Answers
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Related Questions
The substance formed on addition of water to an aldehyde or ketone is called a hydrate or a/an:_______
A) vicinal diol
B) geminal diol
C) acetal
D) ketal
Answers
Answer:
B) geminal diol
Explanation:
Hello,
In this case, considering the attached picture, you can see that the substance resulting from the hydrolysis of an aldehyde or a ketone is a geminal diol since the two hydroxyl groups are in the same carbon. Such hydrolysis could be carried out in either acidic or basic conditions depending upon the equilibrium constant.
Regards.
The insoluble salts below are put into 0.10 M hydrochloric acid solution. Do you expect their solubility to be more, less, or about the same as in a pure water solution?
1. Zinc sulfide
2. Silver chloride
3. Lead iodide
4. Silver hydroxide
Answers
Answer:
1. Zinc sulfide : about the same solubility, no common ion is found.
2. Silver chloride : less solubility due to the presence of chloride ions provided by the 0.10 M hydrochloric acid.
3. Lead iodide : about the same solubility, no common ion is found.
4. Silver hydroxide : about the same solubility, no common ion is found.
Explanation:
Hello,
In this case, we first must remember that adding a common ion (which is related with the dissolving solid) decreases the solubility of the insoluble solid due to the fact Le Chatelier's principle states the reaction will shift leftwards (reactants) to reestablish equilibrium, therefore, we have:
1. Zinc sulfide : about the same solubility, no common ion is found.
2. Silver chloride : less solubility due to the presence of chloride ions provided by the 0.10 M hydrochloric acid.
3. Lead iodide : about the same solubility, no common ion is found.
4. Silver hydroxide : about the same solubility, no common ion is found.
Best regards.
Using the standard reduction potentials Ni2+(aq) + 2 e‑Ni(s) ‑0.25 volt Fe3+(aq) + e‑Fe2+(aq) +0.77 volt Calculate the value of E°cell for the cell with the following reaction. Ni2+(aq) + 2 Fe2+(aq) →Ni(s) + 2 Fe3+(aq)
Answers
Answer:
The correct answer is - 1.02 V
Explanation:
From the reduction-oxidation reaction:
Ni²⁺(aq) + 2 Fe²⁺(aq) → Ni(s) + 2 Fe³⁺(aq)
Ni²⁺ is reduced to Ni(s) while Fe²⁺ is oxidized to Fe³⁺. Thus, the half reactions are:
Reduction (cathode) : Ni²⁺(aq) + 2 e‑ → Ni(s) Eº= ‑0.25 V
Oxidation (anode) : 2 x (Fe²⁺ → Fe³⁺ + e-)(aq) Eº= -0.77 V
-------------------------------------
Ni²⁺(aq) + 2 Fe²⁺(aq) → Ni(s) + 2 Fe³⁺(aq)
In order to calculate the Eºcell, we have to add the reduction potential of the reaction in cathode (reduction) to the oxidation potential of the anode (oxidation):
Eºcell= Eºr + Eºo= (-0.25 V) + (-0.77 V) = - 1.02 V
Write a balanced chemical equation for the base hydrolysis of methyl butanoate with NaOH. (Use either molecular formulas or condensed structural formulas, but be consistent in your equation.)
Answers
Explanation:
C5H10O2 + NaOH = C2H5COONa + C2H5OH
your result are : sodium propanoate and ethanol
A balanced chemical equation represents atoms and their numbers with their charge. The balanced equation for base hydrolysis is C₅H₁₀O₂ + NaOH → C₂H₅COONa + C₂H₅OH.
What is hydrolysis?Base hydrolysis is the splitting of the ester linkage by the basic molecule. As the result the acidic ester portion makes the salt, and also alcohol is produced as the by-product.
The base hydrolysis of methyl butanoate is shown as,
C₅H₁₀O₂ + NaOH → C₂H₅COONa + C₂H₅OH
Here, sodium propanoate and ethanol are produced by the splitting of methyl butanoate in the presence of the base (NaOH).
Therefore, C₅H₁₀O₂ + NaOH → C₂H₅COONa + C₂H₅OH is balanced reaction.
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One way the U.S. Environmental Protection Agency (EPA) tests for chloride contaminants in water is by titrating a sample of silver nitrate solution. Any chloride anions in solution will combine with the silver cations to produce bright white silver chloride precipitate. Suppose an EPA chemist tests a sample of groundwater known to be contaminated with nickel(II) chloride, which would react with silver nitrate solution like this:
Answers
Answer:
6.5 mg/L.
