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.
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.
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.
Calculate the solubility (in mol/L) of Fe(OH)3 (Ksp = 4.0 x 10^-38) in each of the following situations:
(A) Pure water (assume that the pH is 7.0 and remains constant).
(B) A solution buffered at pH = 5.0.
(C) A solution buffered at pH = 11.0.
Answer:
(A) 1.962x10^-10 M solubility in pure water
(B) 4.0 x 10^-33 M solubility
(C) 4.0 x 10^-27 M solubility
Explanation:
(A) Fe(OH)3 would give (Fe3+) and (3OH-)
Ksp = [Fe^3+][OH-]^3 = 4.0 x 10^-38
Let y = [Fe^3+]
Let 3y = [OH-]
4x10^-38 = (y)(3y)^3
4x10^-38 = 27y^4
y^4 = 4x10^-38 ÷ 27
y^4 = 1.481 x 10^-39
y = 1.962x10^-10 M solubility in pure water
(B) pH = 5.0
5.0 = - log [OH-]
-5.0 = log [OH-]
[OH-] = 10^-5.0 = 1.0 x 10^-5 M
So, Ksp = [Fe^3+][OH-]^3 = 4.0 x 10^-38
[Fe^3+][1.0 x 10^-5] = 4.0 x 10^-38
[Fe^3+] = 4.0 x 10^-38 ÷ 1.0 x 10^-5
= 4.0 x 10^-33 M solubility
(C) pH = 11.0
11.0 = - log [OH-]
-11.0 = log [OH-]
[OH-] = 10^-11.0 = 1.0 x 10^-11 M
So, Ksp = [Fe^3+][OH-]^3 = 4.0 x 10^-38
[Fe^3+][1.0 x 10^-11] = 4.0 x 10^-38
[Fe^3+] = 4.0 x 10^-38 ÷ 1.0 x 10^-11
= 4.0 x 10^-27 M solubility
A buffer is prepared containing 0.75 M NH3 and 0.20 M NH4 . Calculate the pH of the buffer using the Kb for NH3. g
Answer:
pH=8.676
Explanation:
Given:
0.75 M [tex]NH_{3}[/tex]
0.20 M [tex]NH_{4}[/tex]
The objective is to calculate the pH of the buffer using the kb for [tex]NH_3[/tex]
Formula used:
[tex]pOH=pka+log\frac{[salt]}{[base]}\\[/tex]
pH=14-pOH
Solution:
On substituting salt=0.75 and base=0.20 in the formula
[tex]pOH=-log(1.77*10^{-5})+log\frac{0.75}{0.20}\\ =4.75+0.5740\\ =5.324[/tex]
pH=14-pOH
On substituting the pOH value in the above expression,
pH=14-5.324
Therefore,
pH=8.676
Given 0.60 mol CO2, 0.30 mol CO, and 0.10 mol H20, what is the partial pressure of the CO if the total pressure of the mixture was 0.80 atm?
Answer:
Explanation:
/ means divided by
* means multiply
1. formula is
partial pressure = no of moles(gas 1)/ no of moles(total)
0.30 mol CO/0.60 mol CO2 + 0.30 mol CO + 0.10 mol H20 ->
.3/(.6+.3+.1) =
.3/1 =
.3 =
partial pressure of CO
2.
.3 * .8 atm = .24
khanacademy
quizlet
The partial pressure of the CO is 0.24 atm if the total pressure of the mixture was 0.80 atm.
Dalton's Law of Partial pressureDalton's Law of partial pressure states that the total pressure exerted by non reacting gaseous mixture at a constant temperature and given volume is equal to the sum of partial pressure of all gases.
Dalton's Law of partial pressure using mole fraction of gas
Partial pressure of carbon monoxide (CO) = Mole fraction of carbon monoxide (CO) × Total pressure
Now, we have to find the first mole fraction of CO
Mole fraction of carbon monoxide (CO) = [tex]\frac{\text{moles of solute}}{\text{total moles of solute}}[/tex]
= [tex]\frac{\text{moles of CO}}{\text{moles of CO}_2 + \text{moles of CO} + \text{moles of H}_{2}O}[/tex]
= [tex]\frac{0.30}{0.60 + 0.30 + 0.10}[/tex]
= [tex]\frac{0.30}{1}[/tex]
= 0.3
Now, put the value in above equation, we get that
Partial pressure of carbon monoxide (CO)
= Mole fraction of carbon monoxide (CO) × Total pressure
= 0.3 × 0.8
= 0.24 atm
Thus, the partial pressure of the CO is 0.24 atm is the total pressure of the mixture was 0.80 atm.
Learn more about the Dalton's Law of partial Pressure here: https://brainly.com/question/14119417
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Suppose a 48. L reaction vessel is filled with 1.6 mol of Br2 and 1.6 mol of OCl2. What can you say about the composition of the mixture in the vessel at equilibrium
Answer:
There will be very little of BrOCl BrCl
Explanation:
Based on the equilibrium:
Br2(g) + OCl2(g) ⇄ BrOCl(g) + BrCl(g)
The equilibrium constant, Kc, is:
Kc = 1.58x10⁻⁵ = [BrOCl] [BrCl] / [Br2] [OCl2]
As Kc is <<< 1, in equilibrium, the concentration of products will remain lower regard to the concentration of the reactants. That means, right answer is;
There will be very little of BrOCl BrClAmmonium phosphate is an important ingredient in many solid fertilizers. It can be made by reacting aqueous phosphoric acid with liquid ammonia. Calculate the moles of phosphoric acid needed to produce 1.80 mol of ammonium phosphate. Be sure your answer has a unit symbol, if necessary, and round it to the correct number of significant digits.
Answer:
Explanation:
The reaction is
H3Po4+3NH3\to→ (NH4)3PO4
Given,7.10g NH3=7.10g/molar mass of NH3
=7.10g/(17.031g/mol)
=0.416mol
From the reaction
3 mol ammonia reacted and produced 1 mole of ammoniam phosphate
So,One mole ammonia reacted and produced 1/ 3 mole ammonium phosphate.
And Also,0.416 mole ammonium reacted and produced (1/3)0.416=0.138 mole ammonium phosphate.
Hence 0.138mole=0.138mole*149.08 g/mole
=20.573gm ammonium phosphate produced.
Hence 20.573g of ammonium phosphate is produced by the reaction of 7.10 g of ammonia.
A sample of calcium fluoride was decomposed into the constituent elements. Write a balanced chemical equation for the decomposition reaction. If the sample produced 294 mg of calcium, how many g of fluorine were formed
Answer:
A sample of calcium fluoride was decomposed into the constituent elements. Write a balanced chemical equation for the decomposition reaction. If the sample produced 294 mg of calcium, how many g of fluorine was formed
Explanation:
The balanced chemical equation for the decomposition of calcium fluoride is shown below:
[tex]CaF_2(s)->Ca(s)+F_2(g)[/tex]
The sample produced 294 g of calcium then, how many grams of fluorine is formed?
From the balanced chemical equation,
1 mol of CaF2 forms 1mol of calcium and 1 mol of fluorine.
That is:
40g of calcium and 38.0 g of fluorine are formed.
then,
If 294 g of calcium is formed then how many grams of fluorine is formed?
[tex]294g Ca * 38g F2 / 40g Ca\\=279.3 g F_2[/tex]
Hence, 279.3 g of fluorine will be formed.
