Answers
find the given attachment
Related Questions
Classify the following unbalanced chemical reaction Fe(OH)2(s) + HCl(aq) = FeCl2(aq) + H2O(l)
1. Acid-Base Reaction
2. Precipitation Reaction
3. Oxidation-Reduction Reaction
4. Combustion Reaction
Answers
Answer:
1. Acid-Base Reaction
Explanation:
Fe(OH)2(s) + HCl(aq) = FeCl2(aq) + H2O(l)
base acid
This a reaction between base and acid.
Ferrous hydroxide is an inorganic alkaline compound whereas hydrochloric acid is an acid. The reaction between Fe(OH)₂and HCl is an acid-base reaction. Thus, option 1 is correct.
What is an acid-base reaction?An acid-base reaction is a chemical change that occurs and takes place when the reactant constitutes an acid and a base. They are characterized by the exchange of protons that results in the formation of conjugate bases and acids or salt.
The acid-base chemical reaction is shown as,
Fe(OH)₂(s) + HCl(aq) ⇒ FeCl₂(aq) + H₂O(l)
Here, ferrous hydroxide is a base with hydroxide ions and hydrochloric acid is an acid with hydrogen ions. HCl donates its proton to form water molecules with hydroxide ions of ferrous hydroxide.
Therefore, in option 1. the reaction is an acid-base reaction.
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n an experiment, 39.26 mL of 0.1062 M NaOH solution was required to titrate 37.54 mL of \ v unknown acetic acid solution to a phenolphthalein end point. Calculate the molarity of the acetic acid solution, and the percent (by weight) of acetic acid in the solution (assuming its density to be 1.00 g/mL).
Answers
Answer:
Molarity: 0.111M
% (w/w): 0.666
Explanation:
The reaction of NaOH with acetic acid (CH₃COOH) is:
NaOH + CH₃COOH → CH₃COO⁻Na⁺ + H₂O
where 1 mole of NaOH reacts per mole of acetic acid producing 1 mole of water and 1 mole of sodium acetate.
As 39.26mL ≡ 0.03926L of 0.1062M are required to titrate the solution of acetic acid. Moles are:
0.03926L × (0.1062mol / L) = 4.169x10⁻³ moles of NaOH. As 1 mole of NaOH reacts per mole of acetic acid:
4.169x10⁻³ moles of CH₃COOH.
Molarity is defined as ratio between moles of substance and volume of solution in liters. Thus, molarity of acetic acid solution is:
4.169x10⁻³ moles of CH₃COOH / 0.03754L = 0.111M
As molar mass of acetic acid is 60g/mol, 4.169x10⁻³ moles weights:
4.169x10⁻³ moles × (60g / mol) = 0.2501 g of acetic acid
Now, assuming density of solution as 1.00g/mL, 37.54mL weights 37.54g.
Thus, percent by weight is:
0.2501g CH₃COOH / 37.54g × 100 = 0.666% (w/w)
The molarity of acetic acid is 0.11M and the percent by weight is 0.666%.
How we calculate molarity?Molarity of any solution is used to define their concentration and it will be calculated as:
M = n/V, where
n = moles
V = volume
Molarity of acetic acid will be calculated as:
M₁V₁ = M₂V₂, where
M₁ = molarity of acetic acid = ?
V₁ = volume of acetic acid = 37.54mL = 0.037L
M₂ = molarity of NaOH = 0.1062M
V₂ = volume of NaOH = 39.26mL = 0.039L
On putting all these values on the above equation we can calculate the molarity as:
M₁ = (0.1062)(39.26) / (37.54) = 0.11M
Now we calculate the moles of acetic acid by using the molarity formula as:
n = 0.11M × 0.037L = 0.00407 moles
Molar mass of acetic acid = 60g/mole
Mass of 0.00407 moles of acetic acid = 4.1x10⁻³ moles×(60g / mol) = 0.2501 g
Density of solution = 1.00 g/mL
So, 37.54mL in 1g/mL = 37.54g/mL
Percent by weight will be calculated as:
%w/w = 0.2501g CH₃COOH / 37.54g × 100 = 0.666% (w/w)
Hence, molarity and %(w/w) of acetic acid is 0.11M and 0.666% respectively.
