Answer:
The answer to this question can be defined as follows:
In question 1, the answer is "Option A".
In question 2, the answer is "[tex]\bold{CH_3COOH}[/tex]".
Explanation:
In the second question, there is mistype error in the choices so the correct answer to this question can be defined as follows:
The product From 1,2-addition as its consequence of 1,4-addition is the result of kinetic regulation by HBr in conjugated dienes.The chemical name of the [tex]CH_3COOH[/tex] is the acetic acid, it is one of the carboxylic acids quite basic. It is a major chemical production factor for use as disposable soft drinks, movies or wood glue, polyethylene terephthalate, and many plastics, fibers, and fabrics. It is also used in the storage of the water and soft drinks in the bottles.Compare strontium with rubidium in terms of the following properties:
a. Atomic radius, number of valence electrons, ionization energy.
b. Strontium is smaller than rubidium.
c. Rubidium is smaller than strontium.
d. Strontium has more valence electrons.
e. Rubidium has more valence electrons.
f. Strontium has a larger ionization energy.
g. Rubidium has a larger ionization energy.
Answer:
Strontium is smaller
Strontium has the higher ionization energy
Strontium has more valence electrons
Explanation:
It must be understood that both elements belong to the same period i.e the same horizontal band of the periodic table
While Rubidium is an alkali metal(group 1) while Strontium is an alkali earth metal(group 2)
Since they are in the same period, periodic trends would be useful in evaluating their properties
In terms of atomic radius, rubidium is larger meaning it has a bigger atomic size
Generally, across the periodic table, atomic radius is expected to decrease and thus Rubidium which is leftmost is expected to have the higher atomic radius
Since strontium belongs to group 2 of the periodic table, it has 2 valence electrons which is more than the single valence electron that rubidium which is in group 1 has
In terms of ionization energy, the atom with the higher number of valence electrons will have the higher ionization energy which is strontium in this case
A reaction mixture at 175 K initially contains 522 torr of NO and 421 torr of O2. At equilibrium, the total pressure in the reaction mixture is 748 torr. Calculate Kp at this temperature. Express your answer to three significant figures.
Answer:
[tex]Kp=0.0386[/tex]
Explanation:
Hello,
In this case, the undergoing chemical reaction is:
[tex]2NO+O_2\rightleftharpoons 2NO_2[/tex]
For which the equilibrium expression is:
[tex]Kp=\frac{p_{NO_2}^2}{p_{NO}^2p_{O_2}}[/tex]
Whereas, at equilibrium, each pressure is computed in terms of the initial pressure and the reaction extent via:
[tex]p_{NO_2}=2x\\p_{NO}=522-2x\\p_{O_2}=421-x[/tex]
And the total pressure:
[tex]p_{eq}=p_{NO_2}+p_{NO}+p_{O_2}\\\\p_{eq}=2x+522-2x+421-x\\\\p_{eq}=943-x[/tex]
Yet it is 748 torr, for which the extent is:
[tex]x=943-p_{eq}=943-748\\\\x=195torr[/tex]
Therefore, Kp turns out:
[tex]Kp=\frac{(2x)^2}{(522-2x)^2(421-x)}\\\\Kp=\frac{(2*195)^2}{(522-2*195)^2(421-195)}\\\\Kp=0.0386[/tex]
Best regards.
Write a balanced equation for the single-replacement oxidation-reduction reaction described, using the smallest possible integer coefficients. The reaction that takes place when chlorine gas combines with aqueous potassium bromide. (Use the lowest possible coefficients. Omit states of matter.)
Answer:
[tex]\rm Cl_2 + 2\; KBr \to Br_2 + 2\; KCl[/tex].
One chlorine molecule reacts with two formula units of (aqueous) potassium bromide to produce one bromine molecule and two formula units of (aqueous) potassium chloride.
Explanation:
Formula for each of the speciesStart by finding the formula for each of the compound.
Both chlorine [tex]\rm Cl[/tex] and bromine [tex]\rm Br[/tex] are group 17 elements (halogens.) Each On the other hand, potassium [tex]\rm K[/tex] is a group 1 element (alkaline metal.) EachTherefore, the ratio between [tex]\rm K[/tex] atoms and [tex]\rm Br[/tex] atoms in potassium bromide is supposed to be one-to-one. That corresponds to the empirical formula [tex]\rm KBr[/tex]. Similarly, the ratio between
The formula for chlorine gas is [tex]\rm Cl_2[/tex], while the formula for bromine gas is [tex]\rm Br_2[/tex].
Balanced equation for the reactionWrite down the equation using these chemical formulas.
