Answer:
1)Ni=1s2, 2s2, 2p6, 3s2, 3p6, 4s0, 3d10 called full-filled
2)Cr=1s2, 2s2, 2p6, 3s2, 3p6, 4s1, 3d5 called half-filled
the pain reliever codeine is a weak base with a kb equal to 1.6 x 10^-6. what is the ph of a 0.05 m aqueous codeine solution
Answer:
[tex]pH=10.45[/tex]
Explanation:
Hello,
In this case, for the dissociation of the given base, we have:
[tex]base\rightleftharpoons OH^-+CA[/tex]
Whereas CA accounts for conjugated acid and OH⁻ for the conjugated base. In such a way, equilibrium expression is:
[tex]Kb=\frac{[OH^-][CA^+]}{[base]}[/tex]
And in terms of the reaction extent [tex]x[/tex] we can write:
[tex]1.6x10^{-6}=\frac{x*x}{0.05M-x}[/tex]
For which the roots are:
[tex]x_1=-0.000284M\\x_2=0.000282M[/tex]
For which clearly the result is the positive root which also equals the concentration of hydroxyl ions and we can compute the pOH:
[tex]pOH=-log([OH^-])=-log(0.000282)\\\\pOH=3.55[/tex]
And the pH:
[tex]pH=14-pOH=14-3.55\\\\pH=10.45[/tex]
Regards.
The pH of the solution is 10.45.
Let us represent codeine with the generic formula BH. We can set up the ICE table as follows;
:B(aq) + H2O(l) ⇄ BH(aq) + OH^-(aq)
I 0.05 0 0
C -x +x +x
E 0.05 - x x x
We know that the Kb of codeine is 1.6 x 10^-6, Hence;
1.6 x 10^-6 = x^2/0.05 - x
1.6 x 10^-6 (0.05 - x ) = x^2
8 x 10^-8 - 1.6 x 10^-6x = x^2
x^2 + 1.6 x 10^-6x - 8 x 10^-8 = 0
x = 0.00028 M
The concentration of hydroxide ions = 0.00028 M
Given that pOH = - log[0.00028 M]
pOH = 3.55
pH + pOH = 14
pH = 14 - 3.55
pH = 10.45
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The recommended application for dicyclanil for an adult sheep is 65 mg/kg of body mass. If dicyclanil is supplied in a spray with a concentration of 50. mg/mL, how many milliliters of the spray are required to treat a 70.-kg adult sheep?
Answer:
91 millilitres
Explanation:
Recommended application = 65mg / Kg
This means 65 mg of dicyclanil per kg (1 kg of body mass).
Concentration = 50 mg / mL
How many millilitres required to treat 70kg adult?
If 65mg = 1 kg
x = 70 mg
x = 70 * 65 = 4550 mg
Concentration = Mass / Volume
50 mg/mL = 4550 / volume
volume = 4550 / 50 = 91 mL
How many moles of gold are equivalent to 1.204 × 1024 atoms? 0.2 0.5 2 5
Answer:
c
Explanation:
How many moles of gold are equivalent to 1.204 × 1024 atoms?
0.2
0.5
2
5
C) 2 Is the correct answer, I took the test and it was correct.
According to the concept of Avogadro's number, there are 2 moles of gold which are equivalent to 1.204×10²⁴ atoms.
What is Avogadro's number?
Avogadro's number is defined as a proportionality factor which relates number of constituent particles with the amount of substance which is present in the sample.
It has a SI unit of reciprocal mole whose numeric value is expressed in reciprocal mole which is a dimensionless number and is called as Avogadro's constant.It relates the volume of a substance with it's average volume occupied by one of it's particles .
According to the definitions, Avogadro's number depend on determined value of mass of one atom of those elements.It bridges the gap between macroscopic and microscopic world by relating amount of substance with number of particles.
Number of atoms can be calculated using Avogadro's number as follows: mass/molar mass×Avogadro's number.Number of moles=number of atoms/Avogadro's number=1.204×10²⁴ /6.023×10²³=1.999≅2
Thus, there are 2 moles of gold which are equivalent to 1.204×10²⁴ atoms.
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If a sample of C-14 initially contains 1.6 mmol of C-14, how many millimoles will be left after 2250 years
Answer: 1.2 millimoles will be left after 2250 years
Explanation:
Expression for rate law for first order kinetics is given by:
[tex]t=\frac{2.303}{k}\log\frac{a}{a-x}[/tex]
where,
k = rate constant
t = age of sample
a = let initial amount of the reactant
a - x = amount left after decay process
a) for completion of half life:
Half life is the amount of time taken by a radioactive material to decay to half of its original value.
