Answer:
The answer is A
Overfishing is not sustainable and ultimately depletes a fishery. Therefore, the correct option is option A among all the given options.
What is overfishing?Overfishing is the activity of capturing more fish than their population can replenish. The population continues to diminish until it is completely wiped off. Bycatch is another phrase that is frequently used in conjunction with overfishing.
This refers to fish captured in nets designed for a certain breed. Dolphins, sea turtles, as well as other non-target fish, for example, are frequently caught in nets designed for anchoveta pollock. Trawling is another kind of overfishing. This is the process of scrapping the ocean floor and scooping up everything within reach within a certain region. Overfishing is not sustainable and ultimately depletes a fishery.
Therefore, the correct option is option A.
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g A 25.00 mL sample of 0.0988 M benzoic acid (HC7H5O2, Ka = 6.5 · 10-5) is a monoprotic acid that is titrated with 0.115 M sodium hydroxide. Calculate the pH at the equivalence point.
Answer:
pH at equivalence point is 8.47
Explanation:
Benzoic acid react with NaOH, thus:
HC₇H₅O₂ + NaOH → C₇H₅O₂⁻ + H₂O + Na⁺
You reach equivalence point when moles of the acid = moles of NaOH.
Moles of benzoic acid are:
0.025L ₓ (0.0988mol / L) = 0.00247 moles
To have 0.00247 moles of NaOH in solution and reach equivalence point you need to add:
0.00247 moles NaOH ₓ (1L / 0.115mol) = 0.0215L of NaOH solution.
Total volume is 0.0465L.
There are produced 0.00247 moles of C₇H₅O₂⁻ and its molarity will be:
0.00247 mol C₇H₅O₂⁻ / 0.0465L = 0.0531M C₇H₅O₂⁻
C₇H₅O₂⁻ is in equilibrium with water, thus:
C₇H₅O₂⁻(aq) + H₂O ⇄ HC₇H₅O₂(aq) + OH⁻(aq)
Where Kb = Kw / Ka = 1x10⁻¹⁴ / 6.5x10⁻⁵ = 1.54x10⁻¹⁰ is:
Kb = 1.54x10⁻¹⁰ = [HC₇H₅O₂] [OH⁻] / [C₇H₅O₂⁻]
The concentrations in equilibrium of the species are:
[HC₇H₅O₂] = X
[OH⁻] = X
[C₇H₅O₂⁻] = 0.0531M - X
Where X represents how much C₇H₅O₂⁻ react, X is reaction coordinate
Replacing in Kb expression:
1.54x10⁻¹⁰ = [HC₇H₅O₂] [OH⁻] / [C₇H₅O₂⁻]
1.54x10⁻¹⁰ = [X] [X] / [0.0531 - X]
8.169x10⁻¹² - 1.54x10⁻¹⁰X = X²
8.169x10⁻¹² - 1.54x10⁻¹⁰X - X² = 0
Solving for X:
X = -2.858x10⁻⁶M → False solution, there is no negative concentrations
X = 2.858x10⁻⁶M → Right solution
As [OH⁻] = X
[OH⁻] = 2.858x10⁻⁶M
pOH is -log [OH⁻]
pOH = 5.54
pH = 14 - pOH
pH = 8.46
pH at equivalence point is 8.47Why does the excess of base used in these eliminations favor the E2 over the E1 mechanism for elimination
Answer:
The base is involved in the rate determining step of an E2 reaction mechanism
Explanation:
Let us get back to the basics. Looking at an E1 reaction, the rate determining step is unimolecular, that is;
Rate = k [Carbocation] since the rate determining step is the formation of a carbonation.
For an E2 reaction however, the reaction is bimolecular hence for the rate determining step we can write;
Rate = k[alkyl halide] [base]
The implication of this is that an excess of either the alkyl halide or base will facilitate an E2 reaction.
Hence, when excess base is used, E2 reaction is favoured since the base is involved in its rate determining step. In an E1 reaction, the base is not involved in the rate determining step hence an excess of the base has no effect on an E1 reaction.
The monomer of poly(vinyl chloride) has the formula C2H3Cl. If there are 1,565 repeat units in a single chain of the polymer, what is the molecular mass (in amu) of that chain? Enter your answer in scientific notation.
Answer:
[tex]\large \boxed{9.780 \times 10^{4}\text{ u}}[/tex]
Explanation:
The molecular mass of a monomer unit is:
C₂H₃Cl = 2×12.01 + 3×1.008 + 35.45 = 24.02 + 3.024 + 35.45 = 62.494 u
For 1565 units,
[tex]\text{Molecular mass} = \text{1565 units} \times \dfrac{\text{62.494 u}}{\text{1 unit }} = \mathbf{9.780 \times 10^{4}}\textbf{ u}\\\\\text{The molecular mass of the chain is $\large \boxed{\mathbf{9.780 \times 10^{4}}\textbf{ u}}$}[/tex]
A solution contains A13+ and Co2+. The addition of 0.3731 L of 1.735 M NaOH results in the complete precipitation of the
ions as Al(OH), and Co(OH)2. The total mass of the precipitate is 22.73 g. Find the masses of Al3+ and Co2+ in the solution.
