Answer:
The mass of NaHCO3 required is 235.22 g
Explanation:
*******
Continuation of Question:
2NaHCO3(s) + H2SO4(aq) → Na2SO4(aq) + 2CO2(g) + 2H2O(l)
You estimate that your acid spill contains about 1.4 mol H2SO4. What mass of NaHCO3 do you need to neutralize the acid?
********\
The question requires us to calculate the mass of NaHCO3 to neutralize the acid.
From the balanced chemical equation;
1 mol of H2SO4 requires 2 mol of NaHCO3
1.4 would require x?
Upon solving for x we have;
x = 1.4 * 2 = 2.8 mol of NaHCO3
The relationship between mass and number of moles is given as;
Mass = Number of moles * Molar mass
Mass = 2.8 mol * 84.007 g/mol
Mass = 235.22 g
A saturated solution was formed when 5.16×10−2 L of argon, at a pressure of 1.0 atm and temperature of 25 ∘C, was dissolved in 1.0 L of water.
Calculate the Henry's law constant for argon. it must be im M/atm
Answer:
The Henry's law constant for argon is [tex]k=2.11*10^{-3}\frac{ M}{atm}[/tex]
Explanation:
Henry's Law indicates that the solubility of a gas in a liquid at a certain temperature is proportional to the partial pressure of the gas on the liquid.
C = k*P
where C is the solubility, P the partial pressure and k is the Henry constant.
So, being the concentration [tex]C=\frac{ngas}{V}[/tex]
where ngas is the number of moles of gas and V is the volume of the solution, you must calculate the number of moles ngas. This is determined by the Ideal Gas Law: P*V=n*R*T where P is the gas pressure, V is the volume that occupies, T is its temperature, R is the ideal gas constant, and n is the number of moles of the gas. So [tex]n=\frac{P*V}{R*T}[/tex]
In this case:
P=PAr= 1 atmV=VAr= 5.16*10⁻² LR=0.082 [tex]\frac{atm*L}{mol*K}[/tex]T=25 °C=298 °KThen:
[tex]n=\frac{1 atm*5.16*10^{-2} L}{0.082 \frac{atm*L}{mol*K} *298K}[/tex]
Solving:
n= 2.11 *10⁻³ moles
So: [tex]C=\frac{ngas}{V}=\frac{2.11*10^{-3} moles}{1 L} =2.11*10^{-3} \frac{moles}{L}= 2.11*10^{-3} M[/tex]
Using Henry's Law and being C=CAr and P =PAr:
2.11*10⁻³ M= k* 1 atm
Solving:
[tex]k=\frac{2.11*10^{-3} M}{1 atm}[/tex]
You get:
[tex]k=2.11*10^{-3}\frac{ M}{atm}[/tex]
The Henry's law constant for argon is [tex]k=2.11*10^{-3}\frac{ M}{atm}[/tex]
The Henry's law constant for argon gas in 1 litre of water is 2.1 × 10⁻³M/atm.
What is Henry's law?Henry's law of gas states that solubility of a gas in any liquid at particular temperature is directly proportional to the partial pressure of the gas.
C∝P
C = kP, where
k = Henry's constant
P = partial pressure of gas
C is the solubility and it is present in the form of concentration and will be calculated as:
C = n/V
n = no. of moles
V = volume
And moles of the gas will be calculated by using the ideal gas equation as:
PV = nRT
n = (1)(5.16×10⁻²) / (0.082)(298) = 2.1 × 10⁻³ moles
And Concentration in liquid will be:
C = 2.1 × 10⁻³mol / 1L = 2.1 × 10⁻³ M
Now we put all these values in the first equation to calculate the value of k as:
k = (2.1 × 10⁻³M) / (1atm) = 2.1 × 10⁻³M/atm
Hence required value of k is 2.1 × 10⁻³M/atm.
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Find the [OH−] of a 0.32 M methylamine (CH3NH2) solution. (The value of Kb for methylamine (CH3NH2) is 4.4×10−4.) Express your answer to two significant figures and include the appropriate units.
