Answers
Answer:
See explanation
Explanation:
We have to remember that theory behind the carbohydrates. Carbohydrates are molecules with several hydroxyl groups in which the main functional group can be an aldehyde or a ketone.
If we have an aldehyde as a main functional group we will have an "aldose". If we have a ketone as a main functional group we will have a "ketose".
We can also, classify the carbohydrates using the number of carbons. So, for example, if we have 5 carbons and a ketone as the main functional group we will have a "keto-pentose". If we have for example 4 carbons and an aldehyde as the main functional group we will have a "tetra-aldose".
In this case, we have an aldohexose, so we will have 6 carbons and an aldehyde as main functional group. So, we can draw a structure with 6 carbons, in carbon 1 we have to put the aldehyde group and in the other carbons we have to put "OH" groups.
See figure 1
I hope it helps!
Related Questions
"The pH of a solution of household ammonia, a 0.950-M solution of NH3, is 11.612. What is Kb" for NH3
Answers
Answer:
Kb = 1.77x10⁻⁵
Explanation:
When NH₃, a weak base, is in equilibrium with waterm the reaction that occurs is:
NH₃(aq) + H₂O(l) ⇄ NH₄⁺(aq) + OH⁻(aq)
And the dissociation constant, Kb, for this equilibrium is:
Kb = [NH₄⁺] [OH⁻] / [NH₃]
To find Kb you need to find the concentration of each species. The equilibrium concentrations are:
[NH₃] = 0.950M - X
[NH₄⁺] = X
[OH⁻] = X
Where X is reaction coordinate.
You can know [OH⁻] and, therefore, X, with pH of the solution, thus:
pH = -log [H⁺] = 11.612
[H⁺] = 2.4434x10⁻¹²
As 1x10⁻¹⁴ = [H⁺] [OH⁻]
1x10⁻¹⁴ / 2.4434x10⁻¹² = [OH⁻]
4.0926x10⁻³ = [OH⁻] = X
Replacing, concentrations of the species are:
[NH₃] = 0.950M - X
[NH₄⁺] = X
[OH⁻] = X
[NH₃] = 0.9459M
[NH₄⁺] = 4.0926x10⁻³M
[OH⁻] = 4.0926x10⁻³M
Replacing in Kb expression:
Kb = [NH₄⁺] [OH⁻] / [NH₃]
Kb = [4.0926x10⁻³M] [4.0926x10⁻³M] / [0.9459M]
Kb = 1.77x10⁻⁵The branch of science which deals with the chemical bond is called Chemistry.
The correct answer to the question is [tex]Kb = 1.77*10^{-5[/tex]
Explanation:
When NH₃, is acts as a weak base it forms an equilibrium with water the reaction occurs is:
[tex]NH_3(aq) + H_2O(l) ---><NH_4^+(aq) + OH^-(aq)[/tex]
The formula we gonna use is as follows:-
[tex]Kb = \frac{[NH_4^+] [OH^-]}{[NH_3]}[/tex]
The data is given in the question is as follows:-
[NH₃] = 0.950M - X [NH₄⁺] = X [OH⁻] = X
Where X stands for reaction coordinate.
After solving the ph of the compound the value is as follows:-
[NH₃] = [tex]0.9459M[/tex] [NH₄⁺] = [tex]4.0926*10^{-3}M[/tex] [OH⁻] = [tex]4.0926*10^{-3}M[/tex]
Putting the value in the formula.
[tex]Kb = \frac{[4.0926*10^{-3}M] [4.0926*10^{-3}M]}{[0.9459M]}[/tex]
After solving the equation the value of Kb is [tex]1.77*10^{-5[/tex]
Hence, the correct answer is [tex]1.77*10^{-5[/tex]
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Which best describes the relationship between heat internal energy, and thermal energy?
O Internal energy is heat that flows, and heat is the part of thermal energy that can be transferred.
O Internal energy is thermal energy that flows, and thermal energy is the part of heat that can be transferred.
O Thermal energy is heat that flows, and heat is the part of internal energy that can be transferred.
O Heat is thermal energy that flows, and thermal energy is the part of internal energy that can be transferred.
