Answer:
4 boiling point elevation
Write the balanced half-reactions and an overall cell reaction below. Be sure to include states of matter.
Answer: Hello your question is incomplete attached below is the missing image
answer:
Reduction half-reaction: Zn(s) ⇒ Zn⁺² (aq) + 2e⁻ ( occurs at anode )
Oxidation half-reaction; CO²⁺(aq) + 2e⁻ ⇒ CO (s) ( occurs at the cathode )
Overall cell reaction ; Zn(s) + CO²⁺(aq) ⇒ Zn⁺² (aq) + CO (s)
Explanation:
stating the standard reduction potentials
E° zn²⁺/zn = -0.76 v
E°Co²⁺ / Co = -0.28 v
since ; -0.76 v < -0.28 v. Zn will be oxidized while Co²⁺ will be reduced .
Reduction half-reaction: Zn(s) ⇒ Zn⁺² (aq) + 2e⁻ ( occurs at anode )
Oxidation half-reaction; CO²⁺(aq) + 2e⁻ ⇒ CO (s) ( occurs at the cathode )
hence the
Overall cell reaction ;
Zn(s) + CO²⁺(aq) ⇒ Zn⁺² (aq) + CO (s)
When naming organic compounds, there are strict rules regarding punctuation.
1. A comma is used to separate two numbers.
2. A hyphen is used to separate a number from a letter.
Rewrite the name of this compound using hyphens and commas as appropriate.
The question is incomplete, the complete question is shown in the image attached to this answer
Answer:
2,3,3-trimethylhexane
Explanation:
IUPAC nomenclature provides a universally acceptable method of naming organic compounds from its structure.
According to this system of nomenclature;
A comma is used to separate two numbers.
A hyphen is used to separate a number from a letter.
Applying these rules, the name of the compound shown in the question should be written as 2,3,3-trimethylhexane.
Given the equation: 2C6H10(l) 17 O2(g) ---> 12 CO2(g) 10 H2O(g) MM( g/mol): 82 32 44 18 If 115 g of C6H10 reacts with 199 g of O2 and 49 g of H2O are formed, what is the percent yield of the reaction
Answer:
74%
Explanation:
Step 1: Write the balanced equation
2 C₆H₁₀(l) + 17 O₂(g) ⇒ 12 CO₂(g) + 10 H₂O(g)
Step 2: Determine the limiting reactant
The theoretical mass ratio (TMR) of C₆H₁₀ to O₂ is 164:544 = 0.301:1.
The experimental mass ratio (EMR) of C₆H₁₀ to O₂ is 115:199 = 0.578:1.
Since EMR > TMR, the limiting reactant is O₂.
Step 3: Calculate the theoretical yield of H₂O
The theoretical mass ratio of O₂ to H₂O 544:180.
199 g O₂ × 180 g H₂O/544 g O₂ = 65.8 g H₂O
Step 4: Calculate the percent yield of H₂O
%yield = (experimental yield/theoretical yield) × 100%
%yield = (49 g/65.8 g) × 100% = 74%
Answer:
Percentage yield of H₂O = 74.24%
Explanation:
The balanced equation for the reaction is given below:
2C₆H₁₀ + 17O₂ —> 12CO₂ + 10H₂O
Next, we shall determine the masses of C₆H₁₀ and O₂ that reacted and the mass of H₂O produced from the balanced equation. This is can be obtained as follow:
Molar mass of C₆H₁₀ = 82 g/mol
Mass of C₆H₁₀ from the balanced equation = 2 × 82 = 164 g
Molar mass of O₂ = 32 g/mol
Mass of O₂ from the balanced equation = 17 × 32 = 544 g
Molar mass of H₂O = 18 g/mol
Mass of H₂O from the balanced equation = 10 × 18 = 180 g
SUMMARY:
From the balanced equation above,
164 g of C₆H₁₀ reacted with 544 g of O₂ to produce 180 g of H₂O.
Next, we shall determine the limiting reactant. This can be obtained as follow:
From the balanced equation above,
164 g of C₆H₁₀ reacted with 544 g of O₂.