Answer:
[tex][OH^-]=0.01165M[/tex]
Explanation:
Hello,
In this case, for the dissociation of methylamine:
[tex]CH_3NH_2(aq)+H_2O(l)\rightleftharpoons CH_3NH_3^+(aq)+OH^-(aq)[/tex]
We can write the basic dissociation constant as:
[tex]Kb=\frac{[CH_3NH_3^+][OH^-]}{[CH_3NH_2]}[/tex]
That in terms of the reaction extent [tex]x[/tex], turns out:
[tex]Kb=\frac{x*x}{[CH_3NH_2]_0-x}[/tex]
[tex]4.4x10^{-4}=\frac{x^2}{0.32M-x}[/tex]
That has the following solution for [tex]x[/tex]:
[tex]x_1=-0.01209M\\x_2=0.01165M[/tex]
Yer 0.01165M is valid only as no negative concentrations are eligible. It means that it is the concentration of hydroxyl ions in the solution:
[tex][OH^-]=0.01165M[/tex]
Best regards.
Draw the Lewis structure for methane (CH4) and ethane (C2H6) in the box below. Then predict which would have the higher boiling point. Finally, explain how you came to that conclusion.
Answer:
Ethane would have a higher boiling point.
Explanation:
In this case, for the lewis structures, we have to keep in mind that all atoms must have 8 electrons (except hydrogen). Additionally, each carbon would have 4 valence electrons, with this in mind, for methane we have to put the hydrogens around the carbon, and with this structure, we will have 8 electrons for the carbon. In ethane, we will have a bond between the carbons, therefore we have to put three hydrogens around each carbon to obtain 8 electrons for each carbon.
Now, the main difference between methane and ethane is an additional carbon. In ethane, we have an additional carbon, therefore due to this additional carbon, we will have more area of interaction for ethane. If we have more area of interaction we have to give more energy to the molecule to convert from liquid to gas, so, the ethane will have a higher boiling point.
I hope it helps!
The Lewis structure shows the valence electrons in a molecule. Ethane will have a higher boiling point than methane.
We can deduce the number of valence electrons in a molecule by drawing the Lewis structure of the molecule. The Lewis structure consists of the symbols of elements in the compound and the valence electrons in the compound.
We know that the higher the molar mass of a compound the greater its boiling point. Looking at the Lewis structures of methane and ethane, we cam see that ethane has a higher molecular mass (more atoms) and consequently a higher boiling point than methane.
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Assume that a nickel weighs exactly 5.038650 g for the sets of weights listed below obtained by a single weighing on the balance below
Answer:
afshkkyfugutuiryfyi
The second-order decomposition of HI has a rate constant of 1.80 · 10-3 M-1s-1. How much HI remains after 27.3 s if the initial concentration of HI is 4.78 M?
Answer: 3.87M of HI remains after 27.3 s
Explanation:
Using the Second order decomposition equation of
1/[H]t =K x t +1/[A]o
Given initial concentration ,[A]o = 4.78M
time, t = 27.3 s
rate of constant , k= 1.80 x 10^-3 M-1s-1
1/[H] t= 1/[A] t= concentration after time, t=?
SOLUTION
1/[A] t =kt +1/[A]o
1/[A] t =(1.80 x 10^-3 (27.3)+1/4.78
0.04914+0.2092=0.2583
1/[A] t =0.2583
[A] t =1/0.2583= 3.87M
Which relationship can be used to aid in the determination of the heat absorbed by bomb calorimeter? 
Answer:
ΔH = [tex]q_{p}[/tex]
Explanation:
In a calorimeter, when there is a complete combustion within the calorimeter, the heat given off in the combustion is used to raise the thermal energy of the water and the calorimeter.
The heat transfer is represented by
[tex]q_{com}[/tex] = [tex]q_{p}[/tex]
where
[tex]q_{p}[/tex] = the internal heat gained by the whole calorimeter mass system, which is the water, as well as the calorimeter itself.
[tex]q_{com}[/tex] = the heat of combustion
Also, we know that the total heat change of the any system is
ΔH = ΔQ + ΔW
where
ΔH = the total heat absorbed by the system
ΔQ = the internal heat absorbed by the system which in this case is [tex]q_{p}[/tex]
ΔW = work done on the system due to a change in volume. Since the volume of the calorimeter system does not change, then ΔW = 0
substituting into the heat change equation
ΔH = [tex]q_{p}[/tex] + 0
==> ΔH = [tex]q_{p}[/tex]
Given a fixed amount of gas in a rigid container (no change in volume), what pressure will the gas exert if the pressure is initially 1.50 atm at 22.0oC, and the temperature is changed to 11.0oC?
