Answer:
0.394 atm
Explanation:
Mathematically, when we increase the pressure of a gas, we are increasing its concentration and when we decrease the pressure, we are decreasing its concentration.l at same temperature
What this means is that pressure and concentration are directly proportional.
Representing concentration by c and pressure by p, we have;
P1/C1 = P2/C2
From the question;
P1 = 0.719 atm
P2 = ?
C1 = 429.7 ppm
C2 = 235.3 ppm
Now, we can rewrite the equation to be;
P1C2/C1 = P2
Substituting the values we have;
0.719 * 235.3/429.7 = 0.394 atm
A student accidentally let some of the vapor escape the beaker. As a result of this error, will the mass of naphthalene you record be too high, too low, or unaffected? Why?
Answer:
too low
Explanation:
If our aim is to recover the naphthalene and measure its mass after separation, then we must not allow any vapour to escape.
Naphthalene is a sublime substance, it can be separated by sublimation. It changes directly from solid to gas. This vapour must be kept securely so that none of it escapes. If part of the naphthalene vapour happens to escape accidentally, then the measured mass of naphthalene will be too low compared to the mass of naphthalene originally present in the mixture.
Which one of the following is most likely to gain electrons when forming an ion, based on the natural tendency of the element?
A Ni
B S
C Na
D Cr
E Be
Answer:
Option B. S
Explanation:
All of the options except sulphur, S is metal.
Metals tend to lose electron in order to form ion. Non metals on the other hand gain electron to form ion.
Sulphur, S has atomic number of 16 with electronic configuration as:
S (16) => 1s² 2s²2p⁶ 3s²3p⁴
From the above illustration, we can see that sulphur needs two more electrons to complete it's octet configuration.
Therefore, sulphur, S will gain two electrons to form ion.
As stated earlier, the rest option given are all metals which will form ion by losing electron(s).
Answer
B) Sulphur (S)
Explanation
Here in the options we have been provided with elements like Nickel (Ni), Sulphur (S), Sodium (Na), Chromium (Cr) and Beryllium (Be) but except for Sulphur all the other ones are metals.
Now, let us understand what is a metal and a non-metal.
Metal- electron donors are called as metal.Non-metal- electron acceptors are called non-metals.So, sulphur being the only non metal will accept electron to complete its octate and to stablize itself and form a Anion.
Now let us also look at the electronic configuration of Sulphur to get the picture more clearly
atomic no. of sulphur would be = 16[tex]S\rightarrow 1s^2\; 2s^2\;2p^6\;3s^2\;3p^4[/tex]
so here the p-subshell is incomplete and is in need of 2 electrons.
Therefore the element which is most likely to gain electrons, forming an Anion will be sulphur.
To learn more about Ion Formation
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Potassium iodide reacts with lead(II) nitrate in the following precipitation reaction: 2 KI(aq) + Pb(NO3)2(aq) → 2 KNO3(aq) + PbI2(s) What minimum volume of 0.400 M potassium iodide solution is required to completely precipitate all of the lead in 310.0 mL of a 0.112 M lead(II) nitrate solution?
Answer:
0.1736 L or 173.6 ml
Explanation:
Number of moles of lead II nitrate is obtained by;
Number of moles = concentration × volume of solution
Concentration= 0.112 M
Volume of solution= 310 ml
n= 0.112 × 310/1000
n= 0.03472 moles
From the reaction equation;
2 moles of potassium iodide reacted with 1 mole of lead II nitrate
x moles of potassium iodide will react with 0.03472 moles of lead II nitrate
x= 2 × 0.03472 moles= 0.06944 moles of potassium iodide
Volume of potassium iodide solution = number of moles/ concentration = 0.06944/ 0.4
Volume of potassium iodide solution= 0.1736 L or 173.6 ml
Liquids A, B, and C are insoluble in one another (i.e., they are immiscible). A, B, and C have densities of 0.780 g/cm3, 1.102 g/cm3 , and 1.040 g/cm3, respectively. Which drawing represents the result of placing all three liquids into the same graduated cylinder?
Answer:
The drawing that represents the result of placing all three liquids into the same graduated cylinder will have the liquid arranged one on top of the other from top to bottom in the order of A, C, B.
Explanation:
The image with the options is not provided in this question, but I can answer this fairly so that you can pick from the question, the correct drawing.
We know that two or more immiscible liquids contained together in a container will always separate in the order of their density from top to bottom, with the densest at the bottom, and the least densest at the top. In this case, liquid A is the least densest, and liquid B is the densest. Liquid A will stay on top, and liquid B will be at the bottom. Liquid C will be in between liquid A and liquid B.
suppose you are titrating vinegar, which is an acetic acid solution
Answer:
0.373 M
Explanation:
The balanced equation for the reaction is given below:
HC2H3O2 + NaOH —> NaC2H3O2 + H2O
From the balanced equation above, the following were obtained:
Mole ratio of the acid, HC2H3O2 (nA) = 1
Mole ratio of the base, NaOH (nB) = 1
Next, we shall write out the data obtained from the question. This include:
Volume of base, NaOH (Vb) = 32.17 mL
Molarity of base, NaOH (Mb) = 0.116 M
Volume of acid, HC2H3O2 (Va) = 10 mL
Molarity of acid, HC2H3O2 (Ma) =..?