A ball is thrown straight up into the air with a speed of 21 m/s. If the ball has a mass of 0.1 kg, how high does the ball go? Acceleration due to gravity is g = 9.8 m/s^2
Answer:
answer = 22.5m
Explanation:
using
[tex]s = \frac{ {v}^{2} - {4}^{2} }{2a} [/tex]
s= (0²-21²)/2(-9.8)
s= -441/19.6
s= 22.5m
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Which equation expresses the solubility product of Zn3(PO4)2?
a. Ksp = [Zn2+][PO43]
b. Ksp = [Zn2+]3 [PO43]2
c. Ksp = 6[Zn2+][PO43]2
d. Ksp = 108[Zn2+][PO43]2
⇒b. Ksp = [Zn2+]3 [PO43]2
Hope It Helps You ✌️
3. A 3.455-g sample of a mixture was analyzed for barium ion by adding a small excess of sulfuric acid to an aqueous solution of the sample. The resultant reaction produced a precipitate of barium sulfate, which was collected by filtration, washed, dried, and weighed. If 0.2815 g of barium sulfate was obtained, what was the mass percentage of barium in the sample
Answer:
[tex]Ba\ percentage\ in\ Mass=4.8\%[/tex]
Explanation:
From the question we are told that:
Mass of mixture [tex]m=3.455g[/tex]
Mass of Barium [tex]m_b=0.2815g[/tex]
Equation of Reaction is given as
[tex]Ba2+ + H2SO4 => BaSO4 + 2 H+[/tex]
Generally the equation for Moles of Barium is mathematically given by
Since
[tex]Moles of Ba^{2+} = Moles of BaSO_4[/tex]
Therefore
[tex]Moles of Ba^{2+} = \frac{mass}{molar mass of BaSO4}[/tex]
[tex]Moles of Ba^{2+} = \frac{0.2815}{233.39}= 0.0012061 mol[/tex]
Generally the equation for Mass of Barium is mathematically given by
[tex]Mass\ of\ Ba^{2+} = Moles * Molar mass of Ba^{2+}[/tex]
[tex]Mass\ of\ Ba^{2+} = 0.0012061 * 137.33 = 0.1656 g[/tex]
Therefore
[tex]Ba\ percentage\ in\ Mass = mass of Ba^{2+}/mass of sample * 100%[/tex]
[tex]Ba\ percentage\ in\ Mass= \frac{0.1656}{ 3.455 }* 100%[/tex]
[tex]Ba\ percentage\ in\ Mass=4.8\%[/tex]
Consider the balanced chemical equation below.
3 A ⟶ C + 4 D
How many moles of D would be produced if 9 moles of A were used?
hope the picture above helps :)
What is the law of multiple proportions?
A. The proportion of elements to compounds is constant.
B. All elements are found in equal proportions in nature.
C. Different compounds may contain the same elements but may have different ratios of those elements.
D. All compounds contain the same elements in the same proportions.
Answer:
I think
(d) All compounds contain the same elements in the same properties
Calculate the mass percent of carbon in the following molecule: Mn3[Mn(CO)4]3
Answer:
21.63 %
Explanation:
The molar mass of Mn₃[Mn(CO)₄]₃ is 665.64 g/mol.
Let's assume we have 1 mol of Mn₃[Mn(CO)₄]₃, if that were the case then we would have 665.64 grams.
There are 12 C moles per Mn₃[Mn(CO)₄]₃, with that in mind we calculate the weight of 12 C moles:
12 mol C * 12 g/mol = 144 gFinally we calculate the mass percent of carbon:
144 g / 665.64 g * 100% = 21.63 %Which procedure could a student use to examine an intensive property of a rectangular block of wood
Find the mass.
Record the length. Measure the volume. Determine the density.
Answer:
density
Explanation:
The procedure that the student could use to examine an intensive property of a rectangular block of wood is to determine its density. Density is intensive because it is the ration between the mass and the volume.
Answer: find the mass option A
Explanation:
A Bronsted-Lowry acid is defined as a substance that ________. A Bronsted-Lowry acid is defined as a substance that ________. increases Ka when placed in H2O increases [OH-] when placed in H2O acts as a proton donor acts as a proton acceptor decreases [H ] when placed in H2O
Answer: A Bronsted-Lowry acid is defined as a substance that acts as a proton donor.
