Answer:
- [tex]AgBr(s)+2NH_3(aq)\rightleftharpoons Ag(NH_3)_2^+(aq)+Br^-(aq)[/tex]
- [tex]K=1.2x10^{-5}[/tex]
Explanation:
Hello,
In this case, by considering the dissolution of silver bromide:
[tex]AgBr(s)\rightleftharpoons Ag^+(aq)+Br^-(aq) \ \ \ Ksp=[Ag^+][Br^-]=7.7x10^{-13}[/tex]
And the formation of the complex:
[tex]Ag^+(aq)+2NH_3(aq)\rightleftharpoons Ag(NH_3)_2^+(aq)\ \ \ Kf=\frac{[Ag(NH_3)_2^+]}{[Ag^+][NH_3]^2}=1.6x10^7[/tex]
We obtain the balanced net ionic equation by adding the aforementioned equations:
[tex]AgBr(s)+Ag^+(aq)+2NH_3(aq)\rightleftharpoons Ag(NH_3)_2^+(aq)+Br^-(aq)+Ag^+(aq)\\\\AgBr(s)+2NH_3(aq)\rightleftharpoons Ag(NH_3)_2^+(aq)+Br^-(aq)[/tex]
Now, the equilibrium constant is obtained by writing the law of mass action for the non-simplified net ionic equation:
[tex]AgBr(s)+Ag^+(aq)+2NH_3(aq)\rightleftharpoons Ag(NH_3)_2^+(aq)+Br^-+Ag^+\\\\K=[Ag^+][Br^-]*\frac{[Ag(NH_3)_2^+]}{[Ag^+][NH_3]^2}[/tex]
So we notice that the equilibrium constant contains the solubility constant and formation constant for the initial reactions:
[tex]K=Ksp*Kf=7.7x10^{-13}*1.6x10^{7}\\\\K=1.2x10^{-5}[/tex]
Best regards.
Analyze: The metallic character of an element is determined by how readily it loses electrons. Elements that lose electrons most easily have the greatest metallic character
A. Which group has the greatest metallic character?
B. Which group has the lowest metallic character?
C. What is the relationship between metallic character and ionization energy?
Answer:
Group 1 or akali metals have the greatest metallic property.
Group 17 has the lowest metallic character.
C. As you move from right to lefton the periodic table, metallic character increases which is the ability to lose electrons. Ionization energy decrease as we move from right to left on the periodic table.
Explanation:
Akali metals in group 1 have the greatest metallic property and they are the most reactive metals. Francium metal on the group has the most metallic characteristics. It is rare and very radioactive. Group 17 has the lowest metallic character. This is because while moving across the period, the number of electrons in the outermost shell increases. This make it difficult for atoms to leave see electrons and become electropositive . Group 17 has the highest tendency of accepting electrons.
Ionization energy is the energy use to remove electron from an atom in gaseous stage. Ionization energy decrease as we move from right to left on the periodic table and metallic character increases as we move from right to left on the periodic table.
Morphine is a well known pain killer but is highly addictive. The lethal dose of morphine varies from person to person based on their body weight and other factors but is somewhere near 70 mg. Calculate the number of millimoles of carbon atoms in 71.891 mg sample of morphine. Report your answer to the third decimal place.
Answer:
0.252 milimoles
Explanation:
To convert mass of a substance to moles it is necessary to use the molar mass of the substance.
The formula of morphine is C₁₇H₁₉NO₃, thus, its molar mass is:
C: 17*12.01g/mol = 204.17g/mol
H: 19*1.01g/mol = 19.19g/mol
N: 1*14g/mol = 14g/mol
O: 3*16g/mol = 48g/mol.
