]N_2[/tex] and [tex]H_2[/tex])

As we know,

Average speed of a molecule,

⇒ [tex]\bar v=\sqrt{\frac{8RT}{\pi M} }[/tex]

Thus, the average speed of [tex]N_2[/tex] will be lower as its molar mass is greater than [tex]H_2[/tex].

Now,

⇒ [tex]Average \ kinetic \ energy = \frac{3}{2} \ KT[/tex] (not depend on molar mass)

Hence, it will be the same.

The other three alternatives aren't connected to the scenario given. So the above is the correct answer.

explain in brief how some bacteria cause tooth cavities​

Answers

Answer:

because some bacteria change the ph of the teeth and leads to tooth cavitiew

The cause of cavities is acid from bacteria dissolving the hard tissues of the teeth (enamel, dentin and cementum). The acid is produced by the bacteria when they break down food debris or sugar on the tooth surface.

When butane reacts with Br2 in the presence of Cl2, both brominated and chlorinated products are obtained. Under such conditions, the usual selectivity of bromination is not observed. In other words, the ratio of 2-bromobutane to 1-bromobutane is very similar to the ratio of 2-chlorobutane to 1-chlorobutane. Can you offer and explanation as to why we do not observe the normal selectivity expected for bromination

Answers

Answer:

Bromine radical formation is carried out in the presence of Br₂ and Cl₂ causing the normal selectivity not to be observed ( this causes the difference in activation energy to be reduced )

Explanation:

Why the normal selectivity expected for bromination is not observed

On the basis of selectivity and applying the Arrhenius equation the greater the difference between the activation energies the more the selectivity.

as seen in the formation of primary and secondary radicals in the Bromine radical formation. this difference is caused mainly by the propagation step ( exothermic ) . But the main reason why the the usual selectivity of bromination is not observed is because it Bromine radical formation is carried out in the presence of Br₂ and Cl₂ ( this causes the difference in activation energy to be reduced )

5. A beam of photons with a minimum energy of 222 kJ/mol can eject electrons from a potassium surface. Estimate the range of wavelengths of light that can be used to cause this phenomenon. Show your calculations with units of measure (dimensional analysis) and briefly explain your reasoning.

Answers

Answer: The range of wavelengths of light that can be used to cause given phenomenon is [tex]8.953 \times 10^{21} m[/tex].

Explanation:

Given: 222 kJ/mol (1 kJ = 1000 J) = 222000 J

Formula used is as follows.

[tex]E = \frac{hc}{\lambda}[/tex]

where,

E = energy

h = Planck's constant = [tex]6.625 \times 10^{-25} Js[/tex]

c = speed of light = [tex]3 \times 10^{8} m/s[/tex]

Substitute the values into above formula as follows.

[tex]E = \frac{hc}{\lambda}\\222000 J = \frac{6.625 \times 10^{-34}Js \times 3 \times 10^{8} m/s}{\lambda}\\\lambda = 8.953 \times 10^{21} m[/tex]

Thus, we can conclude that the range of wavelengths of light that can be used to cause given phenomenon is [tex]8.953 \times 10^{21} m[/tex].

2, classify the following molecules as polar or non polar.
A,CH4 B,CHcl C,Co2 D,H2O2 E,BCl3 F,H2S​

Answers

A. CH4= NON POLAR

B. CH3cl= POLAR

C. CO2= NON POLAR

D.  H2O2= POLAR

E. BCl3= NON POLAR

F. H2S​= SLIGHTLY POLAR

Which statement is true with respect to standard reduction potentials?
SRP values that are greater than zero always represent a reduction reaction.
SRP values that are less than zero always represent a reduction reaction.
Half-reactions with SRP values greater than zero are spontaneous.
Half-reactions with SRP values greater than zero are nonspontaneous.

Answers

Answer:

C). Half-reactions with SRP values greater than zero are spontaneous.

Explanation:

SRPs or Standard Reduction Potentials are characterized as the ability of a probable distinction among the anode and cathode of a usual/standard cell. It aims to examine the capacity of chemicals to reduce themselves.

The third statement asserts a true claim regarding the SRPs(Standard Reduction Potentials) that the 'half-reactions which take place with the SRP possesses the values higher than zero and they are unconstrained.' The other statements are incorrect as they either show the estimation of SRPs more than 0 or display them as being restricted. Thus, option C is the correct answer.