Explanation:
Step one: write out and Balance the chemical reaction in the Question above:
NiCl2 + 2AgNO3 =====> 2AgCl + Ni(NO3)2.
Step two: Calculate or determine the number of moles of AgCl.
So, we are given that the mass of AgCl = 3.6 mg = 3.6 × 10^-3 g. Therefore, the number of moles of AgCl can be calculated as below:
Number of moles AgCl = mass/molar mass = 3.6 × 10^-3 g / 143.32. = 2.5118 × 10^-5 moles.
Step three: Calculate or determine the number of moles of NiCl2.
Thus, the number of moles of NiCl2 = 2.5118 × 10^-5/ 2 = 1.2559 × 10^-5 moles.
Step four: detemine the mass of NiCl2.
Therefore, the mass of NiCl2 = number of moles × molar mass = 1.2559 × 10^-5 moles × 129.6 = 1.6 × 10^-3 g.
Step five: finally, determine the concentration of NiCl2.
1000/ 250 × 1.6 × 10^-3 g. = 6.5 mg/L.
Match the words below to the appropriate blanks in the sentences. Make certain each sentence is complete before submitting your answerβ-1,4- and α-1,6-glycosidicβ-1,4-glycosidicgalactosean unbranchedglucosea branchedfructoseα-1,6-glycosidicAmylose is ......... polymer of ....... units joined by ........ bonds. Amylopectin is ....... polymer of .......units joined by ........ bonds.
Answers
The words given are not clear, so the clear question is as follows:
Match the words below to the appropriate blanks in the sentences. Make certain each sentence is complete before submitting your answer:
A. β-1,4- and α-1,6-glycosidic
B. α-1,4-glycosidic
C. α-1,4-galactose
D. an unbranched glucose
E. a branched fructose
F. α-1,6-glycosidic
Amylose is ......... polymer of ....... units joined by ........ bonds.
Amylopectin is ....... polymer of .......units joined by ........ bonds.
Answer:
D. an unbranched glucose
C. α-1,4-galactose
B. α-1,4-glycosidic
E. a branched fructose
A. β-1,4- and α-1,6-glycosidic
F. α-1,6-glycosidic
Explanation:
Amylose and amylopectin are two types of polysaccharides that can be found in starch granules.
Amylose is linear or unbranched glucose polymer of α-1,4-galactose units that are joined by α-1,4-glycosidic.
Amylopectin is a branched fructose polymer of β-1,4- and α-1,6-glycosidic units joined by α-1,6-glycosidic bonds.
Hence, the correct answers in the sequential order are:
Amylose:
D. an unbranched glucose
C. α-1,4-galactose
B. α-1,4-glycosidic
Amylopectin:
E. a branched fructose
A. β-1,4- and α-1,6-glycosidic
F. α-1,6-glycosidic
To calculate changes in concentration for a system not at equilibrium, the first step is to determine the direction the reaction will proceed. To do so, we calculate Q and compare it to the equilibrium concentration, K. We can then determine that a reaction will shift to the right if:__________
Answers
Answer:
We can then determine that a reaction will shift to the right if Q<K
Explanation:
Comparing Q with K allows to find out the status and evolution of the system:
If the reaction quotient is equal to the equilibrium constant, Qc = Kc, the system has reached chemical equilibrium. If the reaction quotient is greater than the equilibrium constant, Qc> Kc, the system is not in equilibrium and will evolve spontaneously, decreasing the value of Qc until it equals the equilibrium constant. In this way, the concentrations of the products will decrease and the concentrations of the reagents will increase. In other words, the reverse reaction is favored to achieve equilibrium. Then the system will evolve to the left (ie products will be consumed and more reagents will be formed).If the reaction quotient is less than the equilibrium constant, Qc <Kc, the system is not in equilibrium and will evolve spontaneously increasing the value of Qc until it equals the equilibrium constant. This implies that the concentrations of the products will increase and those of the reagents will decrease. In other words, to achieve balance, direct reaction is favored. Then the reaction will shift to the right, that is, reagents will be consumed and more products will be formed.In this case, we can then determine that a reaction will shift to the right if Q<K
While balancing a chemical equation, we change the _____ to balance the number of atoms on each side of the equation.
Answers
Answer:
While balancing a chemical equation, we change the coefficient to balance the number of atoms on each side of the equation
Explanation:
While balancing a chemical equation, we change the coefficient to balance the number of atoms on each side of the equation.