Answers
Answer:
Heat is thermal energy that flows, and thermal energy is the part of internal energy that can be transferred
Explanation:
The authors state in the general procedures that the reaction was monitored by TLC. How would this be done? What would you spot in each lane? How would you know the reaction was done?
Answers
Answer:
Thin Layer Chromatography (TLC) can be used to analyze chemical reactions. During this reaction monitoring, a typical TLC plate would have three spots: the reactant lane, the reaction mixture lane, and a "co-spot" where reaction product would be spotted directly on top of reactant.
The co-spot serves as a reference point and is vital for reactions where reactant and product have similar Rfs, and many other variations of eluent tracking.
To indicate completion of the reaction, the disappearance of a spot (usually the starting reactant) is observed.
Select the oxidation reduction reactions??
Answers
Answer:
Explanation:
1 ) Cl₂ + ZnBr₂ = ZnCl₂ + Br₂
In this reaction , oxidation number of Cl decreases from 0 to -1 so it is reduced and oxidation number of Br increases from -1 to 0 so it is oxidised . Hence this reaction is oxidation - reduction reaction .
2 )
Pb( ClO₄)₂ + 2KI = PbI₂ + 2KClO₄
In this reaction oxidation number of none is changing so it is not an oxidation - reduction reaction.
3 )
CaCO₃ = CaO + CO₂
In this reaction also oxidation number of none is changing so it is not an oxidation - reduction reaction.
So only first reaction is oxidation - reduction reaction.
2nd option is correct.
Which land feature supports the theory of continental drift?
A.canyons B.volcanoes C.coal fields D.oceans
Answers
Answer:
Coals
Explanation:
The land feature that supports the theory of continental drift is ; ( C ) coal fields
Continental drift is the gradual shift in position of the earth tectonic plates ( i.e. gradual shift in the continents in relation to ocean basins) and this due to the heat from the earths' mantle.
Coal fields supports this theory because the it is an area where coal is found in large quantities and mined for commercial purposes. coal fields areas are found as a result of continental drift.
Hence we can conclude that the land feature that supports the theory of continental drift is coal fields
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What is the pH of a solution prepared by dissolving 0.140 g of potassium hydroxide in sufficient pure water to prepare 250.0 ml of solution
Answers
Answer:
pH= 12
Explanation:
Potassium hydroxide (KOH) is a strong base, so it dissociates completely in water by giving OH⁻ anions as follows:
KOH⇒ K⁺ + OH⁻
Since dissociation is complete, it is assumed that the concentration of OH⁻ is equal to the initial concentration of KOH:
[OH⁻]= [KOH]
In order to find the initial concentration of KOH, we have to divide the mass (0.140 g) into the molecular weight of KOH (Mw):
Mw (KOH)= K + O + H = 39 g/mol + 16 g/mol + 1 g/mol = 56 g/mol
moles KOH: mass/Mw= 0.140 g/(56 g/mol) = 2.5 x 10⁻³ moles
The molality of the solution is the number of moles of KOH per liter of solution:
V= 250.0 ml x 1 L/1000 ml= 0.250 L
M = (2.5 x 10⁻³moles)/(0.250 L)= 0.01 M
Now, we calculate pOH:
pOH = -log [OH⁻]= - log [KOH]= -log (0.01) = 2
Finally, we calculate pH from pOH:
pH + pOH = 14
⇒pH = 14 - pOH= 14 -2 = 12
which process is used to produce gases from solutions of salts dissolved in water or another liquid?
A.Electrolysis
B.Metallic bonding
C.Ionic bonding
D. Polar covalent bonding
Answers
Answer:
A.Electrolysis
Explanation:
A.Electrolysis
For example, electrolysis of solution of NaCl in water gives H2 and O2.
An atom of element number 33 (As) is in its ground electronic state. Which one of the following sets quantum numbers could not apply to any of its electrons?
A) n=2 l=1 ml= -1 ms= 1/2
B) n=3 l=0 ml=0 ms= -1/2
C) n=3 l=2 ml=-2 ms= -1/2
D) n=4 l=0 ml=0 ms= -1/2
E) n=4 l=2 ml=1 ms= 1/2
Answers
Answer:
E) n=4 l=2 ml=1 ms= 1/2
Explanation:
Arsenic is a member of group 15 in the periodic table. Its electronic configuration is;
1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p3. Its condensed electronic configuration can be written as [Ar]4s2 3d10 4p3. This electronic configuration shown here can now enable us to consider each option given in the question in order to meaningfully arrive at a logical answer.