Therefore, 115 g of C₆H₁₀ will react to produce = (115 × 544)/164 = 381 g of O₂.
From the calculations made above, we can see that a higher mass (i.e 381 g) of O₂ than what was given (i.e 199 g) is needed to react with 115 g of C₆H₁₀.
Therefore, O₂ is the limiting reactant and C₆H₁₀ is the excess reactant.
Next, we shall determine the theoretical yield of H₂O. This can be obtained by using the limiting reactant as shown below:
From the balanced equation above,
544 g of O₂ reacted to produce 180 g of H₂O.
Therefore, 199 g of O₂ will react to produce = (199 × 180)/544 = 66 g of H₂O.
Thus, the theoretical yield of H₂O is 66 g.
Finally, we shall determine the percentage yield. This can be obtained as follow:
Actual yield of H₂O = 49 g
Theoretical yield of H₂O = 66 g
Percentage yield of H₂O =?
Percentage yield = Actual yield /Theoretical yield × 100
Percentage yield of H₂O = 49/66 × 100
Percentage yield of H₂O = 74.24%
You perform a distillation to separate a mixture of propylbenzene and cyclohexane, and you obtain 2.9949 grams of cyclohexane (density -0.779 g/mL, MW - 84.16 g/mol) and 1.6575 grams of propylbenzene (density = 0.862 g/mL, MW = 120.2 g/mol). What is the volume percent composition of cyclohexane in the mixture?
Answer:
66.67%
Explanation:
From the given information:
mass of cyclohexane = 2.9949 grams
density of cyclohexane = 0.779 g/mL
Recall that:
Density = mass/volume
∴
Volume = mass/density
So, the volume of cyclohexane = 2.9949 g/ 0.779 g/mL
= 3.8445 mL
Also,
mass of propylbenzene = 1.6575 grams
density of propylbenzene = 0.862 g/mL
Volume of propylbenzene = 1.6575 g/ 0.862 g/mL
= 1.9229 mL
The volume % composition of cyclohexane from the mixture is:
[tex]= (\dfrac{v_{cyclohexane}}{v_{cyclohexane}+v_{propylbenzene}})\times 100[/tex]
[tex]= (\dfrac{3.8445}{3.8445+1.9229})\times 100[/tex]
[tex]= (\dfrac{3.8445}{5.7674})\times 100[/tex]
= 66.67%
Enough of a monoprotic weak acid is dissolved in water to produce a 0.0118 M solution. The pH of the resulting solution is 2.32 . Calculate the Ka for the acid.
Answer:
1.94 × 10⁻³
Explanation:
Step 1: Calculate the concentration of H⁺ ions
We will use the definition of pH.
pH = -log [H⁺]
[H⁺] = antilog -pH = antilog -2.32 = 4.79 × 10⁻³ M
Step 2: Calculate the acid dissociation constant (Ka) of the acid
For a monoprotic weak acid, whose concentration (Ca) is 0.0118 M, we can use the following expression.
Ka = [H⁺]²/Ca
Ka = (4.79 × 10⁻³)²/0.0118 = 1.94 × 10⁻³
An aqueous solution contains 0.374 M ammonia (NH3). How many mL of 0.276 M nitric acid would have to be added to 125 mL of this solution in order to prepare a buffer with a pH of 8.970.