The molarity of the acid solution can be obtained as follow:
MaVa/MbVb = nA/nB
Ma x 10 / 0.116 x 32.17 = 1
Cross multiply
Ma x 10 = 0.116 x 32.17
Divide both side by 10
Ma = (0.116 x 32.17) /10
Ma = 0.373 M
Therefore, the concentration of the acetic acid is 0.373 M.
For dinner you make a salad with lettuce, tomatoes, cheese, carrots, and
croutons. Your salad would be classified as a(n)
O A. compound
OB. element
OC. homogeneous mixture
D. heterogeneous mixture
A heterogeneous mixture
The equilibrium constant for the reaction is 1.1 x 106 M. HONO(aq) + CN-(aq) ⇋ HCN(aq) + ONO-(aq) This value indicates that
The given question is incomplete. The complete question is given here :
The equilibrium constant for the reaction is [tex]1.1\times 10^6[/tex] M.
[tex]HONO(aq)+CN^- (aq)\rightleftharpoons HCN(aq)+ONO^-(aq)[/tex]
This value indicates that
A. [tex]CN^-[/tex] is a stronger base than [tex]ONO^-[/tex]
B. HCN is a stronger acid than HONO
C. The conjugate base of HONO is [tex]ONO^-[/tex]
D. The conjugate acid of CN- is HCN
Answer: A. [tex]CN^-[/tex] is a stronger base than [tex]ONO^-[/tex]
Explanation:
Equilibrium constant is the ratio of product of the concentration of products to the product of concentration of reactants.
When [tex]K_{p}>1[/tex]; the reaction is product favoured.
When [tex]K_{p};<1[/tex] ; the reaction is reactant favored.
[tex]When K_{p}=1[/tex]; the reaction is in equilibrium.
As, [tex]K_p>>1[/tex], the reaction will be product favoured and as it is a acid base reaction where [tex]HONO[/tex] acts as acid by donating [tex]H^+[/tex] ions and [tex]CN^-[/tex] acts as base by accepting [tex]H^+[/tex]
Thus [tex]HONO[/tex] is a strong acid thus [tex]ONO^-[/tex] will be a weak conjugate base and [tex]CN^-[/tex] is a strong base which has weak [tex]HCN[/tex] conjugate acid.
Thus the high value of K indicates that [tex]CN^-[/tex] is a stronger base than [tex]ONO^-[/tex]
write the IUPAC name OF THE FOLLOWING COMPOUNDS
Answer:
Explanation:
a) 2 chloro butane
b) 2-3 dimethyl butane
c) 2 bromo 3 nitro pentane
d) 2-3 trimethyl pentane
e) 2-bromo,3-methyl,4-nitro hexane
f) 2-methyl cyclo butane
Sometimes a nuclide is referenced by the name of the element followed by the:______
a. atomic number
b. mass number
c. electrical charge
d. none of the above
Answer:
The correct option is d
Explanation:
Nuclide is synonymous with groups of electrons or protons, that is, a nuclide is the grouping of nucleons.
Considering that catalysts are not consumed in a reaction, how do you think increasing the amount of catalyst would affect the reaction rate for the decomposition of hydrogen peroxide?
a. increase
b. decrease
c. no effect
Answer:
a. increase
Explanation:
Catalysis is the process of increasing the rate of a chemical reaction by adding a substance known as a catalyst, which is not consumed in the catalyzed reaction.
By default, catalysts exists to speed up the rate of reactions. Increasing the amount of catalysts means that there would be an increase in the rate of reaction. The correct option is A.
Ammonia is oxidized with air to form nitric oxide in the first step of the production of nitric acid. Two principal gas-phase reactions occur:
Answer:
4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O
2NO(g) + O₂(g) → 2 NO₂
Explanation:
First of all, we need to consider the reaction for production of ammonia. In this reaction we have as reactants, nitrogen and hydroge.
3H₂ (g) + N₂(g) → 2NH₃ (g)
Afterwards, ammonia reacts to oxygen, to produce NO and H₂O
The equation for the process will be:
4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O
Then, we take the nitric oxide to make it react, to produce NO₂, in order to produce nitric acid, for the final reaction:
2NO(g) + O₂(g) → 2 NO₂
3NO₂(g) + H₂O(g) → 2 HNO₃ (g) + NO(g)
In a combustion chamber, ethane (C2H6) is burned at a rate of 8 kg/h with air that enters the combustion chamber at a rate of 176 kg/h. Determine the percentage of excess air used during this process.
