Answer:
Explanation:
Mn( NO₃ )₂ + 2Na OH = Mn( OH)₂ (s) ↓ + 2Na NO₃
Converting into ions
Mn⁺ + 2 NO₃⁻ + 2 Na⁺ + 2 OH⁻ = Mn( OH)₂ + 2 Na⁻ + 2 NO₃⁻
Cancelling out common terms
Mn⁺ + 2 OH⁻ = Mn( OH)₂
this is net ionic equation required.
2.50 mol NOCl was placed in a 2.50 L reaction vessel at 400ºC. After equilibrium was established, it was found that 28% of the NOCl had dissociated according to the equation 2NOCl(g) 2NO(g) + Cl 2(g). Calculate the equilibrium constant, K c, for the reaction.
Answer:
Explanation:
2NOCl(g) ⇄ 2NO(g) + Cl 2(g)
C ( 1 - .28 ) .28 C .14 C
Kc = [ NO ]² x [ Cl₂ ] / [ NOCl ]²
= (.28 C )² x .14 C / C² ( 1 - .28 )²
= .021173 x C
C = concentration of reactant
= 2.5 / 2/5 = 1 M
Kc = .021173 x 1
= 211.73 x 10⁻⁴ M .
Given:
Number of moles = 2.50 molVolume of solution = 2.5 LAt equilibrium,
Concentration of NO = 0.28 MConcentration of Cl₂ = 0.14 MNow,
The concentration of NOCl will be:
= [tex]\frac{Number \ of \ moles}{Volume \ of \ solution}[/tex]
= [tex]\frac{2.5}{2.5}[/tex]
= [tex]1 \ M[/tex]
At equilibrium,
The concentration of NOCl will be:
= [tex]1-0.28[/tex]
= [tex]0.72 \ M[/tex]
hence,
The equilibrium constant,
→ [tex]K_c =\frac{ [NO]^2 [Cl_2]}{[NOCl]^2}[/tex]
By substituting the values, we get
[tex]= \frac{(0.28)^2\times (0.14)}{(0.72)^2}[/tex]
[tex]= 2.117\times 10^{-2}[/tex]
Thus the above answer is right.
Learn more about equilibrium below:
https://brainly.com/question/14954687
50.0 mL each of 1.0 M HCl and 1.0 M NaOH, at room temperature (20.0 OC) are mixed. The temperature of the resulting NaCl solution increases to 27.5 OC. The density of the resulting NaCl solution is 1.02 g/mL. The specific heat of the resulting NaCl solution is 4.06 J/g OC Calculate the Heat of Neutralization of HCl(aq) and NaOH(aq) in KJ/mol NaCl produced
Answer:
-62.12kJ/mol is heat of neutralization
Explanation:
The neutralization reaction of HCl and NaOH is:
HCl + NaOH → NaCl + H₂O + HEAT
An acid that reacts with a base producing a salt and water
You can find the released heat of the reaction -heat of neutralization- (Released heat per mole of reaction) using the formula:
Q = C×m×ΔT
Where Q is heat, C specific heat of the solution (4.06J/gºC), m its mass of the solution and ΔT change in temperature (27.5ºC-20.0ºC = 7.5ºC).
The mass of the solution can be found with the volume of the solution (50.0mL of HCl solution + 50.0mL of NaOH solution = 100.0mL) and its density (1.02g/mL), as follows:
100.0mL × (1.02g / mL) = 102g of solution.
Replacing, heat produced in the reaction was:
Q = C×m×ΔT
Q = 4.06J/gºC×102g×7.5ºC
Q = 3106J = 3.106kJ of heat are released.
