Answer:
b. Use the periodic table to work out the charge (oxidation number) on the transition metal according to the group in the periodic table
Explanation:
The keyword in this problem us "transition metal". Transition metals are found between the group 2 and group 3 elements. They have the d sub shells and also exhibit variable oxidation numbers (valency).
Among the options, the incorrect option is option B.
This is because transition metals d not have a fixed oxidation number and they cannot be obtained by looking up the group in the periodic table.
The iconic compounds obtain a transition of metal caution and a halon anon. As per the rules the correct name of the compounds should be written as to identify the incorrect one.
Option B use the ability to check and to work out the charges (oxidation number) of the transition metal as per the group given in the table. The problem with the keyword is transition metal.Learn more about the ionic compound containing a transition metal.
brainly.com/question/21578354.
Acetonitrile (CH3CN) is an important industrial chemical. Among other things, it is used to make plastic moldings, which have multiple uses, from car parts to Lego bricks. Which one of the following statements about acetonitrile is not correct?a. Acetonitrile has 16 valence electrons in its Lewis structure. b. Acetonitrile has one triple bond. c. Acetonitrile has one pair of nonbonding electrons. d. All atoms satisfy the octet rule in acetonitrile. e. One carbon atom and the nitrogen atom have nonzero formal charges.
Answer:
One carbon atom and the nitrogen atom have nonzero formal charges.
Explanation:
The compound Acetonitrile has sixteen valence electrons as is easily San from its structure. It contains a carbon nitrogen triple bond with a lone pair of electrons on nitrogen. All atoms satisfy the octet rule and there is no hyper valent atom in the molecule.
The formal charge an carbon and nitrogen is calculated as follows;
No. of valence electron on atom - [non bonded electrons + no. of bonds]
Therefore, for carbon and nitrogen, we have;
formal charge on carbon = 4 - (0 + 4) = 0
formal charge on nitrogen = 5 - (2 + 3) = 0
Hence carbon and nitrogen both possess zero formal charges.
What are the conjugate acid-base pairs in the following chemical reaction? HBr(aq)+ CH3COOH(aq) ⇌ CH3C(OH)2+(aq) + Br-(aq)
Answer:
HBr, CH3C(OH)2 and CH3COOH, Br-
Explanation:
The conjugate acid-base pairs acid reacts with base to form a conjugate acid and conjugate base.
Conjugate acid is formed when a bases receives a proton (H+) and a conjugate base is formed when an acid losses a proton (H+).
From the given equation:
HBr, CH3C(OH)2 and CH3COOH, Br- are conjugate acid-base pair, where HBr is an acid and CH3C(OH)2 is a conjugate acid while CH3COOH and Br- is the conjugate base.
Why can long chain fatty acids can form micelles in solutions with pH > 7 but are insoluble in pH < 5
Answer:
In basic conditions that is ( pH > 7 ), the equilibrium shifts towards right and produces a lot of (-ve) negatively charged fatty acids which are polar, In water, since they have long hydrophobic hydrocarbon part, this form micelles where the hydrocarbon part remain inside the sphere and -coo- group remain outside the sphere due to H-bonding interaction with water.
At ( pH < 5 ) I.e acidic conditions, the equilibrium shift to the left giving neutral molecules which can not have stronger H-bonding interaction .
So micelles cant form as they become insoluble.
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 .
A 50.0 L cylinder of oxygen gas is stored at 150. atm. What volume would the oxygen gas occupy if the cylinder were opened into a hot air balloon (completely deflated) until the final pressure is 735 torr
Answer:
THE VOLUME OF THE OXYGEN GAS AFTER DEFLATION TILL A PRESSURE OF 735 TORR IS ATTAINED IS 7836.99 L
Explanation:
Using Boyle's law,
P1V1 = P2V2
P1 = 150 atm
V1 = 50 L
P2 = 735 Torr
V2 = unknown
We must first convert the pressures into the same SI unit for easy calculation
1torr = 1/760 atm
So converting 735 torr to atm; we have:
1 torr = 1/ 760 atm
735 torr = 735 * 1 / 760 atm
= 0.967 atm
In other words, P2 = 0.957 atm
So rearranging the formula by making V2 the subject of the equation, we have:
V2 = P1 V1 / P2
V2 = 150 * 50 / 0.957
V2 = 7836.99 L
The volume of the oxygen cylinder after deflation to a final pressure of 735 torr or 0.967 atm pressure is 7836.99 L.