A. 301 atm
B. 1.56 atm
C. 0.750 atm
D. 1.44 atm
E. 3.00 atm
Answer:
The pressure the gas will have if the pressure is initially 1.50 atm at 22.0 ° C and the temperature changes at 11.0 ° C is 1.44 atm (option D)
Explanation:
Gay Lussac's law indicates that, as long as the volume of the container containing the gas is constant, as the temperature increases, the gas molecules move more rapidly. Then the number of collisions against the walls increases, that is, the pressure increases. That is, the gas pressure is directly proportional to its temperature.
Gay-Lussac's law can be expressed mathematically as follows:
[tex]\frac{P}{T}=k[/tex]
Where P = pressure, T = temperature, K = Constant
You have a gas that is at a pressure P1 and at a temperature T1. When the temperature varies to a new T2 value, then the pressure will change to P2, and then:
[tex]\frac{P1}{T1}=\frac{P2}{T2}[/tex]
In this case:
P1= 1.50 atmT1= 22 °C= 295 °K (being 0°C= 273 °K)P2= ?T2= 11 °C= 284 KReplacing:
[tex]\frac{1.5 atm}{295 K}=\frac{P2}{284 K}[/tex]
Solving:
[tex]P2= 284 K*\frac{1.5 atm}{295 K}[/tex]
P2=1.44 atm
The pressure the gas will have if the pressure is initially 1.50 atm at 22.0 ° C and the temperature changes at 11.0 ° C is 1.44 atm (option D)
Identify the elements that have the following abbreviated electron configurations.
A) [Ne] 3s23p5.
B) [Ar] 4s23d7.
C) [Xe] 6s1.
Answer:
A) Chlorine (Cl)
B) Cobalt (Co)
C) Caesium (Cs)
Hope this helps.
The abbreviated electron configurations that was given in the question belongs to
Chlorine (Cl)
Cobalt (Co)
Caesium (Cs) respectively.
Electronic configurations can be regarded as the electronic structure, which is the way an electrons is arranged in energy levels towards an atomic nucleus.The electron configurations is very useful when describing the orbitals of an atom in its ground state.To calculate an electron configuration, we can put the periodic table into sections, and this section will represent the atomic orbitals which is the regions that house the electrons. Groups one of the period table and two belongs to s-block, group 3 through 12 belongs to the d-block, while 13 to 18 can be attributed to p-block ,The rows that is found at bottom are the f-blockTherefore, electron configurations explain orbitals of an atom when it is in it's ground state.
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A 400 mL sample of hydrogen gas is collected over water at 20°C and 760 torr the vapor pressure of water at 20°C is 17.5 torr. what volume will the dry hydrogen gas occupy at 20°C and 760 torr?
Answer:
V2 = 17371.43ml
Explanation:
We use Boyles laws
since temperature is constant
P1V1=P2V2
760 x 400 = 17.5 x V2
304000 = 17.5 x V2
V2 = 304000/17.5
V2 = 17371.43ml
The volume will the dry hydrogen gas occupy at the temperature of 20°C and vapor pressure at 760 torrs will be 18 ml.
What is vapor pressure?
The vapor pressure of a liquid is independent of the volume of liquid in the container, whether one liter or thirty liters; both samples will have the same vapor pressure at the same temperature.
The temperature has an exponential connection with vapor pressure, which means that as the temperature rises, the vapor pressure rises as well the equation is -
P1 V1 / T1 = P2 V2 / T1
here, P = pressure
T = temperature
V = volume
substituting the value in the equation,
400 ×760 / 20 = 17.5× V / 20
V = 400× 760 / 20 × 17.5 / 20
V = 18 ml
Therefore the volume of the hydrogen gas remaining at this temperature will be 18 ml.
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If the equilibrium constant of the reaction is 0.85, then which statement is true if the mass of A is 10.5 grams; the density of B is 0.82 g/ml; the concentration of C is 0.64 M; and the concentration of D is 0.38 M.
A(s) + 3 B(l) _____ 2(aq) + D(aq)
Pick the correct statement about this system.