If we look at option E, the data given for that electron is; n=4 l=2 ml=1 ms= 1/2. This refers to an electron in a 4d orbital. In the ground state configuration of arsenic shown above, there is no 4d orbital, hence option E must be the correct answer.
structure and correct name for 5-octyne
Answers
Answer:
Explanation:
(R)-5-octyne-4-ol
C8H140
Correct name for 5-octyne is 3-octane.
Structure is attached below.
3-Octane:Octane is a hydrocarbon and an alkane with the chemical formula C₈H₁₈, and the condensed structural formula CH₃(CH₂)₆CH₃. Octane has many structural isomers that differ by the amount and location of branching in the carbon chain. One of these isomers, 2,2,4-trimethylpentane (commonly called iso-octane) is used as one of the standard values in the octane rating scale.
The structure for 5-octyne is given below.
Correct name for it is 3-octane.
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differentiate between satured and unsatured fats
Answers
Answer:
...
Explanation:
in saturated fats there is no double bond between the acids and are tightly packed and unsaturated fats arent tight and loosely packed/put together
saturated- solid at room temperature
unsaturated= liquid at room temperature
two types of unsaturated fats, Polyunsaturated fats and Monounsaturated fats
Enter an equation for the formation of CaCO3(s) from its elements in their standard states. Enter any reference to carbon as C(s). Express your answer as a chemical equation. Identify all of the phases in your answer.
Answers
Answer:
CaF2 + CO3- ----> CaCO3 + 2 F-
Explanation:
The chemical compounds found on the left side of the date are the reagents and those found on the right are the products, where calcium carbonate appears.
Calcium carbonate is a quaternary salt
A beach has a supply of sand grains composed of calcite, ferromagnesian silicate minerals, and non-ferromagnesian silicate minerals. If it undergoes lots of chemical weathering, which sand grains will be quickly chemically weathered away?
a. Calcite
b. ferromagnesian silicate minerals
c. non-ferromagnesian silicate minerals
Answers
The sand that grained will be quickly chemically weathered away should be option b. ferromagnesian silicate minerals.
What are ferromagnesian silicate minerals?It should be considered as the Silicate minerals where cations of iron and the form of magnesium should be important for the chemical components. It is used for covering up the minerals. Also, calcite should be normal weather via the solution process so it required a lot of water that contains a high amount of carbonic acid.
Hence, the correct option is b.
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A 25.0-mL sample of 0.150 M hydrazoic acid, HN3, is titrated with a 0.150 M NaOH solution. What is the pH after 13.3 mL of base is added? The Ka of hydrazoic acid = 1.9 x 10-5.
Answers
Answer:
pH ≅ 4.80
Explanation:
Given that:
the volume of HN₃ = 25 mL = 0.025 L
Molarity of HN₃ = 0.150 M
number of moles of HN₃ = 0.025 × 0.150
number of moles of HN₃ = 0.00375 mol
Molarity of NaOH = 0.150 M
the volume of NaOH = 13.3 mL = 0.0133
number of moles of NaOH = 0.0133× 0.150
number of moles of NaOH = 0.001995 mol
The chemical equation for the reaction of this process can be written as:
[tex]HN_3 + OH- ---> N^-_{3} + H_2O[/tex]
1 mole of hydrazoic acid react with 1 mole of hydroxide to give nitride ion and water
thus the new number of moles of HN₃ = 0.00375 - 0.001995 = 0.001755 mol
Total volume used in the reaction = 0.025 + 0.0133 = 0.0383 L
Concentration of [tex]HN_3[/tex] = [tex]\dfrac{0.001755}{0.0383}[/tex] = 0.0458 M
Concentration of [tex]N^{-}_3[/tex] = [tex]\dfrac{ 0.001995 }{0.0383}[/tex] = 0.0521 M
GIven that :
Ka = [tex]1.9 x 10^{-5}[/tex]
Thus; it's pKa = 4.72
[tex]pH =4.72 + log(\dfrac{ \ 0.0521}{0.0458})[/tex]
[tex]pH =4.72 + log(1.1376)[/tex]
[tex]pH =4.72 + 0.05598[/tex]
[tex]pH =4.77598[/tex]
pH ≅ 4.80
The pH of the solution 0.150 M hydrazoic acid after 13.3 mL of NaOH base is added is 4.80.
