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 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
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.
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 ;
https://brainly.com/question/16746749
#SPJ2
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]
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
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
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
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
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
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)3.14LProtons,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.
Help me with these please
Answer:
Help you with what hmm I do not know what you are talking about
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.
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.