A. Q < K and reaction shifts left
B. Q > K and reaction shifts left
C. Q > K and reaction shifts right
D. Q = K and reaction does not shift
E. Q < K and reaction shifts right
Answer:
E. Q < K and reaction shifts right
Explanation:
Step 1: Write the balanced equation
A(s) + 3 B(l) ⇄ 2(aq) + D(aq)
Step 2: Calculate the reaction quotient (Q)
The reaction quotient, as the equilibrium constant (K), only includes aqueous and gaseous species.
Q = [C]² × [D]
Q = 0.64² × 0.38
Q = 0.15
Step 3: Compare Q with K and determine in which direction will shift the reaction
Since Q < K, the reaction will shift to the right to attain the equilibrium.
2) 2.5 mol of an ideal gas at 20 oC under 20 atm pressure, was expanded up to 5 atm pressure via; (a) adiabatic reversible and (b) adiabatic irreversible process. Calculate the values of w, q, ΔU, ΔH for each process. (Cv = 5 cal / mol.K ≈ 5/2 R; R ≈ 2 cal / mol.K) (Please find the desired values by making the corresponding derivations
Answer:
a) for adiabatic reversible, ΔU(internal energy is constant) = 0, ΔH = 0(no heat is entering or leaving the surrounding)
workdone (w) = -8442.6 J ≈ -8.443 KJ
heat transferred (q) of the ideal gas = - w
q = 8.443 KJ
b) For ideal gas at adiabatic reversible, Internal energy (ΔU) = 0 and enthalpy (ΔH) = 0
the workdone(w) in the ideal gas= - 4567.5 J ≈ - 4.57 KJ
the heat transfer (q) of an ideal gas = 4.5675 KJ
Explanation:
given
mole of an ideal gas(n) = 2.5 mol
Temperature (T) = 20°C
= (20°C + 273) K = 293 K
Initial pressure of the ideal gas(P₁) = 20 atm
Final pressure of the ideal gas(P₂) = 5 atm.
2) (a)for adiabatic reversible process,
note: adiabatic process is a process by which no heat or mass is transferred between the system and its surrounding.
Work done (w) = nRT ln[tex]\frac{P_{1} }{P_{2} }[/tex]
= 2.5 mol × 8.314 J/mol K × 293 K × ln[tex]\frac{5atm}{20atm}[/tex]
= 6090.01 J × [-1.3863]
= -8442.6 J ≈ -8.443 KJ
So, the work done (w) of ideal gas = -8.443 KJ
For ideal gas at adiabatic reversible, Internal energy (U) = 0 and Enthalpy (H) = 0
From first law of thermodynamics:-
U = q + w
0 = q + w
q = - w
q = - (-8.443 KJ)
q = 8.443 KJ
heat transfer (q) of the ideal gas = 8.443 KJ
(b) For adiabatic irreversible, the temperature T remains constant because the internal energy U depends only on temperature T. Since at constant temperature, the entropy is proportional to the volume, therefore, entropy will increase.
Work done (w) = -nRT(1 - ln[tex]\frac{P_{1} }{P_{2} }[/tex] )
= - 2.5 mol × 8.314 J / mol K× 293 K × [1- (5 atm /20 atm)]
= - 6090.01 J × 0.75
= - 4567.5 J ≈ - 4.57 KJ
∴work done(w) of an ideal gas = - 4.57 KJ
For ideal gas at adiabatic Irreversible, Internal energy (U) = 0 and Enthalpy (H) = 0
From first law of thermodynamics:-
U = q + w
0 = q + w
q = - w
q = - (-4.5675 KJ)
q = 4.5675 KJ
the heat transfer (q) of an ideal gas = 4.5675 KJ
Which is most likely to happen during a precipitation reaction?
A. A solid substance will break down into two new substances that
are gases.
B. An insoluble solid will form when ions in dissolved compounds
switch places.
C. A substance will react with oxygen to form water and carbon
dioxide.
D. A gas will form when positive ions switch places to form new
compounds.
Answer:
I think its B
Explanation:
Precipitation reactions leave a solid behind. The solid is called a precipitate.
Answer:
B
Explanation:
An insoluble solid will form when ions in dissolved compounds switch places.
how are mass and weight affected in chemical reactions?
