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)A stock solution will be prepared by mixing the following chemicals together:
3.0 mL of 0.00200 M KSCN
10.0 mL of 0.200 M Fe(NO3)3
17.0 mL of 0.5 M HNO3
Determine the molar concentration of Fe(NO3)3 in the stock solution.
Answer:
0.067M Fe(NO3)3
Explanation:
A stock solution is a concentrated solution that is diluted to prepare the solutions that you will use.
The volume of the stock solution is 3.0mL + 10.0mL + 17.0mL= 30.0mL.
The ratio between volume of the aliquot (10.0mL) and total volume (30.0mL) is called dilution factor, that is: 30.0mL / 10.0mL = 3
That means the Fe(NO3)3 is diluted 3 times. That means the molar concentration of the stock solution is:
0.200M / 3 =
0.067M Fe(NO3)31. Reaccionan 9.7 Kg de un mineral de níquel al 70% con 8L de una solución de ácido fosfórico al 60% y con una densidad de 1.36g/ml.
Answer:
The reaction produces 201.4 g of hydrogen gas and 12.2 kg of Nickel Phosphate.
Explanation:
English Translation
9.7 Kg of a 70% nickel mineral react with 8L of a 60% phosphoric acid solution and with a density of 1.36g / ml.
Solution
The problem doesn't seen to be complete as it doesn't ask a question in the end. But, we will just calculate the amount of each product expected to cover the grounds.
The balanced chemical reaction between Nickel and Phosphoric acid is given as
3Ni + 2H₃PO₄ → 3H₂ + Ni₃(PO₄)₂
We need to first obtain the limiting reagent, that is, the reagent that is used up during the reaction and is in short supply. This reagent determines the amount of products that will be formed.
Mass of nickel that is present at the start = 70% of 9.7 kg = 6.79 kg
Mass of Phosphoric acid present at the start of the reaction = 60% of (8000 mL × 1.36 g/mL) = 6528 g = 6.528 kg
Converting both of these to number of moles
Number of moles = (mass)/(Molar mass)
For nickel,
Mass = 6.79 kg = 6790 g
Molar mass = 58.6934 g/mol
Number of moles at the start = (6790/58.6934) = 115.7 moles
For Phosphoric acid
Mass = 6528 g
Molar mass = 97.994 g/mol
Number of moles = (6528/97.994) = 66.6 moles
3 moles of Ni reacts with 2 moles of H₃PO₄
From the number of moles present initially, shows that Phosphoric acid is in limited supply and is the limiting reagent.
From the stoichiometric balance of the reaction
2 moles of H₃PO₄ gives 3 moles of H₂
66.6 moles of H₃PO₄ will give (66.6×3/2) of H₂, that is, 99.9 moles of H₂.
Mass of H₂ liberated from the reaction = (Number of moles) × (molar mass) = 99.9 × 2.016 = 201.3984 g = 201.4 g
2 moles of H₃PO₄ gives 1 mole of Ni₃(PO₄)₂
66.6 moles of H₃PO₄ will give (66.6×1/2) of Ni₃(PO₄)₂, that is, 33.3 moles of Ni₃(PO₄)₂.
Mass of Ni₃(PO₄)₂ produced from the reaction = (Number of moles) × (molar mass) = 33.3 × 366.02 = 12,188.466 g = 12.2 kg
Hope this Helps!!!
A red blood cell is placed into each of the following solutions. Indicate whether crenation, hemolysis, or neither will occur. Solution A: 3.75 % (m/v) NaCl Solution B: 1.92 % (m/v) glucose Solution C: distilled H2O Solution D: 9.03 % (m/v) glucose Solution E: 5.0% (m/v) glucose and 0.9% (m/v) NaCl
Answer:
A. The cell will undergo crenation
B. The cell will undergo hemolysis
C. The cell will undergo hemolysis
D. The cell will undergo crenation.
E. The cell will undergo neither crenation nor hemolysis
Explanation:
A hypotonic solution is a solution in which the concentration of solutes is greater inside the cell than outside the cell.
A hypertonic solution is a solution in which the concentration of solutes is greater outside the cell than inside the cell.
An isotonic solution is a solution in which the concentration of solution is the same outside and inside of the cell. A solution with 5% (m/v) glucose and 0.9% (m/v) NaCl is an isotonic solution.
When a red blood cell is placed in a hypotonic solution, it will swell and burst. This is known as hemolysis.