Identify the oxidation half-reaction for this reaction:
Fe(s) + 2HCl(aq) → FeCl2(aq) + H2(g)
A. Fe2+ + 2e → Fe(s)
O B. H2(g) → 2H+ + 2e
O C. Fe(s) → Fe2+ + 2e
O D. 2H+ + 2e → H2(9)

Answers

Answer:

Fe(s)->Fe2+2e-

Explanation:

A.p.e.x

The oxidation half-reaction for the given reaction is Fe(s) → Fe²⁺ + 2e⁻ Hence, Option (C) is correct

What is Oxidation reaction ?

Oxidation reaction is a chemical reaction which can be described as follows ;

Addition of oxygen Removal of hydrogen Loss of ElectronAddition of electronegative atomRemoval of Electropositive element

In the given reaction ;

Fe(s) + 2HCl(aq) → FeCl₂(aq) + H₂(g)

Fe at RHS got converted to Fe²⁺ state at LHS which shows the gain of electron by Fe with in the reaction.

Therefore,

The oxidation half-reaction for the given reaction is Fe(s) → Fe²⁺ + 2e⁻ Hence, Option (C) is correct

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Solutions of Cu2+ turn blue litmus red because of the equilibrium: Cu(H2O)62+(aq) + H2O(l) ↔ Cu(H2O)5(OH)+(aq) + H3O+(aq) for which Ka = 1.0 x 10-8. Calculate the pH of 0.10 M Cu(NO3)2(aq).

Answers

Answer: The pH of 0.10 M [tex]Cu(NO_{3})_{2}(aq)[/tex] is 4.49.

Explanation:

Given: Initial concentration of [tex]Cu(H_{2}O)^{2+}_{6}[/tex] = 0.10 M

[tex]K_{a} = 1.0 \times 10^{-8}[/tex]

Let us assume that amount of [tex]Cu(H_{2}O)^{2+}_{6}[/tex] dissociates is x. So, ICE table for dissociation of  [tex]Cu(H_{2}O)^{2+}_{6}[/tex]  is as follows.

                               [tex]Cu(H_{2}O)^{2+}_{6} \rightleftharpoons [Cu(H_{2}O)_{5}(OH)]^{+} + H_{3}O^{+}[/tex]

Initial:                       0.10 M                       0                       0

Change:                    -x                              +x                      +x

Equilibrium:           (0.10 - x) M                    x                        x

As the value of [tex]K_{a}[/tex] is very small. So, it is assumed that the compound will dissociate very less. Hence, x << 0.10 M.

And, (0.10 - x) will be approximately equal to 0.10 M.

The expression for [tex]K_{a}[/tex] value is as follows.

[tex]K_{a} = \frac{[Cu(H_{2}O)^{2+}_{6}][H_{3}O^{+}]}{[Cu(H_{2}O)^{2+}_{6}]}\\1.0 \times 10^{-8} = \frac{x \times x}{0.10}\\x = 3.2 \times 10^{-5}[/tex]

Hence, [tex][H_{3}O^{+}] = 3.2 \times 10^{-5}[/tex]

Formula to calculate pH is as follows.

[tex]pH = -log [H^{+}][/tex]

Substitute the values into above formula as follows.

[tex]pH = -log [H^{+}]\\= - log (3.2 \times 10^{-5})\\= 4.49[/tex]

Thus, we can conclude that the pH of 0.10 M [tex]Cu(NO_{3})_{2}(aq)[/tex] is 4.49.

PLEASE HELP ASAP MOLES TO MOLECULES

Answers

Answer:

4.77mol is the correct answer

4.77 mol, is the answer:)

Si enfriamos mercurio de 100C. Calcular la cantidad de calor que se debe restar sabiendo que la masa de mercurio es de 1800gr

Answers

Answer:

I do not speak Spanish.

Explanation:

The reversible reaction: 2SO2(g) O2(g) darrow-tn.gif 2SO3(g) has come to equilibrium in a vessel of specific volume at a given temperature. Before the reaction began, the concentrations of the reactants were 0.060 mol/L of SO2 and 0.050 mol/L of O2. After equilibrium is reached, the concentration of SO3 is 0.040 mol/L. What is the equilibrium concentration of O2

Answers

Answer:

[tex][O_2]_{eq}=0.030M[/tex]

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to solve this problem by firstly writing out the mathematical expression for the concentration of oxygen at equilibrium, given the initial one and the change due to the reaction extent:

[tex][O_2]_{eq}=0.050M-x[/tex]

Whereas [tex]x[/tex] can be found considering the equilibrium of SO3:

[tex][SO_3]_{eq}=2x=0.040M[/tex]

Which means:

[tex]x=\frac{0.040M}{2} =0.020M[/tex]

Thus, the equilibrium concentration of oxygen gas turns out:

[tex][O_2]_{eq}=0.050M-0.020M=0.030M[/tex]

Regards!