[tex]t_{\frac{1}{2}}=\frac{0.693}{k}[/tex]
[tex]k=\frac{0.693}{5730}=0.00012years^{-1}[/tex]
b) Amount left after 2250 years
[tex]2250=\frac{2.303}{k}\log\frac{1.6}{a-x}[/tex]
[tex]2250=\frac{2.303}{0.00012}\log\frac{1.6}{a-x}[/tex]
[tex]\log\frac{1.6}{a-x}=0.117[/tex]
[tex]\frac{1.6}{a-x}=1.31[/tex]
[tex]{a-x}=\frac{1.6}{1.31}=1.2[/tex]
Thus 1.2 millimoles will be left after 2250 years
Methanol liquid burns readily in air. One way to represent this equilibrium is: 2 CO2(g) + 4 H2O(g)2 CH3OH(l) + 3 O2(g) We could also write this reaction three other ways, listed below. The equilibrium constants for all of the reactions are related. Write the equilibrium constant for each new reaction in terms of K, the equilibrium constant for the reaction above. 1) CH3OH(l) + 3/2 O2(g) CO2(g) + 2 H2O(g) K1 = 2) CO2(g) + 2 H2O(g) CH3OH(l) + 3/2 O2(g) K2 = 3) 2 CH3OH(l) + 3 O2(g) 2 CO2(g) + 4 H2O(g)
Answer:
Answers are in the explanation
Explanation:
It is possible to obtain K of equilibrium of related reactions knowing the laws:
A + B ⇄ C K₁
C ⇄ A + B K = 1 /K₁
The inverse reaction has the inverse K equilibrium
2A + 2B ⇄ 2C K = K₁²
The multiplication of the coefficients of reaction produce a k powered to the number you are multiplying the coefficients
For the reaction:
2 CO2(g) + 4 H2O(g) ⇄ 2 CH3OH(l) + 3 O2(g) K
1) CH3OH(l) + 3/2 O2(g) ⇄ CO2(g) + 2 H2O(g)
This is the inverse reaction but also the coefficients are dividing in the half, that means:
[tex]K_1 = \frac{1}{k^{1/2}} = (1/K)^{1/2}[/tex]
2) CO2(g) + 2 H2O(g) ⇄ CH3OH(l) + 3/2 O2(g)
Here,the only change is the coefficients are the half of the original reaction:
[tex]K_2 = K^{1/2}[/tex]
3) 2CH3OH(l) + 3 O2(g) ⇄ 2 CO2(g) + 4 H2O(g)
This is the inverse reaction. Thus, you have the inverse K of equilibrium:
[tex]K_3 = \frac{1}{K}[/tex]
Emission of which one of the following leaves both atomic number and mass number unchanged?
(a) positron
(b) neutron
(c) alpha particle
(d) gamma radiation
(e) beta particle
Answer: Gamma Radiation
Explanation:
The emission of Gamma rays does not cause a change in both the atomic and mass number. They are electromagnetic radiation.
The radiations that leaves without changing the atomic mass and atomic number of the particle have been gamma radiations. Thus, option D is correct.
Radiations have been the energy that has been evolved by the particles during energy transitions. The nuclear decay results with the release of the energy from the particle resulting in the change in the atomic mass.
The electromagnetic radiations have been capable of emitting the radiation without changing the mass and atomic number of the element. The gamma radiations have been the electromagnetic radiations. Thus, option D is correct.
For more information about the emissions, refer to the link:
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One hundred fifty joules of heat are removed from a heat reservoir at a temperature of 150 K. What is the entropy change of the reservoir (in J/K)?
Answer:
ΔS surrounding (entropy change of the reservoir) = -1 J/K
Explanation:
Given:
Change in heat (ΔH) = 150 joules
Temperature (T) = 150 K
Find:
ΔS surrounding (entropy change of the reservoir)
Computation:
ΔS surrounding (entropy change of the reservoir) = - ΔH / T
ΔS surrounding (entropy change of the reservoir) = - 150 / 150
ΔS surrounding (entropy change of the reservoir) = -1 J/K
What's the name for the part of Earth made of rock?
A. Geosphere
B. Atmosphere
C. Hydrosphere
D. Biosphere
SUBMIT
Answer:I think it's Geosphere
Explanation:
Answer:
A
Explanation:
Geo means rock, or earth. Hydro means water, Atmosphere is space, and Bio global ecosystem composed of living organisms
Divers often inflate heavy duty balloons attached to salvage items on the sea floor. If a balloon is filled to a volume of 1.20 L at a pressure of 6.25 atm, what is the volume of the balloon when it reaches the surface?
Answer:
7.50 L
Explanation:
The balloon has a volume of 1.20 L (V₁) when the pressure at the sea floor is 6.25 atm (P₁). When it reaches the surface, the pressure is that of the atmosphere, that is, 1.00 atm (P₂). If we consider the gas to behave as an ideal gas and the temperature to be constant, we can calculate the final volume (V₂) using Boyle's law.