Answer:
The correct answer is mass of Al3+ will be 3.23 grams and the mass of Co2+ will be 8.50 grams.
Explanation:
Based on the given information, 0.3731 L of 1.735 M of NaOH is added in a solution resulting in the precipitation of the ions as Al(OH)₃ and Co(OH)₂. Thus, the moles of NaOH will be molarity × V(L) = 1.735 × 0.3731 L = 0.647 moles.
The mass of the precipitate given is 22.73 grams.
Now let us assume that the mass of Al(OH)₃ will be x grams and the mass of Co(OH)₂ will be (22.73-x) grams
Therefore, the moles of Al(OH)₃ will be x grams/78 g/mol and as 3OH⁻ ions are needed so the moles will be 3x/78 mole.
And, the moles of Co(OH)₂ will be (22.73-x)grams/92.94 g/mol and as 2OH⁻ ions are needed so the moles will be 45.46-2x/92.94 moles.
Now the equation will become,
3x/78 + 45.46-2x/92.94 = 0.647 moles
0.03846 x + 0.489 - 0.02152 x = 0.647
0.01694 x + 0.489 = 0.647
0.01694 x = 0.158
x = 0.158/0.01694
x = 9.327 grams
Hence, the mass of Al(OH)₃ is 9.327 grams, and the mass of Al³⁺ will be,
= 9.327 gm/78 g/mol × 27 g/mol = 3.23 grams
Now the mass of Co(OH)₂ will be, (22.73 - 9.327) grams = 13.403 grams
the mass of Co²⁺ will be,
= 13.403 grams / 92.94 g/mol × 58.94 g/mol = 8.50 grams
Calculate the mass of sodium nitrate( NaNO3), in grams, required to produce 128 g of oxygen, using the following equation: 2NaNO3 --> 2NaNO2 O2
Answer:
680g of NaNO3.
Explanation:
The balanced equation for the reaction is given below:
2NaNO3 —> 2NaNO2 + O2
Next, we shall determine the mass of NaNO3 that decomposed and the mass of O2 produced from the balanced equation. This is illustrated below:
Molar mass of NaNO3 = 23 + 14 + (16x3) = 85g/mol
Mass of NaNO3 from the balanced equation = 2 x 85 = 170g
Molar mass of O2 = 16x2 = 32g/mol
Mass of O2 from the balanced equation = 1 x 32 = 32g
From the balanced equation above,
170g of NaNO3 decomposed to produce 32g of O2.
Now, we can obtain the mass of NaNO3 needed to produce 128g of O2 as shown below:
From the balanced equation above,
170g of NaNO3 decomposed to produce 32g of O2.
Therefore, Xg of NaNO3 will decompose to produce 128g of O2 i.e
Xg of NaNO3 = (170 x 128)/32
Xg of NaNO3 = 680g
Therefore, 680g of NaNO3 are needed to produce 128g of O2.
If 0.98 g of an unknown was dissolved in 10.30 g of solvent and the resulting solution has a molality of 0.45 m, what is the molar mass of the unknown
Answer:
211.4g/mol.
Explanation:
Data obtained from the question includes:
Mass of unknown compound = 0.98g
Mass of solvent = 10.30g
Molality = 0.45 M
Next, we shall determine the number of mole of the unknown compound present in the solution.
This can be obtained as follow:
Molality = mole /kg of solvent
Mole of the unknown compound =.?
Mass of solvent = 10.30g = 10.30/1000 = 0.0103Kg
Molality = 0.45 M
Molality = mole /kg of solvent
0.45 = mole /0.0103
Cross multiply
Mole = 0.45 x 0.0103
Mole = 4.635×10¯³ mole
Therefore the mole of the unknown compound that dissolve in solution is 4.635×10¯³ mole
Now, we can obtain the molar mass of the unknown compound as follow:
Mole of the unknown compound = 4.635×10¯³ mole
Mass of unknown compound = 0.98g
Molar mass of the unknown compound =?
Mole = mass /Molar mass
4.635×10¯³ = 0.98 /Molar mass
Cross multiply
4.635×10¯³ x molar mass = 0.98
Divide both side by 4.635×10¯³
Molar mass = 0.98 / 4.635×10¯³
Molar mass = 211.4g/mol.
Therefore, the molar mass of the unknown compound is 211.4g/mol.
The molecular mass of the unknown has been 211.66 g/mol.
Molality can be defined as the moles of the solute per kg of solvent.
Molality can be expressed as:
Molality = [tex]\rm \dfrac{Mass\;of\;solute\;(g)}{molecular\;mass\;of\;solute}\;\times\;\dfrac{1000}{Mass\;of\;solvent\;(g)}[/tex] ......(i)
The given unknown has been the solute.
The mass of solute = 0.98 g.
The mass of solvent = 10.30 g.
The molality of the solution formed has been = 0.45 m.