What is chemical equation?To summarize in chemistry terms, a chemical equation depicts the initial chemicals, or reactants, on the left-hand side and the final compounds, or products, just on right-hand side, divided by an arrow. In the chemical equation, the number of atoms in each element as well as the total charge are the same on opposite of the equation's sides.
Chemical equations are used in chemistry to depict chemical processes by writing the reactants and products in terms of their corresponding chemical formulas. While balancing a chemical equation, we change the coefficient to balance the number of atoms on each side of the equation.
Therefore, while balancing a chemical equation, we change the coefficient to balance the number of atoms on each side of the equation.
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When the nuclide bismuth-210 undergoes alpha decay:
The name of the product nuclide is_____.
The symbol for the product nuclide is_____
Fill in the nuclide symbol for the missing particle in the following nuclear equation.
_____ rightarrow 4He+ 234Th
2 90
Write a balanced nuclear equation for the following:
The nuclide radium-226 undergoes alpha emission.
Answers
Explanation:
An atom undergoes alpha decay by losing a helium atom.
So when bismuth undergoes alpha decay, we have;
²¹⁰₈₃Bi --> ⁴₂He + X
Mass number;
210 = 4 + x
x = 206
Atomic number;
83 = 2 + x
x = 81
The element is Thallium. The symbol is Ti.
For the second part;
X --> ⁴₂He + ²³⁴₉₀Th
Mass number;
x = 4 + 234 = 238
Atomic Number;
x = 2 + 90 = 92
The balanced nuclear equation is;
²³⁸₉₂U --> ⁴₂He + ²³⁴₉₀Th
A 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.
Answers
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.
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Calculate the concentration of H3O+ in a solution that contains 5.5 × 10-5 M OH- at 25°C. Identify the solution as acidic, basic, or neutral.
Answers
Explanation:
To calculate [H3O+] in the solution we must first find the pH from the [ OH-]
That's
pH + pOH = 14
pH = 14 - pOH
To calculate the pOH we use the formula
pOH = - log [OH-]
And [OH-] = 5.5 × 10^-5 M
So we have
pOH = - log 5.5 × 10^ - 5
pOH = 4.26
Since we've found the pOH we can now find the pH
That's
pH = 14 - 4.26
pH = 9.74
Now we can find the concentration of H3O+ in the solution using the formula
pH = - log H3O+
9.74 = - log H3O+
Find the antilog of both sides
H3O+ = 1.8 × 10^ - 10 MThe solution is basic since it's pH lies in the basic region.
Hope this helps you
A chemist fills a reaction vessel with 0.978 g aluminum hydroxide AlOH3 solid, 0.607 M aluminum Al+3 aqueous solution, and 0.396 M hydroxide OH− aqueous solution at a temperature of 25.0°C.
Under these conditions, calculate the reaction free energy ΔG for the following chemical reaction:
Al(OH)3(s) = A1+ (aq) +30H (aq)
Use the thermodynamic information in the ALEKS Data tab. Round your answer to the nearest kilojoule.
KJ
Answers
Answer: [tex]\Delta G^{0}[/tex] = 168.12 kJ
Explanation: Gibbs Free Energy, at any time, is defined as the enthalpy of the system minus product of temperature and entropy of the reaction, i.e.:
[tex]\Delta G^{0} = \Delta H^{0} - T.\Delta S^{0}[/tex]
Enthalpy is defined as internal heat existent in the system. It is calculated as:
[tex]\Delta H^{0} = \Sigma H^{0}_{product} - \Sigma H^{0}_{reagent}[/tex]
Using Enthalpy Formation Table:
[tex]\Delta H^{0} = [3*(-299.9)+(-524.7)] - (-1277)[/tex]
[tex]\Delta H^{0} = 62,6 kJ[/tex]
Entropy is the degree of disorder in the system. It is found by:
[tex]\Delta S^{0} = \Sigma S^{0}_{products} - \Sigma S^{0}_{reagents}[/tex]
Calculating:
[tex]\Delta S^{0} = (-321.7) + 3(-10.8) - 0[/tex]
[tex]\Delta S^{0} = -354.1J[/tex]
And so, Gibbs Free energy will be:
[tex]\Delta G^{0} = \Delta H^{0} - T.\Delta S^{0}[/tex]
[tex]\Delta G^{0} = 62600 - [298.(-354.1)][/tex]
[tex]\Delta G^{0} = 168121.8 J[/tex]
Rounding to the nearest kJ:
[tex]\Delta G^{0}[/tex] = 168.12 kJ
Of the following two gases, which would you predict to diffuse more rapidly? PLZZ HELPP PLZ PLZ PLZ
Answers
Answer:
CO2 will diffuse more rapidly.
