Calculate the amount of ATP in kg that is turned over by a resting human every 24 hours. Assume that a typical human contains ~50g of ATP (Mr 505) and consumes ~8000 kJ of energy in food each day. The energy stored in the terminal anhydride bond of ATP under standard conditions is 30.6 kJmol-1. Assume also that the dietary energy is channeled through ATP with an energy transfer efficiency of ~50%.

Answers

Answer 1

Answer:

The correct answer is 66.35 kilograms.

Explanation:

Based on the data given in the question, the energy consumed by the body of a human being is 50%. Based on the given data, the energy consumed in a day is 8000 kJ, 50 percent is the energy transfer efficiency. Thus, the consumption of total energy is 4000 kJ, and for the transformation of ADP to ATP, the energy involved is 30.6 kJ per mole.  

Hence, the total ATP produced in the process is,  

ATP = 4000 kJ / 30.6 kJ/mol

= 130.7189 mol.  

Thus, with the energy transfer efficiency of 50 percent, the total moles of ATP produced is 130.7 mol.  

The mass of ATP can be calculated by using the formula,  

moles = mass/molecular mass

The molecular mass of ATP is 507.18 g per mol

Now by putting the values we get,  

mass of ATP = 130.7189 mol * 507.18 g/mol

= 66298.011 g or 66.298 kg

It is mentioned that human comprise 50 g of ATP or 0.05 kg of ATP. Therefore, the sum of the available ATP will be.  

= Total production of ATP + Total ATP available

= 66.298 kg + 0.05 kg

= 66.348 kg

Hence, the sum of the ATP that is turned over by a resting human in a day is 66.35 kg.  


Related Questions

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

Answers

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:

1. Which statement describes the particles of an ideal gas, based on the

kinetic molecular theory?*

O There are attractive forces between the particles.

O The particles move in circular paths.

O The collisions between the particles reduce the total energy of the gas.

О

The volume of the gas particles is negligible compared with the total volume of the

gas.

Answers

Answer:the volume of the gas particles is negligible compared with the total volume of the  gas.--D

Explanation:

According to the Kinetic Molecular Theory for ideal gases, it states that

--Gases are composed of larges  molecules which are in constant random motion  in a straight line

--The volume of the gas particles is negligible compared to the total volume in which the gas is contained.

-----The  Attractive and repulsive forces between gas molecules is insignificant ie There are no interactive forces.

----The collisions of the particles  are perfectly elastic and energyis  being transferred between the particles but the total energy remaining constant

From the  statements of the kinetic Molecular theory of ideal gases, it can be seen that the statement which describes the particles of an ideal gas is option D  which is The volume of the gas particles is negligible compared with the total volume of the  gas--- ---This gives the reason why  gases  can be compressed. Since there are no  inter molecular forces between them. The particles of an  ideal gas will move at the same random motion  resulting to high pressures, compressing the gas and making the volume negligible or insignificant.

A stock solution of HNO3 is prepared and found to contain 14.9 M of HNO3. If 25.0 mL of the stock solution is diluted to a final volume of 0.500 L, what is the concentration of the diluted solution

Answers

Answer:

[tex]0.745~M[/tex]

Explanation:

In this case, we have a dilution problem. So, we have to use the dilution equation:

[tex]C_1*V_1=C_2*V_2[/tex]

Now, we have to identify the variables:

[tex]C_1~=~14.9~M[/tex]

[tex]V_1~=~25~mL[/tex]

[tex]C_2~=~?[/tex]

[tex]V_2~=~0.5~L[/tex]

Now, we have different units for the volume, so we have to do the conversion:

[tex]0.5~L\frac{1000~mL}{1~L}=~500~mL[/tex]

Now we can plug the values into the equation:

[tex]C_2=\frac{14.9~M*25~mL}{500~mL}=0.745~M[/tex]

I hope it helps!

A filtration system continuously removes water from a swimming pool, passes the water through filters, and then returns it to the pool. Both pipes are located near the surface of the water. The flow rate is 15 gallons per minute. The water entering the pump is at 0 psig, and the water leaving the pump is at 10 psig.

A. The diameter of the pipe that leaves the pump is 1 inch. How much flow work is done by the water as it leaves the pump and enters the pipe?