Substituting the values in equation (i):
0.45 m = [tex]\rm \dfrac{0.98\;g}{molecular\;mass\;of\;solute}\;\times\;\dfrac{1000}{10.30\;g}[/tex]
0.45 m = [tex]\rm \dfrac{0.98\;g}{molecular\;mass\;of\;solute}\;\times\;97.087[/tex]
[tex]\rm \dfrac{0.98\;g}{molecular\;mass\;of\;solute}[/tex] = [tex]\rm \dfrac{0.45}{97.087}[/tex]
[tex]\rm \dfrac{0.98\;g}{molecular\;mass\;of\;solute}[/tex] = 0.00463
Molecular mass of solute = [tex]\rm \dfrac{0.98}{0.00463}[/tex]
Molecular mass of solute = 211.66 g/mol.
The molecular mass of the unknown has been 211.66 g/mol.
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3A 2B --> 5C If compound A has a molar mass of 159.7 g/mol and compound C has a molar mass of 57.6 g/mole, how many grams of compound C will be produced from 18.24 grams of compound A and excess compound B
Answer:
10.96 grams of compound C will be produced from 18.24 grams of compound A and excess compound B.
Explanation:
3A + 2B ⇒ 5C
By stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction) the following amounts of reagent and products participate in the reaction:
A: 3 molesB: 2 molesC: 5 molesThe excess reagent will be that which is not completely depleted during the reaction.
The amount of product obtained from the reaction will always depend on the amount of limiting reagent in the reaction. Then, being B the excess reagent and therefore A the limiting reagent and knowing that compound A has a molar mass of 159.7 g/mole and compound C has a molar mass of 57.6 g/mole, by stoichiometry the following mass amounts of A and C participate in the reaction:
A: 3 moles* 159.7 g/mole= 479.1 gC: 5 moles* 57.6 g/mole= 288 gThen it is possible to apply the following rule of three: if by stoichiometry of the reaction 479.1 grams of A produce 288 grams of C, 18.24 grams of A, how much mass of C does it produce?
[tex]mass of C=\frac{18.24 grams of A*288 grams of C}{479.1 grams of A}[/tex]
mass of C= 10.96 grams
10.96 grams of compound C will be produced from 18.24 grams of compound A and excess compound B.
A piece of silver with a mass 368 g has a heat capacity of 87.2 J/°C. What is the specific heat of
silver?
A. 0.385 J/g.°C
B. 0.237 J/g.°C
C. 0.184 J/g.°C
D. 1.322 J/g.°C
E. 4.184 J/g.°C
Answer:
B
Explanation:
Heat capacity = mass x specific heat capacity.
(C = mc)
87.2 = 368 x c
= 0.237 J/g.°C
determine the rate of reaction that follows the rate= k[A]^m[B]^n
rate=0.2*3^1*3^2=0.2*3*9=5.4(mol/L)s so the correct answer is C.
what is the polarity of black pepper
Answer:
Polarity in chemistry referred to physical properties of compounds related to solubility, melting and boiling properties.
Polarity of black pepper can be seen when black pepper is sprinkled on water. The balck pepper float on water and get displaced if touched.
It means black pepper is non-polar and have no difference in electronegativity between bonded atoms. Black pepper is so light in weight and non-polar, the surface tension of water keep it floating in the water.
1. What is the net ionic equation for the reaction that occurs when aqueous solutions of AgNO3 and CaCl2 are mixed and a precipitate forms? A. Ca+2(aq) + NO3-(aq) Ca(NO3)2(aq) B. Ag2+(aq) + 2Cl-(aq) AgCl2(s) C. Cl−(aq) + Ag+(aq) ⟶ AgCl(s) D. None of the above because no reaction occurs
Answer:
C. [tex]Cl^−_(_a_q_) + Ag^+_(_a_q_)->AgCl_(_s_)[/tex]
Explanation:
In this question our options are:
A. [tex]Ca^+^2_(_a_q_)+NO_3^-_(_a_q_)->Ca(NO_3)_2_(_a_q_)[/tex]
B. [tex]Ag^2^+_(_a_q_)+2Cl^-_(_a_q_)->AgCl_2_(_s_)[/tex]
C. [tex]Cl^−_(_a_q_) + Ag^+_(_a_q_)->AgCl_(_s_)[/tex]
D. None of the above because no reaction occurs
We have to remember that the ions produced by [tex]AgNO_3[/tex] are:
[tex]Ag^+[/tex] and [tex]NO_3^-[/tex]
And the ions produced by [tex]CaCl_2[/tex] are:
[tex]Ca^+^2[/tex] and [tex]Cl^-[/tex]
Additionally, we will have a double displacement reaction so the compounds produce are:
[tex]AgCl[/tex] and [tex]Ca(NO_3)_2[/tex]
If we remember the solubility rules, all the nitrate salts are soluble and the salts made with silver are not soluble. With this in mind, we will have a solid-state for [tex]AgCl_(_s_)[/tex] and an aqueous state for [tex]Ca(NO_3)_2_(_a_q_)[/tex].
If this is true, the final answer can be B or C. The charge of Ag is +1 so the final answer is C.
I hope it helps!