Answer:
111.95mL of HNO3 are needed to prepare the buffer
Explanation:
We can solve this equation using H-H equation for bases:
pOH = pKb + log [HA+] / [A]
Where pOH is the pOH of the solution
pOH = 14 - pH = 14 - 8.970 = 5.03
pKb is the pKb of NH3 = 4.74
[HA+] could be taken as moles of NH4+
[A] as moles of NH3
The NH3 reacts with nitric acid, HNO3, as follows:
NH3 + HNO3 → NH4+ + NO3-
That means the moles of HNO3 added = X = Moles of NH4+ produced
And moles of NH3 are initial moles NH3 - X
Initial moles of NH3 are:
0.125L * (0.374mol/L) = 0.04675 moles NH3
Replacing in H-H equation:
pOH = pKb + log [HA+] / [A]
5.03 = 4.74 + log [X] / [0.04675-X]
0.29 = log [X] / [0.04675-X]
1.95 = [X] / [0.04675-X]
0.0912 - 1.95X = X
0.0912 = 2.95X
X = 0.0309 moles
We need to add 0.0309 moles of HNO3. From a solution that is 0.276M:
0.0309 moles of HNO3 * (1L / 0.276moles) = 0.11195L of HNO3 are needed
In mL:
111.95mL of HNO3 are needed to prepare the buffer
LION
If 3.0L of helium at 20°C is allowed to expand to 4.4L, with pressure remain the same
Answer:
This question is asking to find the new temperature
The answer for the final temperature is 429.73K
Explanation:
Using Charles law equation as follows:
V1/T1 = V2/T2
Where;
V1 = initial volume (L)
V2 = final volume (L)
T1 = initial temperature (K)
T2 = final temperature (K)
According to this question;
V1 = 3.0L
V2 = 4.4L
T1 = 20°C = 20 +273 = 293K
T2 = ?
Using V1/T1 = V2/T2
3/293 = 4.4/T2
Cross multiply
293 × 4.4 = 3 × T2
1289.2 = 3T2
T2 = 1289.2 ÷ 3
T2 = 429.73K
Which of the following statements is correct concerning the class of reactions to be expected for benzene and cyclooctatetraene?
A) Both substances undergo addition reactions.
B) Both substances undergo substitution reactions.
C) Benzene undergoes addition; cyclooctatetraene undergoes substitution.
D) Benzene undergoes substitution; cyclooctatetraene undergoes addition.
Answer:
Both substances undergo substitution reactions.
Explanation:
Let us go back to the idea of aromaticity. Aromatic substances are said to possess (4n + 2) π electrons according to Huckel rule.
Aromatic substances are unusually stable and the aromatic ring can not be destroyed by addition reactions.
Since both benzene and cyclooctatetraene are both aromatic, they do not undergo addition reactions whereby the aromatic ring is destroyed. They both undergo substitution reaction in which the aromatic ring is maintained.
By process of incineration, a mystery substance is empirically determined to contain 40.00% carbon by weight, 6.67% hydrogen, and 53.33% oxygen. Its molecular weight ranges between 55 and 62 g/mole. a. (6 points) Determine the chemical formula of this substance
Answer:
C₂H₄O₂
Explanation:
Step 1: Divide each percentage by the atomic mass of the element
C: 40.00/12.01 = 3.331
H: 6.67/1.01 = 6.60
O: 53.33/16.00 = 3.333
Step 2: Divide all the numbers by the smallest one
C: 3.331/3.331 = 1
H: 6.60/3.331 ≈ 2
O: 3.333/3.331 ≈ 1
The empirical formula is CH₂O, with a molecular weight of 12 g/mol + 2 × 1 g/mol + 16 g/mol = 30 g/mol. The molecular weight of the compound must be a product of 30, such as 60 (between 55 and 62 g/mol). Since we have to multiply by 2 (30 to 60) to get to the molecular weight of the compound, we also have to multiply the empirical formula by 2 to get the chemical formula of the compound.
CH₂O × 2 = C₂H₄O₂
Which is the electronic configuration for oxygen?
The solvent for an organic reaction is prepared by mixing 70.0 mL of acetone (C3H6O) with 75.0 mL of ethyl acetate (C4H8O2). This mixture is stored at 25.0 ∘C. The vapor pressure and the densities for the two pure components at 25.0 ∘C are given in the following table. What is the vapor pressure of the stored mixture?
Answer:
The answer is "170.9 mm Hg".