A gas at 74°C is heated to 120°C so there is pressure reaches 1.79 ATM. What is its initial pressure?

Answers

Explanation:

here's the answer to your question

A molecule of acetone and a molecule of propyl aldehyde are both made from 3 carbon atoms, 6 hydrogen atoms, and 1 oxygen atom. The molecules differ in their arrangement of atoms. How do formulas for the two compounds compare? Both compounds have the same molecular formula, but have unique structural formulas. Both compounds have unique molecular formulas and structural formulas. Both compounds have the same structural formula, but have unique molecular formulas.

Answers

Explanation:

The structures of both acetone and propanal are shown below:

In the formula of propanal there is -CHO functional group at the end.

In acetone -CO- group is present in the middle that is on the second carbon.

The molecular formula is C3H6O.

Both have same molecular formula but different structural formulas.

Identify the highest energy molecular process that occurs when a molecule absorbs a microwave photon.

Answers

The question is incomplete, the complete question is;

Identify the highest energy molecular process that occurs when a molecule absorbs a microwave photon

a) electronic excitation

b) bond breakage

c) molecular vibration

d) molecular rotation

Answer:

molecular rotation

Explanation:

Microwaves are part of the electromagnetic spectrum. They are lower energy, lower frequency radiation.

When molecules absorb infrared radiation, they transition between the rotational states of the molecule.

Hence, the highest energy molecular process that occurs when a molecule absorbs a microwave photon is molecular rotation.

Determine whether the reaction will be spontaneous at high temperatures only, at low temperatures only, at all temperatures, or no temperatures. (HINT: Use your chemical sense and your real-world knowledge to predict the signs of delta Hrxn & delta Srxn)


4Fe(s) + 3O2(g) ----> 2Fe2O3(s) [rust]


Circle one:High T, Low T, All T, No T

Answers

Answer:

The rusting of iron is spontaneous at low temperatures.

Explanation:

The given chemical reaction is:

4Fe(s) + 3O2(g) ----> 2Fe2O3(s) [rust]

The rusting of iron is a chemical reaction in which iron reacts with oxygen in presence of moisture and forms iron oxide.

This reaction takes place in a faster rate when there is low temperatures in the atmosphere.

When temperature is low, the moisture in the atmosphere is more and hence, rate of rusting is more.

dentify the correct formula for the following ionic compounds. - sodium chloride - magnesium chloride - calcium oxide - lithium phosphide - aluminum sulfide - calcium nitride A. SCl B. LiP 3 C. AlS D. Li 3P E. CaN F. CaO G. Ca 3N 2 H. MgCl 2 I. NaCl J. CaO 2 K. CaN 2 L. LiP M. MnCl 2 N. Al 2S 3 O. AlS 3

Answers

Explanation:

The chemical formula of an ionic compound can be written by using the symbols of the respective cations and anions.

The overall charge on the molecule should be zero.

Hence, the total charge of cations=total charge of anions.

The symbols of the given molecules are shown below:

sodium chloride  ---- NaCl

magnesium chloride ---[tex]MgCl_2[/tex]

calcium oxide ---- CaO

lithium phosphide----[tex]Li_3P[/tex]

aluminum sulfide ----- [tex]Al_2S_3[/tex]

calcium nitride---- [tex]Ca_3N_2[/tex]

During the postabsorptive state, metabolism adjusts to a catabolic state.

a. True
b. False

Answers

Answer:

The postabsorptive state (also called the fasting state) occurs when the food is already digested and absorbed, and it usually occurs overnight, when you sleep (if you skip meals for some days, you will enter in this state).

The catabolic state is the metabolic breakdown of molecules into simpler ones, releasing energy (heat) and utilizable resources.

Now, when you are in a postabsorptive state, the glucose levels start to drop, then the body starts to depend on the glycogen stores, which are catabolized into glucose, this is defined as the start of the postabsorptive state.

So yes, as the postabsorptive states, catabolic processes start to happen, so the statement is true.

A sample of oxygen gas has a volume of 89.6 L at STP. How many moles of oxygen gas are present ?

Answers

Answer:

89,6/22,4 =4(mol)

Explanation:

There are approximately 1.089 moles of oxygen gas present in the sample at STP.