Consider the following reaction where Kc = 6.50×10-3 at 298 K: 2NOBr(g) 2NO(g) + Br2(g) A reaction mixture was found to contain 9.83×10-2 moles of NOBr(g), 5.44×10-2 moles of NO(g), and 4.13×10-2 moles of Br2(g), in a 1.00 liter container. Is the reaction at equilibrium? If not, what direction must it run in order to reach equilibrium? The reaction quotient, Qc
Answer:
This reaction isn't yet at an equilibrium. It must shift in the direction of the reactant (namely [tex]\rm NOBr\; (g)[/tex]) in order to reach an equilibrium.
For this mixture, the reaction quotient is [tex]Q_c = 0.0126[/tex].
Explanation:
A reversible reaction is at equilibrium if and only if its reaction quotient [tex]Q_c[/tex] is equal to the equilibrium constant [tex]K_c[/tex].
Start by calculating the equilibrium quotient [tex]Q_c[/tex] of this reaction. Given the reaction:
[tex]\rm 2\; NOBr\; (g) \rightleftharpoons 2\; NO\; (g) + Br_2\; (g)[/tex].
Let [tex][\mathrm{NOBr\; (g)}][/tex], [tex][\mathrm{NO\; (g)}][/tex], and [tex][\mathrm{Br_2\; (g)}][/tex] denote the concentration of the three species. The formula for the reaction quotient of this system will be:
[tex]\displaystyle Q_c = \frac{[\mathrm{NO\; (g)}]^2 \cdot [\mathrm{Br_2\; (g)}]}{[\mathrm{NOBr\; (g)}]^2}[/tex].
(Note, that in this formula, both [tex][\mathrm{NO\; (g)}][/tex] and [tex][\mathrm{NOBr\; (g)}][/tex] are raised to a power of two. That corresponds to the coefficients in the balanced reaction.)
Calculate the reaction quotient given the concentration of each species:
[tex]\displaystyle Q_c = \frac{[\mathrm{NO\; (g)}]^2 \cdot [\mathrm{Br_2\; (g)}]}{[\mathrm{NOBr\; (g)}]^2} \approx 1.26\times 10^{-2} = 0.0126[/tex].
(Note that the unit is ignored.)
Apparently, [tex]Q_c > K_c[/tex]. Since [tex]Q_c[/tex] and [tex]K_c[/tex] are not equal, this reaction is not at an equilibrium. If external factors like temperature stays the same,
Keep in mind that [tex]Q_c[/tex] denotes a quotient. To reduce the value of a quotient, one may:
reduce the value of the numerator, increase the value of the denominator, orboth.In [tex]Q_c[/tex], that means reducing the concentration of the products while increasing the concentration of the reactants. In other words, the system needs to shift in the direction of the reactants before it could reach an equilibrium.
7. An element's most stable ion forms an ionic compound with chlorine having the formula XCl2. If the ion of element X has a mass of 89 and 36 electrons, what is the identity of the element, and how many neutrons does it have
Answer:
The element is strontium and the number of neutrons it have is 51.
Explanation:
Based on the given information, the ionic compound is,
XCl₂ ⇔ X₂⁺ + 2Cl⁻
X2+ is the ion of the mentioned element
As mentioned in the given question, the number of electrons of the element X is 36 and as seen from the reaction the charge present on the ion is +2. Now the atomic number will be,
No. of electrons = atomic number - charge
36 = atomic number - 2
Atomic number = 38
Based on the periodic table, the atomic number 38 is for strontium element, and the sign of strontium is Sr. Hence, the element X is Sr.
Now based on the given information, the mass number of the element is 89. Now the no. of neutrons will be,
No. of neutrons = mass number - atomic number
= 89 - 38
= 51 neutrons.
Based on the Valence Shell Electron Pair Repulsion Theory (or VSEPR), molecules will arrange to keep the following as far apart as physically possible
a) mobile electrons
b) valence electron pairs
c) inner shell electrons
d) the electrons closest to the nuclei
Answer:
B. Valence Electron Pairs
Explanation:
Valence-shell electron-pair repulsion, or VSEPR, describes the shape of molecules by determining the repulsion of valence electrons. Therefore, our answer is B.
The IE1, for iodine, is 1009 kJ/mol. Calculate the wavelength (in nm) of electromagnetic energy need to ionize an iodine atom. 118.6 743.2 488.3 1042
Answer:
118.6nm
Explanation:
It is possible to calculate wavelength of any energetic process (As an ionization) using:
E = hc / λ (1)
Where E is Energy, h is Planck constant (6.626x10⁻³⁴Js), c speed of light (3x10⁸ms⁻¹) and λ is wavelength In meters.