During a titration, a known concentration of _____ is added to a _____ of an unknown concentration g
Explanation:
The whole process of titration involves finding the concentration of a solution (usually an acid or base) by adding (titrating) it to a solution(acid or base) with a known concentration.
The solution of unknown concentration (the analyte) is usually placed in an flask, while the solution of known concentration (titrant) is placed in a burette and slowly added to the flask.
A compound is found to contain 27.29 % carbon and 72.71 % oxygen by weight. To answer the questions, enter the elements in the order presented above. 1. What is the empirical formula for this compound
Answer:
CO2
Explanation:
Mass of carbon = 27.29
Mass of oxygen = 72.21
Step 1:
We have to first convert these masses to moles
Carbon = 2.29/2.01 = 2.274
Oxygen = 72.71/16 = 4.544
Step 2:
We have to divide the mols by the smallest mol to get simplest ratio of whole number.
The smallest mol is 2.274. we have to divide the mols by this.
2.274/2.274 = 1
4.544/2.274 = 2
Our empirical formula is therefore CO2
how can you prevent frequent landslides from heavy rains
Answer: make a drainage system or make sure that the ground does not absorb the water from the rain and cause a landslide.
Explanation: To prevent frequent landlides you have to make suee there is and area for the rain to go so it does not get stuck in the mud and destabalize the mud
B. Plant vegetation on the slops that expreience landslides.
E. Reduce the sloped where landslides occur.
Hope this helped, please mark brainiest!
A galvanic cell consists of a Cu(s)|Cu2+(aq) half-cell and a Cd(s)|Cd2+(aq) half-cell connected by a salt bridge. Oxidation occurs in the cadmium half-cell. The cell can be represented in standard notation as
Answer:
[tex]Cd(s)|Cd^{2+}(aq) || Cu^{2+}(aq)|Cu(s)[/tex]
Explanation:
A galvanic cell is composed of two electrodes immersed in a suitable electrolyte and connected via a salt bridge. One of the electrodes serves as a cathode where reduction or gain of electrons takes place. The other half cell functions as an anode where oxidation or loss of electrons occurs.
The representation is given by writing the anode on left hand side followed by its ion with its molar concentration. It is followed by a salt bridge. Then the cathodic ion with its molar concentration is written and then the cathode.
As it is given that cadmium acts as anode, it must be on the left hand side and copper must be on right hand side.
[tex]Cd(s)|Cd^{2+}(aq) || Cu^{2+}(aq)|Cu(s)[/tex]
At a certain temperature the equilibrium constant, Kc, equals 0.110 for the reaction: 2 ICl (g) ⇌ I2 (g) +Cl2 (g) What is the equilibrium concentration of ICl if 0.750 mol of I2 and 0.750 mol of Cl2 are initially mixed in a 1.00-L flask?
Answer:
The equilibrium concentration of ICl is 2.26 M
Explanation:
Chemical equilibrium is a state in which no changes are observed as time passes, despite the fact that the substances present continue to react. This is because chemical equilibrium is established when the forward and reverse reaction take place simultaneously at the same rate.
For the study of chemical equilibrium, the so-called equilibrium constant Kc is useful. Being:
aA + bB ⇔ cC + dD
the equilibrium constant Kc is:
[tex]Kc=\frac{[C]^{c} *[D]^{d} }{[A]^{a}*[B]^{b} }[/tex]
That is, the constant Kc is equal to the multiplication of the concentrations of the products raised to their stoichiometric coefficients by the multiplication of the concentrations of the reactants also raised to their stoichiometric coefficients.
For the reaction:
2 ICl (g) ⇌ I₂ (g) +Cl₂ (g)
the constant Kc is:
[tex]Kc=\frac{[I_{2} ]*[Cl_{2} ]}{[ICl]^{2} }[/tex]
Being Kc =0.110 and the concentration being the amount of moles of solute that appear dissolved in each liter of the mixture and being calculated by dividing the moles of the solute by the liters of the solution:
[tex][I_{2} ]=\frac{0.750 moles}{1 L} =0.750 \frac{moles}{L}[/tex][tex][Cl_{2} ]=\frac{0.750 moles}{1 L} =0.750 \frac{moles}{L}[/tex]and replacing in the constant we get:
[tex]0.110=\frac{0.750*0.750}{[ICl]^{2} }[/tex]
Solving, you get the ICl concentration at equilibrium:
[tex][ICl]^{2} =\frac{0.750*0.750}{0.110 }[/tex]
[tex][ICl] =\sqrt{\frac{0.750*0.750}{0.110 }}[/tex]
[ICl]= 2.26 M
The equilibrium concentration of ICl is 2.26 M
When NaOH is added to water, the (OH) = 0.04 M. What is the [H30*]?