[tex]\rm ?\; Cl_2 + ?\; KBr \to ?\;Br_2 + ?\; KCl[/tex].
Start by assuming that the coefficient of compound with the largest number of elements is one. In this particular equation, both [tex]\rm KBr[/tex] and [tex]\rm KCl[/tex] features two elements each.
Assume that the coefficient of [tex]\rm KCl[/tex] is one. Hence:
[tex]\rm ?\; Cl_2 + 1 \; KBr \to ?\;Br_2 + ?\; KCl[/tex].
Note that [tex]\rm KBr[/tex] is the only source of [tex]\rm K[/tex] and [tex]\rm Br[/tex] atoms among the reactants of this reaction.
There would thus be one [tex]\rm K[/tex] atom and one [tex]\rm Br[/tex] atom on the reactant side of the equation.
Because atoms are conserved in a chemical equation, there should be the same number of [tex]\rm K[/tex] and [tex]\rm Br[/tex] atoms on the product side of the equation.
In this reaction, [tex]\rm Br_2[/tex] is the only product with [tex]\rm Br[/tex] atoms.
One [tex]\rm Br[/tex] atom would correspond to [tex]0.5[/tex] units of [tex]\rm Br_2[/tex].
Similarly, in this reaction, [tex]\rm KCl[/tex] is the only product with [tex]\rm K[/tex] atoms.
One [tex]\rm K[/tex] atom would correspond to one formula unit of [tex]\rm KCl[/tex].
Hence:
[tex]\displaystyle \rm ?\; Cl_2 + 1 \; KBr \to \frac{1}{2}\;Br_2 + 1\; KCl[/tex].
Similarly, there should be exactly one [tex]\rm Cl[/tex] atom on either side of this equation. The coefficient of [tex]\rm Cl_2[/tex] should thus be [tex]0.5[/tex]. Hence:
[tex]\displaystyle \rm \frac{1}{2}\; Cl_2 + 1 \; KBr \to \frac{1}{2}\;Br_2 + 1\; KCl[/tex].
That does not meet the requirements, because two of these coefficients are not integers. Multiply all these coefficients by two (the least common multiple- LCM- of these two denominators) to obtain:
[tex]\displaystyle \rm 1\; Cl_2 + 2 \; KBr \to 1\;Br_2 + 2\; KCl[/tex].
A diode has IS = 10−17 A and n = 1.05. (a) What is the diode voltage if the diode current is 70 μA? (b) What is the diode current for VD = 0.1 mV?
Answer:
(a) The diode voltage, [tex]V_D =[/tex] 0.776 V
(b) The diode current, [tex]I_D =[/tex] 3.81 x 10⁻²⁰ A
Explanation:
Given;
saturation current in diode, [tex]I_s[/tex] = 10⁻¹⁷ A
nonideality factor, n = 1.05
(a) the diode voltage
Given diode current, [tex]I_D[/tex] = 70 μA = 7 x 10⁻⁶ A
Diode voltage is calculated as;
[tex]V_D = nV_Tln(1+ \frac{I_D}{I_S} )[/tex]
Where;
[tex]V_T[/tex] is thermal voltage at 25°C = 0.025
[tex]V_D = 1.05 * 0.025 ln(1+ \frac{70*10^{-6}}{1*10^{-17}})\\\\V_D = 0.02625ln(1+ 7*10^{12})\\\\V_D = 0.776 \ V[/tex]
b) the diode current for VD = 0.1 mV
[tex]V_D = nV_Tln(1 +\frac{I_D}{I_S} )\\\\ln(1 +\frac{I_D}{I_S} ) = \frac{V_D}{nV_T} \\\\ln(1 +\frac{I_D}{I_S} ) = \frac{0.1*10^{-3}}{1.05*0.025} \\\\ln(1 +\frac{I_D}{I_S} ) = 0.00381\\\\1 +\frac{I_D}{I_S} = e^{0.00381}\\\\1+ \frac{I_D}{I_S}= 1.00381\\\\ \frac{I_D}{I_S}=1.00381 - 1\\\\ \frac{I_D}{I_S}= 0.00381\\\\I_D = 0.00381(I_S)\\\\I_D = 0.00381(10^{-17})\\\\I_D = 3.81*10^{-20} \ A[/tex]
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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How long should you hold the iron on the hair to heat the strand and set the base ?
A) 5 seconds
B) 15 seconds
C) 30 seconds
D) 1 minute
Which phase change is an example of an exothermic process?
A.
solid to liquid
B.
solid to gas
C.
liquid to solid
D.
liquid to gas
E.
solid to plasma
Reset
Answer:
C
Explanation:
Turning liquid to a solid is like freezing water to ice and requires the water to LOSE (release) heat causing an exothermic reaction.