Explanation:
[tex]\text{Mass of acetone = volume} \times density[/tex]
[tex]= 70.0 \times 0.791\\\\ = 55.37\ g\\[/tex]
[tex]\text{Moles of acetone} = \frac{mass}{molar\ mass}\\\\[/tex]
[tex]=\frac{55.37}{58.08}\\\\ = 0.9533\ mol[/tex]
[tex]\text{Mass of ethyl acetate = volume} \times density[/tex]
[tex]= 73.0 \times 0.900\\\\ = 65.7\ g[/tex]
[tex]\text{Moles of ethyl acetate = mass} \times\ molar\ mass[/tex]
[tex]= \frac{65.7}{88.105} \\\\= 0.7457\ mol[/tex]
[tex]\text{Mole fraction of acetone x(acetone)} = \frac{0.9533}{(0.9533 + 0.7457)}\ = 0.5611\\\\[/tex] [tex]\text{Mole fraction of ethyl acetate x(ethyl acetate)} =\frac{0.7457}{(0.9533 + 0.7457) }= 0.4389[/tex]
Applying Raoult's law: [tex]\text{Vapor pressure = x(acetone)P(acetone) + x(ethyl acetate)P(ethyl acetate)}\\\\= 0.5611 \times 230.0 + 0.4389 \times 95.38\\\\ = 170.9\ mm \ Hg\\[/tex]
The solvent for an organic reaction is prepared by mixing 70.0 mL of acetone (C3H6O) with 75.0 mL of ethyl acetate (C4H8O2).
The vapor pressure of the stored mixture is: 170.03 mmHg
In the given information, there is some information that is still missing.
The parameters that we are being given include:
The volume of acetone = 70.0 mLThe volume of ethyl acetate = 75.0 mLThe standard temperature for the mixture = 25° CThe first step we need to take is to determine the mass and number of moles of each compound (i.e. for acetone and ethyl acetate)
For us to do that:
We need the density of acetone and ethyl acetate, which is not given:
Assuming that at a standard condition of vapour pressure:
230 mmHg of acetone has a density of 0.791 g/mL95.38 mmHg of ethyl acetate has a density of 0.900 g/mLThen;
Using the relation:
[tex]\mathbf{Density = \dfrac{Mass}{volume}}[/tex]
Mass of acetone = Density of acetone × volume of acetone
Mass of acetone = 0.791 g/mL × 70.0 mL
Mass of acetone = 55.37 g
Mass of ethyl acetate = Density of ethyl acetate × volume of ethyl acetate
Mass of ethyl acetate = 0.900 g/mL × 75.0 mL
Mass of ethyl acetate = 67.5 g
At standard conditions;
For acetone, molar mass = 58.08 g/molFor ethyl acetate, molar mass = 88.11 g/molNow, using the formula for calculating the numbers of moles which can be expressed as:
[tex]\mathbf{Number \ of \ moles = \dfrac{mass}{molar \ mass}}[/tex]
For acetone:
[tex]\mathbf{Number \ of \ moles = \dfrac{55.37 \ g}{58.08 \ g/mol}}[/tex]
[tex]\mathbf{Number \ of \ moles =0.95334 \ mol}[/tex]
For ethyl acetate:
[tex]\mathbf{Number \ of \ moles = \dfrac{67.5 \ g}{88.11 \ g/mol}}[/tex]
[tex]\mathbf{Number \ of \ moles =0.76609 \ mol}[/tex]
Now, we will determine the mole fraction of each compound.
The mole fraction describes the ratio a certain constituent of a mixture to the total amount of all the constitutent in the mixture.
Using the formula:
[tex]\mathbf{mole \ fraction = \dfrac{n_A}{n_A+n_B+...n_N}}[/tex]
For Acetone:
[tex]\mathbf{mole \ fraction = \dfrac{0.95334}{0.95334+0.76609}}[/tex]
[tex]\mathbf{mole \ fraction =0.5545 }[/tex]
For ethyl acetate:
[tex]\mathbf{mole \ fraction = \dfrac{0.76609}{0.76609+0.95334}}[/tex]
[tex]\mathbf{mole \ fraction =0.4455}[/tex]
Finally, we can compute determine the vapour pressure of the stored mixture using Raoult's Law.