How we calculate the pH?pH of the given solution will be used by using the following equation:
pH = pKa + log[conjugate base] / [weak acid]
Given chemical reaction will be represented as:
HN₃ + OH⁻ → N₃⁻ + H₂O
Moles will be calculated as:
n = M×V, where
M = molarity
V = volume
Moles of 0.150 M hydrazoic acid = (0.150M)(0.025L) = 0.00375 mol
Moles of 0.150 M NaOH = (0.0133)(0.150) = 0.001995 mol
From the above calculation it is clear that moles of hydrazoic acid is present in excess and it will be:
0.00375 - 0.001995 = 0.001755 mol
And 0.001995 mol of N₃⁻ is preduced by the reaction.
Total volume of the solution = 0.025 + 0.0133 = 0.0383 L
To calculate the pH after titration, first we have to calculate the concentration in terms of molarity of N₃⁻ and HN₃ as:
[N₃⁻] = 0.001995 mol / 0.0383 L = 0.0521 M
[HN₃] = 0.001755 mol / 0.0383 L = 0.0458 M
Ka for HN₃ = 1.9 × 10⁻⁵
pKa = -log( 1.9 × 10⁻⁵ ) = 4.72
On putting all these values on the above equation, we get
pH = 4.72 + log (0.0521) / (0.0458)
pH = 4.80
Hence, pH of the solution is 4.80.
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Draw the structure 2 butylbutane
Answers
Answer:
please look at the picture below.
Explanation:
Suppose that you add 26.7 g of an unknown molecular compound to 0.250 kg of benzene, which has a K f of 5.12 oC/m. With the added solute, you find that there is a freezing point depression of 2.74 oC compared to pure benzene. What is the molar mass of the unknown compound
Answers
Answer: The molar mass of the unknown compound is 200 g/mol
Explanation:
Depression in freezing point is given by:
[tex]\Delta T_f=i\times K_f\times m[/tex]
[tex]\Delta T_f=2.74^0C[/tex] = Depression in freezing point
i= vant hoff factor = 1 (for molecular compound)
[tex]K_f[/tex] = freezing point constant = [tex]5.12^0C/m[/tex]
m= molality
[tex]\Delta T_f=i\times K_f\times \frac{\text{mass of solute}}{\text{molar mass of solute}\times \text{weight of solvent in kg}}[/tex]
Weight of solvent (benzene)= 0.250 kg
Molar mass of solute = M g/mol
Mass of solute = 26.7 g
[tex]2.74^0C=1\times 5.12\times \frac{26.7g}{Mg/mol\times 0.250kg}[/tex]
[tex]M=200g/mol[/tex]
Thus the molar mass of the unknown compound is 200 g/mol
The molar mass of an unknown solute compound in the solution has been 199.626 g/mol.
With the addition of the solute to the solution, there has been a depression in the freezing point. The depression in the freezing point can be expressed as:
Depression in freezing point = Van't Hoff factor × Freezing point constant × molality
The molality can be defined as the moles of solute per kg solvent
Molaity = [tex]\rm \dfrac{Mass\;of\;solute\;(g)}{Molecular\;mass\;of\;solute}\;\times\;\dfrac{1}{Mass\;of\;solvent\;(kg)}[/tex]
The depression in freezing point can be given as:
Depression in freezing point = Van't Hoff factor × Freezing point constant × [tex]\rm \dfrac{Mass\;of\;solute\;(g)}{Molecular\;mass\;of\;solute}\;\times\;\dfrac{1}{Mass\;of\;solvent\;(kg)}[/tex] ......(i)
Given, the depression in freezing point = 2.74 [tex]\rm ^\circ C[/tex]
Van't Hoff factor = 1 (Molecular compound)
Freezing point constant (Kf) = 5.12 [tex]\rm ^\circ C[/tex]/m
Mass of solute = 26.7 g
Mass of solvent = 0.250 kg
Substituting the values in equation (i):
2.74 [tex]\rm ^\circ C[/tex] = 1 × 5.12
[tex]\rm \dfrac{2.74}{5.12}[/tex] = [tex]\rm \dfrac{1}{Molecular\;mass\;of\;solute}\;\times\;\dfrac{26.7}{0.250\;kg}[/tex]
0.535 = [tex]\rm \dfrac{1}{Molecular\;mass\;of\;solute}\;\times\;106.8[/tex]
Molecular mass of solute = [tex]\rm \dfrac{106.8}{0.535}[/tex] g/mol
Molecular mass of solute = 199.626 g/mol
The molar mass of an unknown solute compound in the solution has been 199.626 g/mol.