There are 50.0mL ×1M = 50.0mmoles = 0.0500 moles of HCl and NaOH that reacts releasing 3.106kJ of heat. That means heat of neutralization is:
3.106kJ / 0.0500mol of reaction =
-62.12kJ/mol is heat of neutralizationThe - is because heat is released, absorbed heat has a + sign
A meteorologist filled a weather balloon with 3.00L of the inert noble gas helium. The balloon's pressure was 765 torr. The balloon was released to an altitude with a pressure of 530 torr. What was the volume (L) of the weather balloon
Answer:
4.33 L
Explanation:
Step 1: Given data
Initial volume of the balloon (V₁): 3.00 L
Initial pressure of the balloon (P₁): 765 torr
Final volume of the balloon (V₂): ?
Final pressure of the balloon (P₂): 530 torr
Step 2: Calculate the final volume of the balloon
If we consider Helium to behave as an ideal gas, we can calculate the final volume of the balloon using Boyle's law.
[tex]P_1 \times V_1 = P_2 \times V_2\\V_2 = \frac{P_1 \times V_1}{P_2} = \frac{765torr \times 3.00L}{530torr} = 4.33 L[/tex]
The volume of a sample of oxygen is 300mL when the pressure is 1 atm and the temperature is 27 C . At what temperature is the volume 1.00 L and the pressure.500 atm?
Answer:
T2 = 500K
Explanation:
Given data:
P1 = 1atm
V1 = 300ml
T1= 27 + 273 = 300K
T2 = ?
V2 = 1.00ml
P2 = 500atm
Apply combined law:
P1xV1//T1 = P2xV2/T2 ...eq1
Substituting values into eq1:
1 x 300/300 = 500 x 1/T2
Solve for T2:
300T2 = 500 x 300
300T2 = 150000
Divide both sides by the coefficient of T2:
300T2/300 = 150000/300
T2 = 500K
1. In this experiment, the procedure instructs you to dissolve solid potassium hydrogen tartrate (KHT) in two different solvents. What are these two solvents? (2 pts)
Answer:
Water
Explanation:
Solid potassium hydrogen tartrates (KHT) is soluble in water. This is especially at room temperature.
The solvent for KHT is water.
Draw the structure of beta-D-idose in its pyranose form.
Answer:
See figure 1
Explanation:
In this case, we can start with the linear structure of D-Idose. Then, if we have a "D" configuration the "OH" in the last chiral (carbon 5) will be in the right. This carbon will attack carbon 1 and we will produce a cyclic structure with 6 members (pyranose). Additionally, we have to keep in mind that we want the "beta" structure. So, the "OH" on carbon 1 must point up (red arrow). Finally, we will have a cyclic structure with 6 atoms and the "OH" on carbon 1 pointing up.
See figure 1
I hope it helps!
Assume that a compound is a cyclic, planar, completely conjugated ring. Which number of p electrons would make it aromatic?
a) 0 p electrons
b) 2 p electrons
c) 3 p electrons
d) 4 p electrons
e) 32 p electrons
Answer:
option b is correct
2 p electron makes aromatic
Explanation:
An aromatic compound which is cyclic, planar and has a complete conjugate ring must have (4n + 2)pi electrons(Huckel's rule)
Huckel's Standard (4n+2 rule): For a compound to be an aromatic, a particle must have a specific number of pi (electrons with pi bonds, or lone pairs inside p orbitals) inside a shut loop of parallel, adjoining p orbitals. The pi electron tally is characterized by the arrangement of numbers created from 4n+2 where n = zero or any positive whole number (i..e, n = 0, 1, 2, and so forth.). The most widely recognized case in six pi electrons (n = 1) which is found for example in benzene, pyrrole, furan, and pyridine.
where n is the number of pi electrons
where n = 0
(4n +2) pi electrons = 2pi electrons
attached is an example of aromatic which is cyclic, planar and a complete conjugate ring
Answer:
2 p electrons.
Explanation:
For any compound to be considered an aromatic compound it must be cyclic,flat, conjugated and it must obey Huckel's rule that states an aromatic compound must have 4n + 2 pi electrons in it's p orbitals for it to be an aromatic compound.
n can represent an integer from 2, 6,10, 14,........