P₁ × V₁ = P₂ × V₂
V₂ = P₁ × V₁ / P₂
V₂ = 6.25 atm × 1.20 L / 1.00 atm
V₂ = 7.50 L
A 0.753 g sample of a monoprotic acid is dissolved in water and titrated with 0.250 M NaOH. What is the molar mass of the acid if 21.5 mL of the NaOH solution is required to neutralize the sample?
Answer:
[tex]MM_{acid}=140.1g/mol[/tex]
Explanation:
Hello,
In this case, since the acid is monoprotic, we can notice a 1:1 molar ratio between, therefore, for the titration at the equivalence point, we have:
[tex]n_{acid}=n_{base} \\\\V_{acid}M_{acid}=V_{base}M_{base}\\\\n_{acid}=V_{base}M_{base}[/tex]
Thus, solving for the moles of the acid, we obtain:
[tex]n_{acid}=0.0215L*0.250\frac{mol}{L}=5.375x10^{-3}mol[/tex]
Then, by using the mass of the acid, we compute its molar mass:
[tex]MM_{acid}=\frac{0.753g}{5.375x10^{-5}mol} \\\\MM_{acid}=140.1g/mol[/tex]
Regards.
What is buffers and mention its importance?
Answer:
Buffer is the chemical substance that addition of acids and bases, maintaining constant environment,its called Buffer.
Explanation:
Buffers are use in the system to maintain the value of pH, and the contain the pH value is not to change.Buffer maintain the body of pH for the optimal activity,and they are solution of pH constant.Buffer in used in the lab and that to maintain growth of the micro tissues and the culture media.Buffer are used in maintain necessary optimal reaction activity,determine the indicator of solution with pH.Buffer capacity is that concentration to the buffering agent, is the very small increase,buffer capacity to the pH is 32% , of the maximum value of pH.Buffers in a acid regions to the desired of that value to the particular buffer agent.Buffers can be made from that a mixture of the base and acid, buffer can be a wide range of the obtained.Buffers that the pH calculation and they required to performed in the critic acid that the overlap over the buffer range.For the reaction CO2(g) + H2(g)CO(g) + H20(g)
∆H°=41.2 kJ and ∆S°=42.1 J/K
The standard free energy change for the reaction of 1.96 moles of Co2(g) at 289 K, 1 atm would be_________KJ.
This reaction is (reactant, product)___________ favored under standard conditions at 289 K.
Assume that ∆H° and ∆S° are independent of temperature.
Answer:
The ΔG° is 29 kJ and the reaction is favored towards reactant.
Explanation:
Based on the given information, the ΔH°rxn or enthalpy change is 41.2 kJ, the ΔS°rxn or change in entropy is 42.1 J/K or 42.1 * 10⁻³ kJ/K. The temperature given is 289 K. Now the Gibbs Free energy change can be calculated by using the formula,
ΔG° = ΔH°rxn - TΔS°rxn
= 41.2 kJ - 289 K × 42.1 × 10⁻³ kJ/K
= 41.2 kJ - 12.2 kJ
= 29 kJ
As ΔG° of the reaction is positive, therefore, the reaction is favored towards reactant.
When salt is added to water, all of the following happens except? A. The salt breaks into positive chlorine ions and negative sodium icons B. the positive part of the water molecule is attracted to the negative ions C. The negative part of the water molecule is attracted to the positive ions D. The water molecules surround the dissociated ions
Answer:
The salt breaks into positive chlorine ions and negative sodium icons
Explanation:
The question requested for the wrong option in the list. If we look at the option selected, we will notice that sodium ions are positively charged ions since sodium is a metal. Metals produce cations (positive ions) because they loose electrons. Therefore, a sodium ion can never be negatively charged.
Similarly, chlorine is a highly electronegative nonmetal. It gains electrons in an ionic bond. Hence chlorine ions can not be positive.
Which of the following do we need to know in order to calculate pH during an acid-base titration of a strong monoprotic acid with a strong monoprotic base? Select all that apply
a. the concentration of the acid
b. the concentration of the base titrant
c. the initial volume of the acid solution
d. the volume of the titrant used
Answer:
the volume of the titrant used
Explanation:
Acid-base titrations are usually depicted on special graphs referred to as titration curve. A titration curve is a graph that contains a plot of the volume of the titrant as the independent variable and the pH of the system as the dependent variable.
Hence, a titration curve is a graphical plot showing the pH of the analyte solution plotted against the volume of the titrant as the reaction is in progress. The titration curve is drawn by plotting data obtained during a titration, that is, volume of the titrant added (plotted on the x-axis) and pH of the system (plotted on the y-axis).
