Answer:
V2 = 4.48L
Explanation:
using charles law
V1/T1=V2/T2
3.4/283=V2/373
0.012=V2/373
V2= 0.012 x 373
V2 = 4.48L
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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The Handbook of Chemistry and Physics gives solubilities of the following compounds in grams per 100 mL water. Because these compounds are only slightly soluble, assume that the volume does not change on dissolution and calculate the solubility product for each.
(a) BaSeO4, 0.0118 g/100 mL
(b) Ba(BrO3)2 H20, 0.30 g/100 mL
(c) NH4MgAsO4-6H20, 0.038 g/100 mL
(d) La2(MoOs)3, 0.00179 g/100 mL
Answer:
(a) [tex]Ksp=4.50x10^{-7}[/tex]
(b) [tex]Ksp=1.55x10^{-6}[/tex]
(c) [tex]Ksp=2.27x10^{-12}[/tex]
(d) [tex]Ksp=1.05x10^{-22}[/tex]
Explanation:
Hello,
In this case, given the solubility of each salt, we can compute their molar solubilities by using the molar masses. Afterwards, by using the mole ratio between ions, we can compute the concentration of each dissolved and therefore the solubility product:
(a) [tex]BaSeO_4(s)\rightleftharpoons Ba^{2+}(aq)+SeO_4^{2-}(aq)[/tex]
[tex]Molar\ solubility=\frac{0.0188g}{100mL} *\frac{1mol}{280.3g}*\frac{1000mL}{1L}=6.7x10^{-4}\frac{mol}{L}[/tex]
In such a way, as barium and selenate ions are in 1:1 molar ratio, they have the same concentration, for which the solubility product turns out:
[tex]Ksp=[Ba^{2+}][SeO_4^{2-}]=(6.7x10^{-4}\frac{mol}{L} )^2\\\\Ksp=4.50x10^{-7}[/tex]
(B) [tex]Ba(BrO_3)_2(s)\rightleftharpoons Ba^{2+}(aq)+2BrO_3^{-}(aq)[/tex]
[tex]Molar\ solubility=\frac{0.30g}{100mL} *\frac{1mol}{411.15g}*\frac{1000mL}{1L}=7.30x10^{-3}\frac{mol}{L}[/tex]
In such a way, as barium and bromate ions are in 1:2 molar ratio, bromate ions have twice the concentration of barium ions, for which the solubility product turns out:
[tex]Ksp=[Ba^{2+}][BrO_3^-]^2=(7.30x10^{-3}\frac{mol}{L})(3.65x10^{-3}\frac{mol}{L})^2\\\\Ksp=1.55x10^{-6}[/tex]
(C) [tex]NH_4MgAsO_4(s)\rightleftharpoons NH_4^+(aq)+Mg^{2+}(aq)+AsO_4^{3-}(aq)[/tex]
[tex]Molar\ solubility=\frac{0.038g}{100mL} *\frac{1mol}{289.35g}*\frac{1000mL}{1L}=1.31x10^{-4}\frac{mol}{L}[/tex]
In such a way, as ammonium, magnesium and arsenate ions are in 1:1:1 molar ratio, they have the same concentrations, for which the solubility product turns out:
[tex]Ksp=[NH_4^+][Mg^{2+}][AsO_4^{3-}]^2=(1.31x10^{-4}\frac{mol}{L})^3\\\\Ksp=2.27x10^{-12}[/tex]
(D) [tex]La_2(MoOs)_3(s)\rightleftharpoons 2La^{3+}(aq)+3MoOs^{2-}(aq)[/tex]
[tex]Molar\ solubility=\frac{0.00179g}{100mL} *\frac{1mol}{1136.38g}*\frac{1000mL}{1L}=1.58x10^{-5}\frac{mol}{L}[/tex]
In such a way, as the involved ions are in 2:3 molar ratio, La ion is twice the molar solubility and MoOs ion is three times it, for which the solubility product turns out:
[tex]Ksp=[La^{3+}]^2[MoOs^{-2}]^3=(2*1.58x10^{-5}\frac{mol}{L})^2(3*1.58x10^{-5}\frac{mol}{L})^3\\\\Ksp=1.05x10^{-22}[/tex]
Best regards.
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
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%
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.
