Answer:
The answer is "It takes 1,70 mol of ethanol".
Explanation:
To make acetic acid, we must first write the balanced reaction that occurs of ethanol with oxygen
The response is balanced:
[tex]CH_3CH_2OH+O_2\to CH_3COOH+H_2O[/tex]
1 mol of ethanol creates 1 mol of According the equilibrium Ethanol moles, therefore, required 1.70 mol of water = 1.70 mol
H2SO4 ????????????????
Explanation:
Sulfuric Acid (H2SO4) is a strong mineral acid that has is colorless when pure. This chemical is used as a chemical intermediate to manufacture other chemicals and cleaning metal surfaces. The formula for sulfuric acid is H2SO4. The molar mass of sulfuric acid is 98.07848 g mol.
The specific heat capacity of lead is 0.13 J/g-K. How much heat (in J) is required to raise the temperature of 15 g of lead from 22 °C to 37 °C? a. 5.8 × 10-4 J b. 0.13 J c. 29 J d. 2.0 J e. -0.13 J
Answer:
c. 29 J
Explanation:
Step 1: Given data
Specific heat capacity of Pb (c): 0.13 J/g.K (= 0.13 J/g.°C)Mass of Pb (m): 15 gInitial temperature: 22 °CFinal temperature: 37 °CStep 2: Calculate the temperature change
ΔT = 37 °C - 22 °C = 15 °C
Step 3: Calculate the heat (Q) required to raise the temperature of the lead piece
We will use the following expression.
Q = c × m × ΔT
Q = 0.13 J/g.°C × 15 g × 15 °C = 29 J
Sodium acetate is produced by the reaction of baking soda and vinegar. The resultant solution is then heated until it becomes saturated and allowed to cool. As a result, the solution has become supercooled. Upon addition of a small seed crystal, the solution temperature increases as sodium acetate trihydrate crystallizes. Its molar enthalpy of fusion is 35.9 kJ/mol. How much thermal energy would be released by 276.0 g of sodium acetate trihydrate (molar mass
Answer: The thermal energy that would be released by 276.0g of sodium acetate trihydrate is 71.8kJ.
Explanation:
Supercooling is the process of lowering the temperature a liquid below its freezing point, without it becoming solid. A liquid below its freezing point will crystallize in the presence of a seed crystal because it serves as a structure for formation of crystals. From the question,
The given mass of sodium acetate trihydrate
(CH3COONa.3H2O)= 276.0g
Molar mass of sodium acetate
trihydrate= 136.08g/mol
Thermal heat of fusion of sodium acetate
trihydrate = 35.9 kJ/mol
From the given mass the number of moles present= 276.0/ 136.08
= 2.0moles
Therefore the heat (thermal) energy of the given mass of sodium acetate
trihydrate = 2.0 × 35.9
= 71.8kJ
Therefore, upon addition of a small seed crystal, the solution temperature increases as sodium acetate trihydrate crystallizes.
For the following reaction, 4.77 grams of carbon (graphite) are allowed to react with 16.4 grams of oxygen gas.
carbon (graphite) (s) + oxygen (g) → carbon dioxide (g)
1. What is the maximum amount of carbon dioxide that can be formed?
2. What is the FORMULA for the limiting reagent?
3. What mass of the excess reagent remains after the reaction is complete?
Answer:
1. 17.5 g of CO₂
2. The limiting reactant is carbon (graphite), and its formula is C(graphite)
3. 3.7 g of O₂
Explanation:
First, we have to write the chemical equation for the reaction. For this, we have to know the chemical formula of each reactant and product:
Reactants: carbon(graphite) ⇒ C(graphite) ; oxygen gas ⇒ O₂(g)Products: carbon dioxide ⇒ CO₂(g)Thus, we write the chemical equation:
C(graphite) + O₂(g) → CO₂(g)
The equation is already balanced because it has the same number of C and O atoms on both sides. Thus, we can see that 1 mol of C(graphite) reacts with 1 mol of O₂ and produce 1 mol of CO₂ (mole-to-mole reaction).