Answer:
How the chemical reacts
Explanation:
Calculate the heat absorbed by a sample of water that has a mass of 45.00 g when the temperature increases from 21.0oC to 38.5 oC. (s=4.184 J/g.o C)
Answer:
The heat absorbed by the sample of water is 3,294.9 J
Explanation:
Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.
The sensible heat of a body is the amount of heat received or transferred by a body when it undergoes a temperature variation (Δt) without there being a change of physical state (solid, liquid or gaseous). Its mathematical expression is:
Q = c * m * ΔT
Where Q is the heat exchanged by a body of mass m, made up of a specific heat substance c and where ΔT is the temperature variation.
In this case:
Q=?m= 45 gc= 4.184 [tex]\frac{J}{g*C}[/tex]ΔT= Tfinal - Tinitial= 38.5 C - 21 C= 17.5 CReplacing:
Q= 4.184 [tex]\frac{J}{g*C}[/tex] * 45 g* 17.5 C
Solving:
Q=3,294.9 J
The heat absorbed by the sample of water is 3,294.9 J
Which of the following solutions would have the highest pH? Assume that they are all 0.10 M in acid at 25°C. The acid is followed by its Ka value.
a. HCHO2, 1.8 x 10-4
b. HF, 3.5 x 10-4
c. HClO2, 1.1 x 10-2
d. HCN, 4.9 x 10-10
e. HNO2, 4.6 x 10-4
Answer:
[tex]HCN~~Ka=4.9x10^-^1^0[/tex]
Explanation:
In this case, we have to remember the relationship between the Ka value and the pH. We can use the general reaction for any acid with his Ka value expression:
[tex]HA~->~H^+~+~A^-[/tex] [tex]Ka=\frac{[H^+][A^-]}{[HA]}[/tex]
In the Ka expression, we have a proportional relationship between Ka and the concentration of [tex]H^+[/tex]. Therefore, if we have a higher Ka value we will have a smaller pH (lets keep in mind that with a higher
So, if we have to find the higher pH value we need to search the smaller Ka value in this case [tex]HCN~~Ka=4.9x10^-^1^0[/tex].
I hope helps!
HCN has the highest pH among all the acids listed in the question.
The Ka is called the acid dissociation constant. It shows the extent to which an acid is ionized in water. The pH shows the hydrogen ion concentration of water. The higher the Ka, the higher the hydrogen ion concentration and the lower the pH.
Hence, HCN has the lowest Ka and the lowest hydrogen ion concentration. Therefore, HCN has the highest pH among all the acids listed in the question.
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Review the reversible reactions given, along with the associated equilibrium constant Kat room temperature. In each case, determine whether the forward or reverse reaction is favored.
CH3COOH → CH3C00^- + H^+
Ka=1.8 x 10^-5
AgCl → Ag^+ + Cl^-
Ksp=1.6 x 10^-10
Al(OH)3 → Al^3+ + 3OH^-
Ksp=3.7 x 10^-15
A+B → C
K=4.9 x 10^3
Answer:
The answers to your questions are given below
Explanation:
The following data were obtained from the question:
CH3COOH → CH3C00^- + H^+
Equilibrium constant, Ka = 1.8 x 10^-5
AgCl → Ag^+ + Cl^-
Equilibrium constant, Ksp = 1.6 x 10^-10
Al(OH)3 → Al^3+ + 3OH^-
Equilibrium constant, Ksp = 3.7 x 10^-15
A+B → C
Equilibrium constant, K = 4.9 x 10^3
When the value of the equilibrium constant is grater than 1, it shows that the concentration of product is higher than that of the reactant and it implies that the forward reaction is favored.
When the value of the equilibrium constant is 1, it shows that the the concentration of the product and reactant are the same. Therefore neither the forward nor the reverse reaction is favored.
When the value of the equilibrium constant is lesser than 1, it shows that the concentration of the reactant is higher than the concentration of the product. Therefore, the reversed reaction is favored.
Now, we shall the question given above as follow:
A. CH3COOH → CH3C00^- + H^+
Equilibrium constant, Ka = 1.8 x 10^-5
Since the value of the equilibrium constant is lesser than 1, it means that the reverse reaction is favored.