Explanation:
A substance that is able to donate a proton or hydrogen ion to another substance is a Bronsted-Lowry acid.
For example, HCl is a Bronsted-Lowry acid as it dissociates to give a hydrogen ion.
[tex]HCl \rightleftharpoons H^{+} + Cl^{-}[/tex]
Thus, we can conclude that a Bronsted-Lowry acid is defined as a substance that acts as a proton donor.
a laser emits light with a frequency of 4.69 x 10 to the 14th power s - 1 calculate the wavelength of this light.
Answer:
6.40x10^-7
Explanation:
answer with work is attached.
Which of the following elements is the largest in size
a. O
b. He
c. K
d. H
Answer:
C. K
i took this class before
Answer: The largest element is K
Explanation: As K has the largest radius among O,He and H
Suppose you perform a titration of an unknown weak acid solution. You start with 4.00 mL of the weak acid and find that it takes 14.2 mL of 0.0500 M NaOH to reach the equivalence point. What is the concentration of the unknown weak acid solution
Answer:
0.1775 M
Explanation:
The reaction that takes place is:
HA + NaOH → NaA + H₂OWhere HA is the unknown weak acid.
At the equivalence point all HA moles are converted by NaOH. First we calculate how many NaOH moles reacted, using the given concentration and volume:
0.0500 M NaOH * 14.2 mL = 0.71 mmol NaOHThat means that in 4.00 mL of the weak acid solution, there were 0.71 weak acid mmoles. With that in mind we can now calculate the concentration:
0.71 mmol HA / 4.00 mL = 0.1775 Mgive one important characteristic of the particles of matter very sort answer give
Answer:
The important characteristics of the particles of matter are – They are very, very small. They have spaces between them. They are constantly moving.
Answer:
They are very very small
Which of the following two periods contain the same number of element?
A. 2 & 3
B.3 & 4
c. 4 & 6
D. 2 & 4
Answer:
4 and 6
Explanation:
Period 4 has 18 elements and so does period 6.
In the titration of NaOH with vinegar, a student overshot the endpoint (i.e. added too much NaOH solution). How will this affect the calculate percent acetic acid in the vinegar
Answer:
Overshooting the endpoint leads to a percent acetic acid in vinegar higher than the correct value.
Explanation:
When too much NaOH solution is added, the resulting number of equivalent hydronium (OH⁻) will be higher than what it is in reality. This would directly lead to the number of acetic acid moles in the vinegar being found higher than normal.
In other words, overshooting the endpoint leads to a percent acetic acid in vinegar higher than the correct value.Pls pls pls help me
Predict the products from reaction of 2-hexyne with the following reagents: (a) 2 equiv Br2 (b) 1 equiv HBr (c) Excess HBr (d) Li in NH3 (e) H2O, H2SO4, HgSO4
Answer:
See explanation and image attached
Explanation:
The reactions of the alkynes involved are shown in the image attached to this answer.
First of all, the reaction of two equivalents of bromine with the alkyne converts it to a saturated compound as shown. One equivalent of HBr converts the alkyne to alkene while excess HBr completely reduces the compound to a saturated compound.
Li/NH3 reduces the alkyne to an alkeneby anti addition to the triple bond.
Reaction of the alkyne with H2O, H2SO4, HgSO4 converts it to an aldehyde as shown.
g in the following three compounds(1,2,3) arrange their relative reactivity towards the reagent CH3Cl / AlCl3. Justify your order
Answer:
3 > 2> 1
Explanation:
Aromatic compounds undergo electrophilic substitution reaction with several electrophiles.
Some substituted benzenes are more reactive towards electrophilic aromatic substitution than unsubstituted benzene.
Certain groups of substituents increase the ease with which an aromatic compound undergoes aromatic substitution.
If we look at the compounds closely, we will notice that only toluene leads to easy reaction with CH3Cl / AlCl3. Thus is due to the +I inductive effect of -CH3 which stabilizes the negatively charged intermediate produced in the reaction.