When placed in a hypertonic solution, a red blood cell will lose water and shrivel. This is is known as cremation.
When a red blood cell is placed in an isotonic solution, neither hemolysis or crenation occurs as there is no net movement of water across the cell's membrane.
A: 3.75 % (m/v) NaCl Solution is a hypertonic solution. The cell will undergo crenation
B: 1.92 % (m/v) glucose Solution is hypotonic. The cell will undergo hemolysis
C: Distilled H2O Solution is a hypotonic solution. The cell will undergo hemolysis
D: 9.03 % (m/v) glucose Solution is a hypertonic solution. The cell will undergo crenation
E: 5.0% (m/v) glucose and 0.9% (m/v) NaCl are both isotonic solutions. The cell will undergo neither hemolysis not crenation.
Solution A (3.75% NaCl): Crenation
Solution B (1.92% glucose): HemolysisSolution C (Distilled H2O): HemolysisSolution D (9.03% glucose): CrenationSolution E (5.0% glucose and 0.9% NaCl): CrenationWhat is the SolutionTo find out if a red blood cell will shrink, burst, or stay the same when placed in a solution, we have to think about how concentrated the solution is compared to the red blood cell.
A red blood cell has a normal concentration of about 0. 9% salt or 0. 3% sugar Solutions that have more concentrated substances are called hypertonic, while solutions that have less concentrated substances are called hypotonic.
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A scientist mixed 25.00 mL of 2.00 M KOH with 25.00 mL of 2.00 M HBr. The temperature of the mixed solution rose from 22.7 oC to 31.9 oC. Calculate the enthalpy change for the reaction in kJ/mol HBr, assuming that the calorimeter loses negligible heat, that the volumes are additive, and that the solution density is 1.00 g/mL, and that its specific heat is 4.184 J/g.oC.
Answer:
38.493 KJ/mol
Explanation:
Equation of reaction; HBr + KOH ---> KBr + H2O
Heat evolved = mass * specific heat capacity * temperature rise
Mass of solution = density * volume
Mass = 1.00 g/ml*50 ml = 50g
Temperature rise = 31.9 - 22.7 = 9.2 °C
Heat evolved = 50 * 4.184 * 9.2 = 1924.64 J
From the equation of reaction, 1 mole of HBr reacts with 1mole of KOH to produce 1 mole of H20
Number of moles of HBr involved in the reaction = molar concentration * volume (L)
Molar concentration = 2.0 M, volume = 25 ml = 0.025 L
Number of moles = 2.0 M * 0.025 L= 0.05 moles
Therefore, 0.05 moles of HBr reacts with 0.05 moles of KOH to produce 0.05 moles of H20
Enthalpy change per mole of HBr = 1924.64 J/0.05 moles = 38492.8 J/mol = 38.493 KJ/mol
what is a mitochondrion
Explanation:
Mitochondria (sing. mitochondria) are organelles, or parts of the eukaryote cell. They are in the cytoplasm, not the nucleus. They make the most cell supply of adenosine triphosphate (ATP), a molecule that cells use as an energy source. ... This means that mitochondria are known as '' the powerhouse of the cell'' or ''cell strength".
Good Luck, and have a great day..
a) A molecule of DNA contains 140 A bases, how many T bases will it contain? Plz help ❤️
Answer:
140 T or thymine base
Explanation:
Adenine pairs with Thymine in DNA thus the number of adenine will always equal number of thymine (unless some sort of mutation), therefore in this problem you have 140 A so you have 140 T as well. Remember: Adenine (A) and Thymine(T) is equal, & Cytosine (C) and Guanine (G) is equal
Consider the following specific heats of metals. Metal Specific Heat Aluminum 0.897 J/(g°C) Magnesium 1.02 J/(g°C) Lithium 3.58 J/(g°C) Silver 0.237 J/(g°C) Gold 0.129 J/(g°C) If the same amount of heat is added to 25.0 g of each of the metals, which are all at the same initial temperature, which metal will have the lowest temperature?
Answer:
Lithium
Explanation:
The specific geat capacity of a substance is the energy required to raise 1 unit of that substance by one degree.
Heat energy (Q) = mc∇t
Q = heat energy
M = mass of the substance
c = specific heat capacity
∇t = change in temperature of the substance
Generally, increase in the specific heat capacity will lead to a lower final temperature likewise decrease in the specific heat capacity will lead to increase the final temperature of the substance.
From the data above, we can take just two specific heat capacity and test this theory.