Raoult's Law posits that the constituent of a partial pressure in a mixture of a liquid is proportional to the mole fraction of that constituent in the mixture provided the temperature is constant.
∴ For the stored mixture = Vapor pressure of acetone + vapour pressure of ethyl acetate.
where:
Vapour pressure of the solution = (mole fraction × vapor pressure) of solventFor acetone;
Vapor pressure = 0.5545 × 230 mmHg
Vapour pressure = 127.54 mmHg
For ethyl acetate:
Vapour pressure = 0.4455 × 95.38 mmHg
Vapour pressure ==42.49 mmHg
Thus, the vapor pressure of the stored mixture is
= (127.54 + 42.49 ) mmHg
= 170.03 mmHg
Therefore, we can conclude that the vapour pressure of the stored mixture is 170.03 mmHg
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A solution of the primary standard potassium hydrogen phthalate (KHP), KHC8H4O4 , was prepared by dissolving 0.4877 g of KHP in about 50 mL of water. Titration of the KHP solution with a KOH solution of unknown concentration required 38.91 mL to reach a phenolphthalein end point. What is the concentration of the KOH solution
Answer:
The concentration of KOH is 0.06137 M
Explanation:
Step 1: Data given
Molar mass of KHP = 204.22 g/mol
Mass of KHP = 0.4877 grams
Volume of water = 50 mL
Volume of KOH solution = 38.91 mL
Step 2: The balanced equation
C8H5KO4 + KOH ⇒ C8H4K2O4+ H2O
Step 3: Calculate number of moles of KHP
Moles = Mass / molar mass
Moles KHP = 0.4877 grams / 204.22 g/mol
Moles KHP = 0.002388 moles
Step 4: Calculate moles of KOH
For 1 mol KHP we need 1 mol KOH to produce 1 mol C8H4K2O4 and 1 mol H2O
For 0.002388 moles KHP we need 0.002388 moles KOH
Step 5: Calculate the concentration of KOH
Concentration = moles / volume
Concentration of KOH = 0.002388 moles / 0.03891 L
Concentration of KOH = 0.06137 M
The concentration of KOH is 0.06137 M
A quantity of ideal gas requires 800 kJ to raise the temperature of the gas by 10.0 K when the gas is maintained at constant volume. The same quantity of gas requires 900 kJ to raise the temperature of the gas by 10.0 K when the gas is maintained at constant pressure. What is the adiabatic gas constant of this gas
Answer:
[tex]\gamma=1.125[/tex]
Explanation:
From the question we are told that:
Initial Heat [tex]Q_1=800kJ[/tex]
initial Temperature [tex]T_1=10.0K[/tex]
Final Heat [tex]Q_2=800kJ[/tex]
Final Temperature [tex]T_2=10.0K[/tex]
Generally the equation for Adiabatic constant is mathematically given by
[tex]\gamma=\frac{Cp}{Cv}[/tex]
Since
Equation for Heat [tex]dQ=nCdT[/tex]
Where
[tex]n_1=n_2\\\\T_1=T_2[/tex]
Therefore
[tex]Q_1=Cv\\\\Cv=800[/tex]
And
[tex]Cp=900[/tex]
Therefore
[tex]\gamma=\frac{900}{800}\\\\\gamma=\frac{9}{8}[/tex]
[tex]\gamma=1.125[/tex]
Wet helium gas is placed into a balloon at 24.4 degrees Celsius and a pressure of 765.3 mm Hg. What volume (in L) does the dry gas occupy if the water vapor pressure is 24.3 torr and the mass of dried helium gas in the balloon is 0.498 g
Answer:
Hence the Volume of Gas = 3.04 L.
Explanation:
pressure of dry gas = 765.3 - 24.3 = 741 mmhg
Temperature of gas = 24.4+273.15 = 297.55 k
No of mol of gas = 0.498/4 = 0.1245 mol
R = gas constant = 0.0821 l.atm.k-1.mol-1
From ideal-gas equation
PV = nRT
(741/760) x v = 0.1245 x 0.0821 x 297.55
V = Volume of Gas = 3.04 L
A sample of neon gas occupies 105 L at 27°C under a pressure of
985 torr. What volume would it occupy at standard condition
Answer: Volume occupied by given neon sample at standard condition is 123.84 L.
