Explanation:
The electric force between two charged particles is given by the formula as follows :
[tex]F=k\dfrac{q_1q_2}{r^2}[/tex]
Here,
k is electrostatic constant
[tex]q_1,q_2[/tex] are charges
r is the distance between charges
The above formula shows that the electric force is inversely proportional to the square of the distance between charges. It means that as the distance increases by a factor, the electric force decreases by the square of that factor. Hence, the correct option is (e).
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:
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.
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
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 )
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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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.
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.
A civil engineer designs mostly:
A. building structures.
B. computer parts.
C. new foods.
D. technology that flies.
If the distance between two objects increased, what would happen to the force of gravity between them? It would increase. It would stay the same. It would depend on the speed. It would decrease.
Answer:
it will decrease
Explanation:
force of gravity is inversely proportional to the distance
Answer:
It shall decrease! -w-ll
Explanation:
because of the gravitational pull
Suppose, in an experiment to determine the amount of sodium hypochlorite in bleach, you titrated a 26.34 mL sample of 0.0100 M K I O 3 with a solution of N a 2 S 2 O 3 of unknown concentration. The endpoint was observed to occur at 15.51 mL . How many moles of K I O 3 were titrated
Answer:
0.1 M
Explanation:
The overall balanced reaction equation for the process is;
IO3^- (aq)+ 6H^+(aq) + 6S2O3^2-(aq) → I-(aq) + 3S4O6^2-(aq) + 3H2O(l)
Generally, we must note that;
1 mol of IO3^- require 6 moles of S2O3^2-
Thus;
n (iodate) = n(thiosulfate)/6
C(iodate) x V(iodate) = C(thiosulfate) x V(thiosulfate)/6
Concentration of iodate C(iodate)= 0.0100 M
Volume of iodate= V(iodate)= 26.34 ml
Concentration of thiosulphate= C(thiosulfate)= the unknown
Volume of thiosulphate=V(thiosulfate)= 15.51 ml
Hence;
C(iodate) x V(iodate) × 6/V(thiosulfate) = C(thiosulfate)
0.0100 M × 26.34 ml × 6/15.51 ml = 0.1 M
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
Name the following alkanes, please need answer for f,g,h?!
Answer:
f is =2,2-dimethyl butane
g is = 2,2-dimethyl propane
h is = 3,3-diethyl pentane
Explanation:
please give me brainliest
The correct IUPAC name for the following compound is
Answer:
1-cyclopentylhexan-2-one
Explanation:
1-cyclopentylhexan-2-one
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.
Draw the structure of beta-D-idose in its pyranose form.
Answer:
See figure 1
Explanation:
In this case, we can start with the linear structure of D-Idose. Then, if we have a "D" configuration the "OH" in the last chiral (carbon 5) will be in the right. This carbon will attack carbon 1 and we will produce a cyclic structure with 6 members (pyranose). Additionally, we have to keep in mind that we want the "beta" structure. So, the "OH" on carbon 1 must point up (red arrow). Finally, we will have a cyclic structure with 6 atoms and the "OH" on carbon 1 pointing up.
See figure 1
I hope it helps!
Which of the following statements is not true for an exothermic reaction? Question options: The products have a higher heat content than the reactants. The temperature of the reaction system increases. The temperature of the surroundings increases. Heat passes from the reaction system to the surroundings. The enthalpy change for the reaction is negativ
Answer:
The products have a higher heat content than the reactants.
Explanation:
The statement above is not true for an exothermic reaction because in an exothermic reaction heat is released to the surroundings. This simply means that the total energy of the products is less than that of the reactants.
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%
For carbon: What is the effective nuclear charge? In which orbitals do the valence electrons reside? For silicon: What is the effective nuclear charge? In which orbitals do the valence electrons reside?