B. AgCl → Ag^+ + Cl^-
Equilibrium constant, Ksp = 1.6 x 10^-10
Since the value of the equilibrium constant is lesser than 1, it means that the reverse reaction is favored.
C. Al(OH)3 → Al^3+ + 3OH^-
Equilibrium constant, Ksp = 3.7 x 10^-15
Since the value of the equilibrium constant is lesser than 1, it means that the reverse reaction is favored.
D. A+B → C
Equilibrium constant, K = 4.9 x 10^3
Since the value of the equilibrium constant is greater than 1, it means that the forward reaction is favored.
The reaction conditions are:
A. The reverse reaction is favored.
B. The reverse reaction is favored.
C. The reverse reaction is favored.
D. The forward reaction is favored.
Chemical reaction:
A. [tex]CH_3COOH[/tex] → [tex]CH_3COO^- + H^+[/tex]
Equilibrium constant, Ka = [tex]1.8 * 10^{-5}[/tex]
B. [tex]AgCl[/tex] → [tex]Ag^+ + Cl^-[/tex]
Equilibrium constant, Ksp = [tex]1.6 * 10^{-10}[/tex]
C. [tex]Al(OH)_3[/tex] → [tex]Al^{3+} + 3OH^-[/tex]
Equilibrium constant, Ksp = [tex]3.7 * 10^{-15}[/tex]
D. A+B → C
Equilibrium constant, K = [tex]4.9 * 10^3[/tex]
Conditions for Equilibrium constant:When the value of the equilibrium constant is greater than 1, it shows that the concentration of product is higher than that of the reactant and it implies that the forward reaction is favored.
When the value of the equilibrium constant is 1, it shows that the the concentration of the product and reactant are the same. Therefore neither the forward nor the reverse reaction is favored.
When the value of the equilibrium constant is lesser than 1, it shows that the concentration of the reactant is higher than the concentration of the product. Therefore, the reversed reaction is favored.
Thus, the reactions will be:
A. The reverse reaction is favored.
B. The reverse reaction is favored.
C. The reverse reaction is favored.
D. The forward reaction is favored.
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Determine the volume occupied by 10 mol of helium at
27 ° C and 82 atm
Answer:
3.00 L
Explanation:
PV = nRT
(82 atm × 101325 Pa/atm) V = (10 mol) (8.314 J/mol/K) (27 + 273) K
V = 0.00300 m³
V = 3.00 L
After heating a sample of hydrated CuSO4, the mass of released H2O was found to be 2.0 g. How many moles of H2O were released if the molar mass of H2O is 18.016 g/mol
Answer:
0.1110 mol
Explanation:
Mass = 2g
Molar mass = 18.016 g/mol
moles = ?
These quantities are realted by the following equation;
Moles = Mass / Molar mass
Substituting the values of the quantities and solving for moles, we have;
Moles = 2 / 18.016 = 0.1110 mol
Why was it important to establish the Clean Air Act?
Answer: The Clean Air Act was important because it emphasized cost-effective methods to protect the air; encouraged people to study the effects of dirty air on human health; and created a regulation that makes any activities that pollute the air illegal.
Explanation:
Answer:
Clean Air Act (CAA), U.S. federal law, passed in 1970 and later amended, to prevent air pollution and thereby protect the ozone layer and promote public health. The Clean Air Act (CAA) gave the federal Environmental Protection Agency (EPA) the power it needed to take effective action to fight environmental pollution.
AB2AB2 has a molar solubility of 3.72×10−4 M3.72×10−4 M. What is the value of the solubility product constant for AB2AB2? Express your answer numerically.
Answer:
Ksp = 2.06x10⁻¹⁰
Explanation:
For AB₂. solubility product constant, Ksp, is written as follows:
AB₂(s) ⇄ A²⁺ + 2Br⁻
Ksp = [A²⁺] [Br⁻]²
Molar solubility represents how many moles of AB₂ are soluble per liter of solution. and is obtained from Ksp:
AB₂(s) ⇄ A²⁺ + 2Br⁻
AB₂(s) ⇄ X + 2X
where X are moles that are soluble (Molar solubility)
Ksp = [X] [2X]²
Ksp = 4X³As molar solubility of the salt is 3.72x10⁻⁴M:
Ksp = 4X³
Ksp = 4(3.72x10⁻⁴)³
Ksp = 2.06x10⁻¹⁰Draw a Lewis structure for one important resonance form of HBrO4 (HOBrO3). Include all lone pair electrons in your structure. Do not include formal charges in your structure.