Explanation:
From Graham's law of diffusion, we understood that the rate of diffusion of a gas is inversely proportional to the square root of its density as shown below:
Rate (R) & 1/√Density (d)
R & 1/√d
But, the density of a gas is directly proportional to the relative molecular mass (M) of the gas.
Thus, we can say that the rate of diffusion of a gas is inversely proportional to the square root of the molar mass of the gas. This can be represented mathematically as:
Rate (R) & 1/√Molar mass (M)
R & 1/√M
From the above illustration, we can say that the lighter the gas, the faster the rate of diffusion and the heavier the gas, the slower the rate of diffusion.
Now, to answer the question given above,let us determine the molar mass of Cl2 and CO2.
This is illustrated below:
Molar mass of Cl2 = 2 x 35.5 = 71 g/mol
Molar mass of CO2 = 12 + (2x16) = 12 + 32 = 44 g/mol
Summary
Gas >>>>>> Molar mass
Cl2 >>>>>> 71 g/mol
CO2 >>>>> 44 g/mol
From the illustration above, we can see that CO2 is lighter than Cl2.
Therefore, CO2 will diffuse more rapidly.
Answer: CO2
Explanation:
g Does a reaction occur when aqueous solutions of barium hydroxide and aluminum sulfate are combined
Answers
Answer:
3BaO + Al₂(SO₄)₃ → Al₂O₃+ 3BaSO₄
Explanation:
Yes! A reactiin occurs between barium hydroxide and auminium sulphate.
barium sulfate (BaSO4) and aluminum hydroxide (Al(OH)3) are the products obtained in this reaction.
The reaction is given by the equation below;
3BaO + Al₂(SO₄)₃ → Al₂O₃+ 3BaSO₄
When 1604 J of heat energy is added to 48.9 g of hexane, C6H14, the temperature increases by 14.5 ∘C. Calculate the molar heat capacity of C6H14.
Answers
Answer:
THE MOLAR HEAT CAPACITY OF HEXANE IS 290.027 J/ C
Explanation:
1604 J of heat is added to 48.9 g of hexane
To calculate the molar heat capacity of hexane, it is important to note that the molar heat capacity of a substance is the measure of the amount of heat needed to raise 1 mole of a substance by 1 K.
Since 1604 J of heat = 48.9 g of hexane
Molar mass of hexane = 86 g/mol = 1 mole
then;
1604 J = 48.9 g
x = 86 g
x = 1604 * 86 / 48.9
x = 4205.4 J
Hence, 4205.4 J of heat will be added to 1 mole or 86 g of hexane to raise the temperature by 14.5 C.
In other words,
heat = molar heat capacity * temperature change
molar heat capacity = heat/ temperature change
Molar heat capacity = 4205.4 J / 14.5 C
Molar heat capacity = 290.027 J/C
The molar heat capacity of hexane is 290.027 J/ C
A mixture of 50ml of 0.1M HCOOH and 50ml of 0.05M NaOH is equivalent to
Answers
Answer:
d) a solution that is 0.025M in HCOOH and 0.025M in HCOONa
Explanation:
The reaction of a weak acid (HOOH) with NaOH is as follows:
HCOOH + NaOH → HCOONa + H₂O
Based on the reaction, 1 mole of the acid reacts with 1 mole of the base (Ratio 1:1).
The initial moles of both species are:
HCOOH: 0.050L × (0.1mol / L) = 0.0050 moles of HCOOH
NaOH: 0.050L × (0.05 mol / L) = 0.0025 moles NaOH
After the reaction, all NaOH reacts with HCOOH producing HCOONa (Because moles of NaOH < moles HCOOH).
