Answer:
hlo.......................,
Protons,neutrons and electrons are not considered as------------ *
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.
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.
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.
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
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
Answer:
A. 3.04×10^-19J
Explanation:
Hope this will help you.
what gasous product would you expect when water is drop over calcium carbide
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]
What is a reaction rate?
Answer:
A reaction is the time that is required for a chemical reaction to go essential to completion
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
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/7098974Phosphagens 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
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.
What volume of each solution contains 0.14 mol of KCl? Express your answer using two significant figures.
1.8 M KCl
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=0.14Volume of KCl:-
0.14(22.4)3.14LWhat 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.
Answer:
A
Explanation:
The iron corrodes so it oxidized
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.
Answer:
generally occurs on Ser, Thr, and/or Tyr side chains and to a lesser extent on the His side chain
Critique this statement: Promotion of electrons is accompanied by a release of energy
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 does quantization refer to?
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 ;
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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.
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.
Help me with these please
Answer:
Help you with what hmm I do not know what you are talking about
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)
the rate would decrease because the reactants are being depleted.
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.
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
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?
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!
balance equation of aluminium chloride+ hydrogen
[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]
what is the difference between red phosphorus and white phosphorus?
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
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
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
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]
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
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.
