Label A-F based on the table using C for concentrated and D for dilute.
A_
B_
C_
D_
E_
F_
Answers
Molarity is defined as the number of moles form a certain compound in one liter of solution.
So the higher the molarity the higher the number of moles in one liter of solution, and we say that the concentration is increased. The lower the molarity the lower the number of moles in one liter of solution, and we say that the concentration is decreased.
In a nutshell:
High molarity = concentrated solution
Low molarity = diluted solution
(A) concentrated
(B) dilute
(C) dilute
(D) concentrated
(E) dilute
(F) concentrated
Answer:
A) C
B) D
C) D
D) C
E) D
F) C
hope it helped
ExpLanation:
Related Questions
Which statement about van der Waals forces is true?
a)When the forces are weaker, a substance will have higher volatility.
b)When the forces are stronger, a substance will have lower viscosity.
c)When the forces are weaker, the boiling point of a substance will be higher.
d)When the forces are stronger, the melting point of a substance will be lower.
Answers
Answer:
A
Explanation:
Van der Waals forces are the weak electric forces of attraction between molecules and their strength is dependent on the distance between the molecules. The longer the distance between the molecules the weaker the forces. Weaker Van der Waals forces mean that molecules can easily escape from the liquid - hence meaning higher volatility.
Answer:
A!!!
Explanation:
The solubility of Cd(OH)2 can be increased through formation of the complex ion CdBr2−4 (Kf=5×103). If solid Cd(OH)2 is added to a NaBr solution, what would the initial concentration of NaBr need to be in order to increase the molar solubility of Cd(OH)2 to 1.0×10−3 moles per liter?
Answers
Answer:
Concentration of sodium bromide required = 2.38 M (around 2.4 M)
Explanation:
The equilibrium representing the complex ion formation is:
[tex]Cd^{2+} + 4Br^{-}\rightleftharpoons [CdBr_{4}]^{2-} .....Kf =5*10^{3}[/tex]-----(1)
where K(f) = formation equilibrium
The equilibrium representing the dissolution of Ca(OH)2 is:
[tex]Cd(OH)_{2}\rightleftharpoons Cd^{2+}+2OH^{-}.....Ksp = 2.5*10^{-14}[/tex]---(2)
where Ksp = solubility product
adding Equation (1) and equation(2) gives the net reaction:
[tex]Cd(OH)_{2} + 4Br^{-}\rightleftharpoons [CdBr_{4}]^{2-}+2OH^{-}[/tex]
[tex]K = K_{f}*K_{sp} = 5*10^{3}*2.5*10^{-14}=\frac{[CdBr_{4}^{2-}][OH^{-}]^{2}}{[Br{-}]^{4}}[/tex]
[tex]12.5*10^{-11} =\frac{1*10^{-3} *[2*10^{-3}]^{2}}{[Br-]^{4} }\\[/tex]
[tex][Br-] = 2.38 M[/tex]
The study of chemicals is called chemistry. when the amount of reactant or product gets equal is said to be an equilibrium state.
The correct answer is 2.38M.
What is equilibrium constant?The equilibrium constant of a chemical reaction is the value of its reaction quotient at chemical equilibrium. A state approached by a dynamic chemical system after sufficient time has elapsed at which its composition has no measurable tendency towards further change.The data is given in the question is as follows:-
[tex]Kf =5*10^3\\Ksp =2.5*10^{-14}[/tex]
The reaction in the given question is as follows:-
[tex]Cd(OH)_2 +4Br^- +[CdbBr_4]^{2-} +2OH^-[/tex]
The formula we used to solve the question is as follows:-
[tex]K =\frac{{[cdbr_4^2-]}[OH]^2}{{Br^-}^2}[/tex]After placing the value, the correct answer for the bromine is 2.38M.