204.17 + 19.19 + 14 + 16 = 285.36g/mol
Thus, moles of 71.891 mg = 0.071891g:
0.071891g × (1mol / 285.36g) = 2.5193x10⁻⁴ moles
As 1 mole = 1000 milimoles:
2.5193x10⁻⁴ moles = 0.252 milimoles
Consider the three isomeric alkanes n-hexane,2,3-dimethylbutane, and 2-methylpentane. Which of the following correctly lists these compounds in order of increasing boiling point
a. 2,3-dimethylbutane < 2-methylpentane < n-hexane
b. 2-methylpentane
c. 2-methylpentane < 2,3-dimethylbutane
d. n-hexane < 2-methylpentane < 2,3-dimethylbutane
e. n-hexane < 2,3-dimethylbutane < 2-methylpentane
Answer:
a. 2,3-dimethylbutane < 2-methylpentane < n-hexane
Explanation:
The boiling point of alkanes is highly affected by the degree of branching in the molecule. Branched alkanes generally have a lower boiling point than unbranched alkanes.
The reason for the higher boiling point of unbranched alkanes is because they have greater vanderwaals forces acting between their molecules due to their larger surface area. Recall that branched alkanes have a lesser surface area compared to unbranched alkanes.
n-hexane is an unbranched alkane hence it will have the highest boiling point followed by 2-methyl pentane and lastly 2,3-dimethyl butane. The boiling point continues to decrease as the extent of branching increases.
Design your own experiment: Factors that affect the rate of a reaction
Did anybody do the lab?
Answer:
Reactant concentration, the physical states of the reactants, surface area, temperature and the presence of catalyst.
Vanadium (V) and oxygen (O) form a series of compounds with the following compositions: Mass % V 76.10 67.98 61.42 56.02 Mass % O 23.90 32.02 38.58 43.98 Compound Mass % N 1 33.28 2 39.94 Mass % Si 66.72 60.06 10. What are the relative numbers of atoms of oxygen in the compounds for a given mass of vanadium
Answer:
For every given mass of Vanadium, the relative number of oxygen atoms present or the mole ratio of Oxygen to Vanadium is:
A. 1:1
B. 3:2
C. 2:1
D. 5:2
Note: The question is stated more clearly below:
Vanadium (V) and oxygen (O) form a series of compounds with the following compositions: Mass % V 76.10 67.98 61.42 56.02 Mass % O 23.90 32.02 38.58 43.98 Compound Mass % N 1 33.28 2 39.94.
What are the relative numbers of atoms of oxygen in the compounds for a given mass of vanadium?
Explanation:
Number of moles in 100 g mass = % mass / molar mass
Molar mass of Vanadium, V = 51 g/mol
Molar mass of oxygen atom, O = 16 g/mol
1. Percentage mass of V and O is 76.10% and 23.90% respectively.
Number of moles of each atom;
V = 76.10/51.0 = 1.5 moles
O = 23.9/16 = 1.5 moles
Mole ratio of oxygen to vanadium = 1.5/1.5 = 1 : 1
2. Percentage mass of V and O is 67.98% and 32.02% respectively
Number of moles of each atom:
V = 67.98/51 = 1.33
O = 32.02/16 = 2
Mole ratio of oxygen to vanadium = 2/1.33 = 1.5 : 1 = 3 : 2
3. Percentage mass of V and O is 61.42% and 38.58% respectively
Number of moles of each atom:
V = 61.42/51 = 1.2
O = 38.58/16 = 2.4
Mole ratio of oxygen to vanadium = 2.4/1.2 = 2 : 1
4. Percentage mass of V and O is 56.02% and 43.98% respectively
Number of moles of each atom:
V = 56.02/51 = 1.10
O = 43.98/16 = 2.75
Mole ratio of oxygen to vanadium = 2.75/1.10 = 2.5 : 1 = 5 : 2
Mass of the Vanadium, number of O2 atoms present, or the mole ratio of 1:1 , 3:2 , 2:1 , 5:2 . As Vanadium (V) and oxygen (O) form a series of compounds is given with masses of 76.10 67.98, 23.90 32.02, 33.28 2 39.94, etc.