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WILL GIVE BRAINLIEST!!!!! will give brainliest!!!! ******According to the reaction in question I above, how many grams of solid copper will theoretically be
produced when 14.4 g of aluminum are reacted with 14.4 g of copper (II) sulfate? Which reactant is the
limiting reactant? Show your work. Be sure to include units!
Answers
Answer:
5.73 g Cu
Explanation:
M(CuSO4) = 159.6 g/mol
M(Al) = 27.0 g/mol
M(Cu) = 63.5 g/mol
14.4 g Al * 1 mol/27.0 g = 0.5333 mol Al
14.4 g CuSO4 * 1 mol/159.6 g = 0.0902 mol CuSO4
2 Al + 3CuSO4 -------> 3Cu + Al2(SO4)3
from reaction 2 mol 3 mol
given (0.5333 mol )x 0.0902 mol
needed 0.0601 mol
x= 2*0.0902/3 = 0.0601 mol Al
Al is excess, CuSO4 is a limiting reactant.
2 Al + 3CuSO4 -------> 3Cu + Al2(SO4)3
from reaction 3 mol 3 mol
given 0.0902 mol x mol
x = 0.0902 mol Cu
0.0902 mol Cu * 1 mol/63.5 g Cu = 5.73 g Cu
A gaseous hydride of Nitrogen
contains its own volume of Nitrogen
and twice its volume of Hydrogen
and has vapour density 16. The
formula of the hydride is.
Select one:
a. NH2
b. NH3
c. N3H
• d. N2H4
Answers
Answer:
N2H4
Explanation:
A hydride is a binary compound of hydrogen and another element. Binary compounds contain only two atoms. We have to x-ray the hydrides of nitrogen given in the question in order to make our choice. Remember that we were told that that the hydride contains its own volume of nitrogen and twice its volume of hydrogen.
Now consider the hydride N2H4.
N2H4(g) -----> N2(g) + 2H2(g)
The volume ration of nitrogen gas to hydrogen gas in N2H4 is 1:2.
The molecular mass of the compound is;
N2H4= 2(14) + 4(1)= 28+4= 32
Since
molecular mass= 2 vapour density
Vapour density= molecular mass/2
Vapour density= 32/2
Vapour density = 16
Therefore the hydride of nitrogen referred to in the question is N2H4
When 106 g of water at a temperature of 21.4 °C is mixed with 64.3 g of water at an unknown temperature, the final temperature of the resulting mixture is 46.8 °C. What was the initial temperature of the second sample of water? (The specific heat capacity of liquid water is 4.184 J/g ⋅ K.)
Answers
Answer:
THE INITIAL TEMPERATURE OF THE SECOND SAMPLE IS 4.93 C OR 277.93 K
Explanation:
Mass of first sample of water = 106 g
Initial temp of first sample = 21.4 °C = 21.4 + 273 K = 294.4 K
Mass of second sample = 64.3 g
Final temp of theresulting mixture = 46.8 °C = 46.8 + 273 K = 319.8 K
Specific heat capacity of water = 4.184 J/g K
It is worthy to note that;
Heat gained by the first sample = Heat lost by the second sample
Since heat = mass * specific heat capacity * change in temperature, we have
Mass * specific heat * change in temp of the first sample = Mass * specific heat * change in temp. of the second sample
MC (T2 - T1) = MC (T2-T1)
106 * 4.184 * ( 319.8 - 294.4) = 64.3 * 4.184 * ( 319.8 - T1)
106 * 4.184 * 25.4 = 269.0312 ( 319.8 - T1)
11 265.0016 = 269.0312 (319.8 - T1)
Since the change in temperature = 319.8 -T1
Change in temperature =11265.0016 / 269.0312
Change in temperature = 41.87
Change in temperature = 319.8 -T1
41.87 = 319.8 - T1
T1 = 319.8 - 41.87
T1 = 277.93 K
T1 = 4.93 °C
So therefore, the initial temperature of the sacond sample is 4.73 °C or 277.93 K
Nitrogen is a group 15 element. What does being in this group imply about the structure of the nitrogen atom?
O A. Nitrogen has 15 valence electrons.
OB.
Nitrogen has 15 neutrons.
OC. Nitrogen has 5 valence electrons.
D.
Nitrogen has 5 neutrons.