Now we convert the grams of reactants to moles by using the molecular weight (Mw) of each compound:
Mw(C) = 12 g/mol
moles of C(graphite) = 4.77g/(12 g/mol) = 0.3975 mol
Mw(O₂) = 16 g/mol x 2 = 32 g/mol
moles of O₂ = 16.4 g/(32 g/mol) = 0.5125 mol
Now, we can compare the stoichiometric ratio (given by the moles of reactants in the equation) with the actual ratio (given by the mass of reactants we have):
stoichiometric ratio ⇒ 1 mol C(graphite)/mol O₂
actual ratio ⇒ 0.3975 mol C(graphite)/0.5125 mol O₂
We can see that we need 0.3975 moles of O₂ to react with C(graphite) and we have more moles (0.5125 mol) so the excess reactant is O₂. Thus, the limiting reactant is C(graphite).
The amount of product (CO₂) that is formed is calculated from the amount of limiting reactant. We can see in the chemical equation that 1 mol of CO₂ is produced from 1 mol of C(graphite) ⇒ stoichiometric ratio = 1 mol CO₂/mol C(graphite).
Thus, we multiply the moles of C(graphite) we have by the stoichiometric ratio to calculate the moles of CO₂ produced:
moles of CO₂ = 0.3975 mol C(graphite) x 1 mol CO₂/mol C(graphite) = 0.3975 mol CO₂
Now, we convert the moles of CO₂ to mass by using the Mw:
Mw(CO₂) = 12 g/mol + (16 g/mol x 2) = 44 g/mol
mass of CO₂ = 0.3975 mol CO₂ x 44 g/mol CO₂ = 17.5 g
Therefore, the maximum amount of carbon dioxide (CO₂) formed is 17.5 g.
Since this is a mole-to-mole reaction, the moles of excess reactant that remains after the reaction is complete is calculated as the difference between the moles of excess reactant and limiting reactant:
remaining moles of O₂ = 0.5125 mol - 0.3975 mol = 0.115 mol O₂
Finally, we convert the moles of O₂ to mass with the Mw (32 g/mol) :
mass of O₂ = 0.115 mol O₂ x 32 g/mol = 3.68 g
Therefore, the mass of the excess reagent that remains after the reaction is complete is 3.7 g.
a 67.5 L sample of gas at 27.6 °c and 383.1mm hg expands to 244.2 L at 4.7 °c. what is the new gas pressure.
a 97.8
b 18.0
c 115
d 1.65
Answer:
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What is the correct ratio of carbon to hydrogen to oxygen in glucose (CH1206)?
•12:12:6
•2:1:1
•1:2:1
•6:6:12
Answer:
1:2:1 is the correct ratio of carbon hydrogen to oxygen in glucose.
what are the five main points of kinetic theory of gas?
The kinetic-molecular theory of gases assumes that ideal gas molecules
(1) are constantly moving;
(2) have negligible volume;
(3) have negligible intermolecular forces;
(4) undergo perfectly elastic collisions; and
(5) have an average kinetic energy proportional to the ideal gas's absolute temperature.
The five main postulates of the KMT are as follows:
(1) the particles in a gas are in constant, random motion,
(2) the combined volume of the particles is negligible
(3) the particles exert no forces on one another,
(4) any collisions between the particles are completely elastic.
(5) the average kinetic energy of the particles is proportional to the temperature in kelvins.
Tapeworm is grouped in the phylum Platyhelminthes
Answer:
Tapeworm, also called cestode, any member of the invertebrate class Cestoda (phylum Platyhelminthes), a group of parasitic flatworms containing about 5,000 species. ... Tapeworms also lack a circulatory system and an organ specialized for gas exchange.
What is normality in chemistry?
Answer:
a measure of concentration equal to the gram equivalent weight per liter of solution.
Explanation:
Gram equivalent weight is the measure of the reactive capacity of a molecule. The solute's role in the reaction determines the solution's normality. Normality is also known as the equivalent concentration of a solution.
hope it helped
what characterizes a homogeneous mixture?
Answer:
a mixture that doesn't really show the ingredients or things put into the material or food.
Urea, CH4N2O (s), is manufactured from NH3 (g) and CO2 (g). H2O (l) is another product of this reaction. An experiment is started with 2.6 grams of NH3 (g) added into a reaction vessel with CO2 (g).
Write the balanced equation for this reaction, being sure to include physical states. Based on the balanced equation above, calculate the following:
a. the theoretical yield of urea in grams that can be made from the NH3
b. the actual amount of urea made if the percent yield for this reaction is 34%.