As per No of moles in 100 g mass = % mass / molar mass Mass of Vanadium, V = 51 g/ mol e, Mass of oxygen atom, O = 16 g/mole O = 23.9/16 = 1.5 moles for oxygen to vanadium = 1.5/1.5 = 1 : 1 2. Percentage mass of V and O is 67.98% and 32.02%. Mole ratio of oxygen to vanadium = 2/1.33 = 1.5 : 1 = 3 : 2 3. Percentage mass of V and O is 61.42% and 38.58% Mole ratio of oxygen to vanadium = 2.4/1.2 = 2 : 1 4. Percentage mass of V and O is 56.02% and 43.98%. Mole ratio of oxygen to vanadium = 2.75/1.10 = 2.5 : 1 = 5 : 2Learn more about the Vanadium (V) and oxygen (O).
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The following reaction: NO2 (g) --> NO (g) 1/2 O2 (g) is second-order in the reactant. The rate constant for this reaction is 3.40 L/mol*min. Determine the time needed for the concentration of NO2 to decrease from 2.00 M to 1.50 M.
Answer:
t = 0.049 mins or 2.94 secs
Explanation:
For a simple second order reaction, the integrated law which describes the concentration of reactants at a given time t, is as follows: 1/[A] = 1/[A]o + Kt;
Where [A] is concentration of reactant at time, t, [A]o is initial concentration of A; K is rate constant; t is time at a given instant.
Using the integrated rate law:
I/[NO2]t - 1/[NO2]o = Kt
Where K = 3.40 L/mol/min
[NO2]t = 1.5 mol/L
[N02]o = 2.0 mol/L
t = ?
Making t subject of formula;
t = (1/[NO2]t - 1/[NO2]o) / K
t = (1/1.5 - 1/2.0)/3.40
t = 0.049 mins or 2.94 secs
25.00 mL of a H2SO4 solution with an unknown concentration was titrated to a phenolphthalein endpoint with 28.11 mL of a 0.1311 M NaOH solution. What is the concentration of the H2SO4 solution
Answer:
Concentration of the H₂SO₄ solution is 0.0737 M
Explanation:
Equation of the neutralization reaction between the acid, H₂SO₄, and the base, NaOH, is given below:
H₂SO₄ + 2NaOH -----> Na₂SO₄ + 2H₂O
From the above equation, one mole of acid requires 2 moles of base for complete neutralization which occurs at phenolphthalein endpoint.
mole ratio of acid to base, nA/nB = 1:2
Concentration of the base, Cb = 0.1311 M
Volume of base, Vb, = 28.11 mL
Concentration of acid, Ca = ?
Volume of acid, Va + 25.0 mL
Using the formula, CaVa/CbVb = nA/nB
making Ca subject of the formula, Ca = Cb*Vb*nA/Va*nB
substituting the values into the equation
Ca = (0.1311 * 28.11 * 1) / 25.0 * 2 = 0.0737 M
Therefore, concentration of the H₂SO₄ solution is 0.0737 M
Thermal decomposition of 5.00 metric tons of limestone to lime and carbon dioxide requires 9.00 × 106 kJ of heat. Convert this energy to (a) joules; (b) calories; (c) British thermal units. Give your answers in scientific notation.
Answer:
Take a look at the attachment below
Explanation:
Hope that helps!
Which statement describes both homogeneous mixtures and heterogeneous mixtures?
Answer:
both are the types of mixture and both are impure substances that donot have fixed composition and the composition of constituents is not uniform
Answer:
Their components van be separated by physical processes
Explanation:
Out of the answers im given, it makes the most sense. I would double check before submitting though
Refer to the example about diatomic gases A and B in the text to do problems 20-28.
It was determined that 1 mole of B2 is needed to react with 3 moles of A2.
How many grams in one mole of B2?
__g
Answer:
28g.
Explanation:
Hello,
In this case, considering the statement, we can infer that the monoatomic atomic mass of B is 14 g in one one mole. In such a way, since it is diatomic, we can notice that one mole of B2, is having 28 g of B2, as monoatomic atomic mass is considered twice.