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The constant pressure molar heat capacity of argon, C_{p,m}C
p,m
, is
20.79\text{ J K}^{-1}\text{ mol}^{-1}20.79 J K
−1
mol
−1
at 298\text{ K}298 K. What
will be the value of the constant volume molar heat capacity of argon,
C_{V,m}C
V,m
, at this temperature?
Answers
Answer:
Constant-volume molar heat capacity of argon is 12.47 J K ⁻¹mol⁻¹
Explanation:
Argon is a monoatomic gas that behaves as an ideal gas at 298K.
Using the first law of thermodinamics you can obtain:
Work, Q, for constant pressure molar heat capacity,CP:
CP = (5/2)R
For constant-volume molar heat capacity,CV:
CV = (3/2)R
That means:
2CP/5 = 2CV/3
3/5 = CV / CP
As CP of Argon is 20.79 J K ⁻¹mol⁻¹, CV will be:
3/5 = CV / CP
3/5 = CV / 20.79 J K ⁻¹mol⁻¹
12.47 J K ⁻¹mol⁻¹ = CV
Constant-volume molar heat capacity of argon is 12.47 J K ⁻¹mol⁻¹What is the standard cell potential for the spontaneous voltaic cell formed from the given half-reactions
Answers
Answer:
because it is
For the reaction, 2SO2(g) + O2(g) <--> 2SO3(g), at 450.0 K the equilibrium constant, Kc, has a value of 4.62. A system was charged to give these initial concentrations, [SO3] = 0.254 M, [O2] = 0.00855 M, [SO2] = 0.500 M. In which direction will it go?
Answers
Answer:
To the left.
Explanation:
Step 1: Write the balanced reaction at equilibrium
2 SO₂(g) + O₂(g) ⇄ 2 SO₃(g)
Step 2: Calculate the reaction quotient (Qc)
Qc = [SO₃]² / [SO₂]² × [O₂]
Qc = 0.254² / 0.500² × 0.00855
Qc = 30.2
Step 3: Determine in which direction will proceed the system
Since Qc > Kc, the system will shift to the left to attain the equilibrium.
8. A 25.0 mL sample of an H2SO4 solution is titrated with a 0.186 M NaOH solution. The equivalence point is reached with 12.9 mL of base. The concentration of H2SO4 is ________ M. (Hint: write a balanced chemical equation first!)
Answers
Answer:
0.0480 M
Explanation:
The reaction is ...
H₂SO₄ + 2NaOH ⇒ Na₂SO₄ +2H₂O
That is, 2 moles of NaOH react with each mole of H₂SO₄. Then the molarity of the H₂SO₄ is ...
moles/liter = (0.186 M/2)(12.9 mL)/(25.0 mL) ≈ 0.0480 M
2) Which type movement do pivot joints allow?
Answers
Please let me know if you have more to your question and rate this a brainlist if it’s possible, thanks.