Explanation:
Given: [tex]V_{1}[/tex] = 105 L, [tex]T_{1} = 27^{o}C = (27 + 273) K = 300 K[/tex], [tex]P_{1}[/tex] = 985 torr
At standard conditions,
[tex]T_{2}[/tex] = 273 K, [tex]P_{2}[/tex] = 760 K, [tex]V_{2}[/tex] = ?
Formula used to calculate the volume is as follows.
[tex]\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}[/tex]
Substitute the values into above formula as follows.
[tex]\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}\\\frac{985 torr \times 105 L}{300 K} = \frac{760 torr \times V_{2}}{273 K}\\V_{2} = \frac{94116.75}{760} L\\= 123.84 L[/tex]
Thus, we can conclude that volume occupied by given neon sample at standard condition is 123.84 L.
Adding more than one equivalent of HCl to pent-1-yne will lead to which product:______.
a. 1,2-dichloro-1-butene.
b. 1,1-dichloropentane.
c. 2,2-dichloropentane.
d. 2,2-dichlorobutane.
Answer:
c. 2,2-dichloropentane.
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to firstly draw the structure of the reactant, pent-1-yne:
[tex]CH\equiv C-CH_2-CH_2-CH_2[/tex]
Now, we infer the halogen is added to the carbon atom with the most carbon atoms next to it, in this case, carbon #2, in order to write the following product:
[tex]CH\equiv C-CH_2-CH_2-CH_2+2HCl\rightarrow CH_3- CCl_2-CH_2-CH_2-CH_2[/tex]
Whose name is 2,2-dichloropentane.
Regards!
GM 2 all ,What is an atom define it .Good Day
Answer:
An atom is the smallest particle of an element that can take part in chemical reaction.
Explanation:
hope it will help u Amri
When 3-methyl-1-pentene is treated with in dichloromethane, the major product is 1-bromo-3-methyl-2-pentene.
a. True
b. False
Answer:
True
Explanation:
When Methyl Pentene is introduced in a chemical reaction with dichloromethane then the major product will be bromomethylpentene. There can be small amount of bromo methyl pentene than the amount of methyl pentene introduced for reaction.
Tech A says that hydrocarbons are a result of complete combustion. Tech B says that a catalytic converter creates a chemical reaction, changing carbon monoxide and hydrocarbons to water and carbon dioxide. Who is correct
Answer:
Neither Tech A nor B is correct
Explanation:
Combustion is a chemical reaction that occurs when a chemical molecule(s) interacts quickly with oxygen and produces heat.
When hydrocarbon undergoes a complete combustion reaction, they produce water and CO2.
Tech B is also incorrect because the main purpose of a catalytic converter is to accelerate and speed up the chemical reaction rates, Hence, they are not involved in chemical reaction formation. Catalytic converters are utilized as a control device in exhaust emission to lessen the effect of toxic gas fumes.
2. Write the chemical equation for the reaction NaOH Sodium Hydroxide AgNO3 Silver Nitrate
Answer:
AgNO3 + NaOH = AgOH + NaNO3.
Explanation:
Balancing Strategies: In this reaction, the products are initially NaNO3 + AgOH. However the AgOH would break down into Ag2O and H2O. This would give us NaNO3 + Ag2O + H2O as our products for the overall reaction.
Balancing Strategies: In this reaction, the products are initially NaNO3 + AgOH. However the AgOH would break down into Ag2O and H2O. This would give us NaNO3 + Ag2O + H2O as our products for the overall reaction.However, the equation balanced here is the initial reaction which produces AgOH and NaNO3.