Final moles:
HCOOH: 0.0050 moles - 0.0025 moles (After reaction) = 0.0025 moles
HCOONa: Moles HCOONa = Initial Moles NaOH: 0.0025 moles
As volume of the mixture is 100mL (50 from the acid + 50 from NaOH), molarity of both HCOOH and HCOONa is:
0.0025 moles / 0.100L = 0.025M of both HCOOH and HCOONa
Thus, the initial mixture is equivalent to:
d) a solution that is 0.025M in HCOOH and 0.025M in HCOONa. You have two solutions, both with a concentration of 0.1M. Solution A contains a weak acid with a pKa of 5. ThepH of solution A is 3. Solution B contains a weak acid with a pKa of 9. The pH of solution B is:
Answers
Answer:
pH of solution B is 5
Explanation:
A weak acid, HA, is in equilibrium with water as follows:
HA(aq) + H₂O(l) ⇄ A⁻(aq) + H₃O⁺(aq)
Where Ka (10^-pKa = 1x10⁻⁹) is:
Ka = 1x10⁻⁹ = [A⁻] [H₃O⁺] / [HA]
Where concentrations of this species are equilibrium concentrations
As initial concentration of HA is 0.1M, the equilibrium concentrations of the species are:
[HA] = 0.1M - X
[A⁻] = X
[H₃O⁺] = X
Where X is the amount of HA that reacts until reach the equilibrium, X is reaction coordinate.
Replacing in Ka expression:
1x10⁻⁹ = [A⁻] [H₃O⁺] / [HA]
1x10⁻⁹ = [X] [X] / [0.1 - X]
1x10⁻¹⁰ - 1x10⁻⁹X = X²
1x10⁻¹⁰ - 1x10⁻⁹X - X² = 0
Solving for X:
X = -0.00001 → False solution, there is no negative concentrations.
X = 1x10⁻⁵ → Right solution.
As [H₃O⁺] = X
[H₃O⁺] = 1x10⁻⁵M
And pH = -log[H₃O⁺]
pH = 5
pH of solution B is 5
Because of movements at the Mid-Atlantic Ridge, the Atlantic Ocean widens by about 2.5 centimeters each year. Explain which type of plate boundary causes this motion.
Answers
Answer:
A divergent plate boundary
Explanation:
At a divergent boundary, the plates pull away from each other and generate new crust.
Answer:
Because the ocean becomes larger, this is a divergent plate boundary. Divergent plates cause the ocean floor to expand, making the ocean larger.
Explanation:
PLATO ANSWER
Select the true statement concerning voltaic and electrolytic cells. Select one: a. Voltaic cells involve oxidation-reduction reactions while electrolytic cells involve decomposition reactions. b. Voltaic cells require applied electrical current while electrolytic cells do not. . c. all electrochemical cells, voltaic and electrolytic, must have spontaneous reactions. d. Electrical current drives nonspontaneous reactions in electrolytic cells.
Answers
Answer:
Electrical current drives nonspontaneous reactions in electrolytic cells.
Explanation:
Electrochemical cells are cells that produce electrical energy from chemical energy.
There are two types of electrochemical cells; voltaic cells and electrolytic cells.
A voltaic cell is an electrochemical cell in which electrical energy is produced from spontaneous chemical process while an electrolytic cell is an electrochemical cell where electrical energy is produced from nonspontaneous chemical processes. Current is needed to drive these nonspontaneous chemical processes in an electrolytic cell.
Answer:
electrolytic cells generate electricity through a non-spontaneous reaction while voltaic cells absorb electricity to drive a spontaneous reaction.
Explanation:
Answer via Educere/ Founder's Education
A chemist prepares a solution of sodium chloride by measuring out 25.4 grams of sodium chloride into a 100. mL volumetric flask and filling the flask to the mark with water. Calculate the concentration in mol/L of the chemist's sodium chloride solution. Be sure your answer has the correct number of significant digits.
Answers
Answer:
The concentration in mol/L = 4.342 mol/L
Explanation:
Given that :
mass of sodium chloride = 25.4 grams
Volume of the volumetric flask = 100 mL
We all know that the molar mass of sodium chloride NaCl = 58.5 g/mol
and number of moles = mass/molar mass
The number of moles of sodium chloride = 25.4 g/58.5 g/mol
The number of moles of sodium chloride = 0.434188 mol
The concentration in mol/L = number of mol/ volume of the solution
The concentration in mol/L = 0.434188 mol/ 100 × 10⁻³ L
The concentration in mol/L = 4.34188 mol/L
The concentration in mol/L = 4.342 mol/L
If the theoretical yield of a reaction is 332.5 g and the percent yield for the reaction is 38 percent, what's the actual yield of product in grams? \
A. 8.74 g
B. 12616 g
C. 116.3 g
D. 126.4 g
Answers
Answer: D - 126.4g
Explanation:
% Yield = Actual Yield/Theoretical Yield
38% = Actual Yield/332.5
38/100 = Actual Yield/332.5
(.38)(332.5) = 126.35 g = 126.4 g Actual Yield
Answer:
is D. the correct answer
Explanation:
I'm not sure if it is. Please let me know if I'm mistaking.