Regards.
Spell out the full name of the compound.
Answer:
4–octene.
Explanation:
To name the compound given in the question, we must determine the following:
1. Determine the functional group of the compound and locate its position by giving it the lowest possible count.
2. Locate the longest continuous carbon. This gives the parent name of the compound.
3. Combine the above to obtain the name of the compound.
Now, let us determine the name of the compound bearing in mind the information given above. This is illustrated below:
1. The functional group of the compound is double bond i.e alkene and it located at carbon 4.
2. The longest continuous carbon chain is 8. Since the compound is an alkene, the name becomes octene.
3. Therefore, the name of the compound is:
4–octene.
1-hexanol was prepared by reacting an alkene with either hydroboration-oxidation or oxymercuration-reduction. Draw the structure of the alkene that was used to prepare the alcohol in highest yield. You do not have to consider stereochemistry. Indicate the method of preparation by drawing either BH3 (for hydroboration-oxidation), or Hg (for oxymercuration-reduction), in a separate sketcher. If there is more than one alkene that can be used for a given method, draw all of them. If either hydroboration-oxidation or oxymercuration-reduction can be used, just give the structures for one method. Separate structures with signs from the drop-down menu.
Answer:
Alkene form hexan-1-ol with oxidation in presence of NaOH with highest yield
Explanation:
The solvent was propanone. Which of the three basic colours is most soluble in propanone?
Answer:
Red dye
Explanation:
In the given question, the complete question has not been provided but the propanone is used as a solvent in paper chromatography. The paper chromatography was performed for the black ink in which the black ink got separated in the red, blue and yellow colour.
From these three colours that are red, blue and yellow, the dye which is most soluble in propanone was red as red colour moved the most in the given chromatogram and the dye which travelled the most is most soluble in propanone.
Thus, red dye is the correct answer.
Which of the following would be more reactive than magnesium (Mg)?
A. Calcium (Ca)
B. Potassium (K)
C. Argon (Ar)
D. Beryllium (Be)
Answer:potassium is more reactive than Mg because both lie in the same group and the element potassium has more electropositivity than magnesium
Explanation:
I hope it will help you
Answer: B. Potassium(K)
Explanation:
A student mixes baking soda and vinegar in a glass. Are there any new substances created from this mixture?
Answer:
Explanation:
1. A student mixes baking soda and vinegar in a glass. The results are shown at left. ... Yes I do belive that new substances are being formed because there is a chemical reaction between the baking soda and vinegar turning it into a bubbly substances instead of a powder and liquid.
Yes, there are new substances created from this mixture.
A certain reaction with an activation energy of 155 kJ/mol was run at 495 K and again at 515 K . What is the ratio of f at the higher temperature to f at the lower temperature
Answer:
4.32 is the ratio of f at the higher temperature to f at the lower temperature
Explanation:
Using the sum of Arrhenius equation you can obtain:
ln (f₂/f₁) = Eₐ / R ₓ (1/T₁ - 1/T₂)
Where f represents the rate constant of the reaction at T₁ and T₂ temperatures. Eₐ is the energy activation (155kJ / mol = 155000J/mol) and R is gas constant (8.314J/molK)
Replacing:
ln (f₂/f₁) = 155000J/mol / 8.314J/molK ₓ (1/495K - 1/515)
Where 2 represents the state with the higher temperature and 1 the lower temperature.
ln (f₂/f₁) = 155000J/mol / 8.314J/molK ₓ (1/495K - 1/515)
ln (f₂/f₁) = 1.4626
f₂/f₁ = 4.32
4.32 is the ratio of f at the higher temperature to f at the lower temperature
A piece of wood near a fire is at 23°C. It gains 1,160 joules of heat from the fire and reaches a temperature of 42°C. The specific heat capacity of
wood is 1.716 joules/gram degree Celsius. What is the mass of the piece of wood?