What is the PH of the solution?
Answer:
[H₃O⁺] = 2.5 × 10⁻¹³ M
pH = 12.6
Explanation:
Step 1: Given data
Concentration of OH⁻: 0.04 M
Step 2: Calculate the concentration of H₃O⁺
Let's consider the self-ionization of water reaction.
2 H₂O(l) ⇄ OH⁻(aq) + H₃O⁺(aq)
The ionic product of water is:
Kw = [OH⁻] × [H₃O⁺] = 10⁻¹⁴
[H₃O⁺] = 10⁻¹⁴ / [OH⁻]
[H₃O⁺] = 10⁻¹⁴ / 0.04
[H₃O⁺] = 2.5 × 10⁻¹³ M
Step 3: Calculate the pH
The pH is:
pH = -log [H₃O⁺] = -log 2.5 × 10⁻¹³ = 12.6
In the Lewis structure of AB4 where B is more electronegative than A. Both are main group elements where A has 8 valence electrons and each B has 7 valence electrons.
Required:
a. What is the total number of valence electrons?
b. How many lone pairs are in the molecule?
Answer:
1. 36
2. Two
Explanation:
The Lewis structure shows the valence electrons present in a compound. Usually the valence electrons are shown as dot structures around the symbol of the elements involved in the compound.
For a compound AB4 where B is more electronegative than A and A has 8 electrons in its valence shell, there will be thirty six valence electrons on the outermost shell of the molecule.
There are six electron pair domains present in the molecule, four bond pairs and two lone pairs. The molecule is in a square planar geometry.
Answer: a- 36 valence electrons
b- 14 lone pairs
Explanation:
Valence is equal to A + 4B = 8 + 4(7)
With 4 bonds between A and the 4 B, that is 36 valence minus 8 electrons in those pairs = 28. 28 is 14 lone pairs.
A gas mixture contains 3.50 moles of helium, 5.00 moles of krypton and 7.60 moles of neon. A) What is the mole fraction for each gas
Answer:
.217, .311, and .472, respectively.
Explanation:
The total number of moles of gas is 3.50 + 5.00 + 7.60 = 16.10 (to preserve significant digits).
X of helium=3.50/16.10 = .217
X of krypton=5.00/16.10 = .311
X of neon=7.60/16.10 = .472
4. A 0.100 M solution of NaOH is used to titrate an HCl solution of unknown concentration. To neutralize the solution, an average volume of
the titrant was 38.2 mL. The starting volume of the HCI solution was 20 ml. What's the concentration of the HCI?
O A.0.284 M
B. 3.34 M
C. 0.191 M
D. 0.788 M
Answer:
C. 0.191 M
Explanation:
Our goal for this question, is to calculate the concentration of the HCl solution. For this, in the experiment, a solution of NaOH was used to find the moles of HCl. Therefore, our first step is to know the reaction between HCl and NaOH:
[tex]HCl~+~NaOH~->~NaCl~+~H_2O[/tex]
The "titrant" in this case is the NaOH solution. If we know the concentration of NaOH (0.100M) and the volume of NaOH (38.2 mL=0.0382 L), we can calculate the moles using the molarity equation:
[tex]M=\frac{mol}{L}[/tex]
[tex]0.100~M=\frac{mol}{0.0382~L}[/tex]
[tex]mol=0.100~M*0.0382~L=0.0382~mol~of~NaOH[/tex]
Now, in the reaction, we have a 1:1 molar ratio between HCl and NaOH (1 mol of HCl is consumed for each mole of NaOH added). Therefore we will have the same amount of moles of HCl in the solution:
[tex]0.0382~mol~of~NaOH\frac{1~mol~HCl}{1~mol~NaOH}=0.0382~mol~HCl[/tex]
If we want to calculate the molarity of the HCl solution we have to divide by the litters of HCl used in the experiment (20 mL= 0.02 L):
[tex]\frac{0.0382~mol~HCl}{0.02~L}~=~0.191~M[/tex]
The concentration of the HCl solution is 0.191 M
I hope it helps!
PLEASE HELP!! 40 POINTS
Answer:
1) 6.524779402×10^(-17)
2)521.1g
3)113
Explanation:
Answer: 1) 6.524779402×10^(-17)
2)521.1g
Explanation:
Two elements represents by the letter Q and R atomic number 9 and 12 respectively.