Using the periodic table provided, identify the atomic mass of sodium (Na) . Your answer should have 5 significant figures. Provide your answer below: __ amu
Answer:
Your answer will either be 22.9897 or 22.990 !!
Explanation:
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.
A 45.0 mL sample of 0.020 M acetic acid (HC2H3O2) is titrated with 0.020 M NaOH.? Determine the pH of the solution after adding 35.0 mL of any NaOH. (Ka of acetic acid is 1.8 x 10-5) HC2H3O2 (aq) + NaOH (aq) D NaC2H3O2(aq) + H2O (l) (Hint: Calculate new concentration and ICE table)
Answer:
Explanation:
CH₃COOH + NaOH = CH₃COONa + H₂O .
.02M
CH₃COOH = CH₃COO⁻ + H⁺
C xC xC
Ka = xC . xC / C = x² C
1.8 x 10⁻⁵ = x² . .02
x² = 9 x 10⁻⁴
x = 3 x 10⁻²
= .03
concentration of H⁺ = xC = .03 . .02
= 6 x 10⁻⁴ M , volume = 45 x 10⁻³ L
moles of H⁺ = 6 X 10⁻⁴ x 45 x 10⁻³
= 270 x 10⁻⁷ moles
= 2.7 x 10⁻⁵ moles
concentration of NaOH = .0200 M , volume = 35 x 10⁻³ L
moles of Na OH = 2 X 10⁻² x 35 x 10⁻³
= 70 x 10⁻⁵ moles
=
NaOH is a strong base so it will dissociate fully .
there will be neutralisation reaction between the two .
Net NaOH remaining = (70 - 2.7 ) x 10⁻⁵ moles
= 67.3 x 10⁻⁵ moles of NaOH
Total volume = 45 + 35 = 80 x 10⁻³
concentration of NaOH after neutralisation.= 67.3 x 10⁻⁵ / 80 x 10⁻³ moles / L
= 8.4125 x 10⁻³ moles / L
OH⁻ = 8.4125 x 10⁻³
H⁺ = 10⁻¹⁴ / 8.4125 x 10⁻³
= 1.1887 x 10⁻¹²
pH = - log ( 1.1887 x 10⁻¹² )
= 12 - log 1.1887
= 12 - .075
= 11.925 .
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
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.
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%
Give the major organic products from the oxidation with KMnO4 for the following compounds. Assume an excess of KMnO4.
a) ethylbenzene
b) m-Xylene (1,3- dimethylbenzene)
c) 4-Propyl-3-t-butyltoluene
Answer:
Explanation:
a ) Benzoic acid is formed . In any alkyl benzene derivative , potassium permanganate reacts to form carboxylic acid . It oxidises side chains to carboxylic acid .
C₆H₅CH₃ + 0 = C₆H₅COOH + H₂O
O is provided by KMnO₄
b ) In this reaction isophthalic acid is formed .
C₆H₄(CH₃)₂ +O = C₆H₄(COOH)₂
c)
4-Propyl-3-t-butyltoluene
In this oxidation , three side chains of ring are 1 ) 1-methyl 2 ) 3- butyl 3 ) 4 propyl .
The methyl and 4 - propyl groups are oxidised to di- carboxylic acid and 3 butyl group remains intact ( unoxidised )
You are given 10.00 mL of a solution of an unknown acid. The pH of this solution is exactly 2.18. You determine that the concentration of the unknown acid was 0.2230 M. You also determined that the acid was monoprotic (HA). What is the pKa of your unknown acid
Answer:
[tex]pKa=3.70[/tex]
Explanation:
Hello,
In this case, given the information, we can compute the concentration of hydronium given the pH:
[tex]pH=-log([H^+])\\[/tex]
[tex][H^+]=10^{-pH}=10^{-2.18}=6.61x10^{-3}M[/tex]
Next, given the concentration of the acid and due to the fact it is monoprotic, its dissociation should be:
[tex]HA\rightleftharpoons H^++A^-[/tex]
We can write the law of mass action for equilibrium:
[tex]Ka=\frac{[H^+][A^-]}{[HA]}[/tex]
Thus, due to the stoichiometry, the concentration of hydronium and A⁻ are the same at equilibrium and the concentration of acid is:
[tex][HA]=0.2230M-6.61x10^{-3}M=0.2164M[/tex]
As the concentration of hydronium also equals the reaction extent ([tex]x[/tex]). Thereby, the acid dissociation constant turns out:
[tex]Ka=\frac{(6.61x10^{-3})^2}{0.2164}\\ \\Ka=2.02x10^{-4}[/tex]
And the pKa:
[tex]pKa=-log(Ka)=-log(2.02x10^{-4})\\\\pKa=3.70[/tex]
Regards.