How many milliliters of 7.10 M hydrobromic acid solution should be used to prepare 5.50 L of 0.400 M HBr
Answer:
310 mL
Explanation:
Step 1: Given data
Initial concentration (C₁): 7.10 MInitial volume (V₁): ?Final concentration (C₂): 0.400 MFinal volume (V₂): 5.50 LStep 2: Calculate the initial volume
We have a concentrated HBr solution and we want to prepare a diluted one. We can do so using the dilution rule.
C₁ × V₁ = C₂ × V₂
V₁ = C₂ × V₂ / C₁
V₁ = 0.400 M × 5.50 L / 7.10 M
V₁ = 0.310 L = 310 mL
In the pictured cell, the side containing zinc is the Choose... and the side containing copper is the Choose... . The purpose of the N a 2 S O 4 NaX2SOX4 is to
Answer:
Zinc- anode
Copper- cathode
Sodium sulphate- salt bridge
Explanation:
A galvanic cell is an electrochemical cell in which electrical energy is produced by a spontaneous chemical reaction.
In the pictured galvanic cell, zinc is the anode since it looses electrons according to the reaction; Zn(s) -----> Zn^2+(aq) + 2e
Copper is the cathode as shown here; Cu^2+(aq) + 2e ----> Cu(s)
Sodium sulphate functions as the salt bridge. It keeps the both solutions neutral by ensuring charge balance in the both half cells.
Answer:
zinc=anode
copper=cathode
Explanation:
The front curve of a spectacle lens is called?
Answer:
Corrective lense or just lens.
Explanation:
Calculate the enthalpy change (∆H) for the reaction- N2(g) + 3 F2(g) –––> 2 NF3(g) given the following bond enthalpies: N≡N 945 kJ/mol F–F 155 kJ/mol N–F 283 kJ/mol
Answer:
– 844 kJ/mol.
Explanation:
The following data were obtained from the question:
N2(g) + 3 F2(g) –––> 2 NF3(g)
Enthalpy of N≡N (N2) = 945 kJ/mol
Enthalpy of F–F (F2) = 155 kJ/mol
Enthalpy of N–F3 (NF3) = 283 kJ/mol
Enthalpy change (∆H) =?
Next, we shall determine the enthalpy of reactant.
This is illustrated below:
Enthalpy of reactant (Hr) = 945 + 3(155)
Enthalpy of reactant (Hr) = 945 + 465
Enthalpy of reactant (Hr) = 1410 kJ/mol
Next, we shall determine the enthalpy of the product.
This is illustrated below:
Enthalpy of product (Hp) = 2 x 283
Enthalpy of product (Hp) = 566 kJ/mol
Finally, we shall determine the enthalpy change (∆H) for the reaction as follow:
Enthalpy of reactant (Hr) = 1410 kJ/mol
Enthalpy of product (Hp) = 566 kJ/mol
Enthalpy change (∆H) =?
Enthalpy change (∆H) = Enthalpy of product (Hp) – Enthalpy of reactant (Hr)
Enthalpy change (∆H) = 566 – 1410
Enthalpy change (∆H) = – 844 kJ/mol
Answer:
– 844 kJ/mol.
Explanation:
The following data were obtained from the question:
N2(g) + 3 F2(g) –––> 2 NF3(g)
Enthalpy of N≡N (N2) = 945 kJ/mol
Enthalpy of F–F (F2) = 155 kJ/mol
Enthalpy of N–F3 (NF3) = 283 kJ/mol
Enthalpy change (∆H) =?
Next, we shall determine the enthalpy of reactant.
This is illustrated below:
Enthalpy of reactant (Hr) = 945 + 3(155)
Enthalpy of reactant (Hr) = 945 + 465
Enthalpy of reactant (Hr) = 1410 kJ/mol
Next, we shall determine the enthalpy of the product.
This is illustrated below:
Enthalpy of product (Hp) = 2 x 283
Enthalpy of product (Hp) = 566 kJ/mol
Finally, we shall determine the enthalpy change (∆H) for the reaction as follow:
Enthalpy of reactant (Hr) = 1410 kJ/mol
Enthalpy of product (Hp) = 566 kJ/mol
Enthalpy change (∆H) =?
Enthalpy change (∆H) = Enthalpy of product (Hp) – Enthalpy of reactant (Hr)
Enthalpy change (∆H) = 566 – 1410
Enthalpy change (∆H) = – 844 kJ/mol
Explanation:
A compound (C_9H_9BrO_2) gives the following NMR data. Draw the structure of the compound.
'1^H-NMR: 1.39 ppm, t(3H); 4.38 ppm, q(2H); 7.57 ppm, d(2H); 7.90 ppm, d(2H)
13^C-NMR: 165.73; 131.56; 131.01; 129.84; 127.81; 61.18; 14.18
You do not have to consider stereo chemistry.
You do not have to explicitly draw H atoms.
Do not include lone pairs in your answer.
