Answer:
THE MOLAR MASS OF THE GAS IS 147.78 G/MOLE
Explanation:
Using PV = nRT
n = Mass / molar mass
P = 732.6 mmHg = 1 atm = 760 mmHg
So therefore 732.6 mmHg will be equal to 732.6 / 760 = 0.964 atm
P = 0.964 atm
V = 275 mL = 275 *10 ^-3 L
R = 0.082 Latm/ mol K
T = -28 C = 273 - 28 K = 245 K
mass = 1.95 g
molar mass = unknown
Having known the other variables in the formula, the molar mass of the gas can be obtained.
PV = m R T/ molar mass
Molar mass = m RT / PV
Molar mass = 1.95 * 0.082 * 245 / 0.964 * 275 *10^-3
Molar mass = 39.1755 / 265.1 *10^-3
Molar mass = 39.1755 / 0.2651
Molar mass = 147.78 g/mol
The molar mass of the gas is 147.78 g/mol
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.
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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.
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
g Calculate the concentration of sulfate ion when BaSO4 just begins to precipitate from a solution that is 0.0758 M in Ba2+.
Answer:
1.42 × 10⁻⁹ M
Explanation:
Step 1: Given data
Concentration of Ba²⁺ ([Ba²⁺]): 0.0758 MSolubility product constant of BaSO₄ at 25°C: 1.08 × 10⁻¹⁰Step 2: Write the reaction for the solution of BaSO₄
BaSO₄(s) ⇄ Ba²⁺(aq) + SO₄²⁻(aq)
Step 3: Calculate the concentration of SO₄²⁻
We will use the following expression.
Ksp = 1.08 × 10⁻¹⁰ = [Ba²⁺] × [SO₄²⁻]
[SO₄²⁻] = 1.08 × 10⁻¹⁰ / [Ba²⁺] = 1.08 × 10⁻¹⁰ / 0.0758 = 1.42 × 10⁻⁹ M
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.
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If you weighed out 203 mg of the green chloro complex and dissolved it in 24.14 mL of acidic solvent, the molarity of your stock solution would be 0.0295 M. Using your precise value of mass and volume that you entered above, please enter your calculated value for the concentration of the original green chloro complex stock solution in moles per liter.
Mgreen stock =
Answer:
0.00295M
Explanation:
Mass Concentration = mass/vol
= 0.203 g/ 0.02414 L = 8.409 g/L
But molarity = Mass conc / molar mass
∴ Molar mass(mol/L) = mass conc / molarity
= .84909 / 0.0295
= 285.06 g/mol
If 1 mol of green stock - 285.06g
? mol - 0.203 g
= 0.00071213 g
= 0.00071213 g / .2414L = 0.0095 mol/L.
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]
The pH of a solution prepared by mixing 40.00 mL of 0.10 M NH3 with 50.00 mL of 0.10 M NH4Cl and 30mL of 0.05 M H2SO4 is 5.17. Assume that the volume of the solutions are additive . What would be the Ka for NH4
Answer:
Following are the answer to this question:
Explanation:
The value of pH solution is =5.17 So, the p^{OH}:
[tex]p^{OH}[/tex]=14-56.17
=8.823
The volume of the [tex]NH_{3}[/tex] = 40.00 ml
convert into the liter= 0.040L
The value of the concentrated [tex]NH_{3}[/tex] =0.10 M
The volume of the [tex]NH_{4}Cl[/tex]= 50.00 ml
convert into the liter= 0.050L
The value of concentrated [tex]NH_{4}Cl[/tex]= 0.10 M
The volume of the [tex]H_{2}So_{4}[/tex]= 30 ml
convert into the liter= 0.030L
The value of concentrated [tex]H_2So_4[/tex]=0.05 M
Calculating total volume=(0.40+0.050+0.030)
=0.120 L
calculating the new concentrated value of [tex]NH_3[/tex] = [tex]\frac{0.10\times 0.040}{0.120}= 0.33 \ M[/tex]
calculating the new concentrated value of [tex]NH_4Cl[/tex]= [tex]\frac{0.050\times 0.10}{0.120}= 0.04166 \ M[/tex]calculating the new concentrated value of [tex]H_2So_4= \frac{0.030\times 0.05}{0.120}= 0.0125 \ M[/tex] when 1 mol [tex]H_2So_4[/tex] produced 2 mols [tex]H^{+}[/tex] so, 0.0125 in [tex]H_2So_4[/tex]produced:
[tex]=4 \times (2 \times 0.0125) \ mol H^{+}\\\\= 0.025 mol H^{+}[/tex]
create the ICE table:
[tex]NH_3 \ \ \ \ \ \ \ \ + H^{+} \ \ \ \ \ \ \longrightarrow NH_4^{+}[/tex]
I (m) 0.033(m) 0.025 0.04166
C -0.025 -0.025 + 0.025
E 8.3\times 10^{-3} 0 0.0667
now calculating pH:
when ph= 8.83:
[tex]P^{H}= p^{kb}|+ \log\frac{[NH_4^{+}]}{[NH_3]}\\\\8.83=p^{kb}+\log\frac{0.0667}{8.3 \times 10^{-3}}\\\\p^{kb}=8.83-0.9069\\\\ \ \ \ =7.7231 \\\\\ The P^{kb} \ for \ NH_3 \ is =7.7231\\\\\ The P^{kb} \ for N^{+}H_4=14-7.7231\\\\\ \ \ \ \ \ =6.2769[/tex]
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.
Water (2190 g ) is heated until it just begins to boil. If the water absorbs 5.83×105 J of heat in the process, what was the initial temperature of the water?
Answer:
The initial temperature was [tex]36.4^\circ \:C[/tex]
Explanation:
[tex]\Delta t=\frac{q}{m\cdot C_s}=\frac{5.83\times10^5}{2190\times 4.184}\\\\=63.6^\circ\:C[/tex]
The temperature difference [tex]=100-63.6=36.4^\circ\:C[/tex]
Best Regards!
hen adding a solute to water, the vapor pressure will __________ and the boiling point will __________.
Answer: When a solute is added to water, the vapor pressure will decrease and the boiling point will increase.
Explanation:
When a solute is added to water, a solvent's vapor pressure will decrease because of the displacement of solvent molecules by the solute. i.e. some of the solvent molecules at the surface of the water are replaced by the solute.When a solute is added to water, a solvent's boiling point will increase because water molecules need more energy to produce required pressure to escape the boundary of the liquid , so as the number of particles increase in the liquid it increase the boiling point.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.
Hydrazine, N2H4 , reacts with oxygen to form nitrogen gas and water. N2H4(aq)+O2(g)⟶N2(g)+2H2O(l) If 2.45 g of N2H4 reacts with excess oxygen and produces 0.450 L of N2 , at 295 K and 1.00 atm, what is the percent yield of the reaction?
Answer:
24.15%
Explanation:
According to the given situation the computation of the percent yield of the reaction is shown below:-
PV = NRT = N = [tex]\frac{PV}{RT}[/tex]
Mole of [tex]N_2[/tex] = [tex]\frac{PV}{RT}[/tex]
= [tex]\frac{1\times 0.450}{0.0821\times 295}[/tex]
= [tex]\frac{0.450}{24.2195}[/tex]
= 0.0186
Mole of [tex]N_2H_4 = \frac{2.45}{32}[/tex]
= 0.077
Now, the percentage of yield is
= [tex]\frac{Practical\ yield}{Theoretical\ yield}\times 100[/tex]
= [tex]\frac{0.0186}{0.077}\times 100[/tex]
= 24.15%
Therefore for computing the percentage of yield we simply divide the practical yield by theoretical yield and multiply with 100 so that we can get the result into the percentage form.
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.
When the optically active carboxylic acid below is decarboxylated using the conditions typical in the acetoacetate synthesis, will the ketone product also be optically active?
Answer:
ye, it will be optically active
Explanation:
a compound is said to be optically active if it can optically rotate.
the removal of carboxyl group and release of cabon dioxide from carboxylic acid in acetoacetate synthesis which will result in production of ketone as given the attachment below.
13C NMR is a technique in which the total number of signals represents the number of unique carbon atoms in a molecule. Propose a structure that is consistent with the following data.
a. The IR includes peaks at 1603 and 1495 cm^-1
b. The 13c NMR has a total of 7 signals
c. The compound has one acidic proton.
Answer:
D. Poop Butt.
Explanation:
Based on the given data, we can propose a possible structure that fits the criteria: a. carbonyl group (C=O) and an aromatic ring b. there are seven unique carbon environments. c. Presence of a functional group like a carboxylic acid or phenol .
a. The IR peaks at 1603 [tex]cm^{-1}[/tex] and 1495[tex]cm^{-1}[/tex]suggest the presence of both a carbonyl group (C=O) and an aromatic ring.
b. The 13C NMR having a total of 7 signals indicates that there are seven unique carbon environments in the molecule.
c. Considering the presence of an acidic proton, it suggests the presence of a functional group like a carboxylic acid (COOH) or phenol ([tex]C_6H_5OH[/tex]).
Putting all this information together, a possible structure that fits the data could be benzoic acid ([tex]C_6H_5COOH[/tex]). It contains a benzene ring (giving 6 unique carbon environments), a carbonyl group (giving 1 unique carbon environment), and an acidic proton in the carboxylic acid group. This structure satisfies all the given data.
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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
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
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 major organic product obtained from the following sequence of reactions? PhCH2CHO PhCH2CH2CHO PhCH2CH2COOH PhCH2COOH
Answer:
PhCH2CH2COOH
Explanation:
This is a reaction of PhCH2CH2Br with KCN in the presence of H3O^+. The reaction first leads to the formation of PhCH2CH2CN.
We must recall that part of the properties of nitriles is that they can be converted to carboxylic acids in the presence of H3O^+. This is a common synthetic route for carboxylic acids.
Therefore, when the PhCH2CH2CN is now further reacted with H3O^+, the carboxylic acid PhCH2CH2COOH is formed as the major organic product of the reaction, hence the answer given above.
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.
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
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.
Identify a process that is NOT reversible. A. melting of steel B. freezing water C. melting of ice D. frying an egg E. deposition of carbon dioxide (gas to solid)
A process that is not a reversible reaction is frying an egg.
What are reversible reactions?Reversible reactions are those reactions in which product will again change into the reactant.
Melting of steel and ice are reversible reaction as after cooling again we get the original state of steel and ice.Freezing of water is also reversible reaction as at normal temperature we get the original state of water.Deposition of carbon dioxide is also a reversible reaction.Frying an egg is a non reversible reaction as after frying an egg we didn't get the original egg again.Hence option (D) is correct.
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Hydrogen iodide decomposes according to the equation: 2HI(g) H 2(g) + I 2(g), K c = 0.0156 at 400ºC A 0.660 mol sample of HI was injected into a 2.00 L reaction vessel held at 400ºC. Calculate the concentration of HI at equilibrium.
Answer:
[HI] = 0.264M
Explanation:
Based on the equilibrium:
2HI(g) ⇄ H₂(g) + I₂(g)
It is possible to define Kc of the reaction as the ratio between concentration of products and reactants using coefficients of each compound, thus:
Kc = 0.0156 = [H₂] [I₂] / [HI]²
As initial concentration of HI is 0.660mol / 2.00L = 0.330M, the equlibrium concentrations will be:
[HI] = 0.330M - 2X
[H₂] = X
[I₂] = X
Where X is reaction coefficient.
Replacing in Kc:
0.0156 = [X] [X] / [0.330M - 2X]²
0.0156 = X² / [0.1089 - 1.32X + 4X² ]
0.00169884 - 0.020592 X + 0.0624 X² = X²
0.00169884 - 0.020592 X - 0.9376 X² = 0
Solving for X:
X = - 0.055 → False solution, there is no negative concentrations
X = 0.0330 → Right solution.
Replacing in HI formula:
[HI] = 0.330M - 2×0.033M
[HI] = 0.264MAt 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.
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-
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 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
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