Answer:
37%
Explanation:
From the question, the equation goes does.
C2H6+ (1-x)+a(O2+3.76N2)=bC02 + cH2O + axO2 + 3.76dN2.
Mair=Mair/Rin
( MN)O2 + (MN)N2÷ (MN)O2 + (MN)N2 +(MN)C2H6.
33 . 3.25(1-x) + 28 × 13.16(1-x) ÷ 33 × 3.25(1-x) + 28 × 13.16(1-x). + 30.1
= 176/176+8
X= 0.37
0.37 × 100
X= 37%
A civil engineer designs mostly:
A. building structures.
B. computer parts.
C. new foods.
D. technology that flies.
Provide the reagents necessary to carry out the following conversion. Group of answer choices KMnO4, NaOH,H2O KMnO4, H3O , 75oC H2SO4, heat 1. mCPBA 2. H3O none of these
Answer:
KMnO4,H3O^+,75°C
Explanation:
The conversion of cyclohexene to trans-1,2-cylohexanediol is an oxidation reaction. Alkenes are oxidized in the presence of potassium permanganate and acids to yield the corresponding diols.
These diols may also be called glycols. They are molecules that contain two -OH(hydroxyl) groups per molecule. The reaction closely resembles the addition of the two -OH groups of hydrogen peroxide to an alkene.
The bright color of potassium permanganate disappears in this reaction so it can be used as a test for alkenes.
What is the osmolarity of a 0.20 M solution of KCI?
A) 0.40 Osmol
B) 0.30 Osmol C) 0.20 Osmol D) 0.80 Osmol
E) 0.10 Osmol
Answer:
Osmolarity of solution of KCI = 0.40 osmol
Explanation:
Given:
KCL ⇒ K⁺ + Cl⁻
Find:
Osmolarity of solution of KCI
When M = 0.20 M
Computation:
1 mole of KCL = 2 osmol
1 M of KCl = 2 Osmolarity
So,
Osmolarity of solution of KCI = 2 × 0.20
Osmolarity of solution of KCI = 0.40 osmol
When alkanes react with chlorine in the presence of ultraviolet light, chlorine atoms substitute for one or more alkane hydrogen atoms. What is the number of different chloroalkane compounds that can be formed by the reaction of C2H6 with chlorine?
Answer:
6
Explanation:
Alkanes undergo substitution reaction so the number of replacement reaction hydrogen is 6
A compound decomposes with a half-life of 8.0 s and the half-life is independent of the concentration. How long does it take for the concentration to decrease to one-ninth of its initial value
Answer:
The concentration takes 25.360 seconds to decrease to one-ninth of its initial value.
Explanation:
The decomposition of the compound has an exponential behavior and process can be represented by this linear first-order differential equation:
[tex]\frac{dc}{dt} = -\frac{1}{\tau}\cdot c(t)[/tex]
Where:
[tex]\tau[/tex] - Time constant, measured in seconds.
[tex]c(t)[/tex] - Concentration of the compound as a function of time.
The solution of the differential equation is:
[tex]c(t) = c_{o} \cdot e^{-\frac{t}{\tau} }[/tex]
Where [tex]c_{o}[/tex] is the initial concentration of the compound.
The time is now cleared in the result obtained previously:
[tex]\ln \frac{c(t)}{c_{o}} = -\frac{t}{\tau}[/tex]
[tex]t = -\tau \cdot \ln \frac{c(t)}{c_{o}}[/tex]
Time constant as a function of half-life is:
[tex]\tau = \frac{t_{1/2}}{\ln 2}[/tex]
Where [tex]t_{1/2}[/tex] is the half-life of the composite decomposition, measured in seconds.
If [tex]t_{1/2} = 8\,s[/tex], then:
[tex]\tau = \frac{8\,s}{\ln 2}[/tex]
[tex]\tau \approx 11.542\,s[/tex]
And lastly, given that [tex]\frac{c(t)}{c_{o}} = \frac{1}{9}[/tex] and [tex]\tau \approx 11.542\,s[/tex], the time taken for the concentration to decrease to one-ninth of its initial value is:
[tex]t = -(11.542\,s)\cdot \ln\frac{1}{9}[/tex]
[tex]t \approx 25.360\,s[/tex]
The concentration takes 25.360 seconds to decrease to one-ninth of its initial value.
The addition of 0.242 L of 1.92 M KCl to a solution containing Ag+ and Pb2+ ions is just enough to precipitate all of the ions as AgCl and PbCl2. The total mass of the resulting precipitate is 65.08 g. Find the mass of PbCl2 and AgCl in the precipitate. Calculate the mass of PbCl2 and AgCl in grams.