Answers
Answer:
D. Nitrogen has 5 valence electrons.
Explanation:
Nitrogen is an element in group 5A of the periodic table. Elements in group 5A all contain just 5 valence electrons. (Electrons in the outer shell).
**Elements are organized into these groups in a periodic table based on the number of valence electrons which determines their charge. (Does not apply to transition metals)
Using the following balanced chemical equation 8 H2 + S8à 8 H2S. Determine the mass of the product (molar mass = 34.08g/mol) if you start with 1.35 g of hydrogen and 6.86 g of S8 (Molar mass = 256.5 g/mole).
Answers
Answer: 7.29 g of [tex]H_2S[/tex] will be produced from the given masses of both reactants.
Explanation:
To calculate the moles :
[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]
[tex]{\text{Moles of} H_2}=\frac{1.35g}{2.01g/mol}=0.672moles[/tex]
[tex]\text{Moles of} S_8=\frac{6.86g}{256.5g/mol}=0.0267moles[/tex]
[tex]8H_2+S_8\rightarrow 8H_2S[/tex]
According to stoichiometry :
1 mole of [tex]S_8[/tex] require = 8 moles of [tex]H_2[/tex]
Thus 0.0267 moles of [tex]S_8[/tex] will require=[tex]\frac{8}{1}\times 0.0267=0.214moles[/tex] of [tex]H_2[/tex]
Thus [tex]S_8[/tex] is the limiting reagent as it limits the formation of product and [tex]H_2[/tex] is the excess reagent.
As 1 mole of [tex]S_8[/tex] give = 8 moles of [tex]H_2S[/tex]
Thus 0.0267 moles of [tex]S_8[/tex] give =[tex]\frac{8}{1}\times 0.0267=0.214moles[/tex] of [tex]H_2S[/tex]
Mass of [tex]H_2S=moles\times {\text {Molar mass}}=0.214moles\times 34.08g/mol=7.29g[/tex]
Thus 7.29 g of [tex]H_2S[/tex] will be produced from the given masses of both reactants.
which element causes burning when me mix it with oxygen
Answers
Answer:
Hydrogen peroxide is ans
30
Drag each number to the correct location on the equation. Each number can be used more than once, but not all numbers will be used.
Balance the equation with the correct coefficients.
2
3
4
5
SIO2 +
HF → SiF4 +
H2O
Reset
Next
Answers
Answer:
The balanced equation is given below:
SiO2 + 4HF —> SiF4 + 2H2O
The coefficients are 1, 4, 1, 2
Explanation:
The equation for the reaction is given below:
SiO2 + HF —> SiF4 + H2O
The above equation can be balance by as follow:
There are 4 atoms of F on the right side and 1 atom on the left side. It can be balance by putting 4 in front of HF as shown below:
SiO2 + 4HF —> SiF4 + H2O
Therefore are 4 atoms of H on the left side and 2 atoms on the right side. It can be balance by putting 2 in front of H2O as shown below:
SiO2 + 4HF —> SiF4 + 2H2O
Now the equation is balanced.
What is in period 6 group 8
Answers
Answer:
Osmium
Explanation:
If you take a look at the attached image of a periodic table below, you will see that the element in the 6th period and 8th group is Osmium. Hope this helps!
A solution is made by adding 35.5 mL of concentrated hydrochloric acid ( 37.3 wt% , density 1.19 g/mL1.19 g/mL ) to some water in a volumetric flask, and then adding water to the mark to make exactly 250 mL 250 mL of solution. Calculate the concentration of this solution in molarity.
Answers
Answer:
1.73 M
Explanation:
We must first obtain the concentration of the concentrated acid from the formula;
Co= 10pd/M
Where
Co= concentration of concentrated acid = (the unknown)
p= percentage concentration of concentrated acid= 37.3%
d= density of concentrated acid = 1.19 g/ml
M= Molar mass of the anhydrous acid
Molar mass of anhydrous HCl= 1 +35.5= 36.5 gmol-1
Substituting values;
Co= 10 × 37.3 × 1.19/36.5
Co= 443.87/36.6
Co= 12.16 M
We can now use the dilution formula
CoVo= CdVd
Where;
Co= concentration of concentrated acid= 12.16 M
Vo= volume of concentrated acid = 35.5 ml
Cd= concentration of dilute acid =(the unknown)
Vd= volume of dilute acid = 250ml
Substituting values and making Cd the subject of the formula;
Cd= CoVo/Vd
Cd= 12.16 × 35.5/250
Cd= 1.73 M
CHEMISTRY HELP!
using only the periodic table, determine the charge on the ion that is formed by arsenic.