Find the pH of these buffer solutions using the information provided: 1L solution containing 80g of lactic acid (MW
Answers
Answer:
pH of the solution is 2.0
Explanation:
The lactic acid is a weak acid that is in equilibrium with water as follows:
Lactic acid + H2O ⇄ Lactate + H₃O⁺
And Ka for lactic acid: 1.38x10⁻⁴
Ka = 1.38x10⁻⁴ = [Lactate] [H₃O⁺] / [Lactic acid]
Initial concentration of lactic acid is (MW: 112.06g/mol):
80g * (1mol / 112.06g) / 1L = 0.714M
The equilibrium concentration of the species in the equilibrium are:
[Lactate] = X
[H₃O⁺] = X
[Lactic acid] = 0.714-X
Replacing in Ka expression:
1.38x10⁻⁴ = [X] [X] / [0.714-X]
9.8532x10⁻⁵ - 1.38x10⁻⁴X = X²
9.8532x10⁻⁵ - 1.38x10⁻⁴X - X² = 0
Solving for X:
X = -1.0x10⁻². False solution, there is no negative concentrations
X = 9.86x10⁻³M. Right solution.
As [H₃O⁺] = X
[H₃O⁺] = 9.86x10⁻³M
and pH = -log [H₃O⁺] = -log 9.86x10⁻³M
pH = 2.0
pH of the solution is 2.0Provide the name(s) for the tertiary alcohol(s) with the chemical formula C6H14O that have a 4-carbon chain. Although stereochemistry may be implied in the question, DO NOT consider stereochemistry in your name. Alcohol #1______ Alcohol #2: ______Alcohol #3______
Answers
Answer:
Explanation:
A tertiary alcohol is a compound (an alcohol) in which the carbon atom that has the hydroxyl group (-OH) is also bonded (saturated) to three different carbon atoms.
Based on the question, the only tertiary alcohol that can result from C₆H₁₄O that have a 4-carbon chain is
2-hydroxy-2,3-dimethylbutane
H OH H H
| | | |
H - C - C - C - C - H
| | | |
H CH₃ CH₃ H
From the above, we can see that the carbon atom having the hydroxyl group is also bonded to three other carbon atoms. And since we aren't considering stereochemistry, this is the only tertiary alcohol we can have with a 4-carbon chain
What type of bond would form between two atoms of phosphorus? A. Single covalent bond B. Single ionic bond C. Triple covalent bond D. Double covalent bond
Answers
Answer:
A double bond is formed when two pairs of electrons are shared between the two participating atoms. It is represented by two dashes (=). It is represented by two dashes (=). Double covalent bonds are much stronger than a single bond, but they are less stable
Explanation:
From the following balanced equation, CH4(g)+2O2(g)⟶CO2(g)+2H2O(g) how many grams of H2O can be formed when 1.25g CH4 are combined with 1.25×10^23 molecules O2? Use 6.022×10^23 mol−1 for Avogadro's number.
Answers
Answer:
2.81 g of H2O.
Explanation:
We'll begin by calculating mass of O2 that contains 1.25×10²³ molecules O2.
This can be obtained as follow:
From Avogadro's hypothesis, we understood that 1 mole of any substance contains 6.022×10²³ molecules. This implies that 1 mole of O2 also contains 6.022×10²³ molecules.
1 mole of O2 = 16x2 = 32 g.
Thus 6.022×10²³ molecules is present in 32 g of O2,
Therefore, 1.25×10²³ molecules will be present in =
(1.25×10²³ × 32) / 6.022×10²³ = 6.64 g of O2.
Therefore, 1.25×10²³ molecules present in 6.64 g of O2.
Next, the balanced equation for the reaction. This is given below:
CH4(g) + 2O2(g) —> CO2(g) + 2H2O(g)
Next, we shall determine the masses of CH4 and O2 that reacted and the mass of H2O produced from the balanced equation.
This can be obtained as follow:
Molar mass of CH4 = 12 + (4x1) = 16 g/mol.
Mass of CH4 from the balanced equation = 1 x 16 = 16 g
Molar mass of O2 = 16x2 = 32 g/mol.
Mass of O2 from the balanced equation = 2 x 32 = 64 g
Molar mass of H2O = (2x1) + 16 = 18 g/mol.
