Answer:
[tex]Ksp=2.59x10^{-3}[/tex]
Explanation:
Hello,
In this case, given the 0.0990 moles of the salt are soluble in 1.00 L of water only, we can infer that the molar solubility is 0.099 M. Next, since the dissociation of the salt is:
[tex]AB_3\rightleftharpoons A^{3+}+3B^-[/tex]
The concentrations of the A and B ions in the solution are:
[tex][A]=0.099 \frac{molAB_3}{L}*\frac{1molA}{1molAB_3} =0.0099M[/tex]
[tex][B]=0.099 \frac{molAB_3}{L}*\frac{3molB}{1molAB_3} =0.000.297M[/tex]
Then, as the solubility product is defined as:
[tex]Ksp=[A][B]^3[/tex]
Due to the given dissociation, it turns out:
[tex]Ksp=[0.099M][0.297M]^3\\\\Ksp=2.59x10^{-3}[/tex]
Regards.
. 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:
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
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:__________
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
Of the following two gases, which would you predict to diffuse more rapidly? PLZZ HELPP PLZ PLZ PLZ
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:
A solution is known to contain only one type of cation. Addition of Cl1- ion to the solution had no apparent effect, but addition of (SO4)2- ion resulted in a precipitate. Which cation is present
Answer:
We can have: Calcium, strontium, or barium
Explanation:
In this case, we have to remember the solubility rules for sulfate [tex]SO_4~^-^2[/tex] and the chloride [tex]Cl^-[/tex]:
Sulfate
All sulfate salts are SOLUBLE-EXCEPT those also containing: Calcium, silver, mercury (I), strontium, barium or lead.([tex]Ca^+^2~,Ag^+~,Hg_2^+^2~,Sr^+2~,Ba^+^2~,Pb^+^2[/tex]), which are NOT soluble.
Chloride
All chloride salts as SOLUBLE-EXCEPT those also containing: lead, silver, or mercury (I). ([tex]Pb^+^2~,Ag^+~,Hg_2~^+^2[/tex]), which are NOT soluble.
If we the salt formed a precipitated with the sulfate anion, we will have as possibilities "Calcium, silver, mercury (I), strontium, barium or lead". If We dont have any precipitated with the Chloride anion we can discard "Silver, mercury (I), lead" and our possibilities are:
"Calcium, strontium, or barium".
I hope it helps!
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.
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
2. Which one is the odd one
out and why?
o Water
• Hydrogen
Chlorine
o Aluminum
Answer:
Reaction of Chlorine with Hydrogen Chlorine and Hydrogen mixed together explodes when exposed to sunlight, which produces Hydrogen Chloride. In the dark away from sunlight, no reaction occurs, so light energy is required for a reaction. Cl2 + H2 = 2 HCl Reaction of Chlorine with Non-Metals Chlorine directly combines with most non-metals.
Explanation:
I hope this helps bro
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.
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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Zeros laced at the end of the significant number are...
Answer:
Zeros located at the end of significant figures are significant.
Explanation:
Hope it will help :)
While balancing a chemical equation, we change the _____ to balance the number of atoms on each side of the equation.
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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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
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.
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)
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.
whts the ph of po4 9.78
Answer:
4.22
Explanation:
We know from the question, that the pOH of the solution is 9.78. Now the pOH is defined as -log [OH^-].
If the pOH of a solution is given, one may obtain the pH of such solution from the formula;
pH + pOH =14
Hence we can write;
pH = 14-pOH
pH = 14 - 9.78 = 4.22
Hence the pH of the solution is 4.22.
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
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:
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.
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
g Does a reaction occur when aqueous solutions of barium hydroxide and aluminum sulfate are combined
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
At a constant temperature, a sample of a gas in a balloon that originally had a volume of 5.00 L and pressure of 626 torr has its volume changed to 6.72 L. Calculate the new pressure in torr.
Answer:
466 torr
Explanation:
Step 1: Given data
Initial pressure (P₁): 626 torrInitial volume (V₁): 5.00 LFinal pressure (P₂): ?Final volume (V₂): 6.72 LConstant temperatureStep 2: Calculate the final pressure
Since we have a gas changing at a constant temperature, we can calculate the final pressure using Boyle's law.
P₁ × V₁ = P₂ × V₂
P₂ = P₁ × V₁ / V₂
P₂ = 626 torr × 5.00 L / 6.72 L
P₂ = 466 torr
Arrange the compounds in order of decreasing magnitude of lattice energy:
a. LiBr
b. KI
c. CaO.