Answer:
Mass PbCl₂ = 50.24g
Mass AgCl = 14.84g
Explanation:
The addition of Cl⁻ ions from the KCl solution results in the precipitation of AgCl and PbCl₂ as follows:
Ag⁺ + Cl⁻ → AgCl(s)
Pb²⁺ + 2Cl⁻ → PbCl₂(s)
If we define X as mass of PbCl₂, moles of Cl⁻ from PbCl₂ are:
Xg × (1mol PbCl₂/ 278.1g) × (2moles Cl⁻ / 1 mole PbCl₂) = 0.00719X moles of Cl⁻ from PbCl₂
And mass of AgCl will be 65.08g-X. Moles of Cl⁻ from AgCl is:
(65.08g-Xg) × (1mol AgCl/ 143.32g) × (1mole Cl⁻ / 1 mole AgCl) = 0.45409 - 0.00698X moles of Cl⁻ from AgCl
Moles of Cl⁻ that were added in the KCl solution are:
0.242L × (1.92mol KCl / L) × (1mole Cl⁻ / 1 mole KCl) = 0.46464 moles of Cl⁻ added.
Moles Cl⁻(AgCl) + Moles Cl⁻(PbCl₂) = Moles Cl⁻(added)
0.45409 - 0.00698X moles + (0.00719X moles) = 0.46464 moles
0.45409 + 0.00021X = 0.46464
0.00021X = 0.01055
X = 0.01055 / 0.00021
X = 50.24g
As X = Mass PbCl₂
Mass PbCl₂ = 50.24gAnd mass of AgCl = 65.08 - 50.24
Mass AgCl = 14.84gThe masses of the compounds in the precipitate can be found my knowing
the number of moles of chloride ion contributed by each compound.
The mass of PbCl₂ in the precipitate is approximately 49.24 gThe mass of AgCl in the precipitate is approximately 15.84 gReasons:
The given parameter are;
Volume of KCl solution added = 0.242 L
Concentration of KCl solution = 1.92 M KCl
The ions in the solution to which KCl is added = Ag⁺ and Pb²⁺ ions
Precipitates formed = AgCl and PbCl₂
The mass of the precipitate = 65.08 g
Required:
The mass of PbCl₂ and AgCl in the precipitate
Solution;
Number of moles of chloride ions in a mole of PbCl₂ = 2 moles
Number of moles of chloride ions in a mole of AgCl = 1 mole
Let X represent the mass of PbCl₂ in the precipitate, we have;
The mass of AgCl in the precipitate = 65.08 g - X
[tex]\mathrm{Number \ of \ moles \ of \ PbCl_2} = \dfrac{X \, g}{278.1 \, g} =\mathbf{ \dfrac{X }{278.1}}[/tex]
Number of moles of chloride ions from PbCl₂ is therefore;
[tex]\mathrm{Number \ of \ moles \ of \ Cl^- from \ PbCl_2} =\mathbf{ 2 \times \dfrac{X }{278.1} \ moles \ of \ Cl^-}[/tex]
[tex]\mathrm{Number \ of \ moles \ of \ AgCl \ in \ the \ precipitate} = \dfrac{65.08 -X }{143.32}[/tex]
[tex]\mathrm{Number \ of \ moles \ of \ Cl^- from \ AgCl} = \mathbf{ \dfrac{65.08 -X }{143.32}} \ moles \ of \ Cl^-[/tex]
The number of moles of chloride ions from one mole of KCl = 1 mole
Number of moles of chloride ions from 0.242 L of 1.92 M KCl is therefore;
0.242 L × 1.92 moles/L = 0.46464 moles
Number of moles of chloride ions from KCl = 0.46464 moles
[tex]0.46464 \ moles \ from \ KCl = \overbrace{ \dfrac{ 2 \times X }{278.1} + \dfrac{65.08 -X }{143.32}} \ moles \ in \ PbCl_2 \ and \ AgCl[/tex]
Which gives;
[tex]\displaystyle \frac{192}{896089} \cdot X + \frac{1627}{3583} = \frac{1452}{3125}[/tex]
Therefore;
[tex]\displaystyle X = \frac{\frac{1452}{3125} - \frac{1627}{3583} }{ \frac{192}{896089} } = \frac{105864850549}{2149800000} \approx \mathbf{ 49.24}[/tex]
The mass of PbCl₂ in the precipitate, X ≈ 49.24 g
The mass of AgCl in the precipitate = 65.08 g - 49.24 g ≈ 15.84 g
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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.
. Calculate the final Celsius temperature of sulfur dioxide gas if 50.0 mL of the gas at 20 C and 0.450 atm is heated until the pressure is 0.750 atm. Assume that the volume remains constant.