ОА. 16 g
OB. 29 g
ОC. 36 g
OD. 61 g
Answer:
35.578g or 36g if you round
Explanation:
Q=mc ∆∅ where ∅ is temperature difference
1160= m x 1.716 x (42-23)
m = 1160/ 1.716 x19
m=35.578g
m = 36g to nearest whole number
Answer: C. 36 g
Explanation: I got this right on Edmentum.
Naturally occurring sulfur consists of four isotopes: 32S (31.97207 u, 95.0%); 33S (32.97146 u, 0.76%); 34S (33.96786 u, 4.22%); and 36S (35.96709 u, 0.014%). Calculate the average atomic mass of sulfur in atomic mass units.
Answer:
32.062
Explanation:
The following data were obtained from the question:
Mass of isotope A (32S) = 31.97207 u
Abundance of isotope A (A%) = 95.0%
Mass of isotope B (33S) = 32.97146 u Abundance of isotope B (B%) = 0.76%
Mass of isotope C (34S) = 33.96786 u
Abundance of isotope C (C%) = 4.22%
Mass of isotope D (36S) = 35.96709 u Abundance of isotope D (D%) = 0.014%
Average atomic mass of S =..?
The average atomic mass of sulphur, S can be obtained as follow:
Average atomic mass = [(Mass of A x A%)/100] + [(Mass of B x B%)/100] + [(Mass of C x C%)/100] + [(Mass of D x D%)/100]
Average atomic mass of sulphur =
[(31.97207 x 95)/100] + [(32.97146 x 0.76)/100] + [(33.96786 x 4.22)/100] + [(35.96709 x 0.014)/100]
= 30.373 + 0.251 + 1.433 + 0.005
= 32.062
Therefore, the average atomic mass of sulphur is 32.062
The simplest carboxylic acid is called *
O Formaldehye
O formic acid
acetic acid
O
acetone
The most common isotopic forms of hydrogen are ordinary hydrogen (1H) and deuterium (2H), which have percent compositions of 99.98% and 0.0115%, respectively. Convert the percent isotopic composition value of 2H to decimal form.
Answer:
0. 000115
Explanation:
A percentage is defined as a ratio with a basis of 100 as total substance. Convert a percentage to decimal implies to divide the percentage in 100 because decimal form has as basis 1.
For the isotopic forms:
1H: 99.98% → As percent.
99.98% / 100 = 0.9998 → As decimal form.
2H: 0.0115% → As percent.
0.0115% / 100 = 0. 000115→ As decimal form.
The percent should be 0. 000115
The calculation is as follows:For the isotopic forms:
1H: 99.98% → As percent.
Now
[tex]99.98\% \div 100[/tex]= 0.9998 → As decimal form.
Now
2H: 0.0115% → As percent.
And,
[tex]0.0115\% \div 100[/tex]= 0. 000115→ As decimal form.
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what are the differences between strong and weak acids?
Strong acids are completely ionised and weak acids are partly ionised
Answer:
Como forman los iones en soluciión
Explanation:
Los ácidos fuertes y las bases fuertes se refieren a especies que se disocian completamente para formar los iones en solución. Por el contrario, los ácidos y bases débiles se ionizan solo parcialmente y la reacción de ionización es reversible.
PdPd has an anomalous electron configuration. Write the observed electron configuration of PdPd. Express your answer in complete form in order of orbital filling. For example, 1s22s21s22s2 should be entered as 1s^22s^2. View Available Hint(s)
Answer:
1s²,2s²,2p⁶,3s²,3p⁶,4s²,3d¹⁰,4p⁶,5s⁰,4d¹⁰.
Explanation:
Palladium is a chemical element with the symbol Pd and atomic number 46.
The electronic configuration is;
[Kr] 4d¹⁰
The full electronic configuration observed for palladium is given as;
1s²,2s²,2p⁶,3s²,3p⁶,4s²,3d¹⁰,4p⁶,5s⁰,4d¹⁰.