As the energy to ionize 1 mole of iodine is 1009kJ, one atom requires:
(1009kJ / mol) ₓ (1mol / 6.022x10²³ atoms) = 1.6755x10⁻²¹kJ / atom. = 1.6755x10⁻¹⁸J
Replacing in (1):
λ = hc / E
λ = 6.626x10⁻³⁴Js*3x10⁸ms⁻¹ / 1.6755x10⁻¹⁸J
λ = 1.186x10⁻⁷m
As 1m = 1x10⁹nm:
1.186x10⁻⁷m ₓ (1x10⁹nm / 1m) =
118.6nmWhat is the systematic name for the given compound? (R)-2-bromobutane (S)-2-bromobutane (R)-2-butyl bromide (S)-2-butyl bromide
Answer:
See explanation below
Explanation:
IUPAC came up with the idea of an unambiguous system of nomenclature for organic compounds. This unambiguous system relates the structure of a compound with its name. Thus, IUPAC has established a worldwide standard for the unambiguous naming of organic compounds. Scientists all over the world can now have a uniform system of nomenclature for compounds in order to facilitate easy communication of scientific information.
The systematic names of the following compounds listed in the question are shown below;
(R)-2- butyl bromide has the systematic name (R)-2-bromobutane
(S)-2-butyl bromide has the systematic name (S)-2-bromobutane
This unified system of nomenclature avoids the confusion created by the use of different trivial names in deferent localities and by various scientific academies. This is a major advantage of the systematic nomenclature.
What is the specific heat of a metal with a mass of 14.0 g, heat of 3.45 kJ and a change in temperature of 3.2 ℃?
Recall that when a reaction is at equilibrium, the forward and reverse reactions occur at the same rate. To illustrate this idea, consider the reaction of A (small, red spheres) and B (large, blue spheres) to form AB.
A+B ⇌ AB
Notice that the reaction never stops. Even after several minutes, there is A and B left unreacted, and the forward and reverse reactions continue to occur. Also note that amounts of each species (i.e., their concentrations) stay the same.
Required:
What is the value of the equilibrium constant for this reaction?
Assume each atom or molecule represents a 1 M concentration of that substance.
Answer:
Equilibrium constant Kc = [x]² / [A - x] [B - x]
Explanation:
The equilibrium constant is defined as the ratio of the concentration of the products to that of the reactants at equilibrium
ie Kc = [products] / [reactants].
The balanced equation of the reaction is given as : A + B ⇄ AB
At the beginning of the reaction,
Initial concentration I = A = 1M
B = 1M
AB = 0M
After a period of time and assuming 'x' to be the concentration of product AB formed, the concentrations become
C = reactant A = [A - x] M
rectant B = [B - x] M
Product AB = [x] [x] M
At equilibrium, the concentrations are,
E = rectant A = [A - x] M
reactant B = [B - x] M
product AB = [x]² M
therefore , the equilibrium constant, Kc = [products]/[reactants]
= [x]² / [A - x] [B - x]
A gas sample is collected in a 0.279 L container at 22.7 °C and 0.764 atm. If the sample has a mass of 0.320 g, what is the identity of the gas? Group of answer choices g
Answer:
HCl
Explanation:
Choices:
CO: 28.01g/mol
NO₂: 46g/mol
CH₄: 16.04g/mol
HCl: 36.4g/mol
CO₂: 44.01g/mol
It is possible to identify a substance finding its molar mass (That is, the ratio between its mass in grams and its moles). It is possible to find the moles of the gas using general ideal gas law:
PV = nRT
Where P is pressure of gas 0.764atm; V its volume, 0.279L; n moles; R gas constant: 0.082atmL/molK and T its absolute temperature, 295.85K (22.7°C + 273.15).
Replacing:
PV = nRT
PV / RT = n
0.764atm*0.279L / 0.082atmL/molKₓ295.85K = n
8.786x10⁻³ = moles of the gas
As the mass of the gas is 0.320g; its molar mass is:
0.320g / 8.786x10⁻³moles = 36.4 g/mol
Based in the group of answer choices, the identity of the gas is:
HCl
Molar mass of substance is the mass of one mole of substance. The gas is identified as HCl as it has 36.4 g/mol of molar mass.
The substance can be identified from its molar mass. The formula to calculate the molar mass is
[tex]\bold {m = \dfrac w{n}}[/tex]
Where,
n - number of moles
w - given mass of the substance
The number of moles can be calculated by the using ideal gas law
PV = nRT
[tex]\bold {n = \dfrac {PV }{ RT }}[/tex]
Where
P - pressure of gas 0.764 atm;
V its volume, 0.279 L;
n moles;
R gas constant: 0.082atmL/molK
T its absolute temperature, 295.85 K (22.7°C + 273.15).
Put the values in the formula,
[tex]\bold {n = \dfrac {0.764\ atm \times , 0.279 }{ 0.082 \times 295.85 }}\\\\\bold {n =8.786x10^-^3}[/tex]
Now, put the values in the molar mass formula,
[tex]\bold {m = \dfrac {0.320g}{ 8.786x10^-^3} = 36.4\ g/mol}[/tex]
Therefore, the gas is identified as HCl as it has 36.4 g/mol of molar mass.