Answer:
For carbon, the effective nuclear charge is 3.25 and the valence electrons will reside in the orbitals 2s^2 and 2p^2
For silicon, the effective nuclear charge is 4.15 and its valance electrons will reside in the orbitals 3s^2 and 3p^2
Explanation:
Carbon
The effective nuclear charge of carbon is 3.25
To get the orbitals in which it’s valence electron reside, let’s write the electronic configuration
The atomic number of carbon is 6
So the configuration will be;
1s^2 2s^2 2p^2
So the valence electrons will reside in the orbitals 2s^2 and 2p^2
For silicon;
It’s effective nuclear charge is +4.15
The electronic configuration of silicon with atomic number 14 is;
1s^2 2s^2 2p^6 3s^2 3p^2
So the valence electrons will reside in the orbitals 3s^2 and 3p^2
Determine whether each of the following salts will form a solution that is acidic, basic, or pH-neutral. Drag the appropriate items to their respective bins.
Al(NO3)3
C2H5NH3NO3
NaClO
RbI
CH3NH3CN
Answer:
Al(NO₃)₃: Acidic.
C₂H₅NH₃NO₃: Acidic.
NaClO: Basic
RbI: pH-neutral
CH₃NH₃CN: Solution basic
Explanation:
The general rules to determine if a solution is acidic, basic or neutral are:
If it is a salt of a strong acid and base, the solution will be pH-neutral. If it is a salt of a strong acid and a weak base, the solution will be acidic due to the hydrolysis of the weak base component (cation). If it is a salt of a strong base and a weak acid, the solution will be basic due to the hydrolysis of the weak acid component (anion).For the salts:
Al(NO₃)₃. The repective acid is HNO₃ (Strong acid) and the base is Al(OH)₃ (Weak base). As the salt comes from strong acid and weak base. SOLUTION ACIDIC
C₂H₅NH₃NO₃. The acid is HNO₃ (Strong acid) and the base C₂H₅NH₃OH (Weak base). SOLUTION ACIDIC.
NaClO. Tha acid is HClO (weak acid), and the base NaOH (Strong base). SOLUTION BASIC.
RbI: The acid is HI (Strong acid) and the base RbOH (Strong base). pH-NEUTRAL
CH₃NH₃CN. The acid is HCN (weak acid; pKb = 4.79) and the base CH₃NH₃OH (weak base; pKa = 10.64). Both weak acid and base will produce each hydrolisis. The lower pK will predominate. That is the weak acid. SOLUTION BASIC
Solution of Al(NO₃)₃ and C₂H₅NH₃NO₃ salts is acidic, NaClO is basic and of RbI & CH₃NH₃cyanide is neutral in nature.
What is pH?pH of any solution tells about the acidity or basicity of the solution, pH of any solution ranges from 0 to 14 and from acidity to basicity.
Al(NO₃)₃ is a salt which is formed by the mixing of strong acid HNO₃ (Nitric acid) and weak base Al(OH)₃, so the resultant solution of the salt is acidic in nature.C₂H₅NH₃NO₃ salt is formed by the mixing of strong acid HNO₃ (Nitric acid) and weak base C₂H₅NH₃OH, so the resultant solution of the salt is acidic in nature.NaClO is a salt of weak acid is HClO and strong base NaOH, so the resultant solution of the salt is basic in nature.RbI salt is formed by the combination of strong acid HI and strong base RbOH, so the resultant solution of the salt is neutral in nature.CH₃NH₃Cyanide is a salt of weak acid hydrogen cyanide and weak base CH₃NH₃OH, so the resultant solution of the salt is neutral in nature.Hence, appropriate differentiation was done above.
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. 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
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)
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.
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 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]
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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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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A multistep reaction can only occur as fast as its slowest step. Therefore, it is the rate law of the slow step that determines the rate law for the overall reaction. Consider the following multistep reaction:
A+B ----- AB(slow)
A+AB-----A2B(fast
....................................................
2A+B ----- A2B(overall)
Based on this mechanism, determine the rate law for the overall reaction.
a) rate = kA2BAB
b) rate = kAB
c) rate = kAAB
d) rate = kA2B
Answer:
b) rate = kAB.
Explanation:
Hello,
In this case, considering the given statement, we can notice that the rate law of the overall reaction will be determined for the slowest step, that is:
[tex]A+B \rightarrow AB\ \ (slow)[/tex]
In such a way, we can infer that the rate law will contain both the concentration of A and B to the first power both, since their stoichiometric coefficients in the chemical equation are both one:
[tex]rate=k[A][B][/tex]
Thereby, answer is b) rate = kAB, that should be better rate = k[A][B] by expressing the concentrations.
Best regards.
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
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.
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.