Rank from largest to smallest.
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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Provide 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______
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 are the correct half reactions for the following reaction: Cu2+ + Mg -> Cu + Mg2+
Answer:
Cu2 + 2Mg-> 2Cu+ Mg2
Explanation:
Balance the equation and make sure both the reactant and the products are the same
Hope it will be helpful
[tex]Cu^{+2} + 2Mg[/tex] -> [tex]2Cu + Mg^+2[/tex] is the correct half-reactions.
What is a balanced equation?A balanced equation is an equation for a chemical reaction in which the number of atoms for each element in the reaction and the total oxidation numbers is the same for both the reactants and the products.
[tex]Cu^{+2} + 2Mg[/tex] -> [tex]2Cu + Mg^+2[/tex] is the correct half-reactions.
Magnesium is oxidized because its oxidation state increased from 0 to +2 while Cu is reduced because its oxidation state decreased from +2 to 0.
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An ice cube at 0.00C with a mass of 8.32g is placed Into 55g of water, initially at 25C. If no heat is lost to the surroundings, what is the final temperature of the entire water sample after all the ice is melted (answer must be in 3 sig figs)
Answer:
The final temperature of the entire water sample after all the ice is melted, is 12,9°C. We should realize that if there is no loss of heat in our system, the sum of lost or gained heat is 0. It is logical to say that the temperature has decreased because the ice gave the water "heat" and cooled it
Thats all i know
Aqueous potassium nitrate (KNO3) and solid silver bromide are formed by the reaction of aqueous potassium bromide and aqueous silver nitrate (AgNO3). Write a balanced chemical equation for this reaction
Answer:
For the mentioned reaction, the balanced chemical equation is:
KBr (aq) + AgNO3 (s) ⇒ KNO3 (aq) + AgBr (s)
The number written in front of the ion, atoms, and molecules in a chemical reaction so that each of the elements on both the sides of reactants and products of the equation gets balanced is known as the stoichiometric coefficient.
From the mentioned balanced equation, the stoichiometric coefficient before KBr is 1, AgNO3 is 1, KNO3 is 1, as well as before AgBr is also 1. Thus, it is clear that 1 mole of potassium bromide reacts with 1 mole of silver nitrate to produce 1 mole of potassium nitrate and 1 mole of silver bromide.
Calculate the molality of a solution containing 141.5 g of glycine (NH2CH2COOH) dissolved in 4.456 kg of H2O
Answer:
0.423 m.
Explanation:
The following data were obtained from the question:
Mass of glycine (NH2CH2COOH) = 141.5 g
Mass of water = 4.456 kg
Molality =.?
Next, we shall determine the number of mole in 141.5 g of glycine (NH2CH2COOH.
This is illustrated below:
Mass of glycine (NH2CH2COOH) = 141.5 g
Molar mass of glycine (NH2CH2COOH) = 14 + (2x1) + 12 + (2x1) + 12 + 16 + 16 + 1 = 75 g/mol
Mole of glycine (NH2CH2COOH) =.?
Mole = mass /Molar mass
Mole of glycine (NH2CH2COOH) = 141.5/75
Mole of glycine (NH2CH2COOH) = 1.887 moles
Finally, we shall determine the molality of the solution as follow:
Molality is simply defined as the mole of solute per kilogram of water. Mathematically it is expressed as:
Molality = mole / mass (kg) of water
With the above formula, we can obtain the molality of the solution as follow:
Mole of glycine (NH2CH2COOH) = 1.887 moles
Mass of water = 4.456 kg
Molality =.?
Molality = mole /mass (kg) of water
Molality =1.887/4.456
Molality = 0.423 m
Therefore, the molality of the solution is 0.423 m
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
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
Determine whether the following statement about equilibrium is true or false.
(a) When a reaction system reaches a state of equilibrium, the concentration of the products is equal to the concentration of the reactants.
(b) When a system is at equilibrium, Keq = 1.
(c) At equilibrium, the rates of the forward reaction and the reverse reaction are equal.
(d) Adding a catalyst to a reaction system will shift the position of equilibrium to the right so there are more products at equilibrium than if there was no catalyst present.
Answer:
(a) when a reaction system reaches a state of equilibrium, the concentration of the products is equal to the concentration of the reactants
Determining whether the statements about equilibrium is True or False
A) The concentration of the products is equal to the concentration of the reactants at equilibrium : TRUE
B) When a system is at equilibrium, Keq = 1 : TRUE
C) The rates of the forward reaction and the reverse reaction are equal at equilibrium : TRUE
D) Adding a catalyst to a reaction system will shift the position of equilibrium to the right : FALSE
Reaction at equilibriumIn a chemical reaction at equilibrium the value of Keq will be equal to 1 because the concentration of the products is equal to the concentration of the reactants in the chemica reaction. Also at equilibrium the rate of forward reaction is same as the rate of reverse reaction.
A catalyst can only affect the rate of reaction and not the amount of product ( yield of reaction).
Hence we can conclude that the answers to your questions are as listed above.
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Indicate the peptides that would result from cleavage by the indicated reagent: a. Gly-Lys-Leu-Ala-Cys-Arg-Ala-Phe by trypsin b. Glu-Ala-Phe-Gly-Ala-Tyr by chymotrypsin
Answer:
a. Gly-Lys + Leu-Ala-Cys-Arg + Ala-Phe
b. Glu-Ala-Phe + Gly-Ala-Tyr
Explanation:
In this case, we have to remember which peptidic bonds can break each protease:
-) Trypsin
It breaks selectively the peptidic bond in the carbonyl group of lysine or arginine.
-) Chymotrypsin
It breaks selectively the peptidic bond in the carbonyl group of phenylalanine, tryptophan, or tyrosine.
With this in mind in "peptide a", the peptidic bonds that would be broken are the ones in the "Lis" and "Arg" (See figure 1).
In "peptide b", the peptidic bond that would be broken is the one in the "Phe" (See figure 2). The second amino acid that can be broken is tyrosine, but this amino acid is placed in the C terminal spot, therefore will not be involved in the hydrolysis.
A sample of argon gas (molar mass 40 g) is at four times the absolute temperature of a sample of hydrogen gas (molar mass 2 g). Find the ratio of the rms speed of the argon molecules to that of the hydrogen. Assume hydrogen molecule has only translational degree of freedom.
Answer:
Ratio of Vrms of argon to Vrms of hydrogen = 0.316 : 1
Explanation:
The root-mean-square speed measures the average speed of particles in a gas, and is given by the following formula:
Vrms = [tex]\sqrt{3RT/M}[/tex]
where R is molar gas constant = 8.3145 J/K.mol, T is temperature in kelvin, M is molar mass of gas in Kg/mol
For argon, M = 40/1000 Kg/mol = 0.04 Kg/mol, T = 4T , R = R
Vrms = √(3 * R *4T)/0.04 = √300RT
For hydrogen; M = 1/1000 Kg/mol = 0.001 Kg/mol, T = T, R = R
Vrms = √(3 * R *T)/0.001 = √3000RT
Ratio of Vrms of argon to that of hydrogen = √300RT / √3000RT = 0.316
Ratio of Vrms of argon to that of hydrogen = 0.316 : 1
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.
Answer:
A. The atomic mass is 243 amu, and the atomic number is 95.
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
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
How does the spontaneity of the process below depend on temperature? PCl5(g)+H2O(g)→POCl3(g)+2HCl(g) ΔH=−126 kJ mol−1, ΔS=146 J K−1mol−
The given question is incomplete, the complete question is:
How does the spontaneity of the process below depend on temperature? PCI5(9)+H2O(g)POCI3(g) +2HCI(g) -126 kJ mol1, AS = 146 J K-'mol1 ΔΗ Select the correct answer below: nonspontaneous at all temperatures spontaneous at all temperatures spontaneous at high temperatures and nonspontaneous at low temperatures spontaneous at low temperatures and nonspontaneous at high temperatures
Answer:
The correct answer is spontaneous at all the temperatures.
Explanation:
Gibbs Free energy is an essential relation that determines the spontaneity of any reaction, that is, ΔG = ΔH - TΔS
When ΔG is less than zero, that is, negative, the reaction is considered to be in spontaneous state. Based on the given information, ΔH = -126 kJ/mol
= -126000 J/mol, it is negative
ΔS = 146 J/K/mol, it is positive
Now, ΔG = ΔH-TΔS
= (-ve) - T (+ve), Thus, when ΔH, is -ve, ΔS is +ve, -TΔS is -ve, the ΔG will be -ve. Therefore, reaction will be spontaneous at all the temperatures.
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
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.
