Answers
Answer:
a.
[tex]Keq=\frac{[HCO_3^-][OH^-]}{[CO_3^{2-}]}[/tex]
b.
[tex]Keq=[O_2]^3[/tex]
c.
[tex]Keq=\frac{[H_3O^+][F^-]}{[HF]}[/tex]
d.
[tex]Keq=\frac{[NH_4^+][OH^-]}{[NH_3]}[/tex]
Explanation:
Hello there!
In this case, for the attached reactions, it turns out possible for us to write the equilibrium expressions by knowing any liquid or solid would be not-included in the equilibrium expression as shown below, with the general form products/reactants:
a.
[tex]Keq=\frac{[HCO_3^-][OH^-]}{[CO_3^{2-}]}[/tex]
b.
[tex]Keq=[O_2]^3[/tex]
c.
[tex]Keq=\frac{[H_3O^+][F^-]}{[HF]}[/tex]
d.
[tex]Keq=\frac{[NH_4^+][OH^-]}{[NH_3]}[/tex]
Regards!
Related Questions
When Hg2+ concentration is 6.35x10^-4 M, the observed cell potential at 298K for an electrochemical cell with the following reaction is 1.499V. What is the Cr3+ concentration?
3H^2+ (aq) + 2Cr(s)= 3Hg(l) + 2Cr^3+(aq)
Answers
Answer:
10.5 × 10^5 M
Explanation:
E°cell = E°cathode - E°anode
E°cell = 0.85 - (-0.74) = 1.59 V
From Nernst's equation;
Ecell = E°cell - 0.0592/n log Q
1.499 = 1.59 - 0.0592/6 log [Cr^3+]/6.35x10^-4
1.499 - 1.59 = - 0.0592/6 log [Cr^3+]/6.35x10^-4
-0.091 = -0.00987 log [Cr^3+]/6.35x10^-4
-0.091/ -0.00987 = log [Cr^3+]/6.35x10^-4
9.22 = log [Cr^3+]/6.35x10^-4
Antilog (9.22) = [Cr^3+]/6.35x10^-4
1.66 × 10^9 = [Cr^3+]/6.35x10^-4
[Cr^3+] = 1.66 × 10^9 × 6.35x10^-4
[Cr^3+] = 10.5 × 10^5 M
Give your familiarity for following terms
1. roasting 2. smelting 4. zone refining 5. polling
Answers
Answer:
The roasting process is a delicate combination of art and science . Roasters are familiar with how the beans look and the smells Well, familiarity with the machine makes things much easier to predict, but the best way is to do many different tests .Well, familiarity with the machine makes things much easier to predict, but the best way is to do many different tests.Smelting is a process of applying heat to ore in order to extract a base metal. It is a form of extractive metallurgy. It is used to extract many metals from their ores, including silver, iron, copper, and other base metals.In zone refining, solutes are segregated at one end of the ingot in order to purify the remainder, or to concentrate the impurities. For example, in the preparation of a transistor or diode semiconductor, an ingot of germanium is first purified by zone refining. In zone refining, solutes are segregated at one end of the ingot in order to purify the remainder, or to concentrate the impurities. ... For example, in the preparation of a transistor or diode semiconductor, an ingot of germanium is first purified by zone refining.Polling is the process where the computer or controlling device waits for an external device to check for its readiness or state, often with low-level hardware. For example, when a printer is connected via a parallel port, the computer waits until the printer has received the next character.Explanation:
hope it heloed
A quantity of 1.435 g of naphthalene , was burned in a constant-volume bomb calorimeter. Consequently, the temperature of the water rose from 20.28oC to 25.95oC If the heat capacity of the bomb plus water was , calculate the heat of combustion of naphthalene on a molar basis; that is, find the molar heat of combustion.
Answers
Answer:
molar heat of combustion = -5156 *10³ kJ/mol
Explanation:
A quantity of 1.435 g of naphthalene , was burned in a constant-volume bomb calorimeter. Consequently, the temperature of the water rose from 20.28oC to 25.95oC If the heat capacity of the bomb plus water was 10.17 kJ/°C, calculate the heat of combustion of naphthalene on a molar basis; that is, find the molar heat of combustion.
Step 1: Data given
Mass of naphthalene = 1.435 grams
Initial temperature of water = 20.28 °C
Final temperature of water = 25.95 °C
heat capacity of the bomb plus water was 10.17 kJ/°C
Molar mass naphtalene = 128.2 g/mol
Step 2:
Qcal = Ccal * ΔT
⇒with Qcal =the heat of combustion
⇒with Ccal = heat capacity of the bomb plus water = 10.17 kJ/°C
⇒with ΔT = the difference in temperature = T2 - T1 = 25.95 - 20.28 = 5.67°C
Qcal = 10.17 kJ/°C * 5.67 °C
Qcal = 57.7 kJ
Step 3: Calculate moles
Moles naphthalene = 1.435 grams / 128.2 g/mol
Moles naphthalene = 0.01119 moles
Step 4: Calculate the molar heat of combustion
molar heat of combustion = Qcal/ moles
molar heat of combustion = -57.7 kJ/ 0.01119 moles
molar heat of combustion = -5156 *10³ kJ/mol
Calculate the mass of Na2S needed if a solution containing 2g of Hg(NO3)2 was added to Na2S solution.
( Hg= 200.59, N= 14, O= 16, Na= 23, S=32)
Answers
Answer:
1.433g of HgS are produced
Explanation:
A Solution Containing 2.0 Grams Of Hg(NO3)2 Was Added To A Solution Containing 2.0 Grams Of Na2S. Calculate the mass of the HgS that was formed (it is a precipitate) according to this reaction:
Based on the reaction:
Na2S + Hg(NO3)2 → HgS + 2NaNO3
To solve this question we need to find the moles of each reactant in order to find the limiting reactant. The moles of limiting reactant = moles of HgS:
Moles Na2S -Molar mass: 78.0452 g/mol-
2.0g * (1mol / 78.0452g) = 0.0256 moles Na2S
Moles Hg(NO3)2 -324.7g/mol-
2.0g * (1mol / 324.7g) = 0.006159 moles Hg(NO3)2
As the reaction is 1:1, and moles of Hg(NO3)2 < moles Na2S
The moles of Hg(NO3)2 = Moles HgS = 0.006159 moles
The mass is:
Mass HgS -Molar mass: 232.66g/mol-:
0.006159 moles * (232.66g/mol) =
1.433g of HgS are produced
Write two balanced half-equations for this redox equation:
2Cu + O2 = 2CuO
Answers
Answer:
2 Cu + O2 → 2 CuO
This is an oxidation-reduction (redox) reaction:
2 Cu0 - 4 e- → 2 CuII
(oxidation)
2 O0 + 4 e- → 2 O-II
(reduction)
Cu is a reducing agent, O2 is an oxidizing agent.
En un recipiente cerrado y rígido se introdujo una mezcla gaseosa a cierta temperatura y las presiones parciales de cada gas son: p(F2) = 2,00 atm, p(BrF) = 1,50 atm y p(BrF3) = 0,0150 atm. A la temperatura que se preparó la mezcla tiene lugar la reacción representada por:
BrF3 (g) ⇌ BrF(g) + F2(g) Kp(T) = 64,0
Elegir la afirmación correcta.
Seleccione una:
Qp > Kp, por lo tanto, las presiones parciales de BrF(g) y F2(g) aumentan hasta alcanzar el equilibrio.
Qp < Kp, por lo tanto, la presión parcial de BrF3(g) disminuye hasta alcanzar el equilibrio.
Qp = Kp, por lo tanto, las presiones parciales de BrF3(g), BrF(g) y F2(g) no cambian.
Qp < Kp, por lo tanto, las presiones parciales de BrF(g) y F2(g) disminuyen hasta alcanzar el equilibrio.
Qp > Kp, por lo tanto, la presión parcial de BrF3(g) aumenta hasta alcanzar el equilibrio.
Answers
Answer:
Qp > Kp, por lo tanto, la presión parcial de BrF₃(g) aumenta hasta alcanzar el equilibrio.
Explanation:
Paso 1: Escribir la ecuación balanceada
BrF₃ (g) ⇌ BrF(g) + F₂(g) Kp(T) = 64,0
Paso 2: Calcular el cociente de reacción (Qp)
Qp = pBrF × pF₂ / pBrF₃
Qp = 1,50 × 2,00 / 0,0150 = 200
Paso 3: Sacar una conclusión
Dado que Qp > Kp, la reacción se desplazará hacia la izquierda para alcanzar el equilibrio, es decir, la presión parcial de BrF₃(g) aumenta hasta alcanzar el equilibrio.
determine the mass in grams of 3.75 x 10^21 atoms of zinc. (the mass of one mole of zinc is 65.39 g)
Answers
Answer: The mass in [tex]3.75 \times 10^{21}[/tex] atoms of zinc is 0.405 g.
Explanation:
Given: Atoms of zinc = [tex]3.75 \times 10^{21}[/tex]
It is known that 1 mole of every substance contains [tex]6.022 \times 10^{23}[/tex] atoms. So, the number of moles in given number of atoms is as follows.
[tex]Moles = \frac{3.75 \times 10^{21}}{6.022 \times 10^{23}}\\= 0.622 \times 10^{-2}\\= 0.0062 mol[/tex]
As moles is the mass of a substance divided by its molar mass. So, mass of zinc (molar mass = 65.39 g/mol) is calculated as follows.
[tex]Moles = \frac{mass}{molar mass}\\0.0062 mol = \frac{mass}{65.39 g}\\mass = 0.405 g[/tex]
Thus, we can conclude that the mass in [tex]3.75 \times 10^{21}[/tex] atoms of zinc is 0.405 g.
Name the following aldehyde PLEASE PLEASE HELP
Answers
Answer:
Explanation:
Answer is D 2,5-dimethylheptanal
You should accern the lowest possible number close to the parent name
D
Question 2
1 pts
How many moles of carbon tetrachloride are present in
18.Og of CC14?
Answers
i think .117018
now to get to the 20charaeactrs minimum.
How many moles of carbon dioxide at Stp will fit in a 50 liter container?
Answers
Answer:
If the gas is at STP, THE 1 mole is 22.4 liters.
Explanation:
9. Which property is not important when selecting a material to use as a light bulb filament?
d
high melting point
high vapor pressure
high ductility
Answers
Answer:
high melting point
Explanation:
The filament of a bulb is often heated to very high temperatures as the bulb is in operation.
Many times, electric bulbs may have to be on for a whole day and they may reach temperatures that are outrageously high in the process.
The material of the filament must have a very high melting point so that it doesn't melt while the bulb is still in operation.
Answer:
Actually the answer is High Vapor pressure
Explanation:
PLEASE HELP QUICKLY
The diagram shows the potential energy changes for a reaction pathway. (10 points)
Part 1: Does the diagram illustrate an endothermic or an exothermic reaction? Give reasons in support of your answer.
Part 2: Describe how you can determine the total change in enthalpy and activation energy from the diagram and if each is positive or negative.
Answers
The activation energy of the reaction is the difference between the highest point on the reaction profile and the energy of the reactants.
What is a potential energy diagram?A potential energy diagram or a reaction profile shows us the energ change between the reactants and the products.
As we look at the reaction profile, we observe that the products have a greater energy than the reactants hence the reaction is endothermic. The enthalpy chamgeis obtained by subtracting the energy of the products from the energy of the reactants.
The activation energy of the reaction is the difference between the highest point on the reaction profile and the energy of the reactants.
Learn more about energy profile: https://brainly.com/question/11256472
From the dropdowns, identify whether the compound contains ionic bonds, covalent bonds, or both. a) CBr4 [ Select ] b) copper(II) sulfate [ Select ] c) N2O3 [ Select ] d) phosphorous trichloride
Answers
Answer:
a) Covalent bonds
b) Covalent and ionic bonds
c) Covalent bonds
d) Covalent bonds
Explanation:
Metals and non-metals form ionic bonds (electrons are transferred), whereas nonmetals and nonmetals form covalent bonds.
Identify whether the compound contains ionic bonds, covalent bonds, or both.
a) CBr₄. C and Br are nonmetals. Thus, they form covalent bonds.
b) copper(II) sulfate. Sulfate contains S and O (nonmetals), which are bonded through covalent bonds. Sulfate is bonded to copper (metal) through an ionic bond.
c) N₂O₃. N and O are nonmetals. Thus, they form covalent bonds.
d) phosphorous trichloride. P and Cl are nonmetals. Thus, they form covalent bonds.
What quantity of sodium azide in grams is required to fill a 56.0 liters air bag with nitrogen gas at 1.00 atm and exactly 0 °C:
2 NaN3 is) 2Na (s) + 3N2 (8)
Answers
Answer:
108.6 g
Explanation:
2NaN₃(s) → 2Na(s) + 3N₂(g)First we use the PV=nRT formula to calculate the number of nitrogen moles:
P = 1.00 atmV = 56.0 Ln = ?R = 0.082 atm·L·mol⁻¹·K⁻¹T = 0 °C ⇒ 0 + 273.2 = 273.2 KInputting the data:
1.00 atm * 56.0 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 273.2 Kn = 2.5 molThen we convert 2.5 moles of N₂ into moles of NaN₃, using the stoichiometric coefficients of the balanced reaction:
2.5 mol N₂ * [tex]\frac{2molNaN_3}{3molN_2}[/tex] = 1.67 mol NaN₃Finally we convert 1.67 moles of NaN₃ into grams, using its molar mass:
1.67 mol * 65 g/mol = 108.6 gThe pKa of an acid can be determined through _____ (reduction, titration, filtration) with a strong base.
Gradually increase the volume of the base, stopping _____ (before, as, after) the equivalence point is reached.
The pKa of the acid is equal to the pH at the _____ (equivalence point, midway volume to the equivalence point, maximum volume).
Answers
Answer:
The pKa of an acid can be determined through titration with a strong base.
Gradually increase the volume of the base, stopping before the equivalence point is reached.
The pKa of the acid is equal to the pH at the midway volume to the equivalence point.
Explanation:
An acid HA dissociates in water as follows:
HA ⇄ H⁺ + A⁻ Ka
So, it produces hydrogen ions (H⁺) and a conjugate base (A⁻). The concentrations of HA, H⁺ and A⁻ at equilibrium determine the constant Ka. The pKa is calculated as:
pKa = -log Ka
The relationship between the pH of the solution and the pKa of the acid is described by the Henderson-Hasselbalch equation:
pH = pKa + log ([A⁻]/[HA])
The pKa can be experimentally determined by acid-base titration, in which a strong base is added to the acid solution. As the base is added, the acid HA is neutralized and the conjugate base A⁻ is formed. Thus, the concentration of the acid ([HA]) increases and the concentration of the conjugate base ([A⁻] decreases. The equivalence point is reached when the total amount of acid is neutralized with the added base. Before reaching the equivalence point, at the halfway point, half of the acid is neutralized and converted into the conjugate base. Thus:
[A-] = [HA] ⇒ log [A-]/[HA] = log 1 = 0 ⇒ pH = pKa
We measure the pH at that point and it is equal to the pKa of the acid.
Amateur radio operators in the United States can transmit on several bands. One of those bands consists of radio waves with a wavelength near 160m. Calculate the frequency of these radio waves on the freauensy of these radio waves.
Answers
Answer:
1.875 × 10^6 Hz
Explanation:
From the wave formula;
V= λf
λ= wavelength of the radio waves
f= frequency of the radio waves
f= v/λ
Since radio waves is an electromagnetic wave, it possesses the speed of light which is 3 × 10^8 m/s
f= 3 × 10^8 m/s/160 m
f= 1.875 × 10^6 Hz
Consider the following chemical equilibrium: Now write an equation below that shows how to calculate from for this reaction at an absolute temperature . You can assume is comfortably above room temperature. If you include any common physical constants in your equation be sure you use their standard symbols, found in the ALEKS Calculator.
Answers
Answer:
Kp=Kc *(RT)+-3
Explanation:
The relation between Kp and Kc is given below:
Where,
Kp is the pressure equilibrium constant
Kc is the molar equilibrium constant
R is gas constant , R = 0.082057 L atm.mol⁻¹K⁻¹
T is the temperature in Kelvins
Δn = (No. of moles of gaseous products)-(No. of moles of gaseous reactants)
For the first equilibrium reaction:
Δn = (0)-(2+1) = -3
Thus, Kp is:
Kp=Kc *(RT)+-3
What is the cell potential of an electrochemical cell that has the half-reactions shown below?
Ag⁺ + e⁻ → Ag
Fe → Fe³⁺ + 3e⁻
Answers
Answer:
E°(Ag⁺/Fe°) = 0.836 volt
Explanation:
3Ag⁺ + 3e⁻ => Ag°; E° = +0.800 volt
Fe° => Fe⁺³ + 3e⁻ ; E° = -0.036 volt
_________________________________
Fe°(s) + 3Ag⁺(aq) => Fe⁺³(aq) + 3Ag°(s) ...
E°(Ag⁺/Fe°) = E°(Ag⁺) - E°(Fe°) = 0.800v - ( -0.036v) = 0.836 volt
What is the largest number
(of a single species) a specific area can support?
A. Population
B. Unlimited factor
C. Carrying capacity
D. Niche
Answers
Answer:
carrying capacity
Explanation:
Thus, the carrying capacity is the maximum number of individuals of a species that an environment can support. Population size decreases above carrying capacity due to a range of factors depending on the species concerned, but can include insufficient space, food supply, or sunlight.
If a student drops 2.3g pieces of magnesium into a flask of hydrochloric acid, this reaction occurs: Mg + 2HCl= MgCl2 + H2
How many liters of hydrogen can be produced at a pressure of 2 atm and a temperature of 298 K
Answers
Answer:
1.2 L
Explanation:
Step 1: Write the balanced equation
Mg + 2 HCl ⇒ MgCl₂ + H₂
Step 2: Calculate the moles corresponding to 2.3 g of Mg
The molar mass of Mg is 24.31 g/mol.
2.3 g × 1 mol/24.31 g = 0.095 mol
Step 3: Calculate the moles of H₂ produced
0.095 mol Mg × 1 mol H₂/1 mol Mg = 0.095 mol H₂
Step 4: Calculate the volume occupied by the hydrogen
We will use the ideal gas equation.
P × V = n × R × T
V = n × R × T/P
V = 0.095 mol × (0.0821 atm.L/mol.K) × 298 K/2 atm = 1.2 L
In lipid bilayers, there is an order-to-disorder transition similar to the melting of a crystal. Comparing a lipid bilayer that is made up mostly of palmitoleic acid to one made up mostly of palmitic acid, which would have a higher melting transition temperature
Answers
Answer:
The lipid bilayer made up of Palmitic acid will have a higher melting transition temperature
Explanation:
The one with a higher melting transition temperature is the lipid layer with a higher melting temperature
Melting temperature of palmitoleic acid = -0.5°C
Melting temperature of palmitic acid = 62.9°C
Hence the lipid bilayer made up of Palmitic acid will have a higher melting transition temperature
What is the formula of the compound Pentasilicon trioxide ?
Answers
Answer: the molecular formula of trioxide is ClOClO3 or Cl2O4
hope its helps you.
keep smiling be happy stay safe
Pls pls help me me pls
Answers
Answer:
Danger
Explanation:
which straight-chain alkane would you predict to be the most viscous? all are liquids exhbiting the general bonding pattern ch3-(ch2)n-ch3
Answers
The question is incomplete, the complete question his;
Which straight chain alkane below would you predict to be the most viscous? Why? All are liquids exhibiting the general bonding pattern CH3-(CH2)n-CH3
C9H20
C10H22
C5H12
C6H14
C12H26
Answer:
C12H26
Explanation:
Generally, the viscosity of a liquid increases with increase in molecular mass of the substance.
Liquids of high molecular mass do not flow easily. This means that they posses high viscosity.
Thus, since C12H26 has the highest molecular mass among the options given in the question, C12H26 exhibits the greatest viscosity.
Aqueous hydrochloric acid will react with solid sodium hydroxide to produce aqueous sodium chloride and liquid water . Suppose 35. g of hydrochloric acid is mixed with 73.0 g of sodium hydroxide. Calculate the minimum mass of hydrochloric acid that could be left over by the chemical reaction. Round your answer to significant digits.
Answers
Answer:
No mass of HCl could be left over by the chemical reaction because is the limting reactant and it is all consumed.
Explanation:
Our reactants are: HCl and NaOH
Products are: NaCl and H₂O
This is a neutralization reaction that can also be called an acid base reaction, an acid and a base react to produce water and a neutral salt, in this case where we have strong acid and base.
Ratio is 1:1. We convert mass to moles:
35 g . 1 mol / 36.45 g = 0.960 moles of HCl
73 g . 1 mol / 40 g = 1.82 moles of NaOH
As ratio is 1:1, for 0.960 moles of HCl we need 0.960 moles of NaOH and for 1.82 moles of NaOH, we need 1.82 moles of acid.
As we only have 0.960 moles of HCl and we need 1.82 moles, no acid remains after the reaction goes complete. HCl is the limiting reactant, so the acid, it is all consumed.
Under certain conditions, the substance mercury(II) oxide can be broken down to form mercury and oxygen. If 32.2 grams of mercury(II) oxide react to form 29.8 grams of mercury, how many grams of oxygen must simultaneously be formed
Answers
Explanation:
This is a decomposition reaction. Firstly, you will want to write the chemical equation out and balance it.
[tex]2Hg_2O->4Hg+O_2[/tex] (The -> is supposed to be an arrow, sorry!)
We see that there's only 1mol of Oxygen made in the products, we can do some simple math to solve for the amount of grams of Oxygen produced according to the amount of the reactant (Hg2O).
[tex]32.2gHg_2O*\frac{1molHg_2O}{417.18gHg_2O}*\frac{1molO_2}{2molHg_2O}*\frac{32gO_2}{1molO_2}[/tex]
I want to break this down, just in case:
The 417.18gHg2O is the molecular mass of the molecule (so I doubled Hg and added 16 to it to get this number).
As we can see in the chemical equation, 1mol Hg2O produces 2mol O because Oxygen is a diatomic molecule (so there will always be two of it when it's by itself).
And finally, in 1mol O2 there are 32g of O2.
** When you do math like this, always make sure that all of your units cancel out except for the units you're looking for. For example, here we're looking for the grams of Oxygen, so after everything else cancels out, we should only have grams O2.
So, 1.23gO2 should be your answer.
Please please help me
Answers
chemistry help!
This mysterious gas has a volume of 4.35 L and a pressure of 1.20 atm. If the pressure is changed to 0.95 atm, what will the new volume be? How do i set up this problem and solve it?
Answers
Refer to the attachment.
Hope this helps you...
Rank the following compounds in order of decreasing boiling point: sodium chloride (NaCl), methane (CH4), and iodomethane (CH3I). Rank from highest to lowest boiling point.
Answers
Answer:
CH4< CH4I< NaCl
Explanation
NaCl has the boiling point of 1,413°C ( 2,575°F )
CH3I has a boiling point of 42°C ( 107°F )
CH4 has the boiling point of -161.6°C ( -258.9°F )
17. The density of a population would influence which limiting factor?
O niche
O growth rate
O weather
O space
Answers
Answer:
The answer is growth rate
Explanation:
it will help you