Assuming we have a
Mass = 25g
Heat energy applied (Q) = 1 J
Initial temperature (T1) = 10°C
Final temperature (T2) = ?
Q = mc∇t
Q = mc (T2 - T1)
For Lithium, specific heat capacity = 3.58J/g°C
1 = 25 × 3.58 (T2 - 10)
Solve for T2
1 = 89.5 (T2 - 10)
1 = 89.5T2 - 895
89.5T2 = 896
T2 = 896 / 89.5
T2 = 10.011°C
For Magnesium (Mg) specific heat capacity = 1.02J/g°C
Q = mc∇t
1 = 25 × 1.02 × (T2 - 10)
1 = 25.5 (T2 - 10)
1 = 25.5T2 - 255
Solve for T2
25.5T2 = 256
T2 = 10.039°C
Notice the trend that decrease in the specific heat capacity leads to increase in the final temperature.
Try and continue for the elements and see how it works.
A laser is used in eye surgery to weld a detached retina back into place. The wavelength of the laser beam is 503 nm, while the power is 1.4 W. During surgery, the laser beam is turned on for 0.070 s. During this time, how many photons are emitted by the laser?
Answer:
Number of proton emmitted by laser=[tex]2.48*10^17proton[/tex]
Explanation:
Energy is the ability to cause change; power is directly proportional to energy and its the rate energy is utilized.
Power=energy/time.
First we need to calculate the total energy used which is equal to the total power utilized.
E(total)= P( total) = 1.4W × 0.070 s =[tex]0.098J[/tex]
CHECK THE ATTACHMENT FOR THE REMAINING DETAILED CALCULATION
How many moles of CO2 are produced when 84 0 mol O2 completely react?
Answer:
Explanation:boom
Which is true regarding a water molecule?
Answer:
Has many answers, but one is that it consists of small polar v shaped molecules with a molecular formula H20.
Explanation:
Water molecules consists of 2 hydrogen atoms bonded with on oxygen atom. Each molecule is electrically neutral but polar, with the center of positive and negative charges located in different places.
Each hydrogen atom has a nucleus consisting of a single positively-charged proton surrounded by a 'cloud' of a single negatively-charged electron and the oxygen atom has a nucleus consisting of eight positively-charged protons and eight uncharged neutrons surrounded by a 'cloud' of eight negatively-charged electrons.
Hoped this helped!
What is Keq for the reaction 2HCl(9) = H2(g) + Cl2(g)?
Answer:
Keq= [(Cl2) (H2)] / (HCl)^2
Explanation:
Equilibrium Constant, Keq, is written as products/reactants.
So it's going to be Keq= [(Cl2) (H2)] / (HCl)^2
An ethylene glycol solution contains 21.4 g of ethylene glycol (C2H6O2) in 97.6 mL of water.
Required:
a. Determine the freezing point of the solution. Express you answer in degrees Celsius. (Assume a density of 1.00 g/mL for water.)
b. Compute the boiling point of the solution. (Assume a density of 1.00 g/mL for water.)
Answer:
a. TFinal = -6.57°C
b. Tfinal = 101.80°C
Explanation:
When a solute is added to a solvent producing an ideal solution, the freezing point of the solution decreases with regard to pure solvent. Also, boiling point increases with regard to pure solvent.
The formulas are:
Freezing point:
ΔT = Kf×m×i
Where Kf is freezeing point depression constant of water (1.86°C/m), m is molality of solution and i is van't Hoff factor (1 for ethylene glycol).
Boiling point:
ΔT = Kb×m×i
Where K is freezeing point depression constant of water (0.51°C/m), m is molality of solution and i is van't Hoff factor (1 for ethylene glycol).
Moles of 21.4g of ethylene glycol (Molar mass: 62.07g/mol) are:
21.4g C₂H₆O₂ ₓ (1mol / 62.07g) = 0.345 moles
And kg of 97.6mL of water = 97.6g are 0.0976kg. Molality of the solution is:
0.345mol / 0.0976kg = 3.5325m
Replacing in the formulas:
a. Freezing point:
ΔT = 1.86C/m×3.5325m×1
ΔT = 6.57°C
0°C - Tfinal = 6.57°C
TFinal = -6.57°Cb. Boiling point:
ΔT = 0.51°C/m×3.5325m×1
ΔT = 1.80°C
Tfinal - 100°C = 1.80°C
Tfinal = 101.80°C
which statements describe how chemical formulas, such as H2O, represent compounds?