The lone pair is actually in a pure 2p orbital perpendicular to the ring, which means they count as π electrons.
Answer to the best of your ability please
Answer:
The answer to your question is given below.
Explanation:
To draw the structure of 2–methyl–1–butanamine, we following must be observed:
1. The functional group of the compound is amine –NH2.
2. The functional group is located at carbon 1.
3. The longest continuous carbon chain is carbon 4 i.e butane. Since the functional group is amine, the –e at the end of the butane is replaced with
–amine, making the name to be butanamine.
4. Methyl, CH3 is located at carbon 2.
5. Combine the above to get the structure of 2–methyl–1–butaamine.
Please see attached photo for the structure of 2–methyl–1–butanamine
Sodium pentothal is a short-acting barbiturate derivative used as a general anesthetic and known in popular culture as truth serum. It is synthesized like other barbiturates, but uses thiourea, (H2N)2C=S, in place of urea. The mechanism involves the following steps:
1. Ethoxide ion deprotonates malonic ester, forming enolate anion 1;
2. Enolate anion 1 acts as a nucleophile in an SN2 reaction with ethyl bromide, forming alkylated intermediate 2;
3. Ethoxide ion deprotonates alkylated intermediate 2, forming enolate anion 3;
4. Enolate anion 3 acts as a nucleophile in an SN2 reaction with 2-bromopentane, forming alkylated intermediate 4;
5. Alkylated intermediate 4 reacts with thiourea to form tetrahedral intermediate 5;
6. Tetrahedral intermediate 5 collapses, expelling ethoxide ion and forming intermediate 6;
7. Intermediate 6 reacts with sodium hydroxide to form sodium pentothal.
Draw the mechanism out on a separate sheet of paper and then draw the structure of enolate anion 1. You do not have to consider stereochemistry. You do not have to explicitly draw H atoms. Do not include lone pairs in your answer. They will not be considered in the grading.
Answer:
Figure 1
Explanation:
In this case, we have to ester with a "malonic synthesis" in which we have to add a strong (ethoxide) to produce an enolate ion that would form a new C-C bond with an alkyl halide (ethyl bromide and bromo pentane). Then a "nucleophilic acyl substitution reaction" takes place to add thiourea, in this step two ethanol groups are eliminated to form a cyclic structure. Finally, an "elimination reaction" happen by the addition of sodium hydroxide generating a double bond and a negative charge in the sulfur atom that is neutralized with the positive charge of sodium.
See figure 1 for the total mechanism
I hope it helps!
What is the balanced oxidation half-reaction for the following reaction? Cu2+(aq) + Fe(s) → Cu(s) + Fe2+(aq) Question 8 options: There is no reaction Cu2+(aq) + 2e– → Cu(s) Fe2+(aq) + 2e– → Fe(s) Cu(s) → Cu2+(aq) + 2e– Fe(s) → Fe2+(aq) + 2e–
Answer:
Fe(s) → Fe2+(aq) + 2e-
Explanation:
Cu2+(aq) + Fe(s) → Cu(s) + Fe2+(aq)
Oxidation is the loss of electrons. When the oxidation number of an element increases, that means there is a loss of electrons and that element is being oxidized.
The oxidation half equation in this reaction is;
Fe(s) → Fe2+(aq)
The loss of electron is represented in the product side and is given by;
Fe(s) → Fe2+(aq) + 2e-
The Lucas test has _______ results based on the type of alcohol present because the reaction involves a _________, which is ________ stable for tertiary alcohols compared to primary alcohols. Therefore, tertiary alcohols react ________ primary alcohols.
Answer:
1) positive
2) carbocation
3) most stable
4) faster
Explanation:
A common test for the presence of alcohols can be achieved using the Lucas reagent. Lucas reagent is a mixture of concentrated hydrochloric acid and zinc chloride.