Hence, the correct answer is 2.38M
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Given the following balanced equation, determine the rate of reaction with respect to [Cl2]. If the rate of disappearance of Cl2 is 4.24 × 10–2 M/s, what is the rate of formation of NO? 2 NO(g) + Cl2(g) → 2 NOCl(g)
Answers
Answer : The rate of formation of [tex]NOCl[/tex] is, [tex]8.48\times 10^{-2}M/s[/tex]
Explanation : Given,
Rate of disappearance of [tex]Cl_2[/tex] = [tex]4.24\times 10^{-2}M/s[/tex]
The given rate of reaction is,
[tex]2NO(g)+Cl_2(g)\rightarrow 2NOCl[/tex]
The expression for rate of reaction :
[tex]\text{Rate of disappearance}=-\frac{1}{2}\frac{d[NO]}{dt}=-\frac{d[Cl_2]}{dt}[/tex]
[tex]\text{Rate of formation}=\frac{1}{2}\frac{d[NOCl]}{dt}[/tex]
From this we conclude that,
[tex]\frac{1}{2}\frac{d[NOCl]}{dt}=-\frac{d[Cl_2]}{dt}[/tex]
[tex]\frac{1}{2}\frac{d[NOCl]}{dt}=-\frac{d[Cl_2]}{dt}[/tex]
[tex]\frac{d[NOCl]}{dt}=2\times \frac{d[Cl_2]}{dt}[/tex]
Now put the value of rate of disappearance of [tex]Cl_2[/tex], we get:
[tex]\frac{d[NOCl]}{dt}=2\times (4.24\times 10^{-2}M/s)=8.48\times 10^{-2}M/s[/tex]
Therefore, the rate of formation of [tex]NOCl[/tex] is, [tex]8.48\times 10^{-2}M/s[/tex]
The rate of reaction decides the direction in which the reaction goes. It decides the rate of flow of conversion.
The correct rate of the reaction is [tex]8.48*10^{-2[/tex]
The rate of the reaction of a given element is as follows:-
Formation =[tex]-\frac{1}{2}\frac{d[NO]}{dt} =-\frac{1}{2} \frac{dCL_2}{dt}[/tex]Disappearance =[tex]\frac{1}{2}\frac{d[NOCL]}{dt}[/tex]After solving it from the equation,:-
[tex]\frac{d[NOCL]}{dt} = 2*\frac{d[CL_2]}{dt}[/tex]
After solving it, the value we get is
[tex]2 * 4.24*10^{-2}\\=8.48*10^{-2[/tex]
Hence, the correct answer is [tex]8.48*10^{-2[/tex]
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Which statement about a methyl functional group is correct? 1) a methyl group consists of a carbon bonded of three hydrogen atoms 2) a methyl group is polar 3) a methyl group may be negatively charged
Answers
Answer:
Explanation:
A methyl group consists of a carbon bonded to three hydrogen atoms.
A methyl functional group consists of a carbon atom bonded to three hydrogen atoms. Therefore, the correct option is option 1.
The methyl functional group is a basic building block in organic chemistry. It is made up of a carbon atom that is linked to three hydrogen atoms ([tex]CH_3[/tex]). The methyl group is frequently abbreviated as "Me."
Because carbon-hydrogen (C-H) bonds have equal electronegativities, methyl groups are nonpolar in nature. The methyl group's nonpolarity makes it comparatively unreactive in many chemical reactions, especially when contrasted to more polar functional groups.
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Enter your answer in the provided box. Consider the reaction H2(g) + Cl2(g) → 2HCl(g)ΔH = −184.6 kJ / mol If 2.00 moles of H2 react with 2.00 moles of Cl2 to form HCl, what is ΔU (in kJ) for this reaction at 1.0 atm and 25°C? Assume the reaction goes to completion.
Answers
Answer : The value of [tex]\Delta E[/tex] of the reaction is, -369.2 KJ
Explanation :
Formula used :
[tex]\Delta E=\Delta H-\Delta n_g\times RT[/tex]
where,
[tex]\Delta E[/tex] = internal energy of the reaction = ?
[tex]\Delta H[/tex] = enthalpy of the reaction = -184.6 KJ/mole = -184600 J/mole
The balanced chemical reaction is,
[tex]H_2(g)+Cl_2(g)\rightarrow 2HCl(g)[/tex]
when the moles of [tex]H_2\text{ and }Cl_2[/tex] are 2 moles then the reaction will be,
[tex]2H_2(g)+2Cl_2(g)\rightarrow 4HCl(g)[/tex]
From the given balanced chemical reaction we conclude that,
[tex]\Delta n_g[/tex] = change in the moles of the reaction = Moles of product - Moles of reactant = 4 - 4 = 0 mole
R = gas constant = 8.314 J/mole.K
T = temperature = [tex]25^oC=273+25=298K[/tex]
Now put all the given values in the above formula, we get:
[tex]\Delta E=(-184600J/mole\times 2mole)-(0mole\times 8.314J/mole.K\times 298K)[/tex]
[tex]\Delta E=-369200J[/tex]
[tex]\Delta E=-369.2KJ[/tex]
Therefore, the value of [tex]\Delta E[/tex] of the reaction is, -369.2 KJ
A certain drug is made from only two ingredients: compound A and compound B. There are 7 milliliters of compound A used for every 5 milliliters of compound B. If a chemist wants to make 1116 milliliters of the drug, how many milliliters of compound A are needed?