Answer:
The final temperature of sulfur dioxide gas is 215.43 C
Explanation:
Gay Lussac's Law establishes the relationship between the temperature and the pressure of a gas when the volume is constant. This law says that if the temperature increases the pressure increases, while if the temperature decreases the pressure decreases. In other words, the pressure and temperature are directly proportional quantities.
Mathematically, the Gay-Lussac law states that, when a gas undergoes a transformation at constant volume, the quotient of the pressure exerted by the temperature of the gas remains constant:
[tex]\frac{P}{T}=k[/tex]
Assuming you have a gas that is at a pressure P1 and at a temperature T1 at the beginning of the experiment, by varying the temperature to a new value T2, then the pressure will change to P2, and it will be true:
[tex]\frac{P1}{T1} =\frac{P2}{T2}[/tex]
The reference temperature is the absolute temperature (in degrees Kelvin)
In this case:
P1= 0.450 atmT1= 20 C= 293.15 K (being 0 C= 273.15 K)P2=0.750 atmT2= ?Replacing:
[tex]\frac{0.450atm}{293.15 K} =\frac{0.750 atm}{T2}[/tex]
Solving:
[tex]T2 =\frac{0.750 atm}{\frac{0.450atm}{293.15 K} }[/tex]
[tex]T2=\frac{0.750 atm}{0.450 atm} *293.15K[/tex]
T2=488.58 K
Being 273.15 K= 0 C, then 488.58 K= 215.43 C
The final temperature of sulfur dioxide gas is 215.43 C
A 27.9 mL sample of 0.289 M dimethylamine, (CH3)2NH, is titrated with 0.286 M hydrobromic acid.
(1) Before the addition of any hydrobromic acid, the pH is___________.
(2) After adding 12.0 mL of hydrobromic acid, the pH is__________.
(3) At the titration midpoint, the pH is___________.
(4) At the equivalence point, the pH is________.
(5) After adding 45.1 mL of hydrobromic acid, the pH is_________.
Answer:
(1) Before the addition of any HBr, the pH is 12.02
(2) After adding 12.0 mL of HBr, the pH is 10.86
(3) At the titration midpoint, the pH is 10.73
(4) At the equivalence point, the pH is 5.79
(5) After adding 45.1 mL of HBr, the pH is 1.18
Explanation:
First of all, we have a weak base:
0 mL of HBr is added(CH₃)₂NH + H₂O ⇄ (CH₃)₂NH₂⁺ + OH⁻ Kb = 5.4×10⁻⁴
0.289 - x x x
Kb = x² / 0.289-x
Kb . 0.289 - Kbx - x²
1.56×10⁻⁴ - 5.4×10⁻⁴x - x²
After the quadratic equation is solved x = 0.01222 → [OH⁻]
- log [OH⁻] = pOH → 1.91
pH = 12.02 (14 - pOH)
After adding 12 mL of HBrWe determine the mmoles of H⁺, we add:
0.286 M . 12 mL = 3.432 mmol
We determine the mmoles of base⁻, we have
27.9 mL . 0.289 M = 8.0631 mmol
When the base, react to the protons, we have the protonated base plus water (neutralization reaction)
(CH₃)₂NH + H₃O⁺ ⇄ (CH₃)₂NH₂⁺ + H₂O
8.0631 mm 3.432 mm -
4.6311 mm 3.432 mm
We substract to the dimethylamine mmoles, the protons which are the same amount of protonated base.
[(CH₃)₂NH] → 4.6311 mm / Total volume (27.9 mL + 12 mL) = 0.116 M
[(CH₃)₂NH₂⁺] → 3.432 mm / 39.9 mL = 0.0860 M
We have just made a buffer.
pH = pKa + log (CH₃)₂NH / (CH₃)₂NH₂⁺
pH = 10.73 + log (0.116/0.0860) = 10.86
Equivalence pointmmoles of base = mmoles of acid
Let's find out the volume
0.289 M . 27.9 mL = 0.286 M . volume
volume in Eq. point = 28.2 mL
(CH₃)₂NH + H₃O⁺ ⇄ (CH₃)₂NH₂⁺ + H₂O
8.0631 mm 8.0631mm -
8.0631 mm
We do not have base and protons, we only have the conjugate acid
We calculate the new concentration:
mmoles of conjugated acid / Total volume (initial + eq. point)
[(CH₃)₂NH₂⁺] = 8.0631 mm /(27.9 mL + 28.2 mL) = 0.144 M
(CH₃)₂NH₂⁺ + H₂O ⇄ (CH₃)₂NH + H₃O⁻ Ka = 1.85×10⁻¹¹
0.144 - x x x
[H₃O⁺] = √ (Ka . 0.144) → 1.63×10⁻⁶ M
pH = - log [H₃O⁺] = 5.79
Titration midpoint (28.2 mL/2)This is the point where we add, the half of acid. (14.1 mL)
This is still a buffer area.
mmoles of H₃O⁺ = 4.0326 mmol (0.286M . 14.1mL)
mmoles of base = 8.0631 mmol - 4.0326 mmol
[(CH₃)₂NH] = 4.0305 mm / (27.9 mL + 14.1 mL) = 0.096 M
[(CH₃)₂NH₂⁺] = 4.0326 mm (27.9 mL + 14.1 mL) = 0.096 M
pH = pKa + log (0.096M / 0.096 M)
pH = 10.73 + log 1 = 10.73
Both concentrations are the same, so pH = pKa. This is the maximum buffering capacity.