Answer:
a. 4.41 g of Urea
b. 1.5 g of Urea
Explanation:
To start the problem, we define the reaction:
2NH₃ (g) + CO₂ (g) → CH₄N₂O (s) + H₂O(l)
We only have mass of ammonia, so we assume the carbon dioxide is in excess and ammonia is the limiting reactant:
2.6 g . 1mol / 17g = 0.153 moles of ammonia
Ratio is 2:1. 2 moles of ammonia can produce 1 mol of urea
0.153 moles ammonia may produce, the half of moles
0153 /2 = 0.076 moles of urea
To state the theoretical yield we convert moles to mass:
0.076 mol . 58 g/mol = 4.41 g
That's the 100 % yield reaction
If the percent yield, was 34%:
4.41 g . 0.34 = 1.50 g of urea were produced.
Formula is (Yield produced / Theoretical yield) . 100 → Percent yield
Hãy cho biết giá trị và ý nghĩa của số lượng tử n, l, m, ms khi mô tả trạng thái của electron trong nguyên tử?
Consider the reaction between CaCO3 and HCl. Which of the following could speed up the reaction?
I. Increasing concentration of the HCl
II. Increasing size of the CaCO3 pieces
III. Increasing temperature
a) I and III only
b) I, II, and III
c) I only
d) II and III only
The fact that a beam of particles was deflected in the presence of an electric
or magnetic force led J.J. Thomson to conclude that the particles had a(n)
O A. large mass
B. electric charge
O C. negligible mass
O D. neutral charge
Answer:
electric charge
Explanation:
Charged particles are deflected in an electric or a magnetic field. The particles discovered by J.J. Thomson were charged particles.
When these charged particles are passed through electric and magnetic fields, deflection occurs depending on the nature of the charge.
A positive charge is deflected towards the negative part of an electric field or the south pole of a magnetic field.
A negative charge is selected towards the positive end of an electric field or the north pole of a magnetic field.
A solution has a OH- concentration of 7.7x10-3. What is the pH of this solution?
Answer:
11.9 pH
Explanation:
First, we need to find pOH
To find that, we use the formula -log[OH]
-log[7.7x10^-3] = 2.11351
To find the pH, we'll use this formula: 14 = pH + pOH
14 = pH + 2.11351
Subtract boths sides by 2.11351
14 = pH + 2.11351
-2.11351 -2.11351
pH = 11.88649
If we slowly add a solution of mercury(II) ions to a solution of aqueous halide ions with roughly equal concentrations, a precipitate will form. Explain what the precipitate will consist of initially. g
Water, mercury chloride and nitrogen oxide.
Water, mercury chloride and nitrogen oxide will present in the precipitate when we slowly add a solution of mercury(II) nitrate to a solution of aqueous hydrochloric acid having halide ions both in equal concentrations. The equation of this reaction is Hg2(NO3)2 + 4 HCl ----> 2 HgCl2 + 2 H2O + 2 NO so it is concluded that from this reaction we get precipitate of water, mercury chloride and nitrogen oxide.
https://brainly.com/question/24261598.
chemistry
Which of the following is a characteristic of solid silver?
O A. High electrical conductivity
O B. Brittleness
C. Low melting point
O D. Good insulator
SUBM
Answer:
A. High electrical conductivity
Explanation:
solid silver isn't brittle, it has a high melting point, and its not a good insulator.
)Calculate the molar mass of glucose (C6H12O6)
Answer:
Molar mass = 180 g/mol
Explanation:
Relative Atomic Mass of C = 12
of H = 1
of O =16
Let Molar mass be mm
mm of C6H12O6 = 6(12) + 12(1) + 6(16)
= 72 + 12 + 96 = 180 g/mol
A chemist determines by measurements that moles of bromine liquid participate in a chemical reaction. Calculate the mass of bromine liquid that participates. Round your answer to significant digits.
Answer:
5.20 grams of Br₂
Explanation:
From our previous knowledge;
We understand that:
The number of moles of a given element = mass of the element divided by its molar mass.
Mathematically:
[tex]\mathbf{no \ of \ moles =\dfrac{ mass}{ molar \ mass}}[/tex]
From the given information, let's assume that the 0.065 moles of liquid -bromine partake in the reaction.
From the periodic table, the molar mass of Bromine is = 79.9 g/mol
As such, the mass of liquid that partakes is calculated as:
0.065 mol = mass/ 79.9 g/mol
mass = 0.065 mol × 79.9 g/mol
mass of liquid that partakes in the reaction = 5.20 grams of Br₂
Based on periodic properties, choose the more metallic element from each of the following pairs.