The ion charge is:
a. -3
b. -2
c. -1
d. 0
e. +1
f. +2
g. +3
also what is it for elements lithium and strontium?
Answers
Answer:
A
Explanation:
Arsenic is in the same group as Nitrogen - group 5. They all have 5 valence electrons in their outermost shell. To achieve its most stable state - 8 valence electrons (octet rule - elements are most stable when the entire shell is filled) - arsenic needs to gain 3 electrons. Since electrons have a negative charge, the charge of an As ion would be -3.
Try observing the periodic table and how many valence electrons that each element has. From there, you can determine the charges of the elements lithium and strontium. You can guess, I'll help you with those once you attempt to find the charge of those ions.
In which of these statements are protons, electrons, and neutrons correctly compared?
Quarks are present in protons and neutrons but not in electrons.
Quarks are present in protons, neutrons, and electrons.
Quarks are present in neutrons and electrons but not in protons.
Quarks are present in protons and electrons but not in neutrons
Answers
the second statement is the correct one quarks are needed to balance charges in all subatomic particles such as neutrons, protons and electrons
what happens when you combine Mg2 and NO3
Answers
Answer: they blow up
Explanation: add them together and they will blow up
Answer:
Magnesium nitrate Reactions
Magnesium nitrate has a high affinity towards water. Therefore, heating it results to decompose into magnesium oxide, nitrogen oxides, and oxygen. 2 Mg(NO3)2 → 2 MgO + 4 NO2 + O2.
A rule of thumb is that a reaction rate roughly doubles for every 10 °C increase in temperature. What is the activation energy of a reaction whose rate exactly doubles between 25.0 °C and 35.0 °C
Answers
Answer:
FOR EVERY 10 DEGREE CELSIUS INCREASE IN TEMPERATURE, THE ACTIVATION ENERGY THAT SHOWS THIS IS 52.4 KJ/MOL
Explanation:
From Arrhenius equation, the relationship between the rate constant and the temperature is as shown below:
k = Ae^ -Ea/RT
At initial temperature T1, the initial rate constant is (k1)
At final temperature T2, the final rate constant is k2
For the reaction rate to be doubled, we must double the rate constant which shows that the ratio of k2 / k1 must be equal to 2.
That is, k2 / k1 = 2 (rate is doubled)
Equating this into the Arrhenius equation, we have:
k2 / k1 = Ae^ (-Ea / R ) (1/ T2 - 1/T1)
2 = e^ (-Ea / R) (1 / T2 = 1 / T1)
Taking the natural logarithm of both sides:
ln 2 = - (Ea / R) (1 / T2 - 1 / T1)
Making Ea the subject of the formula, we obtain:
Ea = - (ln 2 R / (1 / T2- 1 / T1))
Let T1 = 25 C = 25 + 273 K = 298 K
T2 = 35 C = 35 + 273 K = 308 K
R = 8.314
So,
Ea = - (ln 2 * 8.314 / ( 1/308 - 1 / 298))
Ea = - (0.693 * 8.314 / 0.00324 - 0.00335)
Ea = - 5.7616 / -0.00011
Ea = 52 378,18 J / mol
So therefore, the activation energy Ea is 52.4 kJ/mol.
Give the full electron configuration for sulfur.
electron configuration:
Answers
Answer:
[Ne] 3s² 3p⁴
Electrons per shell: 2,8,6
Answer : 2,8,6 (Sulphur Atom)
2,8,8 (Sulphur Ion)
How many moles would 1.204x1024 atoms of oxygen be? (Remember Avogadro's number is 6.022x1023
Answers
Answer:
7.25 x 10^47
Explanation:
1.204 x 10^24 moles*6.022 x 10^23 avogadro's number= 7.25 x 10^47
Fishing trawlers in a certain bay catch a large variety of marketable fish along with a species of eel that is toxic. They normally kill the eels and throw them back into the sea. What term is used to refer to the eel? The eel species is called a(n) ______ of the fishing operation.