At 25 oC the solubility of chromium(III) iodate is 2.07 x 10-2 mol/L. Calculate the value of Ksp at this temperature. Give your answer in scientific notation to 2 SIGNIFICANT FIGURES (even though this is strictly incorrect). [a]
Answer:
5.0 × 10⁻⁶
Explanation:
Step 1: Write the balanced equation for the solution of chromium(III) iodate
Cr(IO₃)₃(s) ⇄ Cr³⁺(aq) + 3 IO₃⁻(aq)
Step 2: Calculate the solubility product constant (Ksp)
To relate Ksp and the solubility (S), we will make an ICE chart.
Cr(IO₃)₃(s) ⇄ Cr³⁺(aq) + 3 IO₃⁻(aq)
I 0 0
C +S +3S
E S 3S
The solubility product constant is:
Ksp = [Cr³⁺] × [IO₃⁻]³ = S × (3S)³ = 27 S⁴ = 27 × (2.07 × 10⁻²)⁴ = 5.0 × 10⁻⁶
Low-density polyethylene is formed because _______ polymerization is very unpredictable and difficult to control.
dehydration-condensation
anionic-initiated
radical-initiated
esterification
Answer:
radical-initiated
Explanation:
Radical-initiated polymerization is unpredictable and difficult to control. The reaction proceeds indiscriminately and produces shortened chains, loops, and branches that create holes in the polymer. This reduces its mass to volume ratio.
According to the EPA Lead and Copper Rule (LCR), the action level for Pb in drinking water (the level at which threat to human health requires public notification and action towards mitigation) is 15 ppb. If you were to add enough phosphate to the system
saturated with respect to Pb3(PO4)2(s), would the [Pb2+] be below the action limit?
Answer:
The right answer is "105.17 ppb".
Explanation:
According to the question,
The amount of [tex]Pb^{2+}[/tex] in ppb will be:
= [tex]0.5076\times 10^{-6}\times 207.2\times 106[/tex]
= [tex]105.17 \ ppb[/tex]
Thus, the amount of [tex]Pb^{2+}[/tex] is above action limit.
How many Noble gases we have in Periodic Table???
Answer:
Six
Explanation:
For a different reaction, the plot of the reciprocal of concentration versus time in seconds was linear with a slope of 0.056 M-1 s -1 . If the initial concentration was 2.2 M, calculate the concentration after 100 seconds. Show your work.
Answer:
[tex]C_t=0.165M[/tex]
Explanation:
From the question we are told that:
Slope [tex]K=0.056 M-1 s -1[/tex]
initial Concentration [tex]C_1=2.2M[/tex]
Time [tex]t=100[/tex]
Generally the equation for Raw law is mathematically given by
[tex]\frac{1}{C}_t=kt+\frac{1}{C}_0[/tex]
[tex]\frac{1}{C}_t=0.056*100+\frac{1}{2.2}_0[/tex]
[tex]C_t=0.165M[/tex]
The second-order reaction is the reaction that depends on the reactants of the first or the second-order reaction. The concentration after 100 seconds will be 0.165 M.
What is the specific rate constant?The specific rate constant (k) of the second-order reaction is given in L/mol/s or per M per s. It is the proportionality constant that gives the relation between the concentration and the rate of the reaction.
Given,
Slope (k)= 0.056 per M per s
Initial concentration of the reactant [tex](\rm C_{1})[/tex] = 2.2 M
Time (t) = 100 seconds
The concentration of the reaction after 100 seconds can be given by,
[tex]\rm \dfrac{1}{C_{t}} = kt + \dfrac{1}{C_{1}}[/tex]
Substitute values in the above equation:
[tex]\begin{aligned} \rm \dfrac{1}{C_{t}} &= 0.056 \times 100 + \dfrac{1}{2.02}\\\\&= 0.165 \;\rm M\end{aligned}[/tex]
Therefore, after 100 seconds the concentration is 0.165 M.
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Determine whether each of the examples represents a colligative property or a non-colligative property. boiling point elevation Choose... color Choose... freezing point depression Choose... vapor pressure lowering Choose... density Choose...