Acetic acid and water react to from hydronium cation and acetate anion, like this: HCH3CO2 (aq) + H2O (I) → H3O+(aq) + Ch3CO2-(aq)Imagine 226. mmol of CH3CO2- are added to a flask containing a mixture of HCH3CO2, H2O, H3O + and CH3CO2- at equilibrium, and then answer the following questions. 1. What is the rate of the forward reaction before any HCH3CO2 has been removed from the flask? a. Zero.b. Greater than zero, but less than the rate of the reverse reaction. c. Greater than zero, and equal to the rate of the reverse reaction. d. Greater than zero, and greater than the rate of the reverse reaction. 2. What is the rate of the forward reaction just after the HCH3CO2 has been removed from the flask?a. Zero.b. Greater than zero, but less than the rate of the reverse reaction. c. Greater than zero, and equal to the rate of the reverse reaction. d. Greater than zero, and greater than the rate of the reverse reaction.3. What is the rate of the forward reaction when the system has again reached equilibrium?a. Zero.b. Greater than zero, but less than the rate of the reverse reaction. c. Greater than zero, and equal to the rate of the reverse reaction. d. Greater than zero, and greater than the rate of the reverse reaction.
Answer:
1) Greater than zero, and equal to the rate of the reverse reaction
2) Greater than zero, but less than the rate of the reverse reaction
3) Greater than zero, and equal to the rate of the reverse reaction
Explanation:
A reaction system is said to be in equilibrium when the rate of forward reaction is equal to the rate of reverse reaction.
Before we remove HCH3CO2 from the system, the system was in equilibrium. Recall that when a system is in equilibrium, the rate of forward reaction is equal to the rate of reverse reaction. The rate of reaction is greater than zero because products are being formed as the reactants interact with each other.
When HCH3CO2 is removed from the system, the equilibrium position shifts towards the left hand side hence the rate of reverse reaction is greater than the rate of forward reaction.
When the system attains equilibrium again, the rates of forward and reverse reaction become equal.
All --- is a good conductor of heat
According to the kinetic-molecular theory, gas molecules have
ANSWER:
Part
A. Less energy than molecules of a solid.
B. strong interactions between molecules.
C. little distance between molecule
D. weak interactions between molecules.
Answer:
The choose ( D )
weak interactions between molecules.
Answer:
Explanation:
el a
Enough of a monoprotic acid is dissolved in water to produce a 1.211.21 M solution. The pH of the resulting solution is 2.882.88 . Calculate the Ka for the acid.
Answer:
1.44 × 10⁻⁶
Explanation:
Step 1: Given data
Concentration of the acid (Ca): 1.21 MpH of the solution: 2.88Step 2: Calculate the concentration of H⁺ ions
We will use the definition of pH.
pH = -log [H⁺]
[H⁺] = antilog -pH = antilog -2.88 = 1.32 × 10⁻³ M
Step 3: Calculate the acid dissociation constant of the acid (Ka)
For a weak monoprotic acid, we will use the following expression.
Ka = [H⁺]²/Ca
Ka = (1.32 × 10⁻³)²/1.21 = 1.44 × 10⁻⁶
What is the cell potential of an electrochemical cell that has the half-reactions shown below?
Ag⁺ + e⁻ → Ag
Fe → Fe³⁺ + 3e⁻
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 should be done if a spectrophotometer reports an absorbance that is too high? Select one: Reposition the cuvette in the spectrophotometer. Pour out half the volume of the sample. Restart the spectrophotometer and try again. Dilute the sample. g
Answer:
The sample should be diluted
Explanation:
According to Beer Lambert's law, the absorbance of a sample depends on the concentration of the sample.
Hence, if the concentration of the sample is very high, the spectrophotometer will also report a very high value of absorbance.
When this is the case, the sample should simply be diluted and the readings are taken again using the spectrophotometer.