When we add 45.1 mL of HBrmmoles of acid = 45.1 mL . 0.286 M = 12.8986 mmol
mmoles of base = 8.0631 mmoles
This is an excess of H⁺, so, the new [H⁺] = 12.8986 - 8.0631 / Total vol.
(CH₃)₂NH + H₃O⁺ ⇄ (CH₃)₂NH₂⁺ + H₂O
8.0631 mm 12.8986 mm -
- 4.8355 mm
[H₃O⁺] = 4.8355 mm / (27.9 ml + 45.1 ml)
[H₃O⁺] = 4.8355 mm / 73 mL → 0.0662 M
- log [H₃O⁺] = pH
- log 0.0662 = 1.18 → pH
Calculate the pH of the 1L buffer composed of 500 mL 0.60 M acetic acid plus 500 mL of 0.60 M sodium acetate, after 0.010 mol of NaOH is added (Ka HC2H3O2 = 1.75 x 10-5). Report your answer to the hundredths place.
Answer:
pH = 4.79
Explanation:
The pH of the acetic buffer can be determined using H-H equation:
pH = pKa + log [A⁻] / [HA]
Where pKa is -logKa = 4.76
pH = 4.76 + log [sodium Acetate] / [Acetic Acid]
Where [] can be taken as moles of each specie.
Thus, to find pH of the buffer we need to calculate molesof acetic acid and sodium acetate.
Initial moles:
Initial moles of acetic acid and sodium acetate are:
500mL = 0.500L ₓ (0.60moles / L) = 0.30 moles of both acetic acid and sodium acetate
Moles after reaction:
Now, 0.010 moles of NaOH are added to the buffer reacting with acetic acid, CH₃COOH, producing more acetate ion, as follows:
NaOH + CH₃COOH → CH₃COO⁻ + H₂O
That means after reaction moles of both species are:
Acetic acid: 0.30mol - 0.010mol (Moles that react) = 0.29 moles
Acetate: 0.30mol + 0.010mol (Moles produced) = 0.31 moles
Replacing in H-H equation:
pH = 4.76 + log [0.31] / [0.29]
pH = 4.79
Resonance Structures are ways to represent the bonding in a molecule or ion when a single Lewis structure fails to describe accurately the actual electronic structure. Equivalent resonance structures occur when there are identical patterns of bonding within the molecule or ion. The actual structure is a composite, or resonance hybrid, of the equivalent contributing structures. Draw Lewis structures for thecarbonate ion and for phosphine in which the central atom obeys the octet rule. ... How many equivalent Lewis structures are necessary to describe the bonding in CO32-
Answer:
See explanation
Explanation:
A Lewis structure is also called a dot electron structure. A Lewis structure represents all the valence electrons on atoms in a molecule as dots. Lewis structures can be used to represent molecules in which the central atom obeys the octet rule as well as molecules whose central atom does not obey the octet rule.
Sometimes, one Lewis structure does not suffice in explaining the observed properties of a given chemical specie. In this case, we evoke the idea that the actual structure of the chemical specie lies somewhere between a limited number of bonding extremes called resonance or canonical structures.
The canonical structure of the carbonate ion as well as the lewis structure of phosphine is shown in the image attached to this answer.
What element is primarily used in appliances to make electronic chips
A. Silicon (Si)
B. Nickel (Ni)
C. Copper (Cu)
D. Selenium (Se)
Answer:
Option A
Explanation:
Silicon (Obtained from Sand (SiO2)) is the element that is primarily used in appliances to make electronic chips.
Answer:
A. Silicon (Si)
Explanation:
Silicon (Si) is primarily used as a semiconductor material to make electronic chips.
The decomposition of H2O2 is first order in H2O2 and the rate constant for this reaction is 1.63 x 10-4 s-1. How long will it take for [H2O2] to fall from 0.95 M to 0.33 M?
Answer:
It will take 6486.92 minutes for [H2O2] to fall from 0.95 M to 0.33 M
Explanation:
The order of reaction is defined as the sum of the powers of the concentration terms in the equation. Order of a reaction is given by the number of atoms or molecule whose concentration change during the reaction and determine the rate of reaction.