Answer:
boiling point elevation - colligative property
color - non-colligative property
freezing point depression - colligative property
vapor pressure lowering - colligative property
density - non-colligative property
Explanation:
A colligative property is a property that depends on the number of particles present in the system.
Freezing point depression, boiling point elevation and vapour pressure lowering are all colligative properties of solutions.
Colour and density do not depend on the number of particles present hence they are not colligative properties.
The boiling point elevation, freezing point depression, and vapor pressure lowering are colligative properties. And color and density are non-colligative properties.
Explanation:
The colligative properties are the properties depending upon the number of particles of solute not on the nature of the solute.Example of colligative properties:Vapor pressure loweringElevation boiling pointDepression in freezing pointOsmotic pressureThe non-colligative properties are the properties depending upon the nature of solute and solvent.Example of non-colligative properties :ViscositySurface tensionDensitySolubilitySo, from this, we can conclude that boiling point elevation, freezing point depression, and vapor pressure lowering are colligative properties. And color and density are non-colligative properties.
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11 Explain how you would obtain solid lead carbonate from a mixture of lead carbonate and sodium chloride
Explanation:
Add water, Na2CO3 dissolves, filter, PbCO3 stays in the paper and dissolved Na2CO3 goes through as the solution. Dry the PbCO3 and you have the dry solid.
OR
Add water to dissolve then filter to obtain PbCo3 as you're residue and Na2Co3 as the filtrate. Dry the insoluble PbCo3 between filter papers and you obtain solid PbCo3
Which statement describes the 3d, 4s, and 4p orbitals of Arsenic (As) based on its electronic configuration and position in the periodic table?
The 3d and 4s orbitals are completely filled, and the 4p orbital is partially filled.
The 3d orbital is completely filled, and the 4s and 4p orbitals are partially filled.
The 3d, 4s, and 4p orbitals are completely filled.
The 3d, 4s, and 4p orbitals are partially filled.
Answer:
The 3d and 4s orbitals are completely filled, and the 4p orbital is partially filled.
Explanation:
The correct answer is: The 3d and 4s orbitals are completely filled, and the 4p orbital is partially filled.
The d orbital contains 10 electrons, the s orbital takes 2 electrons and the p orbital takes six electrons.
The orbital in chemistry is defined as a region in space where there is a high probability of finding an electron. There are s, p, d, f orbitals in chemistry which correspond to sharp, principal, diffuse and fundamental.
The electronic configuration of arsenic is 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p3.
From this electronic configuration, we can see that the 4s and 3d orbitals are half filled while the 4p orbital is half filled.
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What do scientist use to form a hypothesis
Answer:
an if/then statement
Explanation:
Evaluate the validity of the following statement: Spontaneous processes are ones that occur quickly and have a low activation energy. A) False. Spontaneous processes always require an input of energy to overcome the activation energy, but always react quickly. B) False. Spontaneous processes can occur slowly, but always have a low activation energy. C) False. Spontaneous reactions can react slowly and can have a high activation energy. D) False. Spontaneous processes always react slowly and always have a high activation energy. E) True.
Answer:
Spontaneous processes are ones that occur quickly and have a low activation energy. - False -
Spontaneous processes always require an input of energy to overcome the activation energy, but always react quickly - False
Spontaneous processes can occur slowly, but always have a low activation- false
Spontaneous reactions can react slowly and can have a high activation energy - True
Spontaneous processes always react slowly and always have a high activation energy- False
Explanation:
A spontaneous reaction is reaction that proceeds on its own without us having to do a thing at all!
A spontaneous reaction may be fast or slow depending on the activation energy of the reaction. A spontaneous reaction having a high activation energy will be slow. However, if the spontaneous reaction has a low activation energy then it will be fast.
We have to note here that a spontaneous reaction proceeds without a prolonged input energy. Sometimes energy may be supplied to the reaction at the beginning for instance in the case of the combustion of hydrocarbons.
So, spontaneous processes are not necessarily fast. Some of them may have a very high activation energy such as in the rusting of iron hence they are slow.