Answer:
ethyl 4-bromobenzoate
Explanation:
In this question, we can start with the Index of Hydrogen Deficiency (I.H.D):
[tex]I.H.D=\frac{2C+2+N-H-X}{2}=\frac{(2*9)+2+0-9-1}{2}~=~5[/tex]
This indicates, that we can have a benzene ring (I.H.D = 4) and a carbonyl group (I.H.D = 1), for a total of 5.
Additionally, in the 1H-NMR info, we have a triplet 1.39 (3H) followed by a doublet 4.38 (2H), this indicates the presence of an ethyl group ([tex]CH_3-CH_2-[/tex]). Also, in the formula, we have 2 oxygens if we have carbonyl group with 2 oxygens we have a high probability to have an ester group.
[tex]O=C-O-CH_2-CH_3[/tex]
Now, if we add this to the benzene ring and the "Br" atom that we have in the formula, we will have ethyl 4-bromobenzoate.
See figures 1 and 2 to further explanations.
I hope it helps!
A laboratory technician combines 35.9 mL of 0.258 M chromium(II) chloride with 35.8 mL 0.338 M potassium hydroxide. How many grams of chromium(II) hydroxide can precipitate
Answer:
0.52 g of chromium(II) hydroxide, Cr(OH)2.
Explanation:
We'll begin by calculating the number of mole of chromium (ii) chloride, CrCl2 in 35.9 mL of 0.258 M chromium(II) chloride solution.
This can be obtained as follow:
Molarity of CrCl2 = 0.258 M
Volume = 35.9 mL = 35.9/1000 = 0.0359 L
Mole of CrCl2 =?
Molarity = mole /Volume
0.258 = mole of CrCl2 /0.0359
Cross multiply
Mole of CrCl2 = 0.258 x 0.0359
Mole of CrCl2 = 0.0093 mole
Next, we shall determine the number of mole of potassium hydroxide, KOH in 35.8 mL 0.338 M potassium hydroxide solution.
This can be obtained as follow:
Molarity of KOH = 0.338 M
Volume = 35.8 mL = 35.8/1000 = 0.0358 L
Mole of KOH =.?
Molarity = mole /Volume
0.338 = mole of KOH /0.0358
Cross multiply
Mole of KOH = 0.338 x 0.0358
Mole of KOH = 0.0121 mole.
Next, we shall write the balanced equation for the reaction. This is given below:
2KOH + CrCl2 → Cr(OH)2 + 2KCl
From the balanced equation above,
2 mole of KOH reacted with 1 mole of CrCl2 to produce 1 mole of Cr(OH)2.
Next, we shall determine the limiting reactant. This can be obtained as follow:
From the balanced equation above,
2 mole of KOH reacted with 1 mole of CrCl2.
Therefore, 0.0121 mole of KOH will react with = (0.0121 x 1)/2 = 0.00605 mole of CrCl2.
From the calculations made above, we can see that only 0.00605 mole out of 0.0093 mole of CrCl2 is needed to react completely with 0.0121 mole of KOH.
Therefore, KOH is the limiting reactant.
Next, we shall determine the number of mole of Cr(OH)2 produced from the reaction.
In this case, we shall be using the limiting reactant because it will give the maximum yield of Cr(OH)2.
The limiting reactant is KOH and the number of mole of Cr(OH)2 produced can be obtained as illustrated below:
From the balanced equation above,
2 mole of KOH reacted to produce 1 mole of Cr(OH)2.
Therefore, Therefore, 0.0121 mole of KOH will react with = (0.0121 x 1)/2 = 0.00605 mole of Cr(OH)2.
Finally, we shall convert 0.00605 mole of Cr(OH)2 to grams.
This is illustrated below:
Mole of Cr(OH)2 = 0.00605 mole
Molar mass of Cr(OH)2 = 52 + 2(16 + 1) = 52 + 2(17) = 86 g/mol
Mass of Cr(OH)2 =..?
Mole = mass /Molar mass
0.00605 = mass of Cr(OH)2/86
Cross multiply
Mass of Cr(OH)2 = 0.00605 x 86
Mass of Cr(OH)2 = 0.52 g
Therefore, 0.52 g of chromium(II) hydroxide, Cr(OH)2 was produced.
You have to prepare a pH 3.65 buffer, and you have the following 0.10M solutions available: HCOOH, CH3COOH, H3PO4, HCOONa, CH3COONa, and NaH2PO4. How many mL of HCOOH and HCOONa would you use to make approximately a liter of the buffer?
Answer:
550mL of HCOOH 0.1M and 450mL of HCOONa 0.1M
Explanation:
It is possible to find the pH of a buffer by using H-H equation:
pH = pKa + log [A⁻]/[HA]
For the formic buffer (HCOOH/HCOONa):
pH = 3.74 + log [HCOONa]/[HCOOH]
As you need a buffer of pH 3.65:
pH = 3.74 + log [HCOONa]/[HCOOH]
3.65 = 3.74 + log [HCOONa]/[HCOOH]
0.81283 = [HCOONa]/[HCOOH] (1)Where [HCOONa]/[HCOOH] can be taken as the moles of each specie.
As molarity of both solutions is 0.10M (0.10mol / L) and you need 1L of solution, total moles of the buffer are:
0.10 moles = [HCOONa] + [HCOOH] (2)Replacing (2) in (1):
0.81283 = 0.10 - [HCOOH] /[HCOOH]
0.81283[HCOOH] = 0.10 - [HCOOH]
1.81283[HCOOH] = 0.10
[HCOOH] = 0.055 molesAnd moles of HCOONa are:
[HCOONa] = 0.1 mol - 0.055mol =
[HCOONa] = 0.045 molesAs concentration of the solutions is 0.1M, the volume you need to add of both solutions is:
HCOOH = 0.055 mol ₓ (1L / 0.1mol) = 0.55L = 550mL of HCOOH 0.1M
HCOONa = 0.045 mol ₓ (1L / 0.1mol) = 0.45L = 450mL of HCOONa 0.1M
The number should be considered like 550mL of HCOOH 0.1M and 450mL of HCOONa 0.1M.
Calculation of mL:Here we used the H-H equation:
pH = pKa + log [A⁻]/[HA]
Now
For the formic buffer (HCOOH/HCOONa):
So,
pH = 3.74 + log [HCOONa]/[HCOOH]
Now
need a buffer of pH 3.65:
So,
pH = 3.74 + log [HCOONa]/[HCOOH]
3.65 = 3.74 + log [HCOONa]/[HCOOH]
0.81283 = [HCOONa]/[HCOOH] (1)
here [HCOONa]/[HCOOH] can be considered as the moles of each specie.
Now the total moles should be
0.10 moles = [HCOONa] + [HCOOH] (2)
Now
0.81283 = 0.10 - [HCOOH] /[HCOOH]
0.81283[HCOOH] = 0.10 - [HCOOH]
1.81283[HCOOH] = 0.10
[HCOOH] = 0.055 moles
And moles of HCOONa should be
[HCOONa] = 0.1 mol - 0.055mol =
[HCOONa] = 0.045 moles
Now
HCOOH = 0.055 mol ₓ (1L / 0.1mol) = 0.55L = 550mL of HCOOH 0.1M
HCOONa = 0.045 mol ₓ (1L / 0.1mol) = 0.45L = 450mL of HCOONa 0.1M
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Using these metal ion/metal standard reduction potentials Cd2+(aq)|Cd(s) Zn2+(aq)|Zn(s) Ni2+(aq)|Ni(s) Cu2+(aq)/Cu(s) -0.40 V -0.76 V ‑0.25 V +0.34 V Calculate the standard cell potential for the cell whose reaction is Ni2+(aq) + Zn(s) →Zn2+(aq)+ Ni(s)
Answer: The standard cell potential for the cell is +0.51 V
Explanation:
Given : [tex]E^0_{Ni^{2+}/Ni}=-0.25V[/tex]
[tex]E^0_{Zn^{2+}/Zn}=-0.76V[/tex]
The given reaction is:
[tex]Ni^{2+}(aq)+Zn(s)\rightarrow Zn^{2+}(aq)+Ni(s)[/tex]
As nickel is undergoing reduction, it acts as cathode and Zinc is undergoing oxidation, so it acts as anode.
[tex]E^0_{cell}=E^0_{cathode}-E^0_{anode}[/tex]
where both [tex]E^0[/tex] are standard reduction potentials.
Thus putting the values we get:
[tex]E^0_{cell}=-0.25-(-0.76)[/tex]
[tex]E^0_{cell}=0.51V[/tex]
Thus the standard cell potential for the cell is +0.51 V
Which ONE of these cations has the same number of unpaired electrons as Fe2+ ? A) Ni2+ B) Fe3+ C) Cr2+ D) Mn2+ E) Co2+
Answer:
Explanation:
Fe2+ Has 4 unpaired electrons.
By method of elimination;
Option A: Ni2+ has two unpaired electrons. so this option is wrong.
Option B: There are 5 unpaired electrons in the Fe3+ ion. so this option is wrong.
Option C: There are 4 unpaired electrons in the Cr2+ ion. so this option is correct.
Option D: There are 5 unpaired electrons in the Mn2+ ion. so this option is wrong.
Option E: There are 3 unpaired electrons in the Co2+ ion. so this option is wrong.
Complete the sentences describing the cell.
a. In the nickel-aluminum galvanic cell, the cathode is ____ .
b. Therefore electrons flow from___ to ____.
c. The ____ electrode loses mass, while the ____ electrode gains mass.
Answer:
a. In the nickel-aluminum galvanic cell, the cathode is nickel electrode.
b. Therefore electrons flow from the aluminium electrode to the nickel electrode.
c. The aluminium electrode loses mass, while the nickel electrode electrode gains mass.