Mass of H2O from the balanced equation = 2 x 18 = 36 g
From the balanced equation above,
16 g of CH4 reacted with 64 g of O2 to produce 36 g if H2O.
Next, we shall determine the limiting reactant.
This can be obtained as follow:
From the balanced equation above,
16 g of CH4 reacted with 64 g of O2.
Therefore, 1.25 g of CH4 will react with = (1.25 x 64)/16 = 5 g of O2.
From the above calculations, we can see that only 5 g out of 6.64 g of O2 is needed to react completely with 1.25 g of CH4.
Therefore, CH4 is the limiting reactant.
Finally, we shall determine the mass of H2O produced from the reaction.
In this case, the limiting reactant will be used because it will give the maximum yield of H2O.
The limiting reactant is CH4 and the mass of H2O produced from the reaction can be obtained as follow:
From the balanced equation above,
16 g of CH4 reacted to produce produce 36 g if H2O.
Therefore, 1.25 g of CH4 will react to produce = (1.25 x 36)/16 = 2.81 g of H2O.
Therefore, 2.81 g of H2O were obtained from the reaction.
The mass in grams of H₂O which can be formed when 1.25g CH₄ are combined with 1.25×10²³ molecules O₂ is 2.8 grams.
What is stoichiometry?Stoichiometry of any reaction tells about the amount of species present before and after the completion of the reaction.
Given chemical reaction is:
CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g)
Moles of CH₄ will b calculate as:
n = W/M, where
W = given mass = 1.25g
M = molar mass = 16g/mol
n = 1.25/16 = 0.078 moles
Molecues of CH₄ in 0.078 moles = 0.078×6.022×10²³ = 0.46×10²³
Given molecules of O₂ = 1.25×10²³
Required molecules of CH₄ is less as compared to the molecules of O₂, so here CH₄ is the limiting reagent and formation of water is depends on it only.
From the stoichiometry of the reaction it is clear that:
1 mole of CH₄ = will produce 2 moles of H₂O
0.078 moles of CH₄ = will produce 2×0.078=0.156 moles of H₂O
Mass of H₂O will be calculated by using its moles as:
W = (0.156)(18) = 2.8g
Hence required mass of H₂O is 2.8g.
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A piece of solid Fe metal is put into an aqueous solution of Cu(NO3)2. Write the net ionic equation for any single-replacement redox reaction that may be predicted. Assume that the oxidation state of in the resulted solution is 2 . (Use the lowest possible coefficients for the reaction. Use the pull-down boxes to specify states such as (aq) or (s). If a box is not needed, leave it blank. If no reaction occurs, leave all boxes blank and click on Submit.)
Answers
Answer:
Fe(s) + Cu^2+(aq) ---> Fe^2+(aq) + Cu(s)
Explanation:
The ionic equation shows the actual reaction that took place. It excludes the spectator ions. Spectator ions are ions that do not really participate in the reaction even though they are present in the system.
For the reaction between iron and copper II nitrate, the molecular reaction equation is;
Fe(s) + Cu(NO3)2(aq)----> Fe(NO3)2(aq) +Cu(s)
Ionically;
Fe(s) + Cu^2+(aq) ---> Fe^2+(aq) + Cu(s)
Consider the reaction: C(s) + O2(g)CO2(g) Write the equilibrium constant for this reaction in terms of the equilibrium constants, Ka and Kb, for reactions a and b below: a.) C(s) + 1/2 O2(g) CO(g) Ka b.) CO(g) + 1/2 O2(g) CO2(g) Kb
Answers
Answer:
A. Ka = [CO2] / [C] [O2]^1/2
B. Kb = [CO2] / [CO] [O2]^1/2
Explanation:
Equilibrium constant is simply defined as the ratio of the concentration of the products raised to their coefficient to the concentration of the reactants raised to their coefficient.