Match the words in the left column to the appropriate blanks in the sentences on the right.
Between Sr and Sb, the more metallic element is ______
Between \rm Sr and \rm Sb, the more metallic element is _______
Between As and Bi, the more metallic element is ______
Between \rm As and \rm Bi, the more metallic element is _______
Between Cl and O, the more metallic element is ______
Between \rm Cl and \rm O, the more metallic element is ______
Between S and As, the more metallic element is ______
Between \rm S and \rm As, the more metallic element is _______
Answer:
Sr is the more metallic element
Bi is the more metallic element
O is the more metallic element
As is the more metallic element
Explanation:
One thing should be clear; metallic character increases down the group but decreases across the period.
Hence, as we move across the period, elements become less metallic. As we move down the group elements become more metallic.
This is the basis upon which decisions were made about the metallic character of each of the elements listed above.
Name the following cycloalkane:
H3C-
CH2CH3
A. 1-ethyl-4-methylcyclohexane
B. 4-ethyl-1-methylcyclohexane
C. 1-methyl-4-ethylcyclohexane
Answer:
A
Explanation:
1-ethyl-4-methylcyclohexane
i believe it is A
hope that helped :)
In a quantitative analysis, a methanol (CH3OH) contaminated water sample was titrated with 0.0021 mol L- potassium permanganate (KMnO4). 50.00 mL samples of the water to be tested were acidified by sulfuric acid, then titrated with the permanganate solution. The results are shown below. Burette reading, ml 1st titration 2nd titration 3rd titration 4th titration Final volume 12.40 19.60 26.60 17.25 Initial volume 4.45 12.50 19.60 10.15 Titre 7.95 7.10 7.00 7.10 The complete equation for the redox titration reaction is: 4MnO4- + 12H+ + 5CH3OH → 4Mn2+ + 11H2O + 5HCOOH a. [5] Calculate the concentration of the methanol in mol L-1.
In a REDOX titration, one specie is oxidized while the other is reduced. The concentration of methanol is 0.012 mol L-1. Methanol is the oxidizing agent while permanganate is the reducing agent.
The average titre value is; [tex]\frac{7.95 + 7.10 + 7.00 + 7.10}{4}[/tex] = 7.29 mL
Equation of the reaction is:
[tex]4MnO4- + 12H+ + 5CH3OH ----> 4Mn2+ + 11H2O + 5HCOOH[/tex]
Concentration of oxidizing agent = CA = ?
Concentration of reducing agent = CB = 0.0021 mol L-1
Volume of oxidizing agent = VA= 7.29 mL
Volume of reducing agent = VB = 50.00 mL
Number of moles of oxidizing agent NA = 4
Number of moles of reducing agent NB = 5
Note that NA and NB are obtained from the balanced reaction equation
CAVA/CBVB = NA/NB
CAVANB = CBVBNA
CA = CBVBNA/VANB
CA = 0.0021 mol L-1 * 50.00 mL * 4/7.29 mL * 5
CA= 0.012 mol L-1
For a comprehensive definition of redox titration see
https://brainly.com/question/24018439
For each reaction, write the chemical formulae of the oxidized reactants.
a. ZnCl2 (aq) + 2Na(s) → Zn(s) + 2NaCl(aq)
b. Al(s) + FeBrz (aq) → AlBrz (aq) + Fe(s)
c. FeSO4 (aq) + Zn (s) → Fe(s) + ZnSO4(aq)
Answer:
a. Na(s); b. Al(s); c. Zn(s)
Explanation:
Let's consider the following redox reactions.
a. ZnCl₂ (aq) + 2 Na(s) → Zn(s) + 2 NaCl(aq)
Na is oxidized because its oxidation number increases from 0 to +1 (in NaCl) whereas Zn is reduced because its oxidation number decreases from 2+ (in ZnCl₂) to 0.
b. Al(s) + FeBr₃ (aq) → AlBr₃ (aq) + Fe(s)
Al is oxidized because its oxidation number increases from 0 to +3 (in AlBr₃) whereas Fe is reduced because its oxidation number decreases from 3+ (in FeBr₃) to 0.
c. FeSO₄ (aq) + Zn(s) → Fe(s) + ZnSO₄(aq)
Zn is oxidized because its oxidation number increases from 0 to +2 (in ZnSO₄) whereas Fe is reduced because its oxidation number decreases from 2+ (in FeSO₄) to 0.
what is the difference between red phosphorus and white phosphorus?