The reason for for the anomlaous electron configuration is beacuse;
1. Full d orbitals are more stable than partially filled ones.
2. At higher energy levels, the levels are said to be degenerated which means that they have very close energies and then electrons can jump from one orbital to another easily.
The value of ΔG°′ΔG°′ for the conversion of glucose-6-phosphate to fructose-6-phosphate (F6P) is +1.67 kJ/mol+1.67 kJ/mol . If the concentration of glucose-6-phosphate at equilibrium is 2.65 mM2.65 mM , what is the concentration of fructose-6-phosphate? Assume a temperature of 25.0°C25.0°C .
Answer:
The concentration of fructose-6-phosphate F6P ≅ 1.35 mM
Explanation:
Given that:
ΔG°′ is the conversion of glucose-6-phosphate to fructose-6-phosphate (F6P) = +1.67 kJ/mol = 1670 J/mol
concentration of glucose-6-phosphate at equilibrium = 2.65 mM
Assuming temperature = 25.0°C
=( 25 + 273)K
= 298 K
We are to find the concentration of fructose-6-phosphate
Using the relation;
ΔG' = -RT In K_c
where;
R = 8.314 J/K/mol
1670 = - (8.314 × 298 ) In K_c
1670 = -2477.572 × In K_c
1670/ 2477.572 = In K_c
0.67 = In K_c
[tex]K_c = e^{-0.67}[/tex]
[tex]K_c =[/tex] 0.511
Now using the equilibrium constant [tex]K_c[/tex]
[tex]K_c = \dfrac{[F6P]}{[G6P]}[/tex]
[tex]0.511 = \dfrac{[F6P]}{[2.65]}[/tex]
F6P = 0.511 × 2.65
F6P = 1.35415
F6P ≅ 1.35 mM
a) What substances are present in an aqueous buffer composed of HC2H3O2 and C2H3O2 - ?b) What happens when LiOH is added to a buffer composed of HC2H3O2 and C2H3O2 - ? Write a chemical equation for that reaction.c) What happens when HBr is added to this buffer? Write a chemical equation for that reaction.
Answer:
a) HC₂H₃O₂, C₂H₃O₂⁻, H₃O⁺, H₂O, OH⁻
b) HC₂H₃O₂ + LiOH ⇄ H₂O + LiC₂H₃O₂
c) C₂H₃O₂⁻ + HBr ⇄ HC₂H₃O₂ + Br⁻
Explanation:
a) In a HC₂H₃O₂/C₂H₃O₂⁻ buffer system, the following reactions take place:
HC₂H₃O₂ + H₂O ⇄ C₂H₃O₂⁻ + H₃O⁺
C₂H₃O₂⁻ + H₂O ⇄ HC₂H₃O₂ + OH⁻
Thus, the species present are: HC₂H₃O₂, C₂H₃O₂⁻, H₃O⁺, H₂O, OH⁻.
b) When LiOH is added to the buffer system, it is partially neutralized according to the following equation.
HC₂H₃O₂ + LiOH ⇄ H₂O + LiC₂H₃O₂
c) When HBr is added to the buffer system, it is partially neutralized according to the following equation.
C₂H₃O₂⁻ + HBr ⇄ HC₂H₃O₂ + Br⁻
find the mass of h2 produced Binary compounds of alkali metals and hydrogen react with water to produce H2(g). The H2H2 from the reaction of a sample of NaH with an excess of water fills a volume of 0.505 L above the water. The temperature of the gas is 35 ∘C∘C and the total pressure is 755 mmHg
Answer: Mass of hydrogen produced is 0.0376 g.
Explanation:
The reaction equation will be as follows.
[tex]NaH(aq) + H_{2}O(l) \rightarrow H_{2}(g) + NaOH(aq)[/tex]
Now, formula for total pressure will be as follows.