1. Write the electronic configuration of R
2. To what group does Q belongs to in the periodic table
3. Write the formula of the compound formed when Q combines with R
Answer:
The two elements with atomic number 9 and 12 are represented by letter Q and R respectively, where Q represents fluorine atom and R represents magnesium atom.
1. Electronic configuration of R that is magnesium (atomic number 12) is:
1s2 2s2 2p6 3s2
2. Q represents fluorine atom, which belongs to group 17 in periodic table that is the most reactive and lightest member of the group.
3. Q and R that is fluorine and magnesium combinely form magnesium fluoride or MgF2.
Which of the following represents six molecules of water? 6HO 2 6H 2O 1 6H 2O H 6O
Answer:
6H20 represents six molecules of water
Answer:
6H20 represents six molecules of water
Explanation:
How many grams of H2O will be formed when 32.0 g H2 is mixed with 73.0 g of O2 and allowed to react to form water
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Given the following values of pKa, determine which is the weakest base of the answers listed. Acid pKa HClO2 1.95 HClO 7.54 HCOOH 3.74 HF 3.17 HNO2 3.15
Answer:
HClO 7.54
Explanation:
Hypochlorous acid (HClO) is a weakest acid because the pKa value of Hypochlorous acid is very high among the options given in the activity. pKa is a method which is used in order to identify the strength of an acid. The higher the value of pKa of a liquid, lower the strength of an acid while lower the value of pKa of chemical, higher the strength of an acid. In the options, HClO2 is a strong acid due to high lower pKa value.
The Lewis structure of N2H2 shows ________. Group of answer choices a nitrogen-nitrogen single bond each hydrogen has one nonbonding electron pair each nitrogen has one nonbonding electron pair each nitrogen has two nonbonding electron pairs a nitrogen-nitrogen triple bond
Answer:
one bond between nitrogen and hydrogen and a double bond between the nitrogen atoms.
Explanation:
H-N=N-H
Assuming that you start with 21.4 g of ammonia gas and 18.0 g of sodium metal and assuming that the reaction goes to completion, determine the mass (in grams) of each product.
Answer:
[tex]m_{NaNH_2}=30.42gNaNH_2[/tex]
[tex]m_{H_2}=0.783gH_2[/tex]
Explanation:
Hello,
In this case, the reaction between sodium and ammonia is:
[tex]2Na+2NH_3\rightarrow 2NaNH_2+H_2[/tex]
Thus, as we know the initial masses of both sodium and ammonia, we should first identify the limiting reactant, for which we firstly compute the available moles of sodium:
[tex]n_{Na}=18.0gNa*\frac{1molNa}{23.0gNa}=0.783molNa[/tex]
And the moles of sodium consumed by 21.4 g of ammonia (2:2 mole ratio):
[tex]n_{Na}^{\ consumed}=21.4gNH_3*\frac{1molNH_3}{17gNH_3} *\frac{2molNa}{2molNH_3} =1.26molNa[/tex]
In such a way, since less moles of sodium are available than consumed by ammonia, we can say, sodium is the limiting reactant. Furthermore, the mass of both sodium amide (39 g/mol) and hydrogen gas (2 g/mol) that are produced turn out:
[tex]m_{NaNH_2}=0.783molNa*\frac{2molNaNH_2}{2molNa}*\frac{39gNaNH_2}{1molNaNH_2}=30.42gNaNH_2[/tex]
[tex]m_{H_2}=0.783molNa*\frac{1molH_2}{2molNa}*\frac{2gH_2}{1molH_2}=0.783gH_2[/tex]
Best regards.
What subatomic particles surround the nucleus? Question 1 options: protons neutrons atoms electrons
Answer:
Electrons "surround"
Explanation:
Protons and neutrons "make up" the nucleus so they are contained "within" the nucleus meaning that electrons would "surround" the nucleus as they orbit around the nucleus
Answer:
Electrons
Explanation:
Protons and nuetrons are present inside the nucleus of an atom while the electron revolve around the nucleus in different energy levels.