Answer:
White phosphorusRed PhosphorusIt is insoluble in water but soluble in carbon disulphide.It is insoluble in both water and carbon disulphide.It undergoes spontaneous combustion in air.It is relatively
Explanation:
I hope it will help you
FILL IN THE BLANK:
The rate of a reaction is measured by how fast a (Product Or Reactant)
is used up or how fast a
(Reactant Or Product) is formed?
Answer:
the rate of a reaction is measured by how fast a REACTANT is used up or how fast a PRODUCT is formed
instruments that have been soaking in cold sterilization for blank minutes are considered aseptic and can be used during non sterile procedures
a. 60
b. 30
c. 15
d. 10
Answer:
Its C you can check
PLEASE HELP!!
this is on USAtestprep
a)
b)
c)
d)
A sample of gas is held at constant volume. If the number of moles of this sample of gas is doubled and the pressure of this sample of gas is halved, what happens to the absolute temperature of the gas?
Select one
a. The absolute temperature is doubled.
b. The absolute temperature is halved.
c. The absolute temperature is quadrupled.
d. The absolute temperature is quartered.
e. The absolute temperature stays the same.
Answer:
number of moles of gas increases the volume also increases.
1) Recall the two written definitions of an oxidation-reduction reaction provided in our lessons. Which of these definitions is
most inclusive of redox reactions? Explain your answer:
A redox reaction is where the oxidation and reduction reaction takes place at the same time, the oxidation half
reaction involves losing electrons and in the reduction half reaction involves gaining electrons. So in a redox
reaction an electron is lost by the reducing agent.
Explain how the reaction below meets these definitions. Which substance is being oxidized and which is
being reduced?
4Ag(s) + 2H2S(g) + O2(g)
2Ag2S(s) + 2H20(9)
Answer:
A redox reaction in which a change of oxidation number of the reacting species occurs either by oxidation or by reduction which occur simultaneously and to the same extent during the reaction.
Silver atoms are oxidized while oxygen atoms are reduced by a loss of electrons and a gain of electrons respectively.
Explanation:
A redox reaction in which a change of oxidation number of the reacting species occurs either by oxidation or by reduction which occur simultaneously and to the same extent during the reaction.
In a redox reaction,the two reactions occurring simultaneously can be divided into two half reactions; an oxidation half-reaction and a reduction half-reaction.
The oxidation half-reaction involves losing electrons and thus an increase in oxidation number of the species being oxidized. Whereas, the reduction half reaction involves gaining electrons and thus, a reduction innthe oxidation number of the species being reduced.
The species which oxidizes another species is known as an oxidizing agent and isnitself reduced due to its accepting electrons from the species being oxidized. Th reducing agent reduces another species and is itself oxidized as it loses electrons to the oxidized agent.
In the given reaction as shown below:
4 Ag (s) + 2 H₂S (g) + O₂ (g) ---> 2 Ag₂S (s) + 2 H₂0 (g)
The reaction is a redox reaction as a change innthe oxidation number of the reacting species; both oxidation and reduction occurs simultaneously and to the same extent.
The metallic silver atoms, have an oxidation number of zero initially. However, each of the four moles of atoms give up one mole of a electrons each to become oxidized to silver (i) ions, Ag+.
On the other hand, molecular oxygen gas also having oxidation number of zero becomes reduced to oxygen ion, O²-. Each of the two moles of atom in the oxygen gas molecule accept two electrons each donated by the metallic silver atoms to become reduced to oxygen ion, O²-.
The oxidation numbers of hydrogen ion and sulfide ion do not change.
Nucleophilic aromatic substitution involves the formation of a resonance-stabilized carbanion intermediate called a Meisenheimer complex as the nucleophile attacks the ring carbon carrying the eventual leaving group.
a. True
b. False
Answer:
True
Explanation:
Aromatic rings undergo nucleophillic substitution reactions in the presence of a electron withdrawing group which stabilizes the Meisenheimer complex.
When the nucleophile attacks the ring carbon atom carrying the eventual leaving group. A resonance-stabilized carbanion intermediate called a Meisenheimer complex is formed.
Subsequent loss of the leaving group from the intermediate complex yields the product of the reaction.