Answer:
2 Hydrogen One oxygen
Explanation:
A student dissolved 5.00 g of Co(NO3)2 in enough water to make 100. mL of stock solution. He took 4.00 mL of the stock solution and then diluted it with water to give 275. mL of a final solution. How many grams of NO3- ion are there in the final solution?
Answer:
0.136g
Explanation:
A student dissolved 5.00 g of Co(NO3)2 in enough water to make 100. mL of stock solution. He took 4.00 mL of the stock solution and then diluted it with water to give 275. mL of a final solution. How many grams of NO3- ion are there in the final solution?
[tex]Co(NO_3)_2(aq)\rightarrow Co^{2+}(aq)+2NO_3^{-}(aq)[/tex]
Initial mole of Co(NO3)2 [tex]=\frac{mass}{molar mass}[/tex]
[tex]=\frac{5.00}{182.94} \\\\=0.02733mol[/tex]
Mole of Co(NO3)2 in final solution
[tex]=\frac{4.00}{100}\times 0.02733\\\\=0.04\times 0.02733\\\\= 0.001093mol[/tex]
Mole of NO3- in final solution = 2 x Mole of Co(NO3)2
[tex]=2\times 0.001093\\\\=0.002186mol[/tex]
Mass of NO3- in final solution is mole x Molar mass of NO3
[tex]=0.002186\times62.01\\\\=0.136g[/tex]
The final solution contains 0.24 g of nitrate ion.
Number of moles of Co(NO3)2 = 5.00 g/183 g/mol = 0.027 moles
Number of moles = concentration × volume
concentration = Number of moles /volume
Volume of solution = 100 mL or 0.1 L
concentration = 0.027 moles/0.1 L = 0.27 M
Using the dilution formula;
C1V1 = C2V2
C1 = 0.27 M
V1 = 4.00 mL
C2 = ?
V2 = 275. mL
C2 = C1V1/V2
C2 = 0.27 × 4.00/ 275
C2 = 0.0039 M
Number of moles of NO3- ion in Co(NO3)2 = 0.0039 M × 62 g/mol = 0.24 g
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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
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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.
Amylase is the enzyme that controls the breakdown of starch to glucose. Describe how the student could investigate the effect of pH on the breakdown of starch by amylase.
Answer:
Explanation:
You will investigate the breakdown of starch by amylase at different pHs.
The different pHs under investigation will be produced using buffer solutions. Buffer solutions produce a particular pH, and will maintain it if other substances are added.
The amylase will break down the starch.
A series of test tubes containing a mixture of starch and amylase is set up at different pHs.
A sample is removed from the test tubes every 10 seconds to test for the presence of starch. Iodine solution will turn a blue/black colour when starch is present, so when all the starch is broken down, a blue-black colour is no longer produced. The iodine solution will remain orange-brown.
A control experiment must be set up - without the amylase - to make sure that the starch would not break down anyway, in the absence of an enzyme. The result of the control experiment must be negative - the colour must remain blue-black - for results with the enzyme to be valid.
When the starch solution is added:
Start timing immediately.Remove a sample immediately and test it with iodine solution.Sample the starch-amylase mixture continuously, for example every 10 seconds.For each pH investigated, record the time taken for the disappearance of starch, ie when the iodine solution in the spotting tile remains orange-brown.
The time taken for the disappearance of starch is not the rate of reaction.
It will give us an indication of the rate, but is the inverse of the rate - the shorter the time taken, the greater the rate of the reaction.
We can calculate the rate of the reaction by calculating \frac{1}{t}, obtaining a measure of the rate of reaction by dividing one by the time taken for the reaction to occur.
A similar experiment can be carried out to investigate the effect of temperature on amylase activity.
Set up a series of test tubes in the same way and maintain these at different temperatures using a water bath - either electrical or a heated beaker of water.
Depending on the chemical reaction under investigation, you might monitor the reaction in a different way. If investigating the effect of temperature on the breakdown of lipid by lipase, you could monitor pH change - lipids are broken down into fatty acids and glycerol. As the reaction begins, the release of fatty acids will mean that the pH will decrease.