[tex]P_{total} = P_{H_{2}} + P_{H_{2}O}[/tex]
Hence, [tex]P_{H_{2}} = P_{total} - P_{H_{2}O}[/tex]
= 755 mm Hg - 42.23 mm Hg
= 712.77 mm Hg
[tex]P_{H_{2}} = \frac{712.77 \times 1 atm}{760 mm Hg}[/tex]
= 0.937 atm
Now, we will calculate the moles of [tex]H_{2}[/tex] as follows.
[tex]P_{H_{2}}V = nRT[/tex]
[tex]0.937 atm \times 0.505 L = n \times 0.0821 \times 308.15 K[/tex]
n = [tex]\frac{0.473}{25.29}[/tex] mol
= 0.0187 mol
Therefore, mass of [tex]H_{2}[/tex] will be calculated as follows.
[tex]m_{H_{2}} = \frac{0.0187 mol \times 2.0158 g}{1 mol}[/tex]
= 0.0376 g
Thus, we can conclude that mass of hydrogen produced is 0.0376 g.
The complete ionic equation for the reaction of aqueous sodium hydroxide with aqueous nitric acid is
Answer and Explanation:
Sodium hydroxide (NaOH) is a strong base and nitric acid (HNO₃) is a strong acid. That means that they dissociates in water by giving the ions:
NaOH ⇒ Na⁺(ac) + OH⁻(ac)
HNO₃ ⇒ H⁺(ac) + NO₃⁻(ac)
The reaction between an acid and a base is called neutralization. In this case, HNO₃ loses its proton and it is converted in NO₃⁻ (nitrate anion). NaOH loses its hydroxyl anion (OH⁻) by giving Na⁺ cations.
Na⁺ cations with NO₃⁻ anions form the salt NaNO₃ (sodium nitrate); whereas H⁺ and OH⁻ form water molecules. The complete equation is the following:
HNO₃(ac) + NaOH(ac) ⇒ NaNO₃(ac) + H₂O(l)
The ionic equation is:
H⁺(ac) + NO₃⁻(ac) + Na⁺(ac) + OH⁻(ac) ⇄ Na⁺(ac) + NO₃⁻(ac) + H₂O(ac)
If we cancel the repeated ions at both sides of the equation, it gives the following ionic reaction:
H⁺(ac) + OH⁻(ac) ⇄ H₂O(ac)
Prepare a solution that is 0.1 M acetic acid and 0.1 M sodium acetate by measuring out 5.0 mL of the 1.0 M acetic acid solution and 5.0 mL of the 1.0 M sodium acetate solution in a 100 mL graduated cylinder, diluting the 10.0 mL to a final volume of 50.0 mL with deionized water, and then stirring. Pour this solution into a clean, dry 100 mL breaker. By knowing that the Ka for acetic acid is 1.8 x -5 10 , calculate the theoretical pH of the solution.
Answer:
4.74
Explanation:
It is possible to find pH of a buffer (The mixture of a weak acid: Acetic acid, with its conjugate base: Sodium acetate) using Henderson-Hasselbalch equation:
pH = pKa + log₁₀ [A⁻] / [HA]
Where pKa is -log Ka of the weak acid, [A⁻] concentration of the conjugate base and [HA] concentration of the weak acid
pKa of acetic acid is -log 1.8x10⁻⁵ = 4.74
The concentration of both, acetic acid and sodium acetate is 0.1M. Replacing in H-H equation:
pH = pKa + log₁₀ [A⁻] / [HA]
pH = 4.74 + log₁₀ [0.1] / [0.1]
pH = 4.74 + log₁₀ 1
pH = 4.74
Theoretical pH is 4.74
An experiment calls for 10.0 mL of bromine (d = 3.12 g/mL). Since an accurate balance is available, it is decided to measure the bromine by mass. How many grams should be measured out? Multiple Choice 3.21 32.1 3.12 31.2 0.312
Answer:
31.2g
Explanation:
The following data were obtained from the question:
Volume of bromine = 10mL
Density of bromine = 3.12 g/mL
Mass of bromine =...?