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
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
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]
A strontium hydroxide solution is prepared by dissolving 10.60 gg of Sr(OH)2Sr(OH)2 in water to make 47.00 mLmL of solution.What is the molarity of this solution? Express your answer to four significant figures and include the appropriate units.
Answer:
Approximately [tex]1.854\; \rm mol\cdot L^{-1}[/tex].
Explanation:
Note that both figures in the question come with four significant figures. Therefore, the answer should also be rounded to four significant figures. Intermediate results should have more significant figures than that.
Formula mass of strontium hydroxideLook up the relative atomic mass of [tex]\rm Sr[/tex], [tex]\rm O[/tex], and [tex]\rm H[/tex] on a modern periodic table. Keep at least four significant figures in each of these atomic mass data.
[tex]\rm Sr[/tex]: [tex]87.62[/tex].[tex]\rm O[/tex]: [tex]15.999[/tex].[tex]\rm H[/tex]: [tex]1.008[/tex].Calculate the formula mass of [tex]\rm Sr(OH)_2[/tex]:
[tex]M\left(\rm Sr(OH)_2\right) = 87.62 + 2\times (15.999 + 1.008) = 121.634\; \rm g \cdot mol^{-1}[/tex].
Number of moles of strontium hydroxide in the solution[tex]M\left(\rm Sr(OH)_2\right) =121.634\; \rm g \cdot mol^{-1}[/tex] means that each mole of [tex]\rm Sr(OH)_2[/tex] formula units have a mass of [tex]121.634\; \rm g[/tex].
The question states that there are [tex]10.60\; \rm g[/tex] of [tex]\rm Sr(OH)_2[/tex] in this solution.
How many moles of [tex]\rm Sr(OH)_2[/tex] formula units would that be?
[tex]\begin{aligned}n\left(\rm Sr(OH)_2\right) &= \frac{m\left(\rm Sr(OH)_2\right)}{M\left(\rm Sr(OH)_2\right)}\\ &= \frac{10.60\; \rm g}{121.634\; \rm g \cdot mol^{-1}} \approx 8.71467\times 10^{-2}\; \rm mol\end{aligned}[/tex].
Molarity of this strontium hydroxide solutionThere are [tex]8.71467\times 10^{-2}\; \rm mol[/tex] of [tex]\rm Sr(OH)_2[/tex] formula units in this [tex]47\; \rm mL[/tex] solution. Convert the unit of volume to liter:
[tex]V = 47\; \rm mL = 0.047\; \rm L[/tex].
The molarity of a solution measures its molar concentration. For this solution:
[tex]\begin{aligned}c\left(\rm Sr(OH)_2\right) &= \frac{n\left(\rm Sr(OH)_2\right)}{V}\\ &= \frac{8.71467\times 10^{-2}\; \rm mol}{0.047\; \rm L} \approx 1.854\; \rm mol \cdot L^{-1}\end{aligned}[/tex].
(Rounded to four significant figures.)
If a bottle of olive oil contains 1.2 kg of olive oil, what is the volume, in milliliters (mL), of the olive oil?
Answer:
1.3 mL
Explanation:
First, get the density of the olive oil, which is 0.917 kg/mL. Then divide the mass by the density:
1.2kg/0.917kg/mL= 1.3086150491 mL. The kg cancel out, leaving us with mL.
It should have 2 significant figures, because 1.2kg has 2 and we are dividing.
The volume of olive oil will be nearly 1300mL or 1.30 L as per the given data.
What is volume?Volume is a measurement of three-dimensional space that is occupied. It is frequently numerically quantified using SI derived units or various imperial units. The definition of length is linked to the definition of volume.
Volume is, at its most basic, a measure of space. The units liters (L) and milliliters (mL) are used to measure the volume of a liquid, also known as capacity.
This measurement is done with graduated cylinders, beakers, and Erlenmeyer flasks.
Here, it is given that mass of olive oil is 1.2kg.
We know that,
Density of olive oil = 0.917kg/l.
Volume = mass/density
Volume = 1.2/0.917.
Volume = 1.30 lit.
Volume = 1300mL.
Thus, the volume of olive oil will be 1300 mL.
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A civil engineer designs mostly:
A. building structures.
B. computer parts.
C. new foods.
D. technology that flies.
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
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.
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.
Explain the term isomers?