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For the iodine trichloride molecule: a. Determine the number of valence electrons for each atom in the molecule b. Draw the Lewis Dot structure c. Describe why the molecule is drawn this way (i.e. any extra rules/steps needed?) d. Show the polarity of each bond and for the molecule by drawing in the dipole +à
Answer:
Explanation:
a. Determine the number of valence electrons for each atom in the molecule
In this case we both atoms are halogens. Therefore we will have 7 electrons for each atom.
b. Draw the Lewis Dot structure
In this case, the formula is [tex]ICl_3[/tex], so the central atom would be "I" and the "Cl" atoms would be placed around "I". See figure 1
c. Describe why the molecule is drawn this way (i.e. any extra rules/steps needed?)
In this specific case, the "I" atom don't follow the octet rule. We will have an expanded octet for iodine (more than 8 electrons).
d. Show the polarity of each bond and for the molecule by drawing in the dipole +d
The negative dipole would be placed in the atom with higher electronegativity, in this case "Cl". The positive dipole would be placed in the atom with low electronegativity, in this case "I".
I hope it helps!
The functional groups in an organic compound can frequently be deduced from its infrared absorption spectrum. A compound containing C, H, and O exhibits intense absorption at 1720 cm-1. No additional information is available. List possible classes for which there is positive evidence.
Relative absorption intensity: (s)=strong, (m)=medium, (w)=weak.
What functional class(es) does the compound belong to?
List only classes for which evidence is given here. Attach no significance to evidence not cited explicitly.
Do not over-interpret exact absorption band positions. None of your inferences should depend on small differences like 10 to 20 cm-1.
a. alkane (List only if no other functional class applies.)
b. alkene h. amine
c. terminal alkyne i. aldehyde or ketone
d. internal alkyne j. carboxylic acid
e. arene k. ester
f. alcohol l. nitrile
g. ether
Answer:
The class of this compound is aldehyde or ketone (i).
Explanation:
Absorption peak at 1720 cm-1 shows the presence of a carbonyl group, possibly an aldehyde or ketone with C=O bond.
Further information on molecular formula would be required for structural elucidation.
At high temperatures one mole of hydrogen gas reacts with one mole of bromine gas to form hydrogen bromide. At a given temperature the equilibrium constant is 57.6. If at the same temperature, a mixture of 4.67 × 10^-3M bromine gas, 2.14 × 10^−3 hydrogen gas, and 2.40 × 10^−2M hydrogen bromide gas is made, then:
a. the system is at equilibrium.
b. the system is far from equilibrium and will shift to form more hydrogen gas.
c. the system is far from equilibrium and will shift to form more hydrogen bromide gas.
d. nothing can be deduced since we do not know whether the reaction is endothermic or exothermic.
e. nothing can be deduced since we do not know whether the equilibrium constant is Kc or Kp.
Answer:
a. the system is at equilibrium.
Explanation:
Hello,
In this case, the undergoing chemical reaction is:
[tex]H_2+Br_2\rightleftharpoons 2HBr[/tex]
Thus, the law of mass action is given by:
[tex]Keq=\frac{[HBr]^2}{[H_2][Br_2]} =57.6[/tex]
Nonetheless, for the given point of 4.67 × 10^-3M bromine gas, 2.14 × 10^−3 hydrogen gas, and 2.40 × 10^−2M hydrogen bromide gas we should compute the reaction quotient in order to know whether the direction of the reaction is to left or to right, thus:
[tex]Q=\frac{[HBr]^2}{[H_2][Br_2]} =\frac{(2.40x10^{-2})^2}{(4.67x10^{-3})(2.14x10^{-3})} \\\\Q=57.6[/tex]
Therefore, since Keq=Q, we say that the system is at equilibrium, for that reason, the answer is a.
Best regards.
Which of the following is an example of a mechanical wave?
O A. A light ray
B. A seismic wave
C. A radio wave
D. An X-ray
Answer:
A seismic wave
Explanation:
It requires a medium for its propagation.
4-Methylphenol, CH3C6H4OH (pKa 10.26), is only slightly soluble in water, but its sodium salt, CH3C6H4O-Na , is quite soluble in water. Describe the solubility of 4-methylphenol 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.
Explanation:
We know that more is the [tex]pK_{a}[/tex] value, weaker will be the acid. Also, an acid completely dissociates into ions in an aqueous base solution when [tex]pK_{a}[/tex] of conjugate acid of base is greater than acid.
4-methylphenol [tex](CH_{3}C_{6}H_{4}OH)[/tex] ([tex]pK_{a} = 10.26[/tex]) is quite soluble in its sodium salt. In NaOH, the dissociation will be as [tex]Na^{+}[/tex] and [tex]OH^{-}[/tex] ions as NaOH is a strong base.