Answer:
The Lewis structure is attached with the answer -
Explanation:
Lewis structure or Lewis dot diagram are diagrams or representation of showing the bonding between different or same atoms of a molecule in any and also shows lone pairs of electrons that may exist in the molecule as dots.
HBrO₄ is bromine oxoacid which is also known as perbromic acid. It is a unstable inorganic compound.
The Lewis structure is attached in form of image with representation of lone pairs of electrons.
Which is a nonpolar molecule?
Answer:
Explanation:
A nonpolar molecule has no separation of charge, so no positive or negative poles are formed. In other words, the electrical charges of nonpolar molecules are evenly distributed across the molecule. Nonpolar molecules tend to dissolve well in nonpolar solvents, which are frequently organic solvents. The answer is hydrogen cyanide.
If the H+ concentration is 0.00001 M, what is the OH- concentration?
Answer:
1.00x10^-9
Explanation:
The reaction, 2 SO3(g) <--> 2 SO2(g) + O2(g) is endothermic. Predict what will happen if the temperature is increased.
Explanation:
This reaction is in equilibrium and would hence obey lechatelier's principle. This principle states that whenever a system at equilibrium undergoes a change, it would react in way so as to annul that change.
Since it is an endothermic reaction, increasing the temperature would cause the reaction to shift towards the right.
This means that it favours product formation and more of the product would be formed.
The accepted value of the number of Liters of gas in a mole is 22.4. List two possible reasons on why our experiment yielded a different value for the number of Liters in a mole of a gas.
Hint: Our experiment was conducted in July, in St. Paul, Minnesota.
Answer:
- Pressure in St. Paul, Minnesota
- Temperature in St. Paul, Minnesota
Explanation:
22.4 L or dm³ is the volume for a gas under Standard pressure and temperature conditions.
It is logically to say, that tempereature value at the day of the experiment was not 273.15 K, which is 32°F
We can say, that the pressure was not 1 atm. St Paul Minnesota has a minimum, but a little height, so the pressure differs by few figures from the standard pressure values.
We also have to mention, that 22.4 L is the value for the Ideal gases at standards conditions. Ideal gases does not exisist on practice, we always talk about real gases. Don't forget the Ideal Gases Law equation:
P . V = n . R . T
Pressure . Volume = number of moles . 0.082 L.atm /mol. K . 273.15K
Number of moles must be 1 at STP, to determine a volume of 22.4L
A student obtained a clean flask. She weighed the flask and stopper on an analytical balance and found the total mass to be 34.232 g. She then filled the flask with water and found the new mass to be 60.167 g. The temperature of the water was measured to be
Answer:
25.99mL is the volume internal volume of the flask
Explanation:
To complete the question:
The temperature of the water was measured to be 21ºC. Use this data to find the internal volume of the stoppered flask
The flask was filled with water, that means the internal volume of the flask is equal to the volume that the water occupies.
To find the volume of the water you need to find the mass and by the use of density of water at 21ºC (0.997992g/mL), you can find the volume of the flask, thus:
Mass water = Mass filled flask - Mass of clean flask
Mass water = 60.167g - 34.232g
Mass water = 25.935g of water.
To convert this mass to volume:
25.935g × (1mL / 0.997992g) =
25.99mL is the volume internal volume of the flaskA sample of N2(g) was collected over water at 25 oC and 730 torr in a container with a volume of 340 mL. The vapor pressure of water at 25 oC is 23.76 torr. What mass of N2 was collected
Answer:
0.36 g of N2.
Explanation:
The following data were obtained from the question:
Temperature (T) = 25 °C
Volume (V) = 340 mL
Measured pressure = 730 torr
Vapour pressure = 23.76 torr
Mass of N2 =..?
First, we shall determine the true pressure of N2. This can be obtained as follow:
Measured pressure = 730 torr
Vapor pressure = 23.76 torr
True pressure =..?
True pressure = measured pressure – vapor pressure
True pressure = 730 – 23.76
True pressure = 706.24 torr.