Answer:
Isomers are molecules that have the same molecular method, however have a unique association of the atoms in space. That excludes any extraordinary preparations which can be sincerely because of the molecule rotating as an entire, or rotating about precise bonds.
Stote 4 ways in which excesine alcohol conscuption is
harmful to humans
Answer:
An addiction could occur, maybe an overdose?, this could lead to death and maybe you would do unreasonable things which could get you fined or arrested.
Explanation:
Answer:
Excessive alcohol is harmful because you could get addicted.Alcohol can affect your nervous system.Your sugar levels will not be good.Parts of your body and organs will become inflamed.You can get a larger amount of muscle cramps.Also you will not be able to get enough vitamins in your body.Accidents that lead to deaths could occur.You would do crazy actions with things such as theft or breaking into a house which could get you fined or arrested.Too much alcohol can lead to high blood pressure, disease and even strokes.You can have birth defectsWith excessive alcohol you can get osteoporosis.You can also get your immune system weakened.Finally, alcohol can lead to cancer.Hope this helped,
Kavitha
what bsic difference is between NMR and MS spectroscopic techniques?
Answer:
The Nuclear magnetic resonance is the process this technique does not use radiation.
The ms is an sensitive technology can be a massive number and small sample of the blood.
Explanation:
The Nuclear magnetic resonance we look at the both side of that coin.
The technique provides that fatty acid composition and various including amino acids.
These are contain the complementary these biomarkers, that are suitable for all kinds of studies. there are many types of research:-
(1) A powerful tool metabolic (2) A versatile tool research (3) Quick analysis (4) Low cost analysis.
The MS is an extremely sensitive technology using a very small number of the blood.
(1) Powerful techniques (2) Highly method (3) Large number of metabolites (4)Small sample volume
MS can be fine mapping metabolic pathways to sign analytical strategy.
Which of the following is a property of salts? Undergo combustion Do not make ionic bonds easily Do not conduct electricity as solids Formed due to reaction of acid with water
Answer:
Do not conduct electricity as solids.
Explanation:
Hello,
In this case, we should remember that salts are formed when an acid and base react in order to yield the salt and water due to the ions exchange during neutralization chemical reactions. For instance, when hydrochloric acid (acid) reacts with potassium hydroxide (base), sodium chloride (salt) and water are yielded via:
[tex]HCl+NaOH\rightarrow NaCl+H_2O[/tex]
Moreover, it is widely known that salts are formed by electrovalent/ionic bonds which involves electron transfer so the metallic atom becomes positively charged (cation) whereas the non-metallic atom becomes negatively charged (anion) once the electrons are received so it can conduct electricity when dissolved in water yet not when solid since electron transfer is facilitated by the aqueous media, otherwise, ions remain together. Thereby, answer is do not conduct electricity as solids.
Regards.
Answer:
c
Explanation:
A base solution contains 0.400 mol of OH–. The base solution is neutralized by 43.4 mL of sulfuric acid. What is the molarity of the sulfuric acid solution?
Answer:
Molarity of the sulfuric acid solution is 4.61M
Explanation:
The neutralization of a base of OH⁻ with sulfuric acid, H₂SO₄, occurs as follows:
2 OH⁻ + H₂SO₄ → 2H₂O + SO₄²⁻
That means, 2 moles of base react with 1 mole of sulfuric acid.
If you add 0.400 moles of OH⁻, moles of sulfuric acid you need to neutralize this amount of OH⁻ are:
0.400 moles OH⁻ ₓ (1 mole H₂SO₄ / 2 moles OH⁻) = 0.200 moles of H₂SO₄
As you add 43.4mL = 0.0434L of sulfuric acid to neutralize this solution, molarity (Ratio between moles and liters) is:
0.200 moles H₂SO₄ / 0.0434L = 4.61M
Molarity of the sulfuric acid solution is 4.61MThe heat of vaporization of 1-pentanol is 55.5 kJ/mol, and its entropy of vaporization is 148 J/K.mol. What is the approximate boiling point of 1-pentanol? 100 oC 375 oC 0 oC 25 oC
Answer:
Approximately 100 °C.
Explanation:
Hello,
In this case, since the entropy of vaporization is computed in terms of the heat of vaporization and the temperature as:
[tex]\Delta S_{vap}=\frac{\Delta H_{vap}}{T}[/tex]
We can solve for the temperature as follows:
[tex]T=\frac{\Delta H_{vap}}{\Delta S_{vap}}[/tex]
Thus, with the proper units, we obtain:
[tex]T=\frac{55500J/mol}{148J/(mol*K)} =375K\\\\T=102 \°C[/tex]
Hence, answer is approximately 100 °C.
Best regards.