In first order reaction;
[tex]In \dfrac{a}{a_o-x}= k_1 t[/tex]
where;
a = concentration at time t
[tex]a_o[/tex] = initial concentration
and k = constant.
[tex]In (\dfrac{0.33}{0.95})= -1.63 \times 10^{-4} \times t[/tex]
[tex]-1.05736933 = -1.63 \times 10^{-4} \times t[/tex]
[tex]t = \dfrac{-1.05736933}{ -1.63 \times 10^{-4} }[/tex]
t = 6486.92 minutes
A sample of an unknown gas effuses in 11.1 min. An equal volume of H2 in the same apparatus at the same temperature and pressure effuses in 2.42 min. What is the molar mass of the unknown gas
Answer:
Molar mass of the gas is 0.0961 g/mol
Explanation:
The effusion rate of an unknown gas = 11.1 min
rate of [tex]H_{2}[/tex] effusion = 2.42 min
molar mass of hydrogen = 1 x 2 = 2 g/m
molar mas of unknown gas = ?
From Graham's law of diffusion and effusion, the rate of effusion and diffusion is inversely proportional to the square root of its molar mass.
from
[tex]\frac{R_{g} }{R_{h} }[/tex] = [tex]\sqrt{\frac{M_{h} }{M_{g} } }[/tex]
where
[tex]R_{h}[/tex] = rate of effusion of hydrogen gas
[tex]R_{g}[/tex] = rate of effusion of unknown gas
[tex]M_{h}[/tex] = molar mass of H2 gas
[tex]M_{g}[/tex] = molar mass of unknown gas
substituting values, we have
[tex]\frac{11.1 }{2.42 }[/tex] = [tex]\sqrt{\frac{2 }{M_{g} } }[/tex]
4.587 = [tex]\sqrt{\frac{2 }{M_{g} } }[/tex]
[tex]\sqrt{M_{g} }[/tex] = [tex]\sqrt{2}[/tex]/4.587
[tex]\sqrt{M_{g} }[/tex] = 0.31
[tex]M_{g}[/tex] = [tex]0.31^{2}[/tex] = 0.0961 g/mol
The molar mass of the unknown gas will be "0.0961 g/mol".
Given:
Effusion rate of unknown gas,
[tex]R_g = 11.1 \ min[/tex]Effusion rate of [tex]H_2[/tex],
[tex]R_h = 2.42 \ min[/tex]Molar mass of hydrogen,
[tex]M_h = 1\times 2[/tex][tex]= 2 \ g/m[/tex]
According to the Graham's law, we get
→ [tex]\frac{R_g}{R_h} = \sqrt{\frac{M_h}{M_g} }[/tex]
By substituting the values, we get
→ [tex]\frac{11.1}{2.42} = \sqrt{\frac{2}{M_g} }[/tex]
→ [tex]4.587=\sqrt{\frac{2}{M_g} }[/tex]
→ [tex]\sqrt{M_g} = \sqrt{\frac{2}{4.587} }[/tex]
[tex]\sqrt{M_g} = 0.31[/tex]
[tex]M_g = 0.0961 \ g/mol[/tex]
Thus the above solution is right.
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14. Which group of diamagnetic transition metals exhibits trends in density and melting points that don't match the same trends seen in
other groups?
A. Group 3
B. Group 12
C. Group 7
D. Group 11
Answer:
Group 12
Explanation:
Group 12 transition metals are diamagnetic. They behave properties that distinguish them. They naturally have twelve electrons hence their outermost shell is fully filled.
Transition metals have high densities which increases down the group. However, the increase in density of transition elements of group 12 varies with temperature at a rate that is quite different from other transition elements. Hence the differences in the value of melting points and density changes by only a very small amount as you come down group 12 compared to other groups of transition elements.
Solid MgO has the same crystal structure as NaCl. How many oxide ions surround each Mg * ion as nearest neighbors in MgO? 4 none of these
Answer:
The number of oxide ions as the nearest neighbors of [tex]{\rm{M}}{{\rm{g}}^{{\rm{2 + }}}}[/tex] ions are known to be as six
Explanation:
The regularity of a crystal structure leads to the idea of space lattice.In order to explain this concept, let us consider a crystal of NaCl, It consists of a perfectly regular arrangement of sodium ions and chlorine ions.
If we represent the position of each Na+ in the crystal by a point marked x the result will be a regular three dimensional network of points. This will be the space lattice of Na+ in the crystal NaCl. The symmetry of the combined lattice determined the symmetry of the crystal as a whole.
The space lattice of a crystal may be considered as built up of a three dimensional basic pattern called unit cell. The unit cell is a repeat unit which generates the whole pattern in three dimensions of the unit cell.
In Solid MgO , the crystal structure which is used to predict the properties of the material, have the same structure as that of NaCl.
The obtain the structure of a face centered cubic FCC unit cell where the ions occupy the corner of the cube and the center of each face of the cube.
The number of oxide ions as the nearest neighbors of [tex]{\rm{M}}{{\rm{g}}^{{\rm{2 + }}}}[/tex] ions are known to be as six. As a result of that , the coordination number of [tex]{\rm{M}}{{\rm{g}}^{{\rm{2 + }}}}[/tex] ions is six.
Classify each of these reactions.
1) Ba(ClO3)2(s)--->BaCl2(s)+3O2(g)
2) 2NaCl(aq)+K2S(aq)--->Na2S(aq)+2KCl(aq)
3) CaO(s)+CO2(g)--->CaCO3(s)
4) KOH(aq)+AgCl(aq)---->KCl(aq)+AgOH(s)
5) Ba(OH)2(aq)+2HNO2(aq)--->Ba(NO2)2(aq)+2H2O(l)
Each classify reaction should be either one of this.
a. acid-base neutralization
b. precipitation
c. redox
d. none of the above
Answer:
1. REDOX
2. None of the above
3. Precipitation
4. Preicipitation
5. Acid base neutralization
Explanation:
Reactions where a solid is formed, are named as precipitation. This solid is called precipitated.
Option 4 and 3.
3) CaO (s) + CO₂ (g) → CaCO₃(s)
4) KOH (aq) + AgCl (aq) → KCl (aq) + AgOH(s)
Reactions where water is produced, and you have an acid and a base as reactants, are named as neutralization. You called them acid-base because, the products.
5) Ba(OH)₂ (aq) + 2HNO₂(aq) → Ba(NO₂)₂ (aq) + 2H₂O(l)
Redox, are the reactions where one of the reactans can be oxidized and reduced, when a mole of electrons is released, or gained.
1) Ba(ClO₃)₂ (s) → BaCl₂ (s) + 3O₂(g)
Oxygen from the chlorate is oxidized (increases the oxidation state from -2 to 0) and the chlorine is reduced (decreases the oxidation state from +5 to -1).
2. 2NaCl(aq) + K₂S(aq) Na₂S (aq) + 2KCl (aq)
None of the above
Rank the following substances in order from most soluble in water to least soluble in water: ethane, C2H6; 1-pentanol, C5H11OH; potassium chloride, KCl; and propane, C3H8.
Rank from most to least soluble in water. To rank items as equivalent, overlap them.
Most soluble Least soluble
Answer:
Explanation:
The substances are:
-) Ethane, [tex]C_2H_6[/tex]
-) 1-pentanol, [tex]C_5H_1_1OH[/tex]
-) Potassium chloride, [tex]KCl[/tex]
-) Propane, [tex]C_3H_8[/tex]
For this question, we have to remember the structure of water. Due to the electronegativity difference between oxygen and hydrogen in this structure, we will have the formation of dipoles. The dipoles interact better with net charges, due to this, the Potassium chloride is the compound with highest solubility (due to the formation of a cation and an anion):
[tex]KC~l->~K^~+~Cl^-[/tex]
Then, in 1-pentanol we an "OH". This structure due to the presence of the hydroxyl group can form hydrogen bonds. Therefore, this compound would be the second more soluble.
Finally, the difference between propane and ethane is a carbon. In propane, we have an additional carbon. If we have more carbons we will have more area of interaction. If we have more area we will have more solubility therefore propane is more soluble than ethanol.
In conclusion, the rank from most soluble to least soluble is:
1) Potassium chloride, [tex]KCl[/tex]
2) 1-pentanol, [tex]C_5H_1_1OH[/tex]
3) Propane, [tex]C_3H_8[/tex]
4) Ethane, [tex]C_2H_6[/tex]
I hope it helps!
Order of solubility in water will be:
KCl > C₅H₁₁OH > C₃H₈ > C₂H₆
Solubility in water:Any solvent soluble in water due to its polarity and ability to form hydrogen bonds. The presence of hydrogen bonding between molecules of a substance indicates that the molecules are polar. This means the molecules will be soluble in a polar solvent such as water.
Substances that are given:
Ethane(C₂H₆), 1-pentanol(C₅H₁₁OH), Potassium chloride(KCl) and propane(C₃H₈).
We will look at each compound one by one:
Potassium chloride is an ionic compound, it has ionic interactions between its solubility in water is highest due to the formation of potassium ([tex]K^{+}[/tex]) and ([tex]Cl^{-}[/tex]) ions.In 1-pentanol, there is presence of hydroxyl group thus it can easily form hydrogen bonds with water. Therefore it will be soluble in water and comes after potassium chloride in ranking order.In ethane and propane molecule, there is one extra carbon in case of propane due to which it leads to the more area for interactions therefore more area for interaction leads to more solubility thus propane is more soluble than ethane in water.Order of solubility in water will be:
KCl > C₅H₁₁OH > C₃H₈ > C₂H₆
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