Answer:
1.7 moles of ethanol would be needed.
Explanation:
* Calculate the moles of ethanol needed to produce 1.70mol of water. Be sure your answer has a unit symbol, if necessary, and round it to the correct number of significant digits.
First off, we have to state the equation for the reaction.
So we know that;
ethanol + oxygen → acetic acid + water
This leads us to;
C2H5OH + O2 → CH3COOH + H2O
1 1 1 1
To obtain the moles of ethanol needed to produce 1.70mol of water, we look at the stoichiometry of the reaction above.
1 mol of ethanol produces 1 mole of water
x mol of thanol would produce 1.7 mol of water
Thus we have;
1 = 1
x = 1.7
x = 1.7 moles of ethanol would be needed.
Suppose an industrial quality-control chemist analyzes a sample from a copper processing plant in the following way. He adds powdered iron to a copper(II) sulfate sample from the plant until no more copper will precipitate. He then washes, dries, and weighs the precipitate, and finds that it has a mass of . Calculate the original concentration of copper(II) sulfate in the sample. Round your answer to significant digits.
Answer:
Concentration of Copper (II) Sulfate in the original sample in mol/L = 0.0035 M
Concentration of Copper (II) Sulfate in the original sample in g/L = 0.56 g/L
Explanation:
Complete Question
Fe(s) + CuSO₄(aq) → Cu(s) + FeSO₄(aq)
Suppose an industrial quality-control chemist analyzes a sample from a copper processing plant in the following way. He adds powdered iron to a 400.mL copper (II) sulfate sample from the plant until no more copper will precipitate. He then washes, dries, and weighs the precipitate, and finds that it has a mass of 89.mg. Calculate the original concentration of copper(II) sulfate in the sample. Round your answer to 2 significant figures.
Solution
Noting that the precipitate is Copper as it is the only solid by-product of this reaction.
89 mg of Copper is produced from this reaction.
We convert this into number of moles for further stoichiometric calculations
Mass of Copper = 89 mg = 0.089 g
Molar mass of Copper = 63.546 amu
Number of moles of Copper produced from the reaction = (0.089/63.546) = 0.0014005602 = 0.001401 mole
From the stoichiometric balance of the reaction,
1 mole of Copper is produced from 1 mole of Copper (II) Sulfate
0.001401 mole of Copper will be produced similarly from 0.001401 mole of Copper (II) Sulfate.
Number of moles of Copper (II) Sulfate in the original sample = 0.001401 mole
Concentration of Copper (II) Sulfate in the original sample in mol/L = (Number of moles) ÷ (Volume in L)
Number of moles = 0.001401 mole
Volume in L = (400/1000) = 0.4 L
Concentration of Copper (II) Sulfate in the original sample in mol/L = (0.001401/0.4) = 0.0035025 mol/L = 0.0035 mol/L to 2 s.f.
Concentration in g/L = (Concentration in mol/L) × (Molar Mass)
Concentration in mol/L = 0.0035025 M
Molar mass of Copper (II) Sulfate = 159.609 g/mol
Concentration of Copper (II) Sulfate in the original sample in g/L = 0.0035025 × 159.609 = 0.559 g/L = 0.56 g/L to 2 s.f
Hope this Helps!!!!
The concentration of the original copper solution is 0.035 M.
The equation of the reaction is;
Fe(s) + CuSO4(aq) -------> FeSO4(aq) + Cu(s)
Number of moles of copper obtained = 89 × 10^-3g/63.5 = 0.0014 moles
Since the reaction is 1:1, the number of moles of copper sulfate that reacted is c.
From the question, we are told that the volume of solution is 400.mL or 0.04L.
Hence, the concentration of the solution is; number of moles /volume
= 0.0014 moles/0.04L = 0.035 M
Learn more: https://brainly.com/question/9352088
Missing parts;
Suppose an industrial quality-control chemist analyzes a sample from a copper processing plant in the following way. He adds powdered iron to a 400.mL copper (II) sulfate sample from the plant until no more copper will precipitate. He then washes, dries, and weighs the precipitate, and finds that it has a mass of 89.mg. Calculate the original concentration of copper(II) sulfate in the sample. Round your answer to 2 significant figures.
10. How many grams of NH, are present in 6 moles
of NH,?