Answers
Answer:
First step:
7 ml + 5 ml = 12 ml
Second step:
% of A = 7/12 x 100 = 58.33%
% of B = 5/12 x 100 = 41.67%
Third step:
In 1116 ml
compound A = 1116 x (58.33/100) = 651 ml
compound B = 1116 x (41.67/100) = 465 ml
Explanation:
In the 1st step: with what is given, the total volume is 12 ml
In the 2nd step: Find the percentage of each compound in the drug according to what is given.
In the 3rd step: calculate the volume of each compound separately in the new total volume of 1116 ml using the percentage composition.
volume of compound A will therefore be 651 milliliters
By setting up a proportion based on the ratio of 7 milliliters of A for every 5 milliliters of B, solving the subsequent equations yields that 650 milliliters of compound A are needed to make 1116 milliliters of the drug.
To determine how many milliliters of compound A are needed to make 1116 milliliters of the drug, we can set up a proportion based on the ratio given. With 7 milliliters of compound A used for every 5 milliliters of compound B, we get the following equation:
Compound A / Compound B = 7 / 5
Let's let x be the amount of compound A and y be the amount of compound B needed to make 1116 milliliters of the drug, where:
x + y = 1116 mL
Furthermore, we have:
x / y = 7 / 5
From the second equation, we solve for y:
y = 5/7 x
Substituting this into the first equation:
x + 5/7 x = 1116
Multiplying every term by 7 to clear the fraction, we get:
7x + 5x = 1116 × 7
12x = 7804
Dividing both sides by 12:
x = 7804 / 12
x = 650.333...
We round this to the nearest milliliter, since we are dealing with a measurable quantity. Therefore:
x = 650 mL
The amount of I−3(aq) in a solution can be determined by titration with a solution containing a known concentration of S2O2−3(aq) (thiosulfate ion). The determination is based on the net ionic equation 2S2O2−3(aq)+I3(aq)⟶S4O2−6(aq)+3I−(aq) Given that it requires 29.6 mL of 0.260 M Na2S2O3(aq) to titrate a 30.0 mL sample of I−3(aq), calculate the molarity of I−3(aq) in the solution.
Answers
0.128 M of iodide solution was reacted with thiosulphate.
The equation of the reaction is;
[tex]2S2O2^-3(aq) + I3^-(aq)------->S4O2^-6(aq) + 3I^- (aq)[/tex]
Number of moles of S2O2^-3- = 29.6/1000 × 0.260 M
= 0.0077 moles
Since 2 moles of thiosulphate reacts with 1 mole of iodide
0.0077 moles of thiosulphate reacts with 0.0077 moles × 1 mole/ 2 moles
= 0.00385 moles of iodide.
Since;
Number of moles = concentration × volume
concentration of iodide = Number of moles/volume
Volume of iodide = 30/1000 = 0.03 L
Concentration of iodide = 0.00385 moles/0.03 L
= 0.128 M
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Use tabulated electrode potentials to calculate ∆G° rxn for each reaction at 25 °C in kJ.
(a) Pb2+(aq) + Mg(s) ➝ Pb(s) + Mg2+(aq)
(b) Br2(l) + 2 Cl-(aq) ➝ 2 Br-(aq) + Cl2( g)
(c) MnO2(s) + 4 H+(aq) + Cu(s) ➝ Mn2+(aq) + 2 H2O(l) + Cu2+(aq)
Answers
Answer: (a) [tex]\Delta G^0=-432.25kJ[/tex] , (b) [tex]\Delta G^0=55.96kJ[/tex] and (c) [tex]\Delta G^0=-171.74kJ[/tex]
Explanation: (a) Oxidation half reaction for the given equation is:
[tex]Mg(s)\rightarrow Mg^2^+(aq)+2e^-E^0=2.37V[/tex]
The reduction half equation is:
[tex]Pb^2^+(aq)+2e^-\rightarrow Pb(s)E^0=-0.13V[/tex]
[tex]E^0_c_e_l_l=E^0_r_e_d_u_c_t_i_o_n+E^0_o_x_i_d_a_t_i_o_n[/tex]
[tex]E^0_c_e_l_l=-0.13V+2.37V[/tex]
[tex]E^0_c_e_l_l=2.24V[/tex]
[tex]\Delta G^0=-nFE^0_c_e_l_l[/tex]
where n is the number of moles of electrons transferred and F is faraday constant.