Complete and balance the molecular equation for the reaction of aqueous sodium sulfate, Na2SO4, and aqueous barium nitrate, Ba(NO3)2. Include physical states.molecular equation:Na_{2}SO_{4}(aq) + Ba(NO_{3})_{2}(aq) ->Na2SO4(aq)+Ba(NO3)2(aq)⟶Enter the balanced net ionic equation for this reaction. Include physical states.net ionic equation:
Answer:
1. The balanced molecular equation is given below:
Na2SO4(aq) + Ba(NO3)2(aq) —> BaSO4(s) + 2NaNO3(aq)
2. The net ionic equation is given below:
SO4^2-(aq) + Ba^2+(aq) —> BaSO4(s)
Explanation:
1. The balanced molecular equation
Na2SO4(aq) + Ba(NO3)2(aq) —> BaSO4(s) + NaNO3(aq)
The above equation can be balance as follow:
Na2SO4(aq) + Ba(NO3)2(aq) —> BaSO4(s) + NaNO3(aq)
There are 2 atoms of Na on the left side and 1 atom on the right side. It can be balance by putting 2 in front of NaNO3 as shown below:
Na2SO4(aq) + Ba(NO3)2(aq) —> BaSO4(s) + 2NaNO3(aq)
Now, the equation is balanced.
2. The bal net ionic equation.
This can be obtained as follow:
Na2SO4(aq) + Ba(NO3)2(aq) —>
In solution, Na2SO4 and Ba(NO3)2 will dissociate as follow:
Na2SO4(aq) —> 2Na^+(aq) + SO4^2-(aq)
Ba(NO3)2(aq) —> Ba^2+(aq) + 2NO3^-(aq)
Na2SO4(aq) + Ba(NO3)2(aq) —>
2Na^+(aq) + SO4^2-(aq) + Ba^2+(aq) + 2NO3^-(aq) —> BaSO4(s) + 2Na^+(aq) + 2NO3^-(aq)
Cancel the spectator ions i.e Na^+ and NO3^- to obtain the net ionic equation.
SO4^2-(aq) + Ba^2+(aq) —> BaSO4(s)
How many valence electrons must two atoms share to form a single covalent bond? answers A.2 B.4 C.3 D.1
Answer:
2
Explanation:
A single covalent bond is formed when two electrons are shared between the same two atoms, one electron from each atom.
Answer:
the answer is 2
Explanation:
From the graph of Density vs. Concentration, created in Graph 1, what was the relationship between the concentration of the sugar solution and the density of the sugar solution?
The graph is not given in the question, so, the required graph is attached below:
Answer:
According to the graph, the relationship between the density of the sugar solution and the concentration of the sugar solution is directly proportional to each other as they both are increasing exponentially.
The graph shows that, the density of sugar solution will increase with the increase in concentration of sugar in the solution.
Which solution has the greatest buffer capacity? Select the correct answer below: 1 mole of acid and 1 mole of base in a 1.0 L solution
Answer:
The answer is
Explanation:
1 mole of acid.
Hope this helps....
Have a nice day!!!!
A buffer that is 1 M in acid and base will have the greatest capacity of buffer, and therefore the greatest buffer capacity.
What do you mean by the buffer solution ?A weak acid and the conjugate base of the weak acid, or a weak base and the conjugate acid of the weak base, are combined to form the buffer solution, a water-based solvent solution.
In a biological system, a buffer's keep intracellular and extracellular pH levels within a relatively small range and to withstand pH fluctuations brought on by both internal and external factors.
A buffer is a substance that can withstand a pH shift when acidic or basic substances are added. It may balance out little quantities of additional acid or base, keeping the pH stable.
Thus, 1 M in acid and base solution has the greatest buffer capacity.
To learn more about the buffer solution, follow the link;
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A quantity of liquid methanol, CH 3OH, is introduced into a rigid 3.00-L vessel, the vessel is sealed, and the temperature is raised to 500K. At this temperature, the methanol vaporizes and decomposes according to the reaction CH 3OH(g) CO(g) + 2 H 2(g), K c= 6.90×10 –2. If the concentration of H 2 in the equilibrium mixture is 0.426M, what mass of methanol was initially introduced into the vessel?
Answer:
74.3g of methanol were introduced into the vessel
Explanation:
In the equilibrium:
CH₃OH(g) ⇄ CO(g) + 2H₂(g)
Kc is defined as the ratio between concentrations in equilibrium of :
Kc = 6.90x10⁻² = [CO] [H₂]² / [CH₃OH]
Some methanol added to the vessel will react producing H₂ and CO. And equilibrium concentrations must be:
[CH₃OH] = ? - X
[CO] = X
[H₂] = 2X
Where ? is the initial concentration of methanol
As [H₂] = 2X = 0.426M; X = 0.213M
[CH₃OH] = ? - 0.213M
[CO] = 0.213M
[H₂] = 0.426M
Replacing in Kc to solve equilibrium concentration of methanol:
6.90x10⁻² = [0.213] [0.426]² / [CH₃OH]
[CH₃OH] = 0.560
As:
[CH₃OH] = ? - 0.213M = 0.560M
? = 0.773M
0.7733M was the initial concentration of methanol. As volume of vessel is 3.00L, moles of methanol are:
3.00L * (0.773 mol / L) = 2.319 moles methanol.