B. The water returns to the pool through an opening that is 1.5 inches in diameter, located at the surface of the water, where the pressure is 1 atm. How much work is done by the water as it leaves the pipe and enters the pool?

C. "The system" consists of the water in the pump and in the pipes that transport water between the pump and the pool. Is the system at steady state, equilibrium, both, or neither?

Answers

Answer:

A .    [tex]\mathbf{W = 7133.2 \dfrac{ft. lb_f}{min} }[/tex]

B.   [tex]\mathbf{W = 4245.24 \dfrac{ft. lb_f}{min} }[/tex]

C.  The system is at steady state but not at equilibrium

Explanation:

Given that:

The volumetric flow rate of the water = 15 gallons per minute

The diameter of the pipe that leaves the pump is 1 inch.

A. The objective here is  to determine how much work flow is done by the water as it leaves the pump and enters the pipe

The work flow that is said to be done can  be expressed by the relation :

W = P × V

where;

P = pressure

V = volume

Also the given outlet pressure is the gauge pressure

The pressure in the pump P  is can now be expressed by the relation:

[tex]P_{absolute} = P_{guage} + P_{atmospheric}[/tex]

[tex]P_{absolute}[/tex] =  10 psig + 14.7 psig

[tex]P_{absolute}[/tex] =  24.7 psig

W =  P × V

W =  24.7 psig  ×  15 gal/min

[tex]W = (24.7 \ psig * \dfrac{\frac{lb_f}{in^2}}{psig}) * ( 15 \frac{gal}{min}* \dfrac{0.1337 \ ft^3}{1 \ gal }* \dfrac{144 \ in^2}{1 \ ft^2})[/tex]

[tex]\mathbf{W = 7133.2 \dfrac{ft. lb_f}{min} }[/tex]

Thus ; the rate of flow of work is said to be done by the water at [tex]\mathbf{W = 7133.2 \dfrac{ft. lb_f}{min} }[/tex]

B.

Given that :

The water returns to the pool through an opening that is 1.5 inches in diameter.

where the pressure is 1 atm.

Then ; the rate of work done  by the water as it leaves the pipe and enter the pool is as follows:

W =  P × V

W = 1 atm  ×  15 gal/min

[tex]W = 1 \ atm * ( 15 \frac{gal}{min}* \dfrac{0.1337 \ ft^3}{1 \ gal }* \dfrac{144 \ in^2}{1 \ ft^2})[/tex]

[tex]\mathbf{W = 4245.24 \dfrac{ft. lb_f}{min} }[/tex]

Thus ; the rate of flow of work done by the water leaving the pipe and enters into the pool is  at [tex]\mathbf{W = 4245.24 \dfrac{ft. lb_f}{min} }[/tex]

C.

We can consider the system to be at steady state due to the fact that; the data given for the flow rate and pressure doesn't reflect upon the change in time in the space between the pump and the pool.

On the other-hand the integral factor why the system is not at equilibrium is that :

the pressure leaving the pipe is different from that of the water at the surface of the pool as stated in the question.

Which compound would you expect to be least soluble in water? Explain.
a. CCl4
b. CH3Cl
c. NH3
d. KF

Answers

Answer:  a.CCl4 aka carbon tetrachloride

Explanation:

ionic compounds and polar molecules can be dissolved in water which is a polar solvent.

choice d (KF) is a salt (an ionic compound) and can be dissolved in water /(K+ and F- ions would be formed in water).

choice c (NH3 or ammonia) is a very polar molecule and thus can be dissolved in water(Hydrogen bonding).

choice b (CH3Cl) is slightly polar because the atoms surrounding the central carbon atom are different(3 H atoms and 1 chlorine atom) and can be dissolved in water(Dipole-dipole interaction).

choice a is nonpolar and cannot be dissolved in water.

Which accurately labels the lysosome?

Answers

Answer:

One of the organelles in eukaryotic cells that carry out digestion and waste removal.

Answer:

It's X

Explanation:

A gaseous hydride of Nitrogen
contains its own volume of Nitrogen
and twice its volume of Hydrogen
and has vapour density 16. The
formula of the hydride is.
Select one:
a. NH2
b. NH3
c. N3H
• d. N2H4

Answers

Answer:

N2H4

Explanation:

A hydride is a binary compound of hydrogen and another element. Binary compounds contain only two atoms. We have to x-ray the hydrides of nitrogen given in the question in order to make our choice. Remember that we were told that that the hydride contains its own volume of nitrogen and twice its volume of hydrogen.

Now consider the hydride N2H4.

N2H4(g) -----> N2(g) + 2H2(g)

The volume ration of nitrogen gas to hydrogen gas in N2H4 is 1:2.

The molecular mass of the compound is;

N2H4= 2(14) + 4(1)= 28+4= 32

Since

molecular mass= 2 vapour density

Vapour density= molecular mass/2

Vapour density= 32/2

Vapour density = 16

Therefore the hydride of nitrogen referred to in the question is N2H4

A certain substance X condenses at a temperature of 120.7 degree C. But if a 500, g sample of X is prepared with 55.4 g of urea (NH_2)_2 CO) dissolved in it, the sample is found to have a condensation point of 125.2 degree C instead. Calculate the molal boiling point elevation constant K_b of X. Round your answer to 2 significant digits.

Answers

Answer: The molal boiling point elevation constant [tex]k_b[/tex] of X is [tex]2.4^0C/m[/tex]

Explanation:

Formula used for Elevation in boiling point :

[tex]\Delta T_b=k_b\times m[/tex]

or,

[tex]T_b-T^o_b=i\times k_b\times \frac{w_2\times 1000}{M_2\times w_1}[/tex]

where,

[tex]T_b-T^o_b =(125.2-120.7)^0C=4.5^0C[/tex]

[tex]k_b[/tex] = boiling point constant  = ?

m = molality

[tex]w_2[/tex] = mass of solute (urea) = 55.4 g

[tex]w_1[/tex] = mass of solvent  X =  500 g

[tex]M_2[/tex] = molar mass of solute (urea) = 60 g/mol

Now put all the given values in the above formula, we get:

[tex]4.5^oC=k_b\times \frac{55.4g\times 1000}{60\times 500g}[/tex]

[tex]k_b=2.4^0C/m[/tex]

Thus the molal boiling point elevation constant [tex]k_b[/tex] of X is [tex]2.4^0C/m[/tex]

What is the temperature at which the substance can be both in the solid and the liquid phase?

Answers

Answer: Gas–liquid–solid triple point

The single combination of pressure and temperature at which liquid water, solid ice, and water vapor can coexist in a stable equilibrium occurs at approximately 273.1575 K (0.0075 °C; 32.0135 °F) and a partial vapor pressure of 611.657 pascals (6.11657 mbar; 0.00603659 atm).

Explanation:

It represents the equilibrium between the liquid and gas phases. The point on this curve where the vapor pressure is 1 atm is the normal boiling point of the substance. The vapor-pressure curve ends at the critical point (B), which is at the critical temperature and critical pressure of the substance.

CHEMISTRY HELP!
using only the periodic table, determine the charge on the ion that is formed by arsenic.
The ion charge is:
a. -3
b. -2
c. -1
d. 0
e. +1
f. +2
g. +3

also what is it for elements lithium and strontium?

Answers

Answer:

A

Explanation:

Arsenic is in the same group as Nitrogen - group 5. They all have 5 valence electrons in their outermost shell. To achieve its most stable state - 8 valence electrons (octet rule - elements are most stable when the entire shell is filled) - arsenic needs to gain 3 electrons. Since electrons have a negative charge, the charge of an As ion would be -3.

Try observing the periodic table and how many valence electrons that each element has. From there, you can determine the charges of the elements lithium and strontium. You can guess, I'll help you with those once you attempt to find the charge of those ions.

The half-life of radium-226 is 1590 years. (a) A sample of radium-226 has a mass of 50 mg. Find a formula for the mass of the sample that remains after t years. (b) Find the mass after 500 years correct to the nearest milligram. (c) When will the mass be reduced to 40 mg

Answers

Answer:

Explanation:

a )

m = m₀ [tex]e^{-\lambda t[/tex]

m is mass after time t . original mass is m₀ , λ is disintegration constant

λ = .693 / half life

= .693 / 1590

= .0004358

m = m₀ [tex]e^{- 0.0004358 t}[/tex]

b )

m = 50 x [tex]e^{-.0004358\times 500}[/tex]

= 40.21 mg .

c )

40 = 50 [tex]e^{-.0004358t[/tex]

.8 = [tex]e^{-.0004358t[/tex]

[tex]e^{.0004358t[/tex] = 1.25

.0004358 t = .22314

t = 512 years .

A 1.00 liter solution contains 0.42 moles nitrous acid and 0.32 moles sodium nitrite .
If 0.16 moles of nitric acid are added to this system, indicate whether the following statements are true or false.
(Assume that the volume does not change upon the addition of nitric acid.)
A. The number of moles of HNO2 will decrease.
B. The number of moles of NO2- will remain the same.
C. The equilibrium concentration of H3O+ will increase.
D. The pH will decrease.
E. The ratio of [HNO2] / [NO2-] will increase

Answers

Answer:

E. The ratio of [HNO2] / [NO2-] will increase

D. The pH will decrease.

Explanation:

Nitrous acid ( HNO₂ ) is a weak acid and NaNO₂ is its salt . The mixture makes a buffer solution .

pH = pka + log [ salt] / [ Acid ]

= 3.4 + log .32 / .42

= 3.4 - .118

= 3.282 .

Now .16 moles of nitric acid is added which will react with salt to form acid

HNO₃ + NaNO₂ = HNO₂ + NaNO₃

concentration of nitrous acid will be increased and concentration of sodium nitrite ( salt will decrease )

concentration of nitrous acid = .42 + .16 = .58 M

concentration of salt = .32 - .16 = .16 M

ratio of [HNO₂ ] / NO₂⁻]

= .42 / .32 = 1.3125

ratio of [HNO₂ ] / NO₂⁻] after reaction

= .42 + .16 / .32 - .16

= 58 / 16

= 3.625 .

ratio will increase.

Option E is the answer .

pH after reaction

= 3.4 + log .16 / .58

= 2.84

pH will decrease.

Calculate the percent saturated fat in the total fat in butter

Answers

about 63% of the fat in butter is saturated fat

A 11.0 mLmL sample of 0.30 MHBrMHBr solution is titrated with 0.16 MNaOHMNaOH. Part A What volume of NaOHNaOH is required to reach the equivalence point? Express the volume to two significant figures and include the appropriate units. nothingnothing

Answers

Answer:

21 mL of NaOH is required.

Explanation:

Balanced reaction: [tex]HBr+NaOH\rightarrow NaBr+H_{2}O[/tex]

Number of moles of HBr in 11.0 mL of 0.30 M HBr solution

= [tex](\frac{0.30}{1000}\times 11.0)[/tex] moles = 0.0033 moles

Let's say V mL of 0.16 M NaOH solution is required to reach equivalence point.

So, number of moles of NaOH in V mL of 0.16 M NaOH solution

= [tex](\frac{0.16}{1000}\times V)[/tex] moles = 0.00016V moles

According to balanced equation-

1 mol of HBr is neutralized by 1 mol of NaOH

So, 0.0033 moles of HBr are neutralized by 0.0033 moles of NaOH

Hence, [tex]0.00016V=0.0033[/tex]

           [tex]\Rightarrow V=\frac{0.0033}{0.00016}=21[/tex]

So, 21 mL of NaOH is required.

Name the advantages of coronavirus

Answers

Answer: Positive environmental changes.

Explanation: Without many humans around, the environment has been getting better as more sea life have been spotted in places they haven't been for decades, as well as clearer waters and less rubbish about. Pollution levels have dropped as there are barley any planes in the sky and not many cars about.

Answer:

honestly,i can say that socially being away from people reduces stress

Explanation:

Identify the particle represented by each symbol as an alpha particle, a beta particle, a gamma ray, a positron, a neutron, or a proton.
a. 11P
b. 42He
c. +10e

Answers

Answer:

[tex]_1^{1} {P}[/tex] is symbol for proton emission in the nucleus.

[tex]_2^{4} {He}[/tex] symbolises alpha emission, equivalent to helium atom emission of a radioactive particle

[tex]+_1^{0} {e}[/tex] is the radiation symbol for positiron particle. which occurs when beta + radioactive decay occurs

A certain metal forms a soluble nitrate salt M(NO3)3. Suppose the left half cell of a galvanic cell apparatus is filled with a 3.0mM solution of M(NO3)3 and the right half cell with a 3.0M solution of the same substance. Electrodes made of M are dipped into both solutions and a voltmeter is connected between them. The temperature of the apparatus is held constant at 20.0 C.

Required:
a. Which electrode will be positive?
b. What voltage will the voltmeter show? Assume its positive lead is connected to the positive electrode.

Answers

Answer:

1.The electrode on the right is positive

2. 0.058V

Explanation:

The above cell is a concentration cell.

A concentration cell is an electrolytic cell that is made of two half-cells with the same electrodes, but differs in concentrations of the solutions. A concentration cell functions by diluting the more concentrated solution and concentrating the more dilute solution, creating a voltage as the cell reaches an equilibrium thereby transferring the electrons from the cell with the lower concentration to the cell with the higher concentration.

In the above cell, electrons flow from the left electrode (less concentrated) to the right electrode (more concentrated). Therefore, the right electrode is the positive electrode (cathode).

Part 2: Please, see the attachment below for the calculations.

If the sign for delta G is negative (spontaneous process) and the sign for delta S is positive (more disorder) for both dissolving processes, how could one be endothermic (positive delta H) and one be exothermic (negative delta H)

Answers

Answer: From your question,

One could be exothermic which means that the final enthalpy will be less than the initial enthalpy. H= Hf-Hi(Hf<Hi).

In Endothermic reaction, the entropy is lowered by absorbing energy in the surronding. By so doing, the surronding losses energy and the reaction is not spontaneous.

H is positive and S (entropy) is positive.

Explanation:

Exothermic reaction is the reaction where heat is released In the surronding which lead to increase in the surronding Temperature.

Endothermic reaction is the reaction that absorb heat from the surronding and decrease the surronding Temperature.

Predict the products of the following elimination reaction, and draw the major product formed. Make sure to consider the stereochemistry of the reaction. 3-chloro-3-methylpentane reacts with sodium tertbutoxide in tertbutanol.
Predict the products of the following elimination reaction, and draw the major product formed. Make sure to consider the stereochemistry of the reaction. 3-chloro-3-methylpentane reacts with sodium ethoxide in ethanol.

Answers

Answer:

see explanation below

Explanation:

In the first case, we have a reaction where we have the 3-chloro-3-methylpentane reacting with t-butoxide. The t-butoxide is a very voluminous base, so the strength of substracting a hydrogen atom is reduced. Therefore, the reaction taking place here will be an E2 but instead of substracting the hydrogen from the carbons 2 or 4, it will substract it from the methyl group, cause it has less steric hindrance there and the reaction will go faster.

In the second case, the sodium ethoxide is a strong base, so it will rapidly substract an atom of hydrogen from carbon 2 or 4 to form the (Z) - 3 - methyl - 2- pentene and the substitution product.

Look picture for mechanism and products.

Which of the following obervations would be classified as a physical change? A) Fireworks releasing light B) Antacid fizzing in water C) Steam condensing on a mirror D) Apple turning brown

Answers

Answer:

C) Steam condensing on a mirror

Explanation:

This was just a change in the physical state.

Gold can be separated from sand by panning or by a sluice-box. In panning, water is mixed with the sand and the resulting slurry is swirled in a shallow, saucer-shaped metal pan. In the sluice-box technique, running water is passed over an agitated sand- gold mixture. What physical property and what technique make this separation possible?

Answers

Answer:

Decantation by means of difference in relative densities.

Explanation:

The specific gravity (relative density) of the gold to the soil/sand is the physical property exploited in panning gold. The particles with lower density would float whilst the heavier gold sinks lower to the bottom of the pan by gravity and is decanted off.

I hope this explanation is easy to comprehend.

A rule of thumb is that a reaction rate roughly doubles for every 10 °C increase in temperature. What is the activation energy of a reaction whose rate exactly doubles between 25.0 °C and 35.0 °C

Answers

Answer:

FOR EVERY 10 DEGREE CELSIUS INCREASE IN TEMPERATURE, THE ACTIVATION ENERGY THAT SHOWS THIS IS 52.4 KJ/MOL

Explanation:

From Arrhenius equation, the relationship between the rate constant and the temperature is as shown below:

k = Ae^ -Ea/RT

At initial temperature T1, the initial rate constant is (k1)

At final temperature T2, the final rate constant is k2

For the reaction rate to be doubled, we must double the rate constant which shows that the ratio of k2 / k1 must be equal to 2.

That is, k2 / k1 = 2 (rate is doubled)

Equating this into the Arrhenius equation, we have:

k2 / k1 = Ae^ (-Ea / R ) (1/ T2 - 1/T1)

2 = e^ (-Ea / R) (1 / T2 = 1 / T1)

Taking the natural logarithm of both sides:

ln 2 = - (Ea / R) (1 / T2 - 1 / T1)

Making Ea the subject of the formula, we obtain:

Ea = - (ln 2 R / (1 / T2- 1 / T1))

Let T1 = 25 C = 25 + 273 K = 298 K

T2 = 35 C = 35 + 273 K = 308 K

R = 8.314

So,

Ea = - (ln 2 * 8.314 / ( 1/308 - 1 / 298))

Ea = - (0.693 * 8.314 / 0.00324 - 0.00335)

Ea = - 5.7616 / -0.00011

Ea = 52 378,18 J / mol

So therefore, the activation energy Ea is 52.4 kJ/mol.

A gas has volume of 800.0mL at -23.0°c and 300.0torr. What would the volume of the gas be at 227.0°c and 600.0torr of pressure

Answers

Answer:

Explanation:

use gas law eqation

P1 * V1  / T1 = P2 * V2 /T2

600*V1/227 = 300*800/23

V1 = 300*800*227 / 23*600 = ............ can you solve this and get the answer?

How is excitation in spectroscopy brought about​

Answers

Answer: the exciation of molecules is brount by absorption of energy  in spectroscpy

Explanation:

RUIGA GIRLS
CHEMISTRY FORM 3. 23/06/2020
MR. GICHURU
IZ
1
Narne the elements present in
Common salt
(2 miks)
Hydrated copper (11) Sulphate.
(2 ks)
Sulphuric (VI) acid,
2 Why is a reaction between zinc metal and Nitric acid not suitable for preparing
hydrogen gae in the laboratory
(2 mi)
(1 m)
3.
What is relative atomic mass?
b)
Define 'isotopes
c)Determine the relative atomic mass of element K whose isotople misure occur in
the proportione:
(2 marks)​

Answers

1)
Common salt= sodium, chlorine
Hydrated copper sulfate= copper, sulfur, water
Sulphuric acid=hydrogen, sulfur, oxygen
2)
Idfk
3)
a)Relative atomic mass is the average mass of an elements atoms(total number of protons and neutrons)
b)An isotope is a form of an element that has the same number of protons but different number of neutrons
c)idfk
Btw ur teachers name is hot

Chemical formula for copper gluconate I have 1.4g of Copper gluconate. There is .2g of copper within the copper gluconate. Determine the chemical formula for Copper gluconate with the given information: Copper Gluconate: Cu(C6H11O?)? Cu = 63.55 g/mol H = 12.01 g/mol O = 1.008 g/mol Cu = 63.55 g/mol

Answers

Answer:

The simplest chemical formula of the compound is Cu(C₆H₁₁O₇)₂

Explanation:

Given mass of sample = 1.4 g

mass of copper in the sample = 0.2 g

mass of the gluconate =1.4 - 0.2 = 1.2 g

The mole ratio is determined first using the formula;

mole ratio = reacting mass / atomic mass

atomic mass of copper = 63.55

mass of gluconate, C₆H₁₁O₇ = 12*6 + 1*11 + 16*7 = 195 g/mol

mole ratio ( copper : gluconate) = 0.2/63.55 : 1.4/195

mole ratio ( copper : gluconate) =  0.003 : 0.007

convert to whole number ratios by dividing with the smallest ratio

mole ratio ( copper : gluconate) = 0.003/0.003 : 0.007/0.003

mole ratio ( copper : gluconate) = 1 : 2

Therefore, the simplest chemical formula of the compound is Cu(C₆H₁₁O₇)₂

The compound ClF contains Group of answer choices polar covalent bonds with partial negative charges on the Cl atoms. ionic bonds. nonpolar covalent bonds. polar covalent bonds with partial negative charges on the F atoms.

Answers

Answer:

polar covalent bonds with partial negative charges on the F atoms.

Explanation:

A covalent bond could be polar or nonpolar depending on the relative electro negativity difference between the two bonding atoms. In this case, the bonding atoms are chlorine and fluorine.

In the Pauling's scale, fluorine has an electro negativity value of 3.98 while chlorine has an electro negativity value of 3.16. The difference in electro negativity between the two atoms is about 0.82. This magnitude of electro negativity difference between the two bonding atoms correspond to the existence of a polar covalent bond in the molecule.

The direction of the dipole depends on the relative electro negativity values of the two bonding atoms. Since fluorine is more electronegative than chlorine, the fluorine atom will be partially negative and the chlorine atom will be partially positive accordingly.

The compound ClF (chlorine monofluoride) contains polar covalent bonds with partial negative charges on the F atoms. Therefore, option D is correct.

In ClF, chlorine (Cl) is more electronegative than fluorine. As a result, the shared electrons in the Cl-F bond are pulled closer to the chlorine atom, creating a partial negative charge on the fluorine atoms and a partial positive charge on the chlorine atom.

This polarity in the Cl-F bond gives the molecule an overall polarity, making it a polar molecule. Thank you for pointing out the error, and I apologize for any confusion caused.

Thus, option D is correct.

To learn more about the polar covalent bonds, follow the link:

https://brainly.com/question/28295508

#SPJ6

A galvanic cell at a temperature of 25.0°C is powered by the following redox reaction:

2MnO4^-(aq)+16H+(aq)+5Pb(s)-->2Mn^2+(aq)+8H2O(l)+5Pb^2+(aq)

Suppose the cell is prepared with 1.87 M MnO−4 and 1.37 M H+ in one half-cell and 3.23 M Mn+2 and 6.62 M Pb+2 in the other. Calculate the cell voltage under these conditions. Round your answer to 3 significant digits.

Answers

Answer:

1.63 V

Explanation:

Let us state the reaction equation again for the purpose of clarity;

2MnO4^-(aq)+16H+(aq)+5Pb(s)-->2Mn^2+(aq)+8H2O(l)+5Pb^2+(aq)

The reduction potentials for the two half reaction equations are;

MnO 4 - (aq) + 8H + (aq) + 5e - → Mn2+(aq) + 4H2O(l) Eo=1.51 V

Pb2+(aq) + 2e - → Pb(s) Eo= -0.13 V

E°cell = E°red – E°Ox

E°cell = 1.51 - (-0.13)

E°cell = 1.51 + 0.13

E°cell = 1.64 V

But Q= [Mn^2+]^2 [Pb^2+]^5/[MnO4^-]^2 [H^+]^16

Q= [3.23]^2 [6.62]^5/[1.87]^2 [1.37]^16

Q= 10.43 × 12714.22/3.4969 × 154

Q= 132609.3/538.5226

Q= 246.25

From Nernst equation

E= E° - 0.0592/n log Q

Where n=10

E= 1.64- 0.0592/10 log 246.25

E= 1.64-0.0142

E= 1.63 V

Best example of potential energy?

Answers

Answer:

water stored in a dam

Explanation:

when the water is in dam it is ready to move bit is not moving

The best example would be like a truck pared at the top of a hill or like a person at the top of a slide . Thanks hope this helps

1.Draw the born-Haber lattice energy cycle for sodium chloride. Explain the concept of resonance using the nitrate ion structure.

Answers

Answer:

1. Born Haber cycle is used to calculate enthalpy of formation of an ionic solid

2. Resonance structures are used to represent the bonding in some chemical species.

Explanation:

The Born–Haber cycle is a method popularly known in chemistry used in computing enthalpy. The enthalpy of formation of an ionic solid cannot be measured directly. The lattice enthalpy refers to the enthalpy change involved in the formation of an ionic compound from gaseous ions the process is exothermic process. A Born–Haber cycle works on the principle of Hess's law. It can be used to calculate the lattice enthalpy by comparing the standard enthalpy change of formation of the ionic compound from the elements to the enthalpy required to make gaseous ions from the elements.

Resonance is an idea introduced by Linus Pauling to explain chemical bonding from the valence bond perspective. The idea of resonance affords us the opportunity to describe the bonding in certain molecules by combining several structures called chemical or canonical structures. The real structure of the specie lie somewhere between the structures indicated by the resonance structures. The resonance structures of the nitrate ion are shown in the image attached.

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