The Density of the substance is related to it's mass and volume by the following equation:
Density = Mass /volume
With the above equation, we can calculate the mass of bromine as follow:
Density = Mass /volume
Volume of bromine = 10mL
Density of bromine = 3.12 g/mL
Mass of bromine =...?
Density = Mass /volume
3.12 = Mass /10
Cross multiply
Mass of bromine = 3.12 x 10
Mass of bromine = 31.2g
Therefore, the mass of bromine is 31.2g
chemical equation for potassium sulfate and lead(II) acetate
Answer:
K₂SO₄ + Pb(C₂H₃O₂)₂ →PbSO₄ + 2KC₂H₃O₂
A chemical equation is a symbolic representation of a chemical reaction. The chemical equation for the reaction between potassium sulfate ([tex]K_2SO_4[/tex]) and lead(II) acetate ([tex]Pb(CH_3COO)_2[/tex]) can be written as follows:
[tex]K_2SO_4 + Pb(CH_3COO)_2 = PbSO_4 + 2CH_3COOK[/tex]
A basic chemical equation consists of two main parts: the reactant side (left side) and the product side (right side), separated by an arrow indicating the direction of the reaction. Reactants are substances that undergo a chemical change, while products are substances formed as a result of the reaction.
In this reaction, potassium sulfate reacts with lead(II) acetate to form lead(II) sulfate and potassium acetate. It is important to note that the equation is balanced with stoichiometric coefficients, ensuring that the number of atoms of each element is the same on both sides of the equation.
Therefore, the chemical equation for the reaction between potassium sulfate ([tex]K_2SO_4[/tex]) and lead(II) acetate ([tex]Pb(CH_3COO)_2[/tex]) can be written as follows:
[tex]K_2SO_4 + Pb(CH_3COO)_2 = PbSO_4 + 2CH_3COOK[/tex]
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Copper sulfate is a blue solid that is used to control algae growth. Solutions of copper sulfate that come in contact with the surface of galvanized ( Zinc-plated) steel pails undergo the following reaction that forms copper metal on the zinc surface. How many grams of Zinc would react with 454g (1lb) of copper sulfate (160g/mol)?
CuSO4(aq)+ Zn(s)>>>>Cu(s) + ZnSO4(aq)
Answer:
185.49 grams of Zinc would react with 454g (1lb) of copper sulfate
Explanation:
Yo know the following balanced reaction:
CuSO₄(aq)+ Zn(s) →Cu(s) + ZnSO₄(aq)
You can see that by stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of reagents and products are part of the reaction:
CuSO₄: 1 moleZn: 1 moleCu: 1 moleZnSO₄: 1 moleBeing:
Cu: 63.54 g/moleS: 32 g/moleO: 16 g/moleZn: 65.37 g/molethe molar mass of the compounds participating in the reaction is:
CuSO₄:63.54 g/mole + 32 g/mole + 4*16 g/mole= 159.54 g/mole ≅ 160 g/moleZn: 65.37 g/moleCu: 63.54 g/moleZnSO₄: 65.37 g/mole + 32 g/mole + 4*16 g/mole= 161.37 g/moleThen, by stoichiometry of the reaction, the following amounts of mass of reagent and product participate in the reaction:
CuSO₄: 1 moles* 160 g/mole= 160 gZn: 1 mole* 65.37 g/mole= 65.37 gCu: 1 mole* 63.54 g/mole= 63.54 gZnSO₄: 1 mole* 161.37 g/mole= 161.37 gNow you can apply the following rule of three: if 160 grams of CuSO₄ react with 65.37 grams of Zn by this reaction stoichiometry, 454 grams of CuSO₄ with how much mass of Zn will it react?
[tex]mass of Zn=\frac{454 grams of CuSO_{4} *65.37 grams of Zn}{160 grams of CuSO_{4}}[/tex]
mass of Zn= 185.49 grams
185.49 grams of Zinc would react with 454g (1lb) of copper sulfate