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
what is radiologist
Radiologists are medical doctors that treat injuries using medical imaging (radiology)
Answer:
a person who uses X-rays or other high-energy radiation, especially a doctor specializing in radiology.
Explanation:
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]
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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M(8,7) is the midpoint of rs. The coordinates of s are (9,5) what is the coordinates of r
Answer:
Coordinate or r = (7,9).
Explanation:
Data obtained from the question include the following:
Mid point = (8,7)
Coordinate of S = (9,5)
Coordinate of r =...?
We shall determine the coordinate of r as follow:
Let the coordinate of r be (x2, y2)
Mid point = (x1 + x2)/2 , (y1 + y2)/2
Mid point = (8,7)
Coordinate of S = (9,5)
x1 = 9
y1 = 5
x2 =?
y2 =?
The value of x2 can be obtained as follow:
8 = (x1 + x2)/2
8 = (9 + x2)/2
Cross multiply
9 + x2 = 2 × 8
9 + x2 = 16
Collect like terms
x2 = 16 – 9
x2 = 7
The value of y2 can be obtained as follow:
5 = (y1 + y2)/2
7 = (5 + y2)/2
Cross multiply
5 + y2 = 2 × 7
5 + y2 = 14
Collect like terms
y2 = 14 – 5
y2 = 9
Coordinate of r = (x2, y2)
Coordinate or r = (7,9)
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)
An atom with 19 protons and 18 neutrons is a(n)
A. Isotope of potassium(K)
B. Standard atom of argon(Ar)
C. Standard atom of (K)
D. Isotope of argon (Ar)
Answer:
A
Explanation:
The number of protons indicates the element so we know it's potassium. To get the number of neutrons you subtract the number of protons (19) from the mass number which for potassium is 39.
39-19=20 neutrons
Because you have 18 neutrons then yours would be an isotope.
Answer: A. Isotope of potassium(K)
Explanation: Founders Educere answer.
The displacement of a bromine atom by an amine is a substituion reaction. Write out the mechanism of this reaction (2-->3) Why might you expect that the reaction you have performed, using t-BuNH2, to be much slower than the same reaction using methylamine
Answer:
An alkyl halide can undergo SN2 reaction with an amine
Explanation:
The displacement of a bromine atom by an an amine (step 2---> 3) in the reaction sequence is an example of an SN2 reaction in which the amine is the nucleophile.
The nitrogen atom of the amine which bears a lone pair of electrons functions as the nucleophile and attacks the electrophilic carbon atom of the alkyl halide displacing the bromide and creating a new Carbon-Nitrogen bond. An ammonium intermediate is immediately formed and the reaction is completed by the abstraction of a hydrogen by a base (such as excess amine present in the system).
This reaction is slower with t-BuNH2 because of steric hindrance and steric crowding in the transition state. SN2 reactions are faster with methylamine where the alkyl carbon is easily accessible.
The detailed mechanism of this reaction has been attached to this answer.
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 )
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.
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 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.
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
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.
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.
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.
A civil engineer designs mostly:
A. building structures.
B. computer parts.
C. new foods.
D. technology that flies.
This substituent deactivates the benzene ring towards electrophilic substitution but directs the incoming group chiefly to the ortho and para positions.
A) -F
B) -OCH2CH3
C) -CF3
D) -NHCOCH3
E) -NO2
Answer:
F
Explanation:
Halogens may interact with the benzene ring via inductive or resonance effects. Halogens deactivate the benzene ring by inductive effect rather than by resonance effects.
The lone pairs of electrons present on the halogen atoms may be donated to the ring by resonance, but an opposite effect, the inductive pull (-I inductive effect) of the halogen atoms on electrons away from the benzene ring due to the high electro negativity of the halogens leads to a deactivation of the ring towards electrophilic substitution.
Hence inductive electron withdrawal by the halogen atom predominates over electron donation by resonance effect and the benzene ring g is deactivated towards electrophilic substitution at the ortho and para positions.
33. Hydrocarbons that release pleasant odors are called_________
hydrocarbons. (1 point)
Answer:
Aromatic Hydrocarbons
Explanation:
Aromatic (Pleasant Odour) Hydrocarbons are those having pleasant odours.
Answer:
substituted hydrocarbons
Explanation:
i think
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.
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 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.