Explanation:
Voltaic or galvanic cells are electrochemical cells in which spontaneous oxidation-reduction reactions. The two halves of the redox reaction are separate and electron transfer is required to occur through an external circuit for the redox reaction to take place. That is, one of the metals in one of the half cells is oxidized while the metal of the other half cell is reduced, producing an exchange of electrons through an external circuit. This makes it possible to take advantage of the electric current.
Given:
E ⁰N i ⁺² = − 0.23 V is the standard reduction potential for the nickel ion
E ⁰ A l ⁺³ = − 1.66 V is the standard reduction potential for the aluminum ion
The most negative potentials correspond to more reducing substances. In this case, the aluminum ion is the reducing agent, where oxidation takes place. In the anodic half cell oxidations occur, while in the cathode half cell reductions occur. So the aluminum cell acts as the anode while the nickel cell acts as the cathode.
So a. In the nickel-aluminum galvanic cell, the cathode is nickel electrode.
The metal that is oxidized gives electrons to the metal that is reduced through the outer conductor. Then the electrons flow spontaneously from the anode to the cathode.
Then b. Therefore electrons flow from the aluminium electrode to the nickel electrode.
Ni⁺², being the cathode, accepts electrons, becoming Ni (s) and depositing on the Ni electrodes.
So, c. The aluminium electrode loses mass, while the nickel electrode electrode gains mass.
Suppose a student completes an experiment with an average value of 2.9 mL and a calculated standard deviation of 0.71 mL. What is the minimum value within a 1 SD range of the average
Answer:
The correct answer is 2.2 mL.
Explanation:
Given:
Average: 2.9 mL
SD: 0.71 mL
We can define a 1 SD range in which the value of volume (in mL) will be comprised:
Volume (mL) = Average ± SD = (2.9 ± 0.7) mL
Maximum value= Average + SD= 2.9 + 0.7 mL = 3.6 mL
Minimum value= Average - SD = 2.9 - 0.7 mL = 2.2 mL
Thus, the minimum value within a 1 SD range of the average is 2.2 mL
The minimum value within 1 SD is 2.19 mL
The z score is used to determine by how many standard deviations the raw score is above or below the mean. The z score is given by:
[tex]z=\frac{x-\mu}{\sigma} \\\\where\ x\ is\ raw\ score, \mu=mean,\sigma=standard\ deviation[/tex]
Given that μ = 2.9 mL, σ = 0.71 mL; hence:
The minimum value within 1 SD range = μ ± σ = 2.9 ± 0.71 = (2.19, 3.61)
Therefore the minimum value within 1 SD is 2.19 mL
Find out more at: https://brainly.com/question/23907081
In which pair do both compounds exhibit predominantly ionic bonding? A) KCl and CO2 B) SO2 and BaF2 C) F2 and N2O D) N2O3 and Rb2O E) NaF and SrO
Answer:
E) NaF and SrO
Explanation:
The ionic bonding occurs between atoms with a great difference in electronegativity. This usually happens between a metal and a non-metal.
In which pair do both compounds exhibit predominantly ionic bonding?
A) KCl and CO₂. NO. C and O are non-metals and present covalent bonding.
B) SO₂ and BaF₂. NO. S and O are non-metals and present covalent bonding.
C) F₂ and N₂O. NO. Both compounds contain non-metals and present covalent bonding.
D) N₂O₃ and Rb₂O. NO. N and O are non-metals and present covalent bonding.
E) NaF and SrO. YES. Na and Sr are metals while F and O are non-metals.
A solution contains 2.2 × 10-3 M in Cu2+ and 0.33 M in LiCN. If the Kf for Cu(CN)42- is 1.0 × 1025, how much copper ion remains at equilibrium?