2 moles of electrons are transferred in the cell reaction which is also clear from both the half equations.
[tex]\Delta G^0=-2mol*96485\frac{C}{mol}*2.44V[/tex]
[tex]\Delta G^0=-432252.8J[/tex]
or [tex]\Delta G^0=-432.25kJ[/tex]
(b) Oxidation half reaction for the given equation is:
[tex]2Cl^-(aq)\rightarrow Cl_2(g)+2e^-E^0=-1.36V[/tex]
Reduction half equation is:
[tex]Br_2(l)+2e^-\rightarrow 2Br^-E^0=1.07V[/tex]
[tex]E^0_c_e_l_l=1.07V+(-1.36V)[/tex]
[tex]E^0_c_e_l_l=1.07V-1.36V[/tex]
[tex]E^0_c_e_l_l=-0.29V[/tex]
Now, we can calculate the [tex]\Delta G^0[/tex] same as we did for part a.
[tex]\Delta G^0=-2mol*96485\frac{C}{mol}*(-0.29V)[/tex]
[tex]\Delta G^0=55961.3J[/tex]
or [tex]\Delta G^0=55.96kJ[/tex]
(c) Oxidation half reaction for the given equation is:
[tex]Cu(s)\rightarrow Cu^2^+(aq)+2e^-E^0=-0.34V[/tex]
reduction half equation is:
[tex]MnO_2(s)+4H^+(aq)+2e^-\rightarrow Mn^2^+(aq)+2H_2O(l)E^0=1.23V[/tex]
[tex]E^0_c_e_l_l=1.23V+(-0.34V)[/tex]
[tex]E^0_c_e_l_l=0.89V[/tex]
[tex]\Delta G^0=-2mol*96485\frac{C}{mol}*0.89V[/tex]
[tex]\Delta G^0=-171743.3J[/tex]
or [tex]\Delta G^0=-171.74kJ[/tex]
To calculate the standard Gibbs free energy change (∆G°) for each reaction at 25°C in kJ, we can use tabulated electrode potentials. By writing half-cell reactions and summing the electrode potentials, we can determine the overall reaction's standard potential (E°). Then, using the formula ∆G° = -nFE°, we can calculate the standard Gibbs free energy change.
Explanation:Use tabulated electrode potentials to calculate ∆G° rxn for each reaction at 25 °C in kJ.
a) Pb2+(aq) + Mg(s) ➝ Pb(s) + Mg2+(aq)First, we need to write half-cell reactions for the given equation:Pb2+ + 2e- ➝ Pb (E° = -0.126 V)Mg2+ + 2e- ➝ Mg (E° = -2.37 V)Next, we can sum up the electrode potentials to calculate the overall reaction's standard potential:E° rxn = E°(Pb) - E°(Mg)E° rxn = -0.126 V - (-2.37 V) = 2.244 VFinally, we can use the formula ∆G° = -nFE° to calculate the standard Gibbs free energy change:∆G° = -2F(2.244 V)b) Br2(l) + 2 Cl-(aq) ➝ 2 Br-(aq) + Cl2( g)First, we need to write half-cell reactions for the given equation:Br2 + 2e- ➝ 2Br- (E° = 1.07 V)Cl2 + 2e- ➝ 2Cl- (E° = 1.36 V)Next, we can sum up the electrode potentials to calculate the overall reaction's standard potential:E° rxn = E°(2Br-) - E°(2Cl-)E° rxn = 2(1.07 V) - 2(1.36 V) = -0.640 VFinally, we can use the formula ∆G° = -nFE° to calculate the standard Gibbs free energy change:∆G° = -2F(-0.640 V)c) MnO2(s) + 4 H+(aq) + Cu(s) ➝ Mn2+(aq) + 2 H2O(l) + Cu2+(aq)First, we need to write half-cell reactions for the given equation:MnO2 + 4H+ + 2e- ➝ Mn2+ + 2H2O (E° = 1.23 V)Cu2+ + 2e- ➝ Cu (E° = 0.34 V)Next, we can sum up the electrode potentials to calculate the overall reaction's standard potential:E° rxn = E°(Mn2+) - E°(Cu)E° rxn = 1.23 V - 0.34 V = 0.89 VFinally, we can use the formula ∆G° = -nFE° to calculate the standard Gibbs free energy change:∆G° = -2F(0.89 V)Learn more about Calculating standard Gibbs free energy change here:https://brainly.com/question/34263086
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