Therefore, 4-methylphenol will readily dissolve in NaOH solution.
As, [tex]NaHCO_{3}[/tex] is not a strong base but as 4-Methylphenol forms a sodium salt hence, it will have a low solubility as compared to NaOH.
Whereas [tex]Na_{2}CO_{3}[/tex] is not a base but when dissolved in water it shows basic character as it produces NaOH (strong base) and [tex]H_{2}CO_{3}[/tex] (weak acid). As a result, the solution gets basic. Hence, 4-methylphenol will readily dissolve in [tex]Na_{2}CO_{3}[/tex].
When an unsymmetrical alkene such as propene is treated with N-bromosuccinimide in aqueous dimethyl sulfoxide, the major product has the bromine atom bonded to the less highly substituted carbon atom. Is this Markovnikov or non-Markovnikov orientation
When an unsymmetrical alkene such as propene is treated with N-bromosuccinimide in aqueous dimethyl sulfoxide, the major product has the bromine atom bonded to the less highly substituted carbon atom. This reaction describes a non-Markovnikov orientation.
In the reaction between an unsymmetrical alkene (such as propene) and N-bromosuccinimide (NBS) in the presence of aqueous dimethyl sulfoxide (DMSO), the major product is formed with the bromine atom bonded to the less highly substituted carbon atom of the alkene.
In Markovnikov's addition, the major product is formed by adding the electrophile (in this case, the bromine atom) to the carbon atom with more hydrogen atoms bonded to it. However, the given reaction exhibits non-Markovnikov selectivity, as the bromine atom adds to the less substituted carbon atom.
This non-Markovnikov selectivity can be attributed to the presence of DMSO, which acts as a polar solvent and helps generate a bromine radical (Br•). The radical intermediate can then undergo reaction with the alkene, leading to the observed regioselectivity where the bromine atom adds to the less substituted carbon. This process is known as a radical addition reaction.
Hence, the reaction demonstrates a non-Markovnikov orientation due to the addition of the bromine atom to the less highly substituted carbon atom of the propene molecule.
Learn more about the Markovnikov rule here:
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g a solution is made by mixing 500.0 mL of 0.037980.03798 M Na2sO4 Na2sO4 with 500.0 mL of 0.034280.03428 M NaOH NaOH . Complete the mass balance expressions for the sodium and arsenate species in the final solution.
Answer:
The concentration of the sodium and arsenate ions at the end of the reaction in the final solution
[Na⁺] = 0.05512 M
[HAsO₄²⁻] = 0.00185 M
[AsO₄³⁻] = 0.01714 M
Explanation:
Complete Question
A solution is made by 500.0 mL of 0.03798 M Na₂HAsO₄ with 500.0 mL of 0.03428 M NaOH. Complete the mass balance expressions for the sodium and arsenate species in the final solution.
Na₂HAsO₄ + NaOH → Na₃AsO₄ + H₂O
From the information provided in this question, we can calculate the number of moles of each reactant at the start of the reaction and we then determine which reagent is in excess and which one is the limiting reagent (in short supply and determines the amount of products to be formed)
Concentration in mol/L = (Number of moles) ÷ (Volume in L)
Number of moles = (Concentration in mol/L) × (Number of moles)
For Na₂HAsO₄
Concentration in mol/L = 0.03798 M
Volume in L = (500/1000) = 0.50 L
Number of moles = 0.03798 × 0.5 = 0.01899 mole
For NaOH
Concentration in mol/L = 0.03428 M
Volume in L = (500/1000) = 0.50 L
Number of moles = 0.03428 × 0.5 = 0.01714 mole
Since the NaOH is in short supply, it is evident that it is the limiting reagent and Na₂HAsO₄ is in excess.
Na₂HAsO₄ + NaOH → Na₃AsO₄ + H₂O
0.01899 0.01714 0 0 (At time t=0)
(0.01899 - 0.1714) | 0 → 0.01714 0.01714 (end)
0.00185 | 0 → 0.01714 0.01714 (end)
Hence, at the end of the reaction, the following compounds have the following number of moles
Na₂HAsO₄ = 0.00185 mole
This means Na⁺ has (2×0.00185) = 0.0037 mole at the end of the reaction and (HAsO₄)²⁻ has 0.00185 mole at the end of the reaction
NaOH = 0 mole
Na₃AsO₄ = 0.01714 moles
This means Na⁺ has (3×0.01714) = 0.05142 mole at the end of the reaction and (AsO₄)³⁻ has 0.01714 mole at the end of the reaction
H₂O = 0.01714 moles
So, at the end of the reaction
Na⁺ has 0.0037 + 0.05142 = 0.05512 mole
(HAsO₄)²⁻ has 0.00185 mole
(AsO₄)³⁻ has 0.01714 mole.
And since the Total volume of the reaction setup is now 500 mL + 500 mL = 1000 mL = 1 L
Hence, the concentration of the sodium and arsenate ions at the end of the reaction is
[Na⁺] = 0.05512 M
[HAsO₄²⁻] = 0.00185 M
[AsO₄³⁻] = 0.01714 M
Hope this Helps!!!
Answer:
[tex]\rm [Na^{+}]= \text{0.055 12 mol/L}[/tex]
[tex]\rm [HAsO_{4}^{2-}] + [AsO_{4}^{3-}] + [H_{2}AsO_{4}^{-}] + [H_{3}AsO_{4}] = \text{0.018 99 mol/L}[/tex]
Explanation:
The overall equation for the reaction is
Na₂HAsO₄ + NaOH ⟶ Na₃AsO₄ + H₂O
1. Mass balance for Na
All the Na⁺ comes from the Na₂HAsO₄ and the NaOH.
The mass balance equation for Na is
[tex]\rm c_{Na^{+}} = 2[Na^{+}]_{Na_{2}HAsO_{4}} + [Na^{+}]_{NaOH}[/tex]
At the moment of mixing and before the reaction started, the total volume had doubled, so the concentrations of each component were halved.
[Na₂HAsO₄] = ½ × 0.037 98 =0.018 99 mol·L⁻¹
[NaOH] = ½ × 0.034 28 = 0.017 14 mol·L⁻¹
[tex]\rm c_{Na^{+}} = 2\times 0.01899 + 0.01714 = 0.03798 + 0.01714\\c_{Na^{+}}= \textbf{0.055 12 mol/L}[/tex]
2. Mass balance for arsenate species
All the arsenate species come from the Na₂HAsO₄.
The reactions involved are
HAsO₄²⁻+ OH⁻ ⇌ AsO₄³⁻ + H₂O
HAsO₄²⁻ + H₂O ⇌ H₂AsO₄⁻ + OH⁻
H₂AsO₄⁻ + H₂O ⇌ H₃AsO₄ + OH⁻
The mass balance equation for arsenate species is
[tex]\rm c_{\text{arsenate}} = [HAsO_{4}^{2-}] + [AsO_{4}^{3-}] + [H_{2}AsO_{4}^{-}] + [H_{3}AsO_{4}][/tex]
At the moment of mixing, the concentration of Na₂HAsO₄ had halved.
[Na₂HAsO₄] = ½ × 0.039 78 = 0.018 99 mol·L⁻¹
[tex]\rm [HAsO_{4}^{2-}] + [AsO_{4}^{3-}] + [H_{2}AsO_{4}^{-}] + [H_{3}AsO_{4}] = \textbf{0.018 99 mol/L}[/tex]
What is the temperature at which the substance can be both in the solid and the liquid phase?
Answer: Gas–liquid–solid triple point
The single combination of pressure and temperature at which liquid water, solid ice, and water vapor can coexist in a stable equilibrium occurs at approximately 273.1575 K (0.0075 °C; 32.0135 °F) and a partial vapor pressure of 611.657 pascals (6.11657 mbar; 0.00603659 atm).
Explanation:
It represents the equilibrium between the liquid and gas phases. The point on this curve where the vapor pressure is 1 atm is the normal boiling point of the substance. The vapor-pressure curve ends at the critical point (B), which is at the critical temperature and critical pressure of the substance.
Magnesium and nitrogen react in a combination reaction to produce magnesium nitride:
3 Mg + N2 → Mg3N2
In a particular experiment, a 8.33-g sample of N2 reacts completely. The mass of Mg consumed is ________ g.
Answer:
21.7 g
Explanation:
Step 1: Write the balanced equation
3 Mg + N₂ → Mg₃N₂
Step 2: Calculate the moles corresponding to 8.33 g of nitrogen
The molar mass of N₂ is 28.01 g/mol.