Answers
Answer:
1. Non-target
2. Bycatch
Explanation:
In the fishing industry, the main aim of the industry is to capture fishes that can be used or eaten and sell. A variety of fishes are captured for this purpose and since they are used therefore are known as Target catch.
But there are some species which has to be discarded because they are toxic and not useful. These non-useful species like eel which gets captured in the net while capturing other fishes are known as Non-target fish.
The eel fish which gets captured is known as bycatch fishes in the fishing operation.
Thus, Non-target and Bycatch are the correct answer.
Answer:
Its just bycatch
Explanation:
The eel species is called a(n) bycatch of the fishing operation.
If 196L of air at 1.0 atm is compressed to 26000ml, what is the new pressure
Answers
Answer:
7.5 atm
Explanation:
Initial pressure P1 = 1.0 ATM
Initial volume V1= 196 L
Final pressure P2= the unknown
Final volume V2= 26000ml or 26 L
From Boyle's law we have;
P1V1= P2V2
P2= P1V1/V2
P2= 1.0 × 196/26
P2 = 7.5 atm
Therefore, as the air is compressed, the pressure increases to 7.5 atm.
For the following reaction, draw the major organic product and select the correct IUPAC name for the organic reactant. If there is more than one major product, both may be drawn.
When drawing hydrogen atoms on a carbon atom, either include all hydrogen atoms or none on that carbon atom, or your structure may be incorrect.
Select the correct IUPAC name for the organic reactant:
a) 2-methylbutene
b) 2-methyl-1-butene
c) 3-methyl-3-butened) 3-methylbutene
Answers
Answer:
The correct IUPAC name for the organic reactant is :
d) 3-methylbutene
Explanation:
Firstly the missing diagram is attached in the diagram below.
The objective of this question is to draw the major organic product and select the correct IUPAC name for the organic reactant. If there is more than one major product, both may be drawn.
From the image attached below; we would see the reaction that occurs between the alkene and the HBr (hydrobromic acid). What really occur in the reaction is that; in the presence of HBr with an alkene compound a secondary 2° carbocation is usually formed. This secondary 2° carbocation formed is usually unstable, so what we called an hydride shift occurs (Markovnikov's product) here to form a stable tertiary 3° carbocation.
The correct IUPAC name for the organic reactant is : 3-methylbutene
Compare the conjugate bases of these three acids. Acid 1: hypochlorous acid , HClO Acid 2: phosphoric acid , H3PO4 Acid 3: hydrogen sulfide , HS- What is the formula for the weakest conjugate base ?
Answers
Answer:
The weakest conjugate is HClO-.
Explanation:
As a general rule, the stronger the Bronsted-Lowry acid, the weaker its conjugate base, and vice versa.
Acid 1: HClO is a strong acid, hence its conjugate base would be weak
Acid 2: H3PO4 is a weak acid, hence its conjugate base would be strong
Acid 3: hydrogen sulphide is also a moderately weak acid with a moderately strong conjugate base.
In order of increasing strengths:
HClO < H2S < H3PO4
Identify the particle represented by each symbol as an alpha particle, a beta particle, a gamma ray, a positron, a neutron, or a proton.
a. 11P
b. 42He
c. +10e
Answers
Answer:
[tex]_1^{1} {P}[/tex] is symbol for proton emission in the nucleus.
[tex]_2^{4} {He}[/tex] symbolises alpha emission, equivalent to helium atom emission of a radioactive particle
[tex]+_1^{0} {e}[/tex] is the radiation symbol for positiron particle. which occurs when beta + radioactive decay occurs
differentiate between sol,aerosol and solid soluti
Answers
Answer:
Sol is a colloidal suspension with solid particles in a liquid. Foam is formed when many gas particles are trapped in a liquid or solid. Aerosol contains small particles of liquid or solid dispersed in a gas. While solid solution contain solid as solute in either solid, liquid or gas.