The activation energy for the decomposition of HI is 183 kJ/mol. At 573 K, the rate constant was measured to be 2.91 x 10^{-6} M/s. At what temperature in Kelvin does the reaction have a rate constant of 0.0760 M/s
Answers
Answer:
[tex]T_2=453.05K[/tex]
Explanation:
Hello,
In this case, the temperature-variable Arrhenius equation is written as:
[tex]\frac{k(T_2)}{k(T_1)}=exp(\frac{Ea}{R}(\frac{1}{T_2}-\frac{1}{T_1} ))[/tex]
Now, for us to solve for the temperature by which the reaction rate constant is 0.0760M/s we proceed as shown below:
[tex]ln(\frac{k(T_2)}{k(T_1)})=\frac{Ea}{R}(\frac{1}{T_2}-\frac{1}{T_1} )\\ln(\frac{0.0760M/s}{0.00000291M/s} )=\frac{183000J/mol}{8.314J/(mol*K)} *(\frac{1}{T_2} -\frac{1}{573K} )\\\frac{1}{T_2} -\frac{1}{573K} =\frac{10.17}{22011.06K^{-1}} \\\\\frac{1}{T_2}=4.62x10^{-4}K^{-1}+\frac{1}{573K}\\\\\frac{1}{T_2}=2.21x10^{-3}K^{-1}\\\\T_2=453.05K[/tex]
Regards.
Based on the properties of the compounds in the interactive, predict whether the given compounds behave as electrolytes or as nonelectrolytes.
1. LioH
2. C4H2O4
3. LiBr
4. HNo3
Answers
Explanation:
Before proceeding we have to understand what electrolytes and non electrolytes are;
An electrolyte is a substance that produces an electrically conducting solution when dissolved. An electrolyte is a compound that can dissociate into ions.
Non electrolytes: A substance whose molecules in solution do not dissociate to ions and thus do not conduct an electric current
Going through the options;
1. LiOH
This is a compound of hat would dissociate into Li+ and OH-. This is an electrolyte.
2. C4H2O4
This is an organic compound. Gnerally organic acids are non electrolytes, with the exception og the acids. This is a nonelectrolyte.
3. LiBr
This is an electrolyte because it would dissociate into Li+ and Br- ions.
4. HNO3
HNO3 is a strong acid. Because it is a strong acid it will dissociate completely into its ions (H+ and NO3-). Therefore we consider HNO3 to be a strong electrolyte.
Do you think you could go a week without causing any chemical reactions?
Answers
yes yes yes yes
yes
yes
yes
yes
yes
Elvira Walks 4 miles to the west from school and stops at the store. She then walks 3 miles south.
What is the Distance and the displacement?
Answers
Answer:
distance = 7 miles
displacement = 5 miles
Explanation:
Distance is a scalar quantity as it takes account of magnitude traveled but not the direction traveled from starting point.
The distance traveled is the sum total of distances moved
distance = 4 + 3 = 7 miles
Displacement however, is a vector and measure the shortest possible distance traveled in a given direction from the starting point.
The path of Elvis' walking forms a right-angle triangle with the hypotenuse being the displacement and the other two sides being the distance traveled west and south.
Using Pythagoras' theorem; c² = a² + b²
where c = hypotenuse and a and b are the other two sides
c² = 4² + 3²
c² = 16 + 9 = 25
√c² = √25
c = 5
Therefore, displacement = 5 miles
2.
Name the following compounds:
a. Rb20
Answers
Answer:
Rubidium oxide
Explanation:
2
What is the main side reaction that competes with elimination when a primary alkyl halide is treated with alcoholic potassium hydroxide, and why does this reaction compete with elimination of a primary alkyl halide but not a tertiary alkyl halide
Answers
Answer:
The main competing reaction when a primary alkyl halide is treated with alcoholic potassium hydroxide is SN2 substitution.
Explanation:
The relative percentage of products of the reaction between an alkyl halide and alcoholic potassium hydroxide generally depends on the structure of the primary alkylhalide. The attacking nucleophile/base in this reaction is the alkoxide ion. Substitution by SN2 mechanism is a major competing reaction in the elimination reaction intended.
A more branched alkyl halide will yield an alkene product due to steric hindrance, similarly, a good nucleophile such as the alkoxide ion may favour SN2 substitution over the intended elimination (E2) reaction.
Both SN2 and E2 are concerted reaction mechanisms. They do not depend on the formation of a carbocation intermediate. Primary alkyl halides generally experience less steric hindrance in the transition state and do not form stable carbocations hence they cannot undergo E1 or SN1 reactions.