Now, we shall obtain the expression for the equilibrium constant for the reaction as follow:
A. Determination of the expression for equilibrium constant Ka.
This is illustrated below:
C(s) + 1/2 O2(g) <==> CO(g)
Ka = [CO2] / [C] [O2]^1/2
B. Determination of the expression for equilibrium constant Kb.
This is illustrated below:
CO(g) + 1/2 O2(g) <==> CO2(g)
Kb = [CO2] / [CO] [O2]^1/2
Find the density if the volume is 15 mL and the mass is 8.6 g. (5 pts)
Find the volume if the density is 2.6 g/mL and the mass is 9.7 g.(5 pts)
Find the mass if the density is 1.6 g/cm3 and the volume is 4.1 cm3 (5 pts)
Find the density if the initial volume of water is 12.8 mL, the final volume is 24.6 mL and the mass of the object is 4.3 g. Make a drawing to show the water displacement using a graduated cylinder. (gdoc, gdraw)
Answers
Answer:
[tex]\large \boxed{\text{0.57 g/mL; 3.7 mL; 6.6 g; 0.366 g/mL}}[/tex]
Explanation:
1. Density from mass and volume
[tex]\text{Density} = \dfrac{\text{mass}}{\text{volume}}\\\\\rho = \dfrac{m}{V}\\\\\rho = \dfrac{\text{8.6 g}}{\text{15 mL}} = \text{0.57 g/mL}\\\text{The density is $\large \boxed{\textbf{0.57 g/mL}}$}[/tex]
2. Volume from density and mass
[tex]V = \text{9.7 g}\times\dfrac{\text{1 mL}}{\text{2.6 g}} = \text{3.7 mL}\\\\\text{The volume is $\large \boxed{\textbf{3.7 mL}}$}[/tex]
3. Mass from density and volume
[tex]\text{Mass} = \text{4.1 cm}^{3} \times \dfrac{\text{1.6 g}}{\text{1 cm}^{3}} = \textbf{6.6 g}\\\\\text{The mass is $\large \boxed{\textbf{6.6 g}}$}[/tex]
4. Density by displacement
Volume of water + object = 24.6 mL
Volume of water = 12.8 mL
Volume of object = 11.8 mL
[tex]\rho = \dfrac{\text{4.3 g}}{\text{11.8 mL}} = \text{0.36 g/mL}\\\text{The density is $\large \boxed{\textbf{0.36 g/mL}}$}[/tex]
Your drawing showing water displacement using a graduated cylinder should resemble the figure below.
Arrange the compounds in order of decreasing magnitude of lattice energy:
a. LiBr
b. KI
c. CaO.
Rank from largest to smallest.
Answers
Answer:
The correct answer is CaO > LiBr > KI.
Explanation:
Lattice energy is directly proportional to the charge and is inversely proportional to the size. The compound LiBr comprises Li+ and Br- ions, KI comprises K+ and I- ions, and CaO comprise Ca²⁺ and O²⁻ ions.
With the increase in the charge, there will be an increase in lattice energy. In the given case, the lattice energy of CaO will be the highest due to the presence of +2 and -2 ions. K⁺ ions are larger than Li⁺ ion, and I⁻ ions are larger than Br⁻ ion.
The distance between Li⁺ and Br⁻ ions in LiBr is less in comparison to the distance between K⁺ and I⁻ ions in KI. As a consequence, the lattice energy of LiBr is greater than KI. Therefore, CaO exhibits the largest lattice energy, while KI the smallest.
Arranging the chemical compounds in order of decreasing magnitude of lattice energy, we have:
c. CaO.
a. LiBr
b. KI
Lattice energy can be defined as a measure of the energy required to dissociate one (1) mole of an ionic compound into its constituent anions and cations, in the gaseous state.
Hence, it is typically used to measure the bond strength of ionic compounds.
Generally, lattice energy is inversely proportional to the size of the ions and directly proportional to their electric charges.
Lithium bromide (LiBr) comprises the following ions:
[tex]Li^+[/tex] and [tex]Br^-[/tex]Potassium iodide (KI) comprises the following ions:
[tex]K^+[/tex] and [tex]I^-[/tex]Calcium oxide (CaO) comprises the following ions:
[tex]Ca^{2+}[/tex] and [tex]O^{2-}[/tex]From the above, we can deduce that there is an increase in the charge possessed by the ionic chemical compounds and as such this would result in an increase in the lattice energy.
In order of decreasing magnitude of lattice energy, the chemical compounds are arranged as:
I. CaO.
II. KI.
III. LiBr.
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243
Am
95
1. The atomic symbol of americium-243 is shown. Which of the following is correct?
• A. The atomic mass is 243 amu, and the atomic number is 95.
B. The atomic mass is 338 amu, and the atomic number is 95.
• C. The atomic mass is 95 amu, and the atomic number is 243.
D. The atomic mass is 243 amu, and the atomic number is 338.
Answers
Answer:
A. The atomic mass is 243 amu, and the atomic number is 95.