Converting 706.24 torr to atm, we have:
760 torr = 1 atm
Therefore,
706.24 torr = 706.24 / 760 = 0.929 atm
Next, we shall convert 340 mL to L. This is illustrated below:
1000 mL = 1 L
Therefore,
340 mL = 340/1000 = 0.34 L
Next, we shall convert 25 °C to Kelvin temperature. This is illustrated below:
Temperature (K) = Temperature (°C) + 273
T(K) = T (°C) + 273
T (°C) = 25 °C
T(K) = 25 °C + 273
T (K) = 298 K
Next, we shall determine the number of mole of N2. This can be obtained as follow:
Pressure (P) = 0.929 atm
Volume (V) = 0.34 L
Temperature (T) = 298 K
Gas constant (R) = 0.0821 atm.L/Kmol
Number of mole (n) =...?
PV = nRT
0.929 x 0.34 = n x 0.0821 x 298
Divide both side by 0.0821 x 298
n = (0.929 x 0.34 ) /(0.0821 x 298)
n = 0.0129 mole
Finally, we shall determine the mass of N2 as shown below:
Mole of N2 = 0.0129 mole
Molar mass of N2 = 2x14 = 28 g/mol
Mass of N2 =.?
Mole = mass /Molar mass
0.0129 = mass of N2/ 28
Cross multiply
Mass of N2 = 0.0129 x 28
Mass of N2 = 0.36 g
Therefore, 0.36 g of N2 was collected.
If D+2 would react with E-1, what do you predict to be the formula?
Answer:
DE2
Explanation: for every one D+2 you need two E-1 because +2=-2
If you are given the molarity of a solution, what additional information would you need to find the weight/weight percent (w/w%)?
Answer:
- The molar mass of the solute, in order to convert from moles of solute to grams of solute.
- The density of solution, to convert from volume of solution to mass of solution.
Explanation:
Hello,
In this case, since molarity is mathematically defined as the moles of solute divided by the volume of solution and the weight/weight percent as the mass of solute divided by the mass of solution, we need:
- The molar mass of the solute, in order to convert from moles of solute to grams of solute.
- The density of solution, to convert from volume of solution to mass of solution.
For instance, if a 1-M solution of HCl has a density of 1.125 g/mL, we can compute the w/w% as follows:
[tex]w/w\%=1\frac{molHCl}{L\ sln}*\frac{36.45gHCl}{1molHCl}*\frac{1L\ sln}{1000mL\ sln}*\frac{1mL\ sln}{1.125g\ sln} *100\%\\\\w/w\%=3.15\%[/tex]
Whereas the first factor corresponds to the molar mass of HCl, the second one the conversion from L to mL of solution and the third one the density to express in terms of grams of solution.
Regards.
For the w/w% of the solution, information about the molecular mass of the solute, and density of the solution has been required.
Molarity can be defined as the moles of the solute per liter of the solution. The molarity can be used for the determination of the weight of the solute, by the information about the molecular weight of the compound.
Thus, for the w/w% of the solution, the weight of the solute has been determined with information about the molecular mass of the solute.
The weight of the solvent has been determined with the density of the solution. The density has been defined as the mass per unit volume.
Thus, for the w/w% of the solution, the weight of the solvent has been determined by the density of the solution.
For more information about the w/w% of the solution, refer to the link:
https://brainly.com/question/12369178
How many moles of aqueous magnesium ions and chloride ions are formed when 0.250 mol of magnesium chloride dissolves in water
Answer:
0.250 mol Mg²⁺
0.500 mol Cl⁻
Explanation:
Magnesium chloride (MgCl₂) dissociates into ions according to the following equilibrium:
MgCl₂ ⇒ Mg²⁺ + 2 Cl⁻
1 mol 1 mol 2 mol
1 mol of Mg²⁺ and 2 moles of Cl⁻ are formed per mole of MgCl₂. If we have 0.250 mol of MgCl₂, the following amounts of ions will be formed:
0.250 mol MgCl₂ x 1 mol Mg²⁺/mol MgCl₂= 0.250 mol Mg²⁺
0.250 mol MgCl₂ x 2 mol Cl⁻/mol MgCl₂= 0.500 mol Cl⁻
Answer:
HEY THE ANSWER ABOVE ME IS RIGHT!! i defientely misclicked my rating :/
5/5 all the way.
Explanation: