An analytical chemist is titrating of a solution of hydrazoic acid with a solution of . The of hydrazoic acid is . Calculate the pH of the acid solution after the chemist has added of the solution to it.

Answers

Answer 1

Answer:

pH = 12.43

Explanation:

...is titrating 212.7 mL of a 0.6800 M solution of hydrazoic acid (HN3) with a 0.2900 M solution of KOH. The p Ka of hydrazoic acid is 4.72. Calculate the pH of the acid solution after the chemist has added 571.6 mL of the KOH solution to it.

To solve this question we need to know that hidrazoic acid reacts with KOH as follows:

HN3 + KOH → KN3 + H2O

Moles KOH:

0.5716L * (0.2900mol /L) =0.1658 moles of KOH

Moles HN3:

0.2127L * (0.6800mol/L) = 0.1446 moles HN3

As the reaction is 1:1, the KOH is in excess. The moles in excess of KOH are:

0.1658 moles - 0.1446 moles =

0.0212 mol KOH

In 212.7mL + 571.6mL = 784.3mL = 0.7843L

The molarity of KOH = [OH-] is:

0.0212 mol KOH / 0.7843L = 0.027M = [OH-]

The pOH is defined as -log [OH-]

pOH = -log 0.027M

pOH = 1.57

pH = 14 - pOH

pH = 12.43


Related Questions

What volume of each solution contains 0.14 mol of KCl? Express your answer using two significant figures.

1.8 M KCl

Answers

Answer:

Solution given:

1 mole of KCl[tex]\rightarrow [/tex]22.4l

1 mole of KCl[tex]\rightarrow [/tex]74.55g

we have

0.14 mole of KCl[tex]\rightarrow [/tex]74.55*0.14=10.347g

74.55g of KCl[tex]\rightarrow [/tex]22.4l

10.347 g of KCl[tex]\rightarrow [/tex]22.4/74.55*10.347=3.11litre

volume of each solution contains 0.14 mol of KCl contain 3.11litre.

1 mol of any gas contains 22.4L of volume at STP

Moles of KCl=0.14

Volume of KCl:-

0.14(22.4)3.14L

0.5008 g of an unknown triprotic acid, H3A, is dissolved in 47.3 mL of water and then titrated with 0.315 M NaOH. It takes 25.72 mL of the NaOH solution to completely neutralize the acid. What is the molar mass of this acid

Answers

Answer:

The molar mass is "185.44 g/mol".

Explanation:

According to the question,

The moles of NaOH will be:

= [tex]\frac{0.315}{1000}\times 25.72[/tex]

= [tex]8.1018\times 10^{-3} \ moles[/tex]

Number of moles of an acid will be:

= [tex]\frac{8.1018\times 10^{-3}}{3}[/tex]

= [tex]2.7006\times 10^{-3} \ moles[/tex]

We know that,

⇒ [tex]Moles = \frac{Mass}{Molar \ mass}[/tex]

hence,

Molar mass of unknown acid will be:

= [tex]\frac{Mass \ in \ g}{Moles}[/tex]

= [tex]\frac{0.5008}{2.7006\times 10^{-3}}[/tex]

= [tex]185.44 \ g/mol[/tex]

what gasous product would you expect when water is drop over calcium carbide​

Answers

Answer:

Ethyn gas ( acetylene gas )

Explanation:

All group II carbides react with water to form ethyn gas apart from beryllium which produces methane gas.

[tex]{ \sf{CaC_{2(s)} + 2H _{2} O_{(l)} → Ca(OH) _{2(s)} + C _{2} H _{2(g)} }}[/tex]

An aqueous solution contains 0.374 M ammonia (NH3). How many mL of 0.276 M nitric acid would have to be added to 125 mL of this solution in order to prepare a buffer with a pH of 8.970.

Answers

Answer:

111.95mL of HNO3 are needed to prepare the buffer

Explanation:

We can solve this equation using H-H equation for bases:

pOH = pKb + log [HA+] / [A]

Where pOH is the pOH of the solution

pOH = 14 - pH = 14 - 8.970 = 5.03

pKb is the pKb of NH3 = 4.74

[HA+] could be taken as moles of NH4+

[A] as moles of NH3

The NH3 reacts with nitric acid, HNO3, as follows:

NH3 + HNO3 → NH4+ + NO3-

That means the moles of HNO3 added = X = Moles of NH4+ produced

And moles of NH3 are initial moles NH3 - X

Initial moles of NH3 are:

0.125L * (0.374mol/L) = 0.04675 moles NH3

Replacing in H-H equation:

pOH = pKb + log [HA+] / [A]

5.03 = 4.74 + log [X] / [0.04675-X]

0.29 = log [X] / [0.04675-X]

1.95 =  [X] / [0.04675-X]

0.0912 - 1.95X = X

0.0912 = 2.95X

X = 0.0309 moles

We need to add 0.0309 moles of HNO3. From a solution that is 0.276M:

0.0309 moles of HNO3 * (1L / 0.276moles) = 0.11195L of HNO3 are needed

In mL:

111.95mL of HNO3 are needed to prepare the buffer

What does quantization refer to?

Answers

Answer:

Quantization is the process of constraining an input from a continuous or otherwise large set of values (such as the real numbers) to a discrete set (such as the integers).

Explanation:

Quantization refers to the situation where an electromagnetic field consists of discrete energy parcels, photons.

What is Quantatization in Chemistry ?

In Chemistry , the concept that a system cannot have any possible energy value but instead is limited to certain specific energy values (states). This states depend on the specific system in question.

Under this system, Energy could be gained or lost only in integral multiples of some smallest unit of energy, a quantum (the smallest possible unit of energy).

Hence, Quantization refers to the situation where an electromagnetic field consists of discrete energy parcels, photons.

Learn more about Quantum here ;

https://brainly.com/question/16746749

#SPJ2

balance equation of aluminium chloride+ hydrogen​

Answers

[tex]\boxed{\sf {AlCl_3\atop Aluminium\:Chloride}+{H_2\atop Hydrogen}\longrightarrow {Al\atop Aluminium}+{HCl\atop Hydrochloric\:acid}}[/tex]

Balanced Equation:-

[tex]\boxed{\sf {2AlCl_3\atop Aluminium\:Chloride}+{3H_2\atop Hydrogen}\longrightarrow {2Al\atop Aluminium}+{6HCl\atop Hydrochloric\:acid}}[/tex]

Phosphagens are a group of molecules that includes creatine phosphate (in vertebrates), and arginine phosphate, lombricine, and phosphoopheline (in invertebrates). These molecules have similar functions in different organisms.

a. True
b. False

Answers

Answer: True

Explanation:

Phosphagens are high energy storage compounds that are usually found in the tissue of animals.

Based on the question, the molecules have similar functions in different organisms such as the fact that they can accept phosphoryl groups from ATP in a situation where the ATP is in excess. Also, they donate phosphoryl groups to ADP in order for the regeneration of ATP.

Critique this statement: Promotion of electrons is accompanied by a release of energy

Answers

Answer: Promotion of electrons is accompanied by a release of energy because of absorption of photon.

Explanation:

Promotion of electrons occurs when an electron accepts or absorbs a photon which leads to it's movement from a lower energy level orbital to a higher energy orbital.

According to Bohr, the electrons was restricted to certain energy levels and was thought to move along certain circular orbits around the nucleus. These energy levels were identified by means of principal quantum number, n. The wave mechanics model of atom does not restrict the electrons to a certain energy levels only. Instead it describes a region around the Nucleus called orbitals where there is a possibility of finding an electron with a certain amount of ENERGY.

The energy levels are composed of one or more orbitals and the distribution of electrons around the nucleus is determined by the number and kind of energy levels that are occupied.

Bohr made an assumption that an electron emits energy in the form of radiation when it moves from a higher to a lower permitted orbit, this produces a line in the atomic emission spectrum. Since the energies of the higher and lower orbits are fixed, the line will be of a particular energy and frequency.

What process occurs during the corrosion of iron?
Answers

A.
Iron is oxidized.

B.
Iron is reduced.

C.
Iron (III) is oxidized.

D.
Iron (III) is reduced.

Answers

Answer:

A

Explanation:

The iron corrodes so it oxidized

A student has accidentally spilled 100.0 mL of 3.0 mol/L nitric acid onto the lab bench. What mass of sodium bicarbonate would the teacher need to sprinkle on this spill to neutralize and clean it up?

Answers

Answer:

25 g

Explanation:

Step 1: Write the balanced equation

HNO₃ + NaHCO₃ ⇒ NaNO₃ + H₂O + CO₂

Step 2: Calculate the reacting moles of HNO₃

100.0 mL of 3.0 mol/L HNO₃ reacted.

0.1000 L × 3.0 mol/L = 0.30 mol

Step 3: Calculate the reacting moles of NaHCO₃

The molar ratio of HNO₃ to NaHCO₃ is 1:1. The reacting moles of NaHCO₃ are 1/1 × 0.30 mol = 0.30 mol.

Step 4: Calculate the mass corresponding to 0.30 moles of NaHCO₃

The molar mass of NaHCO₃ is 84.01 g/mol.

0.30 mol × 84.01 g/mol = 25 g

3. (07.05 LC)
When zinc reacts with hydrochloric acid, it produces hydrogen gas. As the reaction proceeds, why does the rate of production of hydrogen gas decrease? (3 points)

Answers

the rate would decrease because the reactants are being depleted.

Five kg of carbon dioxide (CO2) gas undergoes a process in a well-insulated piston-cylinder assembly from 2 bar, 280 K to 20 bar, 520 K. If the carbon dioxide behaves as an ideal gas, determine the amount of entropy produced, in kJ/K.

Answers

This question is incomplete, the complete question is;

Five kg of carbon dioxide (CO2) gas undergoes a process in a well-insulated piston-cylinder assembly from 2 bar, 280 K to 20 bar, 520 K. If the carbon dioxide behaves as an ideal gas, determine the amount of entropy produced, in kJ/K. Assuming;

a) constant specific heats Cp = 0.939 kJ/Kg K

b) variable specific heats

Answer:

a) the amount of entropy produced is 0.731599 kJ/K

b) the amount of entropy produced is 0.69845 kJ/K

Explanation:

Given the data in the question;

5 kg of carbon dioxide (CO₂) gas undergoes a process in a well-insulated piston-cylinder assembly.

m = 5 kg

Molar mass M = 44.01 g/mol

P₁ = 2 bar, P₂ = 20

T₁ = 280 K, P₂ = 520 K

Since its insulated { q = 0 } ( kinetic and potential energy effects = 0 )

Now,

a) the amount of entropy produced, in kJ/K, Assuming constant specific heats with Cp = 0.939 kJ/Kg K

S[tex]_{Generation[/tex] = m × ((Cp × In( T₂/T₁) - R × In( P₂/p₁ ))

we substitute

S[tex]_{Generation[/tex] = 5 × (( 0.939  × In( 520/280) - 0.1889 × In( 20/2 ))

= 5 × ( 0.5812778 - 0.434958 )

= 5 × 0.1463198

= 0.731599 kJ/K

Therefore, the amount of entropy produced is 0.731599 kJ/K

b) the amount of entropy produced, in kJ/K, Assuming variable specific heats.

Now, from  Table A-23: Ideal Gas Properties of Selected Gases;

T₁,T₂ : s₁⁰ = 211.376 kJ/kmol-K, s₂⁰ = 236.575 kJ/kmol-K

now, s₁ = s₁⁰ / M and s₂ = s₂⁰ / M

we substitute

s₁ = s₁⁰ / M = 211.376 / 44.01  = 4.8029 kJ/kg

s₂ = s₂⁰ / M = 236.575 / 44.01 = 5.37548 kJ/kg

S[tex]_{Generation[/tex] = m × (( s₂ - s₁ ) - R × In( p₂ / p₁ ))

we substitute

S[tex]_{Generation[/tex] = 5 × (( 5.37548 - 4.8029  ) - 0.1880 × In( 20 / 2 ))

= 5 × ( 0.57258 - 0.432885997 )

= 5 × 0.13969

= 0.69845 kJ/K

Therefore, the amount of entropy produced is 0.69845 kJ/K

According to an informal 1992 survey, the drinking water in about one-third of the homes in Chicago had lead levels of about 10 ppb. Dr. Koether lived in Chicago from 1996 to 1998. Assuming she drank 1.4 L of water a day, calculate the total amount of lead in mg (using one decimal place) that she was exposed to over the two years if she lived in a home that had such high levels of lead.

Answers

Answer:

10.2 mg

Explanation:

Step 1: Calculate the total amount of water she drank

1 year has 365 days and she lived in Chicago for 2 years = 2 × 365 days = 730 days.

If she drank 1.4 L of water per day, the total amount of water she drank is:

730 day × 1.4 L/day = 1022 L

Step 2: Calculate the amount of Pb in 1022 L of water

The concentration of Pb is 10 ppb (10 μg/L).

1022 L × 10 μg/L = 10220 μg

Step 3: Convert 10220 μg to milligrams

We will use the conversion factor 1 mg = 1000 μg.

10220 μg × 1 mg/1000 μg = 10.2 mg

Oxygen is composed of three isotopes: oxygen-16, oxygen-17 and oxygen-18 and has an average atomic mass of 15.9982 amu. Oxygen-17 has a mass of 16.988 amu and makes up 0.032% of oxygen. Oxygen-16 has a mass of 15.972 amu and oxygen-18 has a mass of 17.970 amu. What is the percent abundance of oxygen-18?

Answers

Answer:

The percent abundance of oxygen-18 is 1.9066%.

Explanation:

The average atomic mass of oxygen is given by:

[tex] m_{O} = m_{^{16}O}*\%_{16} + m_{^{17}O}*\%_{17} + m_{^{18}O}*\%_{18} [/tex]

Where:

m: is the atomic mass

%: is the percent abundance

Since the sum of the percent abundance of oxygen isotopes must be equal to 1, we have:  

[tex] 1 = \%_{16} + \%_{17} + \%_{18} [/tex]

[tex] 1 = x + 3.2 \cdot 10^{-4} + \%_{18} [/tex]

[tex] \%_{18} = 1 - x - 3.2 \cdot 10^{-4} [/tex]

Hence, the percent abundance of O-18 is:  

[tex] m_{O} = m_{^{16}O}*\%_{16} + m_{^{17}O}*\%_{17} + m_{^{18}O}*\%_{18} [/tex]  

[tex]15.9982 = 15.972*x + 16.988*3.2 \cdot 10^{-4} + 17.970*(1 - 3.2 \cdot 10^{-4} - x)[/tex]

[tex] x = 0.980614 \times 100 = 98.0614 \% [/tex]                                                              

Hence, the percent abundance of oxygen-18 is:

[tex]\%_{18} = (1 - 3.2 \cdot 10^{-4} - 0.980614) \times 100 = 1.9066 \%[/tex]                      

Therefore, the percent abundance of oxygen-18 is 1.9066%.

I hope it helps you!                                                      

Help me with these please

Answers

Answer:

Help you with what hmm I do not know what you are talking about

All of the following are TRUE for activities and activity coefficients, except: A) activity for a chemical species is the product of concentration and activity coefficient. B) the activity coefficient corrects for non-ideal behavior due to ionic strength. C) as ionic strength increases, the value of the activity coefficient increases. D) for ions, the activity coefficient approaches unity as the ionic strength approaches 0. E) the activity coefficient for neutral molecules is approximately unity.

Answers

Answer:

C) as ionic strength increases, the value of the activity coefficient increases.

Explanation:

The effective concentration of ions available for reactions is known as the activity of the ion.

The activity coefficient important in chemistry because it accounts for the deviation of a solution from ideal behaviour.

The activity of a chemical species is defines as the product of concentration and activity coefficient.

Following the Debye–Hückel limiting law; log γ = −0.509z2I1/2. The ionic strength of a solution tends to increase as the activity coefficient (γ) of the ion decreases.

Phosphorylation of enzymes:_______.
a. always increases their activity.
b. generally occurs on Ser, Thr, and/or Tyr side chains and to a lesser extent on the His side chain.
c. is irreversible.
d. is one of only five known covalent forms of regulation.

Answers

Answer:

generally occurs on Ser, Thr, and/or Tyr side chains and to a lesser extent on the His side chain

If 1 mol of ferric oxide reacts with 3 moles of carbon monoxide to yield 2 mols of iron and 3 mols of carbon dioxide, how much CO will be needed to completely react with 50.26 g of ferric oxide?

Answers

Answer:

26.4g

Explanation:

The balanced chemical equation as stated in this question is given as follows:

Fe2O3 + 3CO → 2Fe + 3CO2

According to this balanced equation, 3 moles of carbon monoxide (CO) will react with 1 mole of Ferric oxide (Fe2O3).

We need to change 50.26 g of ferric oxide to moles by using the formula;

mole = mass/molar mass

Molar mass of Fe2O3 = 56(2) + 16(3)

= 112 + 48

= 160g/mol

mole = 50.26/160

mole = 0.314mol of Fe2O3

If 3 moles of carbon monoxide (CO) will react with 1 mole of Ferric oxide (Fe2O3).

Hence, 0.314 mol of Fe2O3 will completely react with (0.314 × 3) mol of CO

0.314 × 3 = 0.94 mol of CO

molar mass of CO = 12 + 16 = 28g/mol

mole = mass/molar mass

mass = mole × M.M

mass = 0.94 × 28

mass = 26.4g of CO

what is the difference between red phosphorus and white phosphorus?​

Answers

Answer:

White phosphorusRed PhosphorusIt is insoluble in water but soluble in carbon disulphide.It is insoluble in both water and carbon disulphide.It undergoes spontaneous combustion in air.It is relatively

Explanation:

I hope it will help you




Read the scales of this balance.
The unknown sample has a mass of:
11.2 g
01.012 kg
1.220 g
O 1.200 g

Answers

Answer:

and I'll call you when the party's over

quiet when I'm come in home

when I'm all alone

Answer:

Explanation:

Don't you know too much already?

I'll only hurt you if you let me

Call me friend but keep me closer (call me back)

And I'll call you when the party's over

the ability for carbon to form long chain or rings is

Answers

califactual. thsi is correct bcuz carbon takes 20 minutes to dissolve and ring making it a factual

which primitive organic molecule was essential to form lipid bilayer?
a)protenoid
b)phospholipid
c)autocatalytic RNA
d)aminoacids​

Answers

Answer:

c) autocatalytic RNA is the primitive organic molecules was essential to form lipid bilayer.

g consider the following pair of aqueous solutions. which pair will result in the formation of a precipitate? give the formula for the precipitate in the blank. write none if no precipitate forms. a) libr and nh4no3 b) kcl and pb(ch3coo)2

Answers

Answer:

kcl and pb(ch3coo)2

The precipitate is PbCl2

Explanation:

Let us take the options provided one after the other;

In the first case, we have;

LiBr(aq) + NH4NO3(aq) ----> LiNO3(aq) + NH4Br(aq)

You can see that no precipitate is formed here.

In the second case;

2KCl(aq) + Pb(CH3COO)2(aq) ----> PbCl2(s) + 2CH3COOK(aq)

The precipitate here is PbCl2.

Protons,neutrons and electrons are not considered as------------ *

Answers

Electrons are a type of subatomic particle with a negative charge. Protons are a type of subatomic particle with a positive charge. ... Neutrons are a type of subatomic particle with no charge (they are neutral). Like protons, neutrons are bound into the atom's nucleus as a result of the strong nuclear force.

PLEASE HELP ASAP
Use the equation below to answer the following questions.
2Al(s) + 3Cu(NO3)2(aq) 3Cu(s) + 2Al(NO3)3(aq)

Determine the oxidation state of the atoms in the equation's reactants and products: (6 points)

Oxidation state of Al in reactant:
in product:

Oxidation state of Cu in reactant:
in product:

Oxidation state of N in reactant:
in product:

Oxidation state of O in reactant:
in product:

Explain why this is a redox reaction.

Thank you!

Answers

Answer:

hlo.......................,

What is the energy of a photon emitted with a wavelength of 654 nm?
O A. 3.04 x 10^-19 J
O B. 1.01 * 10^-27 J
O C. 1.30 x 10^-22 J
O D. 4.33 * 10^-22 J ​

Answers

Answer:

A. 3.04×10^-19J

Explanation:

Hope this will help you.

Using the following equation for the combustion of octane calculate the heat associated with the formation of 100.0 g of carbon dioxide. The molar mass of octane is 114.33 g/mole.

2C8H18 + 25O2 → 16 CO2 + 18 H2O

ΔH°rxn = -11018 kJ

Answers

Answer:

The right solution is "-602.69 KJ heat".

Explanation:

According to the question,

The 100.0 g of carbon dioxide:

= [tex]\frac{100.0 \ g}{114.33\ g/mole}[/tex]

= [tex]0.8747 \ moles[/tex]

We know that 16 moles of [tex]CO_2[/tex] formation associates with -11018 kJ of heat, then

0.8747 moles [tex]CO_2[/tex] formation associates with,

= [tex]-\frac{0.8747}{16}\times 11018 \ KJ \ of \ heat[/tex]

= [tex]-0.0547\times 11018[/tex]

= [tex]-602.69 \ KJ \ heat[/tex]

What is a reaction rate?

Answers

Answer:

A reaction is the time that is required for a chemical reaction to go essential to completion

What is the mass of carbon in 69.00 mg of co2

Answers

Answer:

18.82 mg

Explanation:

From the given information:

The molar mass of CO2  is calculated as follow

= (12 + (16 ×2))

= 44

The mass of carbon is determined by dividing the mass no of carbon from co2 by the molar mass of CO2,  followed by multiplying it by 69.00 mg

= [tex](\dfrac{12}{44}\times 69 )[/tex]

=(0.2727 × 69 )

= 18.82 mg

Kevin's supervisor, Jill, has asked for an update on today's sales. Jill is pretty busy moving back and forth between different store locations. How can Kevin most effectively deliver an update to her? a) Send a detailed email Send a detailed text message Oc) Book a one-hour meeting for tomorrow morning O d) Call with a quick update

Answers

Kevin can effectively deliver an update by sending a detailed EMAIL to Jill

Email, which means electronic mail is a technological advanced way of passing information from persons to persons without physical contact. Sending emails are also official ways of passing vital information regarding business, work to and fro.

According to this question, Jill is a very busy supervisor who hardly. The best way for Kevin to deliver any update concerning the store he is managing is to send Jill an updated email that can even be assessed outside work hours. Learn more: https://brainly.com/question/7098974
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