Consider the following reaction where Kc = 1.29×10-2 at 600 K: COCl2 (g) CO (g) + Cl2 (g) A reaction mixture was found to contain 0.104 moles of COCl2 (g), 4.66×10-2 moles of CO (g), and 3.76×10-2 moles of Cl2 (g), in a 1.00 liter container. Indicate True (T) or False (F) for each of the following:
1. In order to reach equilibrium COCl2(g) must be consumed.
A. True B. False
2. In order to reach equilibrium Kc must increase.
A. True B. False
3. In order to reach equilibrium CO must be consumed.
A. True B. False
4. Qc is greater than Kc.
A. True B. False
5. The reaction is at equilibrium. No further reaction will occur.
A. True B. False
Answers
Answer:
1. In order to reach equilibrium COCl₂(g) must be consumed.
B. False
2. In order to reach equilibrium Kc must increase.
B. False .
3. In order to reach equilibrium CO must be consumed.
A. True.
4. Qc is greater than Kc.
A. True
5. The reaction is at equilibrium. No further reaction will occur.
B. False.
Explanation:
Based on the reaction:
COCl₂(g) → CO (g) + Cl₂(g)
And Kc is defined as:
Kc = 1.29x10⁻² = [CO] [Cl₂] / [COCl₂]
Molar concentrations of each species are:
[COCl₂] = 0.104 moles of COCl₂ / 1L = 0.104M
[CO] = 4.66×10⁻² moles of CO / 1L = 4.66×10⁻²M
[Cl₂] = 3.76×10⁻² moles of Cl₂ / 1L = 3.76×10⁻²M
Replacing in Kc formula:
4.66×10⁻²M × 3.76×10⁻²M / 0.104M = 1.68x10⁻²
As the concentrations are not in equilibrium, 1.68x10⁻² is defined as the reaction quotient, Qc.
As Qc > Kc, the reaction will shift to the left producing more COCl₂ and consuming CO and Cl₂. Thus
1. In order to reach equilibrium COCl₂(g) must be consumed.
B. False
2. In order to reach equilibrium Kc must increase.
B. False . Kc is a constant that never change.
3. In order to reach equilibrium CO must be consumed.
A. True.
4. Qc is greater than Kc.
A. True
5. The reaction is at equilibrium. No further reaction will occur.
B. False. The reaction is in equilibrium when Qc = Kc
3. Scientific methods may include three steps of study as listed below. Explain each step in detail with a complete content related sentence. (ref: p.12-16) a. Hypothesis b. Theory c. Scientific law
Answers
Answer:
Hypothesis is an assumption or idea about a particular topic or argument. An hypothesis should be one which is able to be tested and measurable to determine its authenticity.
A theory is an explanation of a scientific observation which has undergone series of experiments and is reproducible in any part of the world.
A law is simply a rule which gives an in depth explanation of a scientific finding. If new findings emerge the law could be changed or modified.
The half-life of radium-226 is 1590 years. (a) A sample of radium-226 has a mass of 50 mg. Find a formula for the mass of the sample that remains after t years. (b) Find the mass after 500 years correct to the nearest milligram. (c) When will the mass be reduced to 40 mg
Answers
Answer:
Explanation:
a )
m = m₀ [tex]e^{-\lambda t[/tex]
m is mass after time t . original mass is m₀ , λ is disintegration constant
λ = .693 / half life
= .693 / 1590
= .0004358
m = m₀ [tex]e^{- 0.0004358 t}[/tex]
b )
m = 50 x [tex]e^{-.0004358\times 500}[/tex]
= 40.21 mg .
c )
40 = 50 [tex]e^{-.0004358t[/tex]
.8 = [tex]e^{-.0004358t[/tex]
[tex]e^{.0004358t[/tex] = 1.25
.0004358 t = .22314
t = 512 years .
please help!!!! Chem question
Answers
Answer : The net ionic equation will be,
[tex]Ba^{2+}(aq)+SO_4^{2-}(aq)\rightarrow BaSO_4(s)[/tex]
Explanation :
In the net ionic equations, we are not include the spectator ions in the equations.
Spectator ions : The ions present on reactant and product side which do not participate in a reactions. The same ions present on both the sides.
The given balanced ionic equation will be,
[tex]Ba(OH)_2(aq)+H_2SO_4(aq)\rightarrow 2H_2O(aq)+BaSO_4(s)[/tex]
The ionic equation in separated aqueous solution will be,
[tex]Ba^{2+}(aq)+2OH^-(aq)+2H^{+}(aq)+SO_4^{2-}(aq)\rightarrow BaSO_4(s)+2H^+(aq)+2OH^{-}(aq)[/tex]
In this equation, [tex]H^+\text{ and }OH^-[/tex] are the spectator ions.