Using molar mass of methanol (32.04g/mol), initial mass of methanol added was:
2.319 moles * (32.04g / mol) =
74.3g of methanol were introduced into the vesselA 1.0 L buffer solution is 0.250 M HC2H3O2 and 0.050 M LiC2H3O2. Which of the following actions will destroy the buffer?
A. adding 0.050 moles of NaOH
B. adding 0.050 moles of LiC2H3O2
C. adding 0.050 moles of HC2H3O2
D. adding 0.050 moles of HCl
E. None of the above will destroy the buffer.
Answer:
D
Explanation:
Addition of 0.05 M HCl, will react with all of the C2H3O2- from LiAc which will give 0.05 M more HAc. So there will be no Acetate ion left to make the solution buffer. Hence, the correct option for the this question is d, which is adding 0.050 moles of HCl.
The action that destroys the buffer is option c. adding 0.050 moles of HCl.
What is acid buffer?It is a solution of a weak acid and salt.
Here, The buffer will destroy at the time when either HC2H3O2 or NaC2H3O2 should not be present in the solution.
The addition of equal moles of HCl finishly reacts with equal moles of NaC2H3O2. Due to this, there will be only acid in the solution.
Since
moles of HC2H3O2 = 1*0.250 = 0.250
moles of NaC2H3O2 = 1*0.050 = 0.050.
moles of HCl is added = 0.050
Now
The reaction between HCl and NaC2H3O2
[tex]HCl + NaC_2H_3O_2 \rightarrow HC_2H_3O_2 + NaCl[/tex]
Now
BCA table is
NaC2H3O2 HCl HC2H3O2
Before 0.050 0.050 0.250
Change -0.050 -0.050 +0.050
After 0 0 0.300
Now, the solution contains the acid (HC2H3O2 ) only.
Therefore addition of 0.050 moles of HCl will destroy the buffer.
Learn more about moles here: https://brainly.com/question/24817060
Enter the balanced chemical equation for the reaction of each of the following carboxylic acids with KOH.Part Aacetic acidExpress your answer as a chemical equation. Assume that there is no dissociation (i.e., enter only whole compounds, not ions).Part B2-methylbutanoic acid (CH3CH2CH(CH3)COOH)Express your answer as a chemical equation. Assume that there is no dissociation (i.e., enter only whole compounds, not ions).Part C4-chlorobenzoic acid (ClC6H4COOH)Express your answer as a chemical equation. Assume that there is no dissociation (i.e., enter only whole compounds, not ions).
Answer:
Explanation:
Answer in attached file .
Calculate the molality of a solution containing 141.5 g of glycine (NH2CH2COOH) dissolved in 4.456 kg of H2O
Answer:
0.423 m.
Explanation:
The following data were obtained from the question:
Mass of glycine (NH2CH2COOH) = 141.5 g
Mass of water = 4.456 kg
Molality =.?
Next, we shall determine the number of mole in 141.5 g of glycine (NH2CH2COOH.
This is illustrated below:
Mass of glycine (NH2CH2COOH) = 141.5 g
Molar mass of glycine (NH2CH2COOH) = 14 + (2x1) + 12 + (2x1) + 12 + 16 + 16 + 1 = 75 g/mol
Mole of glycine (NH2CH2COOH) =.?
Mole = mass /Molar mass
Mole of glycine (NH2CH2COOH) = 141.5/75
Mole of glycine (NH2CH2COOH) = 1.887 moles
Finally, we shall determine the molality of the solution as follow:
Molality is simply defined as the mole of solute per kilogram of water. Mathematically it is expressed as:
Molality = mole / mass (kg) of water
With the above formula, we can obtain the molality of the solution as follow:
Mole of glycine (NH2CH2COOH) = 1.887 moles
Mass of water = 4.456 kg
Molality =.?
Molality = mole /mass (kg) of water
Molality =1.887/4.456
Molality = 0.423 m
Therefore, the molality of the solution is 0.423 m