Answer:
90.08784 grams
Explanation:
idk
The aluminum in a package containing 75 ft2 of kitchen foil weighs approximately 12 ounces. Aluminum has a density of 2.70 g/cm3 . What is the approximate thickness of the aluminum foil in millimeters?(1 ounce = 28.4g)
Answer:
18130 mm
Explanation:
Now we have a lot of unit conversions to do in order to correctly answer this question. We shall do these conversions gradually.
First we convert the weight in ounce to grams.
If 1 ounce = 28.4g
12 ounces = 12×28.4 = 340.8 g
Next we convert the area of aluminum from ft2 to m2
1ft2= 0.0929 m2
75 ft2= 75 × 0.0929= 6.9675m2
Now density of aluminum= 2.70 gcm-3
Density= mass/volume
But volume= area× thickness
Density= mass/ area × thickness
Density × area × thickness= mass
Thickness= mass/ density × area
Thickness= 340.8g / 2.70gcm-3 × 6.9675m2
Thickness= 340.8/18.8
Thickness= 18.13 m
Since 1000 milimeters make 1 metre
Thickness= 18130 mm
Ni
Express your answer in condensed form in the order of orbital filling as a string without blank space between orbitals. For example, [He]2s22p2 should be entered as [He]2s^22p^2.
Answer:
[Ar]3d^84s^2
Explanation:
From the question given, we are asked to write the condensed form of electronic configuration of nickel, Ni.
To do this, we simply write the symbol of the noble gas element before Ni in a squared bracket followed by the remaining electrons to make up the atomic number of Ni.
This is illustrated below:
The atomic number of Ni is 28.
The noble gas before Ni is Argon, Ar.
Therefore, the condensed electronic configuration of Ni is written as:
Ni(28) => [Ar]3d^84s^2
Answer:
[Ar] 4s^23d^8
Explanation:
A 20.0-mL sample of lake water was acidified with nitric acid and treated with excess KSCN to form a red complex (KSCN itself is colorless). The solution was then diluted to 50.0-mL and put in a 1.00 cm pathlength cell, where it yielded an absorbance of 0.345. For comparison, a 5.0-mL reference sample of 4.80 x 10-4 M Fe3 was treated with HNO3 and KSCN and diluted to 50.0 mL. The reference solution was also placed in a 1.00-cm cell and gave an absorbance of 0.512. What is the concentration of Fe3 in Jordan Lake
Answer:
8.09x10⁻⁵M of Fe³⁺
Explanation:
Using Lambert-Beer law, the absorbance of a sample is proportional to its concentration.
In the problem, the Fe³⁺ is reacting with KSCN to produce Fe(SCN)₃ -The red complex-
The concentration of Fe³⁺ in the reference sample is:
4.80x10⁻⁴M Fe³⁺ × (5.0mL / 50.0mL) = 4.80x10⁻⁵M Fe³⁺
Because reference sample was diluted from 5.0mL to 50.0mL.
That means a solution of 4.80x10⁻⁵M Fe³⁺ gives an absorbance of 0.512
Now, as the sample of the lake gives an absorbance of 0.345, its concentration is:
0.345 × (4.80x10⁻⁵M Fe³⁺ / 0.512) = 3.23x10⁻⁵M.
As the solution was diluted from 20.0mL to 50.0mL, the concentration of Fe³⁺ in Jordan lake is:
3.23x10⁻⁵M Fe³⁺ × (50.0mL / 20.0mL) = 8.09x10⁻⁵M of Fe³⁺
The concentration of Fe³⁺ in Jordan Lake is = 8.09* 10⁻⁵ M
According to Lambert-Beer law ;The rate of absorbance of a sample is directly proportional to concentration of the sample
The reaction that produces a red complex
Fe³⁺ + KScN ----> Fe ( SCN )₃ ( red complex )
First step: Determine the Concentration of Fe³⁺ in reference sample
= 4.80x10⁻⁴ * ( 5.0 / 50.0 ) = 4.80 * 10⁻⁵M
reference sample was diluted from 5.0 mL to 50.0 mL
∴ Concentration of 4.80 * 10⁻⁵M has an absorbance = 0.512
Given that Lake sample absorbance = 0.345
Next step : Determine the concentration of the lake sample
Concentration of lake sample :
= absorbance of lake sample * ( conc of reference sample / absorbance )
= 0.345 * ( 4.80* 10⁻⁵ / 0.512 ) = 3.23* 10⁻⁵M.