By removing the spectator ions from the balanced ionic equation, we get the net ionic equation.
The net ionic equation will be,
[tex]Ba^{2+}(aq)+SO_4^{2-}(aq)\rightarrow BaSO_4(s)[/tex]
A chemist titrates of a butanoic acid solution with solution at . Calculate the pH at equivalence. The of butanoic acid is__________ .Round your answer to decimal places.
Note for advanced students: you may assume the total volume of the solution equals the initial volume plus the volume of solution added.
Answers
Answer:
pH = 8.75
Explanation:
100.0mL of a 0.8108M of a butanoic acid (HC₃H₇CO₂, pKa 4.82) solution is titrated with 0.0520M KOH.
The reaction is:
HC₃H₇CO₂ + KOH → H₂O + KC₃H₇CO
Moles of butanoic acid are:
0.1000L × (0.8108mol / L) = 0.08108 moles of butanoic
For a complete reaction, volume of KOH must be added is the volume in which 0.08108 moles of KOH are added, that is:
0.08108 mol × (L / 0.0520mol) = 1.56L of KOH.
Total volume in equilibrium is 1.56L + 0.10L = 1.66L
That means concentration of butanoic acid is:
0.08108 mol / 1.66L = 0.04884M HC₃H₇CO₂
At equivalence point, there is just C₃H₇CO⁻ in solution
Kb of butanoic acid is:
C₃H₇CO⁻ + H₂O ⇄ HC₃H₇CO₂ + OH⁻
Kb = Kw / Ka
Ka = 10^-pKa
Ka = 1.51x10⁻⁵
Kb = 1x10⁻¹⁴ / 1.51x10⁻⁵ = 6.61x10⁻¹⁰
The equilibrium of Kb is:
Kb = 6.61x10⁻¹⁰ = [HC₃H₇CO₂] [OH⁻] / [C₃H₇CO⁻]
As at equivalence point there is just C₃H₇CO⁻, the equilibrium concentrations are:
[C₃H₇CO⁻] = 0.04884M - X
[HC₃H₇CO₂] = X
[OH⁻] = X
Replacing in Kb:
6.61x10⁻¹⁰ = X² / [0.04884M - X]
0 = X² + 6.61x10⁻¹⁰X - 3.23x10⁻¹¹
Solving for X:
X = -5.68x10⁻⁶ → False solution. There is no negative concentrations
X = 5.683x10⁻⁶ → Right solution.
As [OH⁻] = X, [OH⁻] = 5.683x10⁻⁶.
pOH = - log [OH⁻]
pOH = 5.245
pH = 14 - pOH
pH = 8.75An unknown compound, B, has the molecular formula C7H12. On catalytic hydrogenation 1 mol of B absorbs 2 mol of hydrogen and yields 2-methylhexane. B has significant IR absorption band at about 3300 and 2200 cm-1. Which compound best represents B?
a. 5-methyl-1,3-hexadiene
b. 5-methyl-1-hexyne
c. 3-methyl-1-hexyne
d. 5-methyl-2-hexyne
e. 2-methyl-1,5-hexadiene
Answers
Answer:
B and D
Explanation:
If we use the info given we have a band a 3300 cm-1 and 2200 cm-1 this indicates that we have an alkyne functional group. Additionally, the hydrogenation of the unknown molecule will consume two moles of hydrogens this fits with the 2 pi bonds in the alkyne functional group. So, we can discard "a" and "e". The product of this hydrogenation is 2-methylhexane therefore we can discard c because the methyl group is placed on carbon 3. Structures b and d can work.
See figure 1
I hope it helps!
Which metal can replace another metal in a reaction
Answers
Answer:
The products of the reaction are aqueous magnesium nitrate and solid copper metal. This subcategory of single-replacement reactions is called a metal replacement reaction because it is a metal that is being replaced (zinc)
Explanation:
The products of the reaction are aqueous magnesium nitrate and solid copper metal. This subcategory of single-replacement reactions is called a metal replacement reaction because it is a metal that is being replaced (zinc)