At STP (Standard Temperature and Pressure), the conditions are defined as follows:

Temperature (T) = 0 degrees Celsius = 273.15 Kelvin

Pressure (P) = 1 atmosphere (atm) = 101.325 kPa = 1013.25 hPa

Now, to find the number of moles of oxygen gas (O2) present in the sample, we can use the ideal gas law:

PV = nRT

Where:

P = pressure (in atm)

V = volume (in liters)

n = number of moles

R = ideal gas constant = 0.0821 L.atm/(mol.K)

T = temperature (in Kelvin)

Given:

V = 89.6 L (volume at STP)

T = 273.15 K (STP temperature)

Let's plug in the values and solve for n (number of moles):

n = PV / RT

n = (1 atm) × (89.6 L) / (0.0821 L.atm/(mol.K) × 273.15 K)

n = 1.089 moles

So, there are approximately 1.089 moles of oxygen gas present in the sample at STP.

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Poly(ethylene terephthalate) (PET), which has glass transition (Tg) and crystalline melting (Tm) temperature of 69 and 267 °C, respectively, can exist in a number of different states depending upon temperature and thermal history. Thus, it is possible to prepare materials that are semicrystalline with amorphous regions that are either glassy or rubbery and amorphous materials that are glassy, rubbery or melts. Consider a sample of PET cooled rapidly from 300 °C (state A) to room temperature. The resulting material is rigid and perfectly transparent (state B). The sample is then heated to 100 °C and maintained at this temperature, during which time is gradually becomes translucent (state C). It is then cooled to room temperature, where it is again observed to be translucent (state D).

Answers

Answer:

Following are the solution to the given points:

Explanation:

For point A:

The sample cooking (PET) is between 300°C and room temperature.Now in nature, the substance is exceedingly stiff.Samples of PET up to 100°C were heated and stayed on equal footing.Now it has cooled off the same sample below 100° C and we may see how it is again TRASNEPARENT in nature.

For point B:

In point 3, the mixture was added to 100°C, which implies that the granular material flows and deforms, enabling it to become elongated. This is termed solid-state crystalline such that grains are flexible, but this material contaminates numerous little crystalline that has spheres when we cool down in point  4 polymers. It forms therefore an unstructured solid, which then in point 4 is higher in particles and less pliable in orderly atoms.

For point C:

In point 2, the specimen gets forced at room temperature to organize a huge molecule in an ordinary and crystal fashion and therefore is transparent due to highly crystalline atoms in point 2 of the PET sample.

In point 4, however, we notice how amorphous, firm but not crystalline develops. It's why light tends to disperse over many cereal limits, since many microscopic crystallines, therefore dispersion, PET in point 4 is translucent.

it takes 513 kj to remove a mole of electrons from the atoms at the surface of a piece of metal. how much energy does it take to remove a single electron from n atom at the surface of the metal

Answers

Answer:

The right solution is "[tex]8.5\times 10^{-19} \ joule[/tex]".

Explanation:

As we know,

1 mole electron = [tex]6.023\times 10^{23} \ no. \ of \ electrons[/tex]

Total energy = [tex]513 \ KJ[/tex]

                     = [tex]513\times 1000 \ joule[/tex]

For single electron,

The amount of energy will be:

= [tex]\frac{513\times 1000}{(6.023\times 10^{23})}[/tex]

= [tex]8.5\times 10^{-19} \ joule[/tex]

Balance the following chemical equation.

CCl4 -> ___ C+ ___ Cl2

Answers

Answer:

Explanation:

CCl4 => C + 2Cl2

cual es la masa atomica del hidrogeno

Answers

El hidrógeno es el elemento químico de número atómico 1, representado por el símbolo H. Con una masa atómica de 1.00784 u  ​ es el más ligero de la tabla periódica de los elementos. Por lo general, se presenta en su forma molecular, formando el gas diatómico H₂ en condiciones normales.

Use the following key to classify each of the elements below in its elemental form:
A. Discrete atoms .. C. Metallic lattice
B. Molecules ... D. Extended, three-dimensional network
1. Magnesium
2. Nitrogen ...
3. Lithium
4. Potassium ...

Answers

Answer:

Magnesium - Metallic lattice

Nitrogen - Molecules

Lithium - Metallic lattice

Potassium - Metallic lattice

Explanation:

Metals exist in metallic lattices. In this lattice, metal ions are held together with a sea of electrons by strong electrostatic forces.

All metals possess this metallic lattice, hence; potassium, lithium and magnesium all consist of metal lattices.

Nitrogen is a nonmetal and consists of molecules of N2.

What is the major product in this reaction

Answers

Answer:

I think option A is right answer

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