Final step : Determine the concentration of Fe³⁺ in Jordan lake
= 3.23 * 10⁻⁵ * ( 50.0mL / 20.0mL) = 8.09* 10⁻⁵ M
Note : Solution was diluted from 20.0 mL to 50.0 mL
Hence we can conclude that The concentration of Fe³⁺ in Jordan Lake is = 8.09* 10⁻⁵ M .
Learn more : https://brainly.com/question/24287054
There are __________ moles of N atoms present in a 2.0 g C8H10O2N4.
Answer:
[tex]n_N=0.041molN[/tex]
Explanation:
Hello,
In this case, for this mole-mass relationship, we are able to compute the moles of nitrogen atoms by firstly obtaining the moles of the given compound, considering its molar mass that is 194 g/mol:
[tex]n_{C_8H_{10}O_2N_4}=2.0gC_8H_{10}O_2N_4*\frac{1molC_8H_{10}O_2N_4}{194gC_8H_{10}O_2N_4} =0.01molC_8H_{10}O_2N_4[/tex]
Then, by knowing that one mole of the given compound has four moles of nitrogen atoms, we apply the following relationship:
[tex]n_N=0.01molC_8H_{10}O_2N_4*\frac{4molN}{1molC_8H_{10}O_2N_4} \\\\n_N=0.041molN[/tex]
Best regards.
C12H22O11 + 12O2 ---> 12CO2 + 11H2O
there are 10.0 g of sucrose and 10.0 g of oxygen reacting. Which is the limiting reagent?
Answer:
Oxygen is the limiting reactant.
Explanation:
Based on the reaction:
C₁₂H₂₂O₁₁ + 12O₂ → 12CO₂ + 11H₂O
1 mole of sucrose reacts with 12 moles of oxygen to produce 12 moles of CO₂ and 11 moles of H₂O.
10.0g of sucrose (Molar mass: 342.3g /mol) are:
10.0g C₁₂H₂₂O₁₁ × (1mole / 342.3g) = 0.0292 moles of C₁₂H₂₂O₁₁
And moles of 10.0g of oxygen (Molar mass: 32g/mol) are:
10.0g O₂ × (1mole / 32g) = 0.3125 moles of O₂
For a complete reaction of 0.0292 moles of C₁₂H₂₂O₁₁ you need (knowing 12 moles of oxygen react per mole of sucrose):
0.0292 moles of C₁₂H₂₂O₁₁ × (12 moles O₂ / 1 mole C₁₂H₂₂O₁₁) = 0.3504 moles of O₂
As you have just 0.3125 moles of O₂, oxygen is the limiting reactant.
reasons for good care on computer
answer
1)maximise your software efficiency
2)Prevention against viruses and malware
3)Early detection of problematic issues
4)prevent data loss
5)Speed up your computer
what is the correct ionic equation, including all coefficients, charges, and phases for the following sets of reactants? Assume that the contribution of protons from H2SO4 is near 100%.
Ba(OH)2(aq)+H2SO4(aq) —>
help, I have no clue
Answer:
Ba(OH)2(aq)+H2SO4(aq) gives us 2BaH+H2O
Explanation:
Which of the following structures in the human body has the highest level of organization
Answer:
The brain
Explanation:
With all those instructions the body recqures to respond to it must be so
Hope it helps
which best describes a mixture.
A it has a single composition and it has a set of characteristics
B it can have different compositions but it has a set of charactaristics that does not change
C it has a single composition but it has a set of characteristics that does change
D it can have different compositions and it has a set of characteristics that does change
Answer:
B) It can have different compositions, but it has a set of characteristics that does not change.
Explanation:
On e d g e n u i t y
I believe the answer is d lmk if im wrong or right
Calculate the percentage of the void space out of the total volume occupied by 1 mole of water molecules. The density of water is assumed to be 1.0 g/mL that is 1.0 g/cm3. The molar mass of water is 18.0 g/mol. The atomic radius of hydrogen is 37 pm and of oxygen is 73 pm. The formula for the volume of a sphere is 4/3(r3
Answer:
The percentage of the void space out of the total volume occupied is 93.11%
Explanation:
A mole of water contains 2 atoms of hydrogen and 1 atom of oxygen.
To calculate the volume of a mole of water, we calculate 2 times the volume of the hydrogen atom and 1 times the volume of the oxygen atom
Let's calculate this one after the other.
For the hydrogen, formula for the volume will be
[tex]V_{hydrogen[/tex] = 2 × 4/3 × π × [tex]r_{H}^{3}[/tex]
where [tex]r_{H}^{3}[/tex] = 37 pm which is read as 37 picometer (1 picometer = 10^-12 m) = 37 × [tex]10^{-12}[/tex] meters
Volume of the hydrogen = 8/3 × (37 × 10^-12)^3 = 4.05 * 10^-31
we multiply this by the avogadro's number = 6.02 * 10^23
= 4.05 * 10^-31 * 6.02 * 10^23 = 2.6 * 10^-8 m^3
We do same for thr oxygen, but this time we do not multiply the volume of the oxygen by 2 as we have only one atom of oxygen
Volume of oxygen = 4/3 * π * (73 * 10^-12) ^3 * avogadro's number = 9.81 * 10^-7 m^3
adding both volumes together, we have 1.24 * 10^-6 m^3 or simply 1.24 ml ( 0.01 m = 1 ml)
Dividing the molar mass of one mole of water by its density, we can get the volume of 1 mole of water
= (18g/mol)/(1 g/ml) = 18 ml/mol
Now we proceed to calculate the volume of void = Total volume - volume of molecule = 18 - 1.24 = 16.76 ml
Now, the percentage of void = volume of void/total volume * 100%
= 16.76/18 * 100% = 93.11%
4-Nitrophenol, NO2C6H4OH (pKa 7.15), is only slightly soluble in water, but its sodium salt, NO2C6H4O-Na+, is quite soluble in water. Describe the solubility of 4-nitrophenol in solutions of sodium hydroxide, sodium bicarbonate (NaHCO3), and sodium carbonate (Na2CO3). The pKa values for the conjugate acids of sodium hydroxide, sodium bicarbonate (NaHCO3), and sodium carbonate (Na2CO3) are 15.7, 6.36, and 10.33, respectively. Aqueous NaOH: _________ Aqueous NaHCO3: _________ Aqueous Na2CO3: _________
Answer:
Aqueous NaOH: soluble
Aqueous NaHCO₃: insoluble
Aqueous Na₂CO₃: soluble
Explanation:
The organic acid is insoluble. Its salt (ionic) is soluble.
The important principle is:
If you have two acids in a flask, the stronger acid (smaller pKₐ) will protonate the weaker one. The stronger acid will become ionic and therefore more soluble.
1. In NaOH
Let's write the formula for 4-nitrobenzoic acid as HA.
The equation for the reaction is
HA + OH⁻ ⇌ A⁻ + H₂O
pKₐ: 7.15 15.7
HA is the stronger acid. It will protonate the hydroxide ion and be converted to the soluble 4-nitrobenzoate ion.
4-Nitrophenol is soluble in NaOH.
2. In NaHCO₃
HA + HCO₃⁻ ⇌ A⁻ + H₂CO₃
pKₐ: 7.15 6.36
HCO₃⁻ is the stronger acid. It will protonate 4-nitrophenol.
4-Nitrobenzoic acid is insoluble in NaHCO₃.
3. In Na₂CO₃
HA + CO₃²⁻ ⇌ A⁻ + H₂CO₃
pKₐ: 7.15 10.33
HA is the stronger acid. It will protonate the carbonate ion.
4-Nitrophenol is soluble in Na₂CO₃.
Propane (C3H8) burns in a combustion reaction. How many grams of C3H8 are needed to produce 80.3 mols CO2 ?
Answer:
1177.88g of C3H8
Explanation:
We'll begin by writing the balanced equation for the reaction. This is given below:
C3H8 + 5O2 —> 3CO2 + 4H2O
Next we shall determine the number of mole of C3H8 required to produce 80.3 moles of CO2. This is illustrated below:
From the balanced equation above,
1 mole of C3H8 reacted to produce 3 moles of CO2.
Therefore, Xmol of C3H8 will react to produce 80.3 moles of CO2 i.e
Xmol of C3H8 = 80.3/3
Xmol of C3H8 = 26.77 moles
Finally, we shall convert 26.77 moles of C3H8 to grams.
Molar mass of C3H8 = (3x12) + (8x1) = 44g/mol
Mole of C3H8 = 26.77 moles
Mass of C3H8 =..?
Mass = mole x molar mass
Mass of C3H8 = 26.77 x 44
Mass of C3H8 = 1177.88g
Therefore, 1177.88g of C3H8 are needed for the reaction
One of the reagents below gives predominantly 1,2 addition (direct addition) while the other gives predominantly 1,4 addition (conjugate addition). a) Which major organic product is the result of 1,2 addition? ---Select--- b) Draw the skeletal structure of major organic product A
Complete Question
The complete question is shown on the first uploaded image
Answer:
a
The correct option is reagent B
b
The skeletal structure of major organic product A is shown on the third uploaded image
Explanation:
The mechanism of the reaction for A and B are shown on the second the second reaction and looking at this we can see that the reagent that predominately gives 1,2 addition is reagent B
Which best describes thermal energy? It is the difference between internal energies of two or more substances. It is the sum of internal energies of two or more substances. It is the portion of internal energy that can be transferred from one substance to another. It is the portion of potential energy that can be transferred from one substance to another.
Answer:
It is the portion of internal energy that can be transferred from one substance to another.
Thermal energy is the portion of internal energy that can be transferred from one substance to another.
What is thermal energy?Thermal energy is the energy an object posses which is as a result of particles movement within it.
It is also the internal energy system in a state of thermodynamic equilibrium which is as a result of its temperature. Thermal energy cannot be concert to useful work easily.
Therefore, thermal energy is the portion of internal energy that can be transferred from one substance to another.
Learn more about thermal energy from the link below.
https://brainly.com/question/19666326
which statements describe how chemical formulas, such as H2O, represent compounds?
Answer:
2 Hydrogen One oxygen
Explanation:
ch3-ch2-ch-ch(cl)-ch=o IUPAC name
Answer:
2-chloropentanal
Answer:
2-chloropentanal
Explanation:
ch3-ch2-ch-ch(cl)-ch=o IUPAC name
H H H H
H - C - C - C - C - C = O
H H H Cl
So as can be seen 2 as the Chlorine is on the second carbon.
Chloro because of the chlorine.
Pent because there's 5 carbon
al because there's an aldehydes
Aldehyde = −CHO
2-chloropentanal
In E. coli, the enzyme hexokinase catalyzes the reaction: Glucose + ATP → glucose 6-phosphate + ADP The equilibrium constant, Keq, is 7.8 x 102. In the living E. coli cells, [ATP] = 7.9 mM; [ADP] = 1.04 mM, [glucose] = 2 mM, [glucose 6-phosphate] = 1 mM. Determine if the reaction is at equilibrium. If the reaction is not at equilibrium, determine which side the reaction favors in living E. coli cells.
Answer:
Explanation:
Glucose + ATP → glucose 6-phosphate + ADP The equilibrium constant, Keq, is 7.8 x 102.
In the living E. coli cells,
[ATP] = 7.9 mM;
[ADP] = 1.04 mM,
[glucose] = 2 mM,
[glucose 6-phosphate] = 1 mM.
Determine if the reaction is at equilibrium. If the reaction is not at equilibrium, determine which side the reaction favors in living E. coli cells.
The reaction is given as
Glucose + ATP → glucose 6-phosphate + ADP
Now reaction quotient for given equation above is
[tex]q=\frac{[\text {glucose 6-phosphate}][ADP]}{[Glucose][ATP]}[/tex]
[tex]q=\frac{(1mm)\times (1.04 mm)}{(7.9mm)\times (2mm)} \\\\=6.582\times 10^{-2}[/tex]
so,
[tex]q<<K_e_q[/tex] ⇒ following this criteria the reaction will go towards the right direction ( that is forward reaction is favorable until q = Keq
After recrystallizing an impure sample with isopropanol, you isolate your product by filtration. What solvent do you use to wash your crystals? Room temperature distilled water Room temperature isopropanol Ice cold distilled water Ice cold isopropanol
Answer:
The correct answer is ice cold isopropanol.
Explanation:
Any compound in the initial stage is first dissolved in any suitable solvent and is heated for a certain duration for the process of recrystallization. Afterward, the compound is kept at room temperature so that it gets cooled gradually. In the process, the impurities remain dissolved in the solvent and the pure compound gets separated in the form of a precipitate.
Post all this, the filtration of the pure compound is done and is then washed with the cold solvent, which was initially used to dissolve the compound. Therefore, the appropriate solvent to use in the process is ice-cold isopropanol.
A water tank measures 24in.×48in.×12in. Find the capacity of the water tank in cubic feet. Do not include units in your answer.
Answer: 8 (feet)
Explanation:
24 inches = 2 feet
48 inches = 4 feet
12 inches = 1 foot
To find volume you do Base * Width * Height
2*4*1 = 8
Hope this helps!
The correct answer is 8 (feet).
How to calculate ?
24 inches = 2 feet48 inches = 4 feet12 inches = 1 footTo find volume the method is Base * Width * Height
Therefore, 2*4*1 = 8
Hence, the capacity of the water tank in cubic feet is 8 feet.learn more about capacity below,
https://brainly.com/question/15747509
#SPJ2
Unscramble the following words to form a complete
sentence about the cycles of nature:
limited is through environment Matter recycled the on Earth is and
Answer:
recycled is limited through enviroment and matter on earth
Explanation:
What is the balanced form of the chemical equation shown below?
Na2SO4(aq) + Sr(NO3)2(aq) → SrSO4(s) + NaNO3(aq)
A. Na2SO4(aq) + Sr(NO3)2(aq)
SISO4(s) + NaNO3(aq)
B. NaSO4(aq) + SINO3(aq) → SSO4(s) + NaNO3(aq)
C. Na2SO4(aq) → SrSO4(s) + 2NaNO3(aq)
Ο Ο
D. Na2SO4(aq) + 2Sr(NO3)2(aq) → 2SSO4(s) + 2NaNO3(aq)
Answer:
C. Na₂SO₄(aq) + Sr(NO₃)₂(aq) → SrSO₄(s) + 2 NaNO₃(aq)
Explanation:
Based on the equation:
Na₂SO₄(aq) + Sr(NO₃)₂(aq) → SrSO₄(s) + NaNO₃(aq)
As you can see, sulfate ions (SO₄) are been replaced for nitrate ions (NO₃). That is a double replacement reaction and is a very important information because 2 NO₃ ions in Sr(NO₃)₂ are producing 1 NO₃ ion. To balance NO₃:
Na₂SO₄(aq) + Sr(NO₃)₂(aq) → SrSO₄(s) + 2 NaNO₃(aq)
1 SO₄ ion in Na₂SO₄ produce 1 SO₄ ion in SrSO₄. And Na and Sr metals are balanced yet. Thus, the balanced form of this chemical equation is:
Na₂SO₄(aq) + Sr(NO₃)₂(aq) → SrSO₄(s) + 2 NaNO₃(aq)In the activity, click on the Keq and ΔG∘ quantities to observe how they are related. Calculate ΔG∘using this relationship and the equilibrium constant (Keq) obtained in Part A at T=298K:Keq=1.24×1020Express the Gibbs free energy (ΔG∘) in joules to three significant figures.
Answer: The Gibbs free energy of the reaction is -114629.4 J
Explanation:
To calculate the Gibbs free energy of the reaction, we use the equation:
[tex]\Delta G^o=-RT\ln K_{eq}[/tex]
where,
[tex]\Delta G^o[/tex] = Gibbs free energy of the reaction = ?
R = Gas constant = [tex]8.314 J/K.mol[/tex]
T = temperature of the reaction = 298 K
[tex]K_{eq}[/tex] = equilibrium constant of the reaction = [tex]1.24\times 10^{20}[/tex]
Putting values in above equation, we get:
[tex]\Delta G^o=-(8.314J/mol.K\times 298K\times \ln (1.24\times 10^{20}))\\\\\Delta G^o=-114629.4J[/tex]
Hence, the Gibbs free energy of the reaction is -114629.4 J
A blood sample is left on a phlebotomy tray for 4 hours before it is delivered to the laboratory. Which group of tests could be performed:
Fractionation of Crude Oil Select the correct ranking of the following alkanes according to the height reached in a fractionating column, highest first: butane, heptadecane, dodecane, ethane, decane Select the correct ranking of the following alkanes according to the height reached in a fractionating column, highest first:
butane, heptadecane, dodecane, ethane, decane
A. ethane > butane > decane > dodecane > heptadecane
B. heptadecane > > dodecane > decane butane > ethane
C. ethane > butane > decane> heptadecane >
D. dodecane butane > ethane > decane > dodecane > heptadecane
Answer:
A. ethane > butane > decane > dodecane > heptadecane
Explanation:
In fractionating column, crude oil is separated by means of fractional distillation due to the wide range of boiling point of the crude products such as ethane, propane, butane pentane etc.
The product with the least weight rises to top height while the product with highest weight will move down.
For the given hydrocarbon products, the ranking according to their molecular weight, starting with the lighter product to heavier product is
ethane (C2), butane (C4), decane(C10), dodecane (C12), heptadecane(C17).
Thus, the correct ranking, starting with the product that will rise highest is ethane > butane > decane > dodecane > heptadecane
Show work plzzz
Unknown Metal Bar #8
Mass of Unknown Metal bar 11.3g
Length of bar 13.90cm
Width of bar 2.9cm
Thickness of bar 0.081cm
1. Calculate the volume of the bar:
2. Calculate the (experimental) density of the bar:
3. Based on the provided list of (true) densities, what is the possible identity of the Unknown metal?
4. What is the percent difference between the true density of your metal and the calculated density?
= | − | ∗ 100%
Answer:
1= Volume
= Length x breath x height
= 13.90 x 2.9 x 0.081
=3.26511
2= Density = Mass ÷ volume
= 11.3 ÷ 3.26511
= 3.461 (3d.p)
idk the rest because you haven't shown a picture of the rest
Answer:
1. 3.3 cm³; 2. 3.5 g/cm³; 3. barium; 4. 4%
Explanation:
Experimental data:
Mass = 11.3 g
Length = 13.90 cm
Width = 2.9 cm
Thickness = 0.081 cm
Calculations:
1. Volume of bar
V = lwh = 13.90 cm × 2.9 cm × 0.081 cm = 3.3 cm³
2. Experimental density
[tex]\text{Density} = \dfrac{m}{V} = \dfrac{\text{11.3 g}}{\text{3.27 cm}^{3}} = \textbf{3.5 cm}^{\mathbf{3}}[/tex]
3. Identity of metal
The three most likely metals are scandium (3.00 g/cm³), barium (3.59 g/cm³), and yttrium (4.47 g/cm³)
The metal is probably barium.
4. Percent difference
[tex]\begin{array}{rcl}\text{Percent difference}&= &\dfrac{\lvert \text{ True - Calculated}\lvert}{ \text{True}} \times 100 \,\%\\\\& = & \dfrac{\lvert 3.59 - 3.5\lvert}{3.59} \times 100 \, \% \\\\& = & \dfrac{\lvert 0.1\lvert}{3.59} \times 100 \, \%\\ \\& = & 0.04 \times 100 \, \%\\& = & \mathbf{4 \, \%}\\\end{array}\\\text{The percent difference is $\large \boxed{\mathbf{4 \, \%} }$}[/tex]
The breaking buffer that we use this week contains 10mM Tris, pH 8.0, 150mM NaCl. The elution buffer is breaking buffer that also contains 300mM imidazole. Describe how the instructor made the 0.25L elution buffer for all the students this week given 500ml of 1M of Tris (121.1 g/mole) (pH8.0), 750ml of 5M NaCl (MW
Answer:
Explanation:
From the given information ;the objective is to determine how the instructor made the 0.25L elution buffer
0.25 L elution buffer = 250 mL elution butter
The breaking buffer that we use this week contains
10mM Tris = 0.01 M
150mM NaCl = 0.15 M
300mM imidazole. = 0.3 M
The stock concentration of Tris in 1M
Therefore ; by using the formula: [tex]M_1V_1 = M_2 V_2[/tex]; we can determine the volume in the preparation; so;
[tex]1*V_1 = 0.0 1 \ M * 250 \ mL[/tex]
[tex]V_1 = \dfrac{0.0 1 \ M * 250 \ mL}{1 }[/tex]
[tex]V_1 = 2.5 \ mL[/tex]
In NaCl, The amount of stock concentration is 5 M
so; using the same formula; we have:
[tex]5*V_1 = 0.15 \ M * 250 \ mL[/tex]
[tex]V_1 = \dfrac{0. 15 \ M * 250 \ mL}{5 }[/tex]
[tex]V_1 = 7.5 \ mL[/tex]
From Imidazole ; the amount of stock concentration is
[tex]1*V_1 = 0.3 \ M * 250 \ mL[/tex]
[tex]V_1 = \dfrac{0. 3 \ M * 250 \ mL}{1 }[/tex]
[tex]V_1 = 75 \ mL[/tex]
Thus; we can have a table as shown as :
Stock concentration volume to be added Final concentration
1 M of Tris 2.5 mL 10 mM
5 M of NaCl 7.5 mL 150 mM
1 M of Imidazole 75 mL 300 mM
In conclusion. the addition of all the volume make up the 250 mL elution buffer that is equivalent to 0.25 L.
An ideal gas sealed in a rigid 4.86-L cylinder, initially at pressure Pi=10.90 atm, is cooled until the pressure in the cylinder is Pf=1.24 atm. What is the enthalpy change for this process? ΔH =
Answer:
[tex]\Delta H=-11897J[/tex]
Explanation:
Hello,
In this case, it is widely known that for isochoric processes, the change in the enthalpy is computed by:
[tex]\Delta H=\Delta U+V\Delta P[/tex]
Whereas the change in the internal energy is computed by:
[tex]\Delta U=nCv\Delta T[/tex]
So we compute the initial and final temperatures for one mole of the ideal gas:
[tex]T_1= \frac{P_1V}{nR}=\frac{10.90atm*4.86L}{0.082*n}=\frac{646.02K }{n} \\\\T_2= \frac{P_2V}{nR}=\frac{1.24atm*4.86L}{0.082*n}=\frac{73.49K }{n}[/tex]
Next, the change in the internal energy, since the volume-constant specific heat could be assumed as ³/₂R:
[tex]\Delta U=1mol*\frac{3}{2} (8.314\frac{J}{mol*K} )*(73.49K-646.02K )=-7140J[/tex]
Then, the volume-pressure product in Joules:
[tex]V\Delta P=4.86L*\frac{1m^3}{1000L} *(1.24atm-10.90atm)*\frac{101325Pa}{1atm} \\\\V\Delta P=-4756.96J[/tex]
Finally, the change in the enthalpy for the process:
[tex]\Delta H=-7140J-4757J\\\\\Delta H=-11897J[/tex]
Best regards.
The change in enthalpy is 70.42J
Data;
Volume of the gas = 4.86LInitial Pressure = 10.90 atmFinal Pressure = 1.24 atmChange in Enthalpy = ?Change in EnthalpyThe change of enthalpy is calculated as
[tex]\delta H = \delta V + \delta nRT\\\delta n = 0\\\delta H = \delta U \\[/tex]
The volume change is negligible
The change in enthalpy here is equal to change in internal energy over ΔE
[tex]\delta H = \delta U = nCv\delta T\\\delta H = \frac{3}{2}(nR\delta T)\\\delta H = \frac{3}{2}\{\delta PV)\\ \delta H = \frac{3}{2}[(10.90-1.24)*4.86] \\\delta H = 70.42J[/tex]
The change in enthalpy is 70.42J
Learn more on change in enthalpy here;
https://brainly.com/question/14047927
At 25.0 °C the Henry's Law constant for sulfur hexafluoride (SP) gas in water is 2.4x 10 M/atm Calculate the mass in grams of SFo, gas that can be dissolved in S25. ml. of water at 25.0 C and a SF, partial pressure of 1.90 atm Be sure your answer has the correct number of significant digits.
Complete Question
The complete question is shown on the first uploaded image
Answer:
The mass is [tex]m = 0.0349 \ g[/tex]
Explanation:
From the question we are told that
The Henry's Law constant is [tex]k = 2.4 *10^{10} M/atm[/tex]
The volume of water is [tex]V = 525 \ ml = 0.525 \ L[/tex]
The partial pressure is [tex]P = 1.90 \ atm[/tex]
The temperature is [tex]T = 25 ^oC[/tex]
Henry's law is mathematically represented as
[tex]C = P * k[/tex]
Where C is the concentration of sulfur hexafluoride(SP)
substituting value
[tex]C = 1.90 * 2.4*10^{-4}[/tex]
[tex]C = 4.56*10^{-4} \ M[/tex]
The number of moles of SP is mathematically represented as
[tex]n = C * V[/tex]
substituting value
[tex]n = 0.525 * 4.56*10^{-4}[/tex]
[tex]n = 2.39 *10^{-4} \ moles[/tex]
The mass of SP that dissolved is
[tex]m = n * Z[/tex]
Where Z is the molar mass of SP which has a constant value of
[tex]Z = 146 g/mole[/tex]
So
[tex]m = 2.394*10^{-4} * 146[/tex]
[tex]m = 0.0349 \ g[/tex]