SN2 substitution cannot occur in a tertiary alkyl halides because the stability of tertiary carbocations favours the formation of a carbocation intermediate. The formation of this carbocation intermediate will lead to an SN1 or E1 mechanism. SN2 reactions is never observed for a tertiary alkyl halide due to steric crowding of the transition state. Also, with strong bases such as the alkoxide ion, elimination becomes the main reaction of tertiary alkyl halides.
A(n) _____ reaction occurs when an acid and a base are present in the same solution.
Answers
Answer:
The answer is Neutralization reaction
It occurs when an acid and a base are present in the same solution and react to form salt and water only
Hope this helps you
Use standard reduction potentials to calculate the equilibrium constant for the reaction: 2Cr3+(aq) + Pb(s)2Cr2+(aq) + Pb2+(aq) Hint: Carry at least 5 significant figures during intermediate calculations to avoid round off error when taking the antilogarithm. Equilibrium constant: G° for this reaction would be _________ than zero. Submit AnswerRetry Entire Group
Answers
Answer:
3.47 ×10^-10
Explanation:
The equation of the reaction is 2Cr3+(aq) + Pb(s)------->2Cr2+(aq) + Pb2+(aq)
A total of two moles of electrons were transferred in the process. The chromium was reduced while the lead was oxidized. Hence the lead species will constitute the oxidation half equation and the chromium will constitute the reduction half equation.
E°cell = E°cathode - E°anode
E°cathode = -0.41 V
E°anode = -0.13 V
E°cell = -0.41 -(-0.13) = -0.28 V
From
E°cell = 0.0592/n log K
n= 2, K= the unknown
-0.28 = 0.0592/2 log K
log K = -0.28/0.0296
log K = -9.4595
K = Antilog ( -9.4595)
K= 3.47 ×10^-10
A 5.00g of X, the product of organic synthesis is obtained in a 1.0 dm3 aqueous solution. Calculate the mass of X that can be extracted from the aqueous solution by a 50cm3 of ethoxy ethane. (KD (X) =40.
Answers
Answer:
mass of X extracted from the aqueous solution by 50 cm³ of ethoxy ethane = 3.33 g
Explanation:
The partition coefficient of X between ethoxy ethane (ether) and water, K is given by the formula
K = concentration of X in ether/concentration of X in water
Partition coefficient, K(X) between ethoxy ethane and water = 40
Concentration of X in ether = mass(g)/volume(dm³)
Mass of X in ether = m g
Volume of ether = 50/1000 dm³ = 0.05 dm³
Concentration of X in ether = (m/0.05) g/dm³
Concentration of X in water = mass(g)/volume(dm³)
Mass of X in water left after extraction with ether = (5 - m) g
Volume of water = 1 dm³
Concentration of X in water = (5 - m/1) g/dm³
Using K = concentration of X in ether/concentration of X in water;
40 = (m/0.05)/(5 - m)
(m/0.05) = 40 × (5 - m)
(m/0.05) = 200 - 40m
m = 0.05 × (200 - 40m)
m = 10 - 2m
3m = 10
m = 10/3
m = 3.33 g of X
Therefore, mass of X extracted from the aqueous solution by 50 cm³ of ethoxy ethane = 3.33 g
Write a net ionic equation to show that benzoic acid, C6H5COOH, behaves as a Brønsted-Lowry acid in water.
Answers
Answer:
H⁺(aq) + H₂O(l) ⇄ H₃O⁺(aq)
Explanation:
According to Brönsted-Lowry acid-base theory, an acid is a substance that donates H⁺. Let's consider the molecular equation showing that benzoic acid is a Brönsted-Lowry acid.
C₆H₅COOH(aq) + H₂O(l) ⇄ C₆H₅COO⁻(aq) + H₃O⁺(aq)
The complete ionic equation includes all the ions and molecular species.
C₆H₅COO⁻(aq) + H⁺(aq) + H₂O(l) ⇄ C₆H₅COO⁻(aq) + H₃O⁺(aq)
The net ionic equation includes only the ions that participate in the reaction and the molecular species.
H⁺(aq) + H₂O(l) ⇄ H₃O⁺(aq)