Answer:
[Cu²⁺] = 2.01x10⁻²⁶
Explanation:
The equilibrium of Cu(CN)₄²⁻ is:
Cu²⁺ + 4CN⁻ ⇄ Cu(CN)₄²⁻
And Kf is defined as:
Kf = 1.0x10²⁵ = [Cu(CN)₄²⁻] / [Cu²⁺] [CN⁻]⁴
As Kf is too high you can assume all Cu²⁺ is converted in Cu(CN)₄²⁻ -Cu²⁺ is limiting reactant-, the new concentrations will be:
[Cu²⁺] = 0
[CN⁻] = 0.33M - 4×2.2x10⁻³ = 0.3212M
[Cu(CN)₄²⁻] = 2.2x10⁻³
Some [Cu²⁺] will be formed and equilibrium concentrations will be:
[Cu²⁺] = X
[CN⁻] = 0.3212M + 4X
[Cu(CN)₄²⁻] = 2.2x10⁻³ - X
Where X is reaction coordinate
Replacing in Kf equation:
1.0x10²⁵ = [2.2x10⁻³ - X] / [X] [0.3212M +4X]⁴
1.0x10²⁵ = [2.2x10⁻³ - X] / 0.0104858X + 0.524288 X² + 9.8304 X³ + 81.92 X⁴ + 256 X⁵
1.04858x10²³X + 5.24288x10²⁴ X² + 9.8304x10²⁵ X³ + 8.192x10²⁶ X⁴ + 2.56x10²⁷ X⁵ = 2.2x10⁻³ - X
1.04858x10²³X + 5.24288x10²⁴ X² + 9.8304x10²⁵ X³ + 8.192x10²⁶ X⁴ + 2.56x10²⁷ X⁵ - 2.2x10⁻³ = 0
Solving for X:
X = 2.01x10⁻²⁶
As
[Cu²⁺] = X
[Cu²⁺] = 2.01x10⁻²⁶11. The mass (in grams) of FeSO4.7H2O required for preparation of 125 mL of 0.90 M
solution is:
(a) 16 g
(b) 25 g
(c) 13 g
(d) 31 g
(e) 43 g
Answer:
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Taking into account the definition of molarity and the molar mass of the compound, the correct answer is option (d): the mass of FeSO₄.7H₂O required for preparation of 125 mL of 0.90 M solution is 31 g.
In first place, you have to know tha molarity is a measure of the concentration of that substance that indicates the number of moles of solute present in the solution.
The molarity of a solution is calculated by dividing the moles of the solute by the volume of the solution.
[tex]molarity=\frac{number of moles of solute}{volume of solution}[/tex]
Molarity is expressed in units [tex]\frac{moles}{liter}[/tex].
In this case, you know:
molarity= 0.90 Mnumber of moles of solute= ?volume= 125 mL= 0.125 L (being 1000 mL=1 L)So, by definition of molarity, the number of moles is calculated as:
[tex]0.90 M=\frac{number of moles of solute}{0.125 L}[/tex]
Solving:
number of moles of solute= 0.90 M× 0.125 L
number of moles of solute= 0.1125 moles
On the other side, molar mass is the mass of one mole of a substance, which can be an element or a compound. In this case, the molar mass of FeSO₄.7H₂O is 277.85 [tex]\frac{g}{mole}[/tex].
Then you can apply the following rule of three: if by definition of molar mass, 1 mole of the compound contains 277.85 g, 0.1125 mole contains how much mass?
[tex]mass=\frac{0.1125 moles*277.85 g}{1 mole}[/tex]
Solving:
mass ≅ 31 g
Finally, the correct answer is option (d): the mass of FeSO₄.7H₂O required for preparation of 125 mL of 0.90 M solution is 31 g.
Learn more about molarity with this example: brainly.com/question/15406534?referrer=searchResults
Chlorine can be prepared in the laboratory by the reaction of manganese dioxide with hydrochloric acid, HCl(aq), as
described by the chemical equation
MnO,(s) + 4 HCl(aq)
MnCl(aq) + 2 H2O(l) + Cl (8)
How much MnO(s) should be added to excess HCl(aq) to obtain 175 mL C12(g) at 25 °C and 715 Torr?
mass of MnO2
Answer:
Explanation:
MnO₂(s) + 4 HCl(aq) = MnCl₂(aq) + 2 H₂O(l) + Cl₂
87 g 22.4 x 10³ mL
volume of given chlorine gas at NTP or at 760 Torr and 273 K
= 175 x ( 273 + 25 ) x 715 / (273 x 760 )
= 179.71 mL
22.4 x 10³ mL of chlorine requires 87 g of MnO₂
179.4 mL of chlorine will require 87 x 179.4 / 22.4 x 10³ g
= 696.77 x 10⁻³ g
= 696.77 mg .
What type of chemist exclusively studies most carbon compounds?
-biochemist
-physical chemist
-organic chemist
-inorganic chemist
Answer:
Organic chemist? I do not know.
Explanation:
Thanks you.
The type of chemist exclusively studies most carbon compounds are organic chemist. Therefore, option C is correct.
What is an organic chemist ?The structure, characteristics, and reactivity of compounds containing carbon are studied by organic chemists. Additionally, they create novel organic materials with distinct features and uses.
Analytical capabilities, communication skills, and numeracy skills are three of the most important soft skills for an organic chemist.
Organic chemists often work in research and development in labs at universities, pharmaceutical, industrial, and biotechnology businesses, as well as government agencies, according to the American Chemical Society.
According to one assessment, organic chemistry is the hardest college course. According to certain statistics, almost one out of every two students in organic chemistry leave the course. The hopes of a medical career come tumbling down for those who fit this description. Organic chemistry is undoubtedly challenging.
Thus, option C is correct.
To learn more about an organic chemist, follow the link;
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