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How much heat is liberated at constant pressure when 1.41 g of potassium metal reacts with 6.52 mL of liquid iodine monochloride (d = 3.24 g/mL)? 2K(s) + ICl(l) → KCl(s) + KI(s)
Answer:
The correct answer is - 13.33 kJ of heat
Explanation:
To know which one is the limiting reagent, determine the number of moles of each reagent in order .
n(K) = mass/atomic weight = 1.41/39 = 0.036 moles
Density of ICl = Mass/Volume
3.24 = Mass/6.52
Mass of ICl = 21.12 g
n(ICl) = mass/molar mass = 21.12/162.35 = 0.130 moles
2 moles of K reacts with 1 mole of ICl
0.036 moles of K will react with = 0.036/2 = 0.018 moles of ICl
since the amount of moles of ICl is more than 0.018, it is in excess and hence K is the limiting reagent. Now, use the balance equation to determine the amount of heat liberated:
2 moles of K gives out -740.71 kJ of heat
1 mole of K will give out = -740.71/2 = 370.36 kJ of heat
0.036 moles of K will give out = 0.036 × 370.36 = 13.33 kJ of heat
Thus, the correct answer is - 13.33 kJ of heat
The amount of heat liberated at constant pressure is -13.33 kJ
The given parameters are:
Mass of potassium metal = 1.41gAmount of liquid iodine monochloride = 6.52 mLStart by calculating the number (n) of moles of each reagent using:
[tex]n = \frac{Mass}{Atomic\ weight }[/tex]
For the potassium metal, we have:
[tex]n_k = \frac{1.41g}{39g/mole}[/tex] ---where 39 is the atomic weight of potassium
[tex]n_k = 0.036\ moles[/tex]
For the liquid iodine monochloride, we start by calculating its mass using:
[tex]Mass = Density \times Volume[/tex]
So, we have:
[tex]Mass = 3.24 \times 6.52[/tex]
[tex]Mass = 21.12g[/tex]
The number of moles is then calculated as:
[tex]n_I=\frac{21.12g}{162.35}[/tex]
[tex]n_I = 0.130\ moles[/tex]
The reaction equation 2K(s) + ICl(l) → KCl(s) + KI(s) means that:
2 moles of potassium reacts with 1 mole of liquid iodine monochloride.
So, 0.036 moles of potassium will react with the following moles of liquid iodine monochloride.
[tex]A = \frac{0.036}{2}[/tex]
[tex]A = 0.018\ moles[/tex]
i.e. 0.036 moles of potassium will react with of liquid iodine monochloride
By comparison: 0.018 moles is less than 0.036 moles
So, the amount of heat liberated at constant pressure is:
[tex]Amount = 0.036 \times -\frac{740.71}{2} kJ[/tex]
[tex]Amount = -13.33 kJ[/tex]
Hence, the amount of heat liberated at constant pressure is -13.33 kJ
Read more about heat at:
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What is the new mass/volume percent (m/v) of a KOH solution that is prepared by diluting 110 mL of a 6.0% (m/v) KOH solution to 330 mL
Answer:
The new mass/volume percent is 2.0% (m/v)
Explanation:
Dilution is a procedure by which the concentration of a solution is decreased, generally with the addition of a diluent. In other words, dilution is a process in which a concentrated solution is always started, to which a greater volume of solvent is added, causing the concentration and volume of the resulting solution to change. But the amount of solute used to prepare the initial solution remains the same.
The calculation of a dilution is made by:
Cinitial. Vinitial = Cfinal. Vfinal
where C indicates concentration and V indicates volume.
In this case:
Cinitial: 6.0% (m/v)Vinitial: 110 mLCfinal: ?Vfinal: 330 mLReplacing:
6.0% (m/v) * 110 mL= Cfinal* 330 mL
Solving:
[tex]Cfinal=\frac{ 6.0 (m/v)*110 mL}{330 mL}[/tex]
Cfinal= 2.0% (m/v)
The new mass/volume percent is 2.0% (m/v)
Calcium carbide, CaC2, reacts with water to form calcium hydroxide and the flammable gas ethyne (acetylene) in the reaction: What mass of ethyne can be produced
Answer:
1 mole of CaC₂ will produce 26g of C₂H₂ or 64.1g of CaC₂ will produce 26g of C₂H₂
Explanation:
Hello,
To solve this question, we'll require a balanced chemical equation of reaction between calcium carbide and water.
Equation of reaction
CaC₂ + 2H₂O → Ca(OH)₂ + C₂H₂
Molar mass of calcium carbide (CaC₂) = 64.1g/mol
Molar mass of water (H₂O) = 18g/mol
Molar mass of calcium hydroxide (Ca(OH)₂) = 74g/mol
Molar mass of ethyne (C₂H₂) = 26g/mol
From the equation of reaction, 1 mole of CaC₂ will produce 1 mole of C₂H₂
1 mole of CaC₂ = mass / molar mass
Mass = 1 × 64.1
Mass = 64.1g
1 mole of C₂H₂ = mass / molar mass
Mass = 1 × 26
Mass = 26g
Therefore, 1 mole of CaC₂ will produce 26g of C₂H₂
Note: this is a hypothetical calculation since we were not given the initial mass of CaC₂ that starts the reaction
A certain radioactive element has a half life of 8694 years. How much of a 8.30 g sample is left after 8323 years
Answer: The amount of sample left after 8323 years is 4.32g
Explanation:
Expression for rate law for first order kinetics is given by:
[tex]t=\frac{2.303}{k}\log\frac{a}{a-x}[/tex]
where,
k = rate constant
t = age of sample
a = let initial amount of the reactant
a - x = amount left after decay process
a) for completion of half life:
Half life is the amount of time taken by a radioactive material to decay to half of its original value.
[tex]t_{\frac{1}{2}}=\frac{0.693}{k}[/tex]
[tex]k=\frac{0.693}{8694years}=7.97\times 10^{-5}years^{-1}[/tex]
b) amount left after 8323 years
[tex]t=\frac{2.303}{7.97\times 10^{-5}}\log\frac{8.30g}{a-x}[/tex]
[tex]8323=\frac{2.303}{7.97\times 10^{-5}}\log\frac{8.30g}{a-x}[/tex]
[tex]0.285=\log\frac{8.30}{a-x}[/tex]
[tex]\frac{8.30}{a-x}=1.92[/tex]
[tex](a-x)=4.32g[/tex]
The amount of sample left after 8323 years is 4.32g
Multiply each half reaction by the correct number, in order to balance charges for the two half reactions. Mg--> mg2++2c Au++e--> Au Please i need the answer quick i couldnt find it anywhere on the web
Answer:
Mg(s) + 2Ag^+(aq) ---->Mg^2+(aq) + 2Ag(s)
Explanation:
The key to balancing redox reaction equations is this; ensure that the number of electrons lost in the oxidation half reaction equation is equal to the number of electrons gained in the reduction half reaction equation. The both half reaction equations can now be combined to give the overall reaction equation.
For the redox reaction under consideration;
Oxidation half equation;
Mg(s) ------> Mg^2+(aq) + 2e
Reduction half equation;
2Ag^+(aq) +2e ----> 2Ag(s)
Overall balanced redox reaction equation;
Mg(s) + 2Ag^+(aq) ---->Mg^2+(aq) + 2Ag(s)
The elements will lose of gain electrons as needed to have an electron configuration that matches a noble gas.
a) true
b) false
Answer:
true
Explanation:
This is true because elements aim to have a full octet of electrons in their outermost (also called valence) shell. Noble gases already have a full valence shell which is why the elements that are not noble gases aim to be like them.
Answer:
a) true
Explanation:
this is the answer coz elements aim to have a full octet of electrons in their outermost (also called valence) shell.
Assuming the same temperature and pressure for each gas, how many milliliters of carbon dioxide are produced from 16 0 mL of CO
2 CO(g) + O2(g) 4, 2 CO2 (g)
Express your answer with the appropriate units.
Answer:
[tex]V_{CO_2}=16.0mL[/tex]
Explanation:
Hello,
In this case, given that the same temperature and pressure is given for all the gases, we can notice that 16.0 mL are related with two moles of carbon monoxide by means of the Avogadro's law which allows us to understand the volume-moles relationship as a directly proportional relationship. In such a way, since in the chemical reaction:
[tex]2CO(g)+O_2(g)\rightarrow 2CO_2(g)[/tex]
We notice two moles of carbon monoxide yield two moles of carbon dioxide, therefore we have the relationship:
[tex]n_{CO}V_{CO}=n_{CO_2}V_{CO_2}[/tex]
Thus, solving for the yielded volume of carbon dioxide we obtain:
[tex]V_{CO_2}=\frac{n_{CO}V_{CO}}{n_{CO_2}} =\frac{2mol*16.0mL}{2mol}\\ \\V_{CO_2}=16.0mL[/tex]
Best regards.
Place ~0.8 g Fe(NO3)3·9 H2O into a 10 mL volumetric flask and dissolve to the volumetric line with 0.1 M HNO3(aq). (Make sure you pre-rinse the volumetric flask with the 0.1 M HNO3(aq) solution before making the solution. Also, 9 waters of hydration (·9H2O) are present in this chemical, and must be included in the formula weight calculation.) Record the exact mass of iron nitrate used and show the calculation for the solution concentration in your observations.
Before lab, create a plan for the dilution of the Fe(NO3)3(aq) solution created above with DI water. This plan should be clearly outlined in your ELN. Create enough 0.0020 M Fe(NO3)3
How do I go about solving for the dilution plan?
Answer:
You can take 1mL of your stock solution in a 100mL volumetric flask and complete to volume.
Explanation:
You need to create a 0.00200M solution of Fe(NO₃)₃. First, you have to obtain the concentration of the first solution you made. That is:
0.8g Fe(NO₃)₃.9H₂O × (1mol / 403.9972g) =
0.0020 moles of Fe(NO₃)₃.9H₂O = Moles of Fe(NO₃)₃
In 10mL = 0.010L:
0.0020 moles of Fe(NO₃)₃ / 0.010L = 0.20M Fe(NO₃)₃
This is the concentration of your stock solution, as you want to obtain a 0.0020M solution, you dilution factor must be:
0.20M / 0.0020M = 100
That means you need to dilute your stock solution 100 times.
You can make this dilution, for example,
taking 1mL of your stock solution in a 100mL volumetric flask completing to volume with the solvent, 0.1M HNO₃(aq).What volume of water is required to dilute 120 cm3 of 10 mol dm–3 sulphuric acid to a concentration of 2 mol dm–3?
Answer:
0.48 dm3 (or 480 cm3)
Explanation:
First find the original no. of moles existing in the sulphuric acid:
no. of moles = volume (in dm3) x concentration
= 120/1000 x 10
= 1.2 mol
Then let the total volume of the diluted acid be v dm3.
Since
Concentration = no. of moles / volume,
so by substituting the given information,
2 = 1.2 / v
v = 0.6 dm3
Hence, the volume of water required
= 0.6 - 120/1000
= 0.48 dm3 (or 480 cm3)
Considering the definition of dilution, 600 cm³ of water is required to dilute 120 cm³ of 10 [tex]\frac{mol}{dm^{3} }[/tex] sulphuric acid to a concentration of 2 [tex]\frac{mol}{dm^{3} }[/tex].
First of all, you have to know that when it is desired to prepare a less concentrated solution from a more concentrated one, it is called dilution.
Dilution is the procedure followed to prepare a less concentrated solution from a more concentrated one and consists of reducing the amount of solute per unit volume of solution. This is accomplished simply by adding more solvent to the solution in the same amount of solute.
In a dilution the amount of solute does not change, but as more solvent is added, the concentration of the solute decreases, as the volume of the solution increases.
A dilution is mathematically expressed as:
Ci×Vi = Cf×Vf
where
Ci: initial concentration Vi: initial volume Cf: final concentration Vf: final volume
In this case, you know:
Ci= 10 [tex]\frac{mol}{dm^{3} }[/tex] Vi= 120 cm³ Cf= 2 [tex]\frac{mol}{dm^{3} }[/tex] Vf= ?Replacing in the definition of dilution:
10[tex]\frac{mol}{dm^{3} }[/tex]× 120 cm³= 2 [tex]\frac{mol}{dm^{3} }[/tex]× Vf
Solving:
Vf= (10[tex]\frac{mol}{dm^{3} }[/tex]× 120 cm³) ÷2 [tex]\frac{mol}{dm^{3} }[/tex]
Vf= 600 cm³
In summary, 600 cm³ of water is required to dilute 120 cm³ of 10 [tex]\frac{mol}{dm^{3} }[/tex] sulphuric acid to a concentration of 2 [tex]\frac{mol}{dm^{3} }[/tex].
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brainly.com/question/20113402?referrer=searchResults brainly.com/question/22762236?referrer=searchResultsWhich of the following types of electromagnetic radiation have higher frequencies than visible light and which have shorter frequencies than visible light?
1. Gamma rays
2. Infrared radiation
3. Ultraviolet liht
4. X-rays
5. Microwaves
6. Radio waves
Answer:
3,4,1 and 6,5,2
Explanation:
In the electromagnetic spectrum the arrangement of the waves in increasing frequencies and decreasing wavelengths are as follows;
Radio waves
Microwaves
Infrared waves
Visible light rays
Ultraviolet rays
X-rays
Gamma rays
(a simple mnemonic is RMIVUXG)
when 6g acetic acid is dissolved in 1000cm3 of solution then how many molecules ionize out of 1000 acetic acid molecules
Answer:
24.8 molecules are ionized from 1000 acetic acid molecules.
Explanation:
Acetic acid, CH₃COOH dissociates in water, thus:
CH₃COOH ⇄ CH₃COO⁻ + H⁺
Ka = 6.3x10⁻⁵ = [CH₃COO⁻] [H⁺] / [CH₃COOH]
That means amount of CH₃COO⁻ (the dissociated form) that are produced is followed by the equilibrium of the weak acid.
The initial molar concentration of acetic acid (Molar mass: 60g/mol) is:
6g ₓ (1mol / 60g) = 0.1 moles acetic acid, in 1000cm³ = 1L.
0.1 moles / L = 0.1M
The 0.1M of acetic acid will dissociate producing X of CH₃COO⁻ and H⁺, thus:
[CH₃COOH] = 0.1M - X
[CH₃COO⁻] = X
[H⁺] = X
Replacing in Ka formula:
6.3x10⁻⁵ = [CH₃COO⁻] [H⁺] / [CH₃COOH]
6.3x10⁻⁵ = [X] [X] / [0.1 - X]
6.3x10⁻⁶ - 6.3x10⁻⁵X = X²
6.3x10⁻⁶ - 6.3x10⁻⁵X - X² = 0
Solving for X
X = - 0.0025 → False solution, there is no negative concentrations.
X = 0.00248M
That means, a 0.1M of acetic acid produce:
[CH₃COO⁻] = X = 0.00248M solution of the ionized form.
In a basis of 1000 molecules:
1000 molecules × (0.00248M / 0.1M) = 24.8
24.8 molecules are ionized from 1000 acetic acid molecules.