[tex]8.33 g \times \frac{1mol}{28.01g} =0.297mol[/tex]
Step 3: Calculate the moles of magnesium that reacts with 0.297 moles of nitrogen
The molar ratio of Mg to N₂ is 3:1. The reacting moles of Mg are 3/1 × 0.297 mol = 0.891 mol
Step 4: Calculate the mass corresponding to 0.891 moles of magnesium
The molar mass of Mg is 24.31 g/mol.
[tex]0.891 mol \times \frac{24.31g}{mol} = 21.7 g[/tex]
Answer:
[tex]m_{Mg}=21.7 g Mg[/tex]
Explanation:
Hello,
In this case, considering the given reaction, we are able to compute the mass of magnesium that is consumed by considering its molar mass (24.31 g/mol), the molar mass of diatomic nitrogen (28.02 g/mol), the initial mass of nitrogen (8.33 g) and the 3:1 molar ratio of magnesium to nitrogen in the reaction.
Hence we compute it by applying the shown below stoichiometric procedure:
[tex]m_{Mg}=8.33 gN_2*\frac{1molN_2}{28.02gN_2} *\frac{3molMg}{1molN_2} *\frac{24.31gMg}{1molMg} \\\\m_{Mg}=21.7 g Mg[/tex]
Regards.
what is the sign of Mercury
Answer:
The answer is Hg.
Explanation:
Symbol for Mercury is Hg.
how many molecules (not moles) of NH3 are produced from 5.25x10^-4 g of H2?
due in a few, please help. will mark as brainliest
Answer:
not 100% but i think its 1.57x10^20
Explanation:
5.25x10^-4g / 2.016g
2.60x10^-4 x 6.022x10^23= 1.56x10^20 molecules
g Identify the type of reaction [(combination (c), decomposition (d), combustion (co), single replacement (sr), double replacement (dr) , or neutralization (n)] and write a balanced equation. If there is no reaction, write NR. Aqueous ammonium phosphate reacts with aqueous calcium sulfide.
Answer: [tex]2(NH_4)_3PO_4(aq)+3CaS(aq)\rightarrow Ca_3(PO_4)_2(aq)+3(NH_4)_2S(s)[/tex] is a double displacement reaction (dr).
Explanation:
A double displacement reaction (dr) is one in which exchange of ions take place. The salts which are soluble in water are designated by symbol (aq) and those which are insoluble in water and remain in solid form are represented by (s) after their chemical formulas.
[tex]2(NH_4)_3PO_4(aq)+3CaS(aq)\rightarrow Ca_3(PO_4)_2(aq)+3(NH_4)_2S(s)[/tex]
Combination reaction (C) is defined as the reaction where substances combine in their elemental state to form a single compound.
Single displacement reaction (sr) is defined as the reaction where more reactive element displaces a less reactive element from its chemical reaction.
Decomposition reaction (d) is defined as the reaction where a single substance breaks down into two or more simpler substances.
Combustion (Co) is a type of chemical reaction in which hydrocarbons burn in the presence of oxygen to form carbon dioxide and water along with heat.
When vinylcyclohexane is treated with in dichloromethane, the major product is (2-bromo ethylidene)cyclohexane . Account for the formation of this product by drawing the structure of the most stable radical intermediate. Include all valence lone pairs in your answer. Include all valence radical electrons in your answer.
Answer:
Explanation:
Vinylcyclohexane is an example of a cyclic hydrocarbon where the vinyl group (-CH=CH₂ ) attaches itself to an end of a cyclohexane in ring form thereby giving rise to a vinylcyclohexane. The vinyl group are ethylene with a reduction in one hydrogen atom given them the name vinyl.
SOo, when vinylcyclohexane is treated with NBS ( i.e N-Bromosuccinimide a chemical reagent used in organic reactions) ; the bromine in the NBS reacts with the cyclohexane thereby giving rise to a allyl radical first. The allyl radical is resonance stabilized radical with an unpaired electron on the allylic carbon . As a result of stabilization ; a more stable substituted cycloalkene is formed as an intermediate .
This stable substituted cycloalkene intermediate then finally react with a bromine ion to give a major product known as ; (2-bromo ethylidene)cyclohexane.
The diagram emphasizing more on the above explanation can be seen in the attached image below
Reactive oxygen species (ROS) are unstable, or reactive, compounds that result from the partial reduction of oxygen. ROS can cause damage to molecules, including membrane lipids and nucleic acids, and may be associated with some diseases. Which of these compounds are reactive oxygen species? Choose all that apply.
a. OH
b. OH-
c. O2-
d. H2O
e. H2O2
f. H-