The reaction of Lucas reagent reacts with alcohols leading to the formation of an alkyl chloride. Since the reaction proceeds via a carbocation mechanism, tertiary alcohols give an immediate reaction. Once a tertiary alcohol is mixed with Lucas reagent, the solution turns cloudy almost immediately indicating an instant positive reaction.
Secondary alcohols may turn cloudy within five minutes of mixing the solutions. Primary alcohols do not significantly react with Lucas reagent obviously because they do not form stable carbocations.
Therefore we can use the Lucas reagent to distinguish between primary, secondary and tertiary alcohols.
If a boy (m = 50kg) at rest on skates is pushed by another boy who exerts a force of 200 N on him and if the first boy's final velocity is 8 m/s, what was the contact time? t= s
Answer:
t = 2 seconds
Explanation:
It is given that,
Mass of a boy, m = 50 kg
Initial speed of boy, u = 0
Final speed of boy, v = 8 m/s
Force exerting by another boy, F = 200 N
Let t is the time of contact. The force acting on an object is given by :
F = ma
a is acceleration
So,
[tex]F=\dfrac{m(v-u)}{t}\\\\t=\dfrac{m(v-u)}{F}\\\\t=\dfrac{50\times 8}{200}\\\\t=2\ s[/tex]
So, the contact time is 2 seconds.
Answer:
t=2 s
Explanation:
The idea that light can act as packets led to what new field of science?
A. Quantum mechanics
B. Nuclear mechanics
C. Electrical mechanics
D. Physical mechanics
What is/are the major organic product(s) of the following reaction, Question 2 options: A) CH3CH2CCH Br B) CH3CCCH3 Br C) CH2CH2 HCCH Br D) HCCCH2CH2Br E) HCCBr
Answer:
CH3CH2C≡CH
Explanation:
The particular reaction under study is known as the alkykation of acetylide ions. An acetylide ion can be alkykated using a suitable alkyl halide. The overall scheme of the reaction is;
CH≡C^- + RX -----> RC≡CH + X^-
This reaction is most effective when primary alkyl halides are used. It involves SN2 substitution of a halide in the alkyl halide by an acetylide ion. Secondary, tertiary or even bulky primary substrates are known to yield alkenes and alkynes owing to elimination by E2 mechanism.
An analytical laboratory balance typically measures mass to the nearest 0.1 mg. You may want to reference (Page) Section 21.6 while completing this problem. Part A What energy change would accompany the loss of 0.1 mg in mass
Answer:
The energy change is [tex]E = 9.0 *10^{9}\ J[/tex]
Explanation:
From the question we are told that
Mass loss is [tex]m_l = 0.1 \ mg = 0.1 *10^{-3} mkg = 0.1 *10^{-6} \ kg[/tex]
Generally the energy change that would accompany this loss is mathematically represented as
[tex]E = m * c^2[/tex]
Where c is the speed of light with values [tex]c = 3.0*10^{8} \ m/s[/tex]
[tex]E = 0.1 *10^{-6} * [3.0 *10^{8}]^2[/tex]
[tex]E = 9.0 *10^{9}\ J[/tex]
A 25.0-mL sample of 0.100M Ba(OH)2(aq) is titrated with 0.125 M HCl(aq).
How many milliliters of the titrant will be needed to reach the equivalence point?
Answer:
20.0
Explanation:
NaOH = (25.0) (0.100m) \ 0.125M = 20.0mL
The following reaction is part of the electron transport chain. Complete the reaction and identify which species is reduced. The abbreviation Q represents coenzyme Q. Use the appropriate abbreviation for the product.
FADH2+Q→
The reactant that is reduced is: _____
Answer:
[tex]FADH_2+Q --> FAD + QH_2[/tex]
The reactant that is reduced is Q.
Explanation:
The complete equation for the reaction is such that:
[tex]FADH_2+Q --> FAD + QH_2[/tex]
Two molecules of H atom is lost from [tex]FADH_2[/tex] and the H atoms are gained by the coenzyme Q. Consequently, [tex]FADH_2[/tex] becomes FAD while Q becomes [tex]QH_2[/tex].
From the definition of oxidation as loss of hydrogen and reduction as the addition of hydrogen, it can be concluded that the FADH2 that lost hydrogen is a reactant that is oxidized while the coenzyme Q that gained hydrogen is a reactant that is reduced in the reaction.
What is the molar concentration of H atoms at equilibrium if the equilibrium concentration of H2 is 0.28 M? Express your answer to two significant figures and include the appropriate units.
Answer:
0.56M
Explanation:
Molar concentration is defined as the ratio between moles of solute and volume in liters of solution.
In a 0.28M H₂ there are 0.28moles of H₂ per liter of solution.
Now, in 1 molecule of H₂ there are 2 atoms of H. Following this idea, in 0.28 moles of H₂ there are 0.28*2 = 0.56 moles of H atoms.
Thus, molar concentration of H atoms in a 0.28M H₂ is 0.56M
An aqueous solution is made by dissolving 29.4 grams of aluminum acetate in 433 grams of water. The molality of aluminum acetate in the solution is
Answer:
0.333 m
Explanation:
Molality (m) is moles of solute over kilograms of solvent.
Convert grams of the solute (aluminum acetate) to moles.
(29.4 g)/(204.11 g/mol) = 0.144 mol
Convert grams of the solvent (water) to kilograms.
433 g = 0.433 kg
Divide the solute by the solvent.
(0.144 mol)/(0.433 kg) = 0.333 m
The molality of the solution is 0.333 m.
A 1.0 kg object absorbs 1,303 J of heat energy and experiences a temperature increase of 5.2∘C. What is the object’s specific heat, in joules per gram-degree celsius? Report your answer with the correct number of significant figures.
Answer:
c = 250.58 J/kg/[tex]^{0}C[/tex]
Explanation:
The specific heat of a substance is the required quantity of heat to increase or decrease the temperature of its unit mas by 1 kelvin.
Q = mcΔθ
where: Q is the quantity of heat absorbed or released, m is the mass of the substance, c is its specific heat and Δθ is the change in temperature of the substance.
Given that; m = 1.0 kg, Q = 1303 J and Δθ = 5.2 [tex]^{0}C[/tex], then;
c = Q ÷ (mΔθ)
= 1303 ÷ (1.0 × 5.2)
= 1303 ÷ 5.2
= 250.58 J/kg/[tex]^{0}C[/tex]
The specific heat of the object is 250.58 J/kg/[tex]^{0}C[/tex].
Answer:
0.25
Explanation:
A newly found element with the symbol J has two naturally occurring isotopes. Isotope one has an atomic mass of 139.905 amu and an abundance of 37.25%. Isotope two has an atomic mass of 141.709 amu and an abundance of 62.75%. Calculate the mass of the element.
Answer:
The mass of the element is 141.03701 amu
Explanation:
The catch here is that it notes a " newly found element. " Otherwise you could just refer to the average atomic mass of the element in the periodic table, and receive your solution in a much faster way.
The first isotope has an atomic mass of 139.905 amu, and a respective percent abundance of 37.25%. The second isotope has an atomic mass of 141.709 amu, and the remaining percent abundance, 100% - 37.25% = 62.75% ( given ). We can calculate the mass of the unknown element by associating each percentage with the mass of their respective isotope, over 100%.
Mass = ( ( 139.905 amu )( 37.25% ) + ( 141.709 amu )( 62.75% ) )/ 100,
Mass = ( ( 5211.46125 ) + ( 8892.23975 ) ) / 100,
Mass = ( 14103.701 ) / 100 = 141.03701 amu
1. Suppose 1.00 g of NaOH is used to prepare 250 mL of an NaOH solution. Compare the expected molarity of this solution to the actual average molarity you measured in the standardization. What do you notice? 2. Do you think the results would have been more accurate if a different type of acid or base were used in the standardization? Why, or why not? 3. There are many different primary standards that could be used in a standardization titration. What are the criteria for a primary standard?
Answer:
See explanation
Explanation:
The calculated concentration of the sodium hydroxide is;
Number of moles= mass/molar mass = 1g/40gmol-1 = 0.025 moles
Concentration= number of moles/volume= 0.025×1000/250 = 0.1 M
This calculated concentration will be different from the molarity of NaOH obtained by standardization with acid. The result will not be more accurate if a different acid is used for the standardization this is because sodium hydroxide is deliquescent and absorbs moisture thereby leading to inaccuracy in the calculated molarity.
Any substance that must be used as a primary standard must not absorb moisture, it must be stable and it must be a substance in its pure form.
2) Os foguetes são utilizados para levar pessoas ao espaço (os astronautas), mas principalmente cargas como, por exemplo, os satélites artificiais, os telescópios espaciais, levar sondas a outros planetas etc. Escreva V(verdadeiro) ou F (falso) em cada afirmação.
( ) Foguetes só levam astronautas ao espaço.
( ) Satélites artificiais servem para ajudar na previsão do clima.
( ) Satélites artificiais "fotografam" o planeta para descobrir queimadas ilegais.
( ) Satélites artificiais permitem vermos jogos ao vivo até do Japão.
( ) Foguetes são movidos com pólvora e dinamite.
Answer:
F, V, V , V, F
Explanation:
1 - "Os foguetes são utilizados para levar pessoas ao espaço (os astronautas), mas principalmente cargas como, por exemplo, os satélites artificiais, os telescópios espaciais, levar sondas a outros planetas etc".
2 - Tipo Meteorologia: utilizados para monitorar o tempo e o clima no planeta Terra, por exemplo, os da série Meteosat.
3 - ...
4 - ...
5 - Usam combustivel solido, liquido, hibridos (solido e liquido), iônica:
Solido:
São sistemas simples que unem os dois propelentes envolvidos em uma massa sólida que, quando inflamada, não para de queimar até o esgotamento completo.
Liquido:
São muito mais complexos e envolvem o bombeamento de quantidades imensas de propelentes para as câmaras de combustão dos motores.
Hibridos:
O propelente sólido – normalmente o combustível – é distribuído ao longo do tanque de maneira homogênea. O propelente líquido ou gasoso "normalmente o oxidante" fica armazenado em tanques.
Podem ser desligados depois de sofrerem ignição, além de permitirem um controle de queima relativamente preciso.
Iônica:
Usando eletricidade (captada por painéis solares ou gerada por reatores atômicos) para ionizar átomos (normalmente gases nobres, como xenônio), e expulsá-los em velocidades altíssimas.
What is the percent yield for a chemical reaction if the actual yield is 36 g and the theorical yield is 45 g.
Answer:
⇒ Percent yield = 80 %
Explanation:
Given:
Actual yield = 36 g
Theoretical yield = 45 g
Find:
Percent yield
Computation:
⇒ Percent yield = [Actual yield / Theoretical yield] 100%
⇒ Percent yield = [36 / 45] 100%
⇒ Percent yield =[0.8] 100%
⇒ Percent yield = 80 %
We wear cotton clothes in summer.
Answer:
we wear cotton clothes because it helps to cool us down and remove the excess heat that causes us to feel hot.
Answer:
[tex]\boxed{\mathrm{view \: explanation}}[/tex]
Explanation:
We wear cotton clothes in the summer beacuse cotton absorbs and removes body moisture caused by the sweat and allows better air circulation than fabric clothes.
Content attribution
QUESTION 2 • 1 POINT
Which anion would bond with K+ in a 1: 1 ratio to form a neutral ionic compound?
The given question is incomplete. The complete question is :
Which anion would bond with K+ in a 1: 1 ratio to form a neutral ionic compound?
a) [tex]O^{2-}[/tex]
b) [tex]F^{-}[/tex]
c) [tex]N^{3-}[/tex]
d) [tex]S^{2-}[/tex]
Answer: b) [tex]F^{-}[/tex]
Explanation:
For formation of a neutral ionic compound, the charges on cation and anion must be balanced. The cation is formed by loss of electrons by metals and anions are formed by gain of electrons by non metals.
Here potassium is having an oxidation state of +1 called as cation and thus is an anion must have an oxidation state of -1 if they have to combine in 1: 1 ratio to give neutral ionic compound.
Thus the anion has to be [tex]F^-[/tex] which combines with [tex]K^+[/tex] in 1: 1 ratio to give [tex]KF[/tex]
At 25 °C, what is the hydroxide ion concentration, [OH−] , in an aqueous solution with a hydrogen ion concentration of [H+]=3.0×10−8 M?
Answer:
[tex][OH^-]=3.33x10^{-7}M[/tex]
Explanation:
Hello,
In this case, for the given concentration of hydronium, we can compute the pH as shown below:
[tex]pH=-log([H^+])=-log(3.0x10^{-8})=7.52[/tex]
Now, given the relationship between pH and pOH we can compute the pOH which is directly related with the concentration of hydroxyl in the solution:
[tex]pOH=14-pH=14-7.52=6.48[/tex]
Then, the concentration of hydroxyl turns out:
[tex][OH^-]=10^{-pOH}=10^{-6.48}[/tex]
[tex][OH^-]=3.33x10^{-7}M[/tex]
Best regards.
A vehicle travels 2345 meter in 35 second toward the evening sun in the West. What is its speed? A. 47 m/s West
Explanation:
Speed = 2345 ÷ 35 = 67m/s
Which of the following solutions would have the highest pH? Assume that they are all 0.10 M in acid at 25°C. The acid is followed by its Ka value.
a. HCHO2, 1.8 x 10-4
b. HF, 3.5 x 10-4
c. HClO2, 1.1 x 10-2
d. HCN, 4.9 x 10-10
e. HNO2, 4.6 x 10-4
Answer:
[tex]HCN~~Ka=4.9x10^-^1^0[/tex]
Explanation:
In this case, we have to remember the relationship between the Ka value and the pH. We can use the general reaction for any acid with his Ka value expression:
[tex]HA~->~H^+~+~A^-[/tex] [tex]Ka=\frac{[H^+][A^-]}{[HA]}[/tex]
In the Ka expression, we have a proportional relationship between Ka and the concentration of [tex]H^+[/tex]. Therefore, if we have a higher Ka value we will have a smaller pH (lets keep in mind that with a higher
So, if we have to find the higher pH value we need to search the smaller Ka value in this case [tex]HCN~~Ka=4.9x10^-^1^0[/tex].
I hope helps!
HCN has the highest pH among all the acids listed in the question.
The Ka is called the acid dissociation constant. It shows the extent to which an acid is ionized in water. The pH shows the hydrogen ion concentration of water. The higher the Ka, the higher the hydrogen ion concentration and the lower the pH.
Hence, HCN has the lowest Ka and the lowest hydrogen ion concentration. Therefore, HCN has the highest pH among all the acids listed in the question.
Learn more: https://brainly.com/question/6505878
The reaction, 2 NO(g) + O2(g) → 2 NO2(g), was found to be first order in each of the two reactants and second order overall. The rate law is therefore
Answer:
[tex]r=k[NO][O_2][/tex]
Explanation:
Hello,
In this case, rate laws allows us to compute how fast a chemical reaction is carried out by means of the change in the concentration of the species affecting the rate. In such a way, since the statement says that the reaction was found to be first order to both nitrogen monoxide and oxygen, it means that their concentrations are powered to first power by separated. It also implies that the overall order is second-order since the specific orders are added (powers properties). Therefore, the rate law is:
[tex]r=k[NO][O_2][/tex]
Whereas k is the rate constant and we find the concentration of the reactants to the first power each one.
Best regards.