Answer:
Check the explanation
Explanation:
Kindly check the attached image below to see the step by step explanation to the question above.
Which of the following would you expect to be elastomers and which thermosetting polymers at room temperature? (a) Linear and highly crystalline polyethylene (b) Phenol-formaldehyde (c) Heavily crosslinked polyisoprene having a glass-transition temperature of 50oC (122oF) (d) Lightly crosslinked polyisoprene having a glass-transition temperature of -60oC (-76oF) (e) Linear and partially amorphous poly(vinyl chloride)
Answer: Please see below for answer
Explanation:A thermosetting polymer is a polymer that consists a of cross-linked structure or heavily network of branched molecules.
Elastomers are loosely cross-linked polymers which are highly flexible and elastic which can make them stretch easily and return to thier original shapes.
a) Linear and highly crystalline polyethylene is a linear thermoplastic it will not be an elastomer or themoset.
.
(b) Phenol-formaldehyde is anetwork structured thermosetting resin which is regarded as the most useful thermosetting resin especially in the production of wood based panels.
(c) Heavily crosslinked polyisoprene having a glass-transition temperature of 50 deg C is a thermoset since it is heavily cross linked above room temperature.
(d) Lightly crosslinked polyisoprene having a glass-transition temperature of -60 deg C is both an elastomer and thermoset since it has both characteristics .
(e) Linear and partially amorphous poly (vinyl chloride) is a thermoplastic and not an elastomer or a thermoset.
In polymers the crystalline order is often set up by a regular arrangement of the chains. Therefore what you would expect to be elastomers and which thermosetting polymers at room temperature are;
a. Linear and crystalline polyethylene would be neither an elastomer nor a thermoset because it is a linear polymer.
b. Phenol-formaldehyde is said too have a network structure called thermosetting polymer because it has a network structure. It would therefore not be an elastomer as it does not have a crosslinked chain structure.
c. Heavily crosslinked polyisoprene having a glass transition temperature of 50°C can be a said to be a thermosetting polymer due to the fact that it is heavily crosslinked. It cannot not be an elastomer because it is heavily crosslinked and room temperature is below its Tg.
d. Linear and partially amorphous poly(vinyl chloride) is said to be neither an elastomer nor a thermoset. If it want to be any of thee both, it must have some crosslinking.
Polymers is said to be crystallized from a solution or upon evaporation of a solvent but it based on the degree of dilution.
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The heart is an example of which level of organization within an organism?
O
cell
organ
tissue
organ system
O
O
How much heat should be transferred when 20.1 g of hydrogen bromide is formed during the reaction of hydrogen gas with liquid bromine?
Answer:
9.1 KJ
Explanation:
We must first put down the reaction equation;
H2(g) + Br2(g) ----> 2HBr(g)
Secondly we find the number of moles of HBr involved;
number of moles of HBr= mass of HBr/ molar mass of HBr
But molar mass of HBr= 80.91 g/mol
Given mass of HBr as given in the question= 20.1g
Hence;
Number of moles of HBr= 20.1 g/80.91g
Number of moles of HBr= 0.25 moles of HBr
Lastly we calculate the heat transferred from the number of moles involved;
If 2 moles of HBr has a heat of formation of 72.80KJ
Then 0.25 moles of HBr will have a heat of formation of 0.25× 72.80/2= 9.1 KJ
Then; 20.1 g of HBr will have a heat of formation of 9.1 KJ
In a laboratory experiment, students synthesized a new compound and found that when 11.09 grams of the compound were dissolved to make 180.9 mL of a diethyl ether solution, the osmotic pressure generated was 3.88 atm at 298 K. The compound was also found to be nonvolatile and a non-electrolyte. What is the molecular weight they determined for this compound
Answer:
The molecular weight of this compound is 386.4 g/mol
Explanation:
Step 1: Data given
MAss of the compound = 11.09 grams
Volume of diethyl ether = 180.9 mL
Osmotic pressure = 3.88 atm
Temperature = 298 K
The compound = nonvolatile and non-electrolyte
Step 2: Calculate molar concentration
π = i*M*R*T
⇒with π = the osmotic pressure = 3.88 atm
⇒with i = the van't Hoff factor = 1
⇒with C = the molar concentration = = TO BE DETERMINED
⇒ with R = the gas constant = 0.08206 L*atm/mol*K
⇒with T = 298 K
C = 3.88 / (0.08206*298)
C = 0.1587 M
Step 3: Calculate moles compound
C = moles / volume
moles = 0.1587 M * 0.1809 L
Moles compound = 0.0287 moles
Step 4: Calculate molecular weight of the compound
Molar mass = mass / moles
Molar mass compound = 11.09 grams / 0.0287 moles
Molar mass compound = 386.4 g/mol
The molecular weight of this compound is 386.4 g/mol
What the first song come to your mind?
Answer: Hey Brother by Avicii
a 0.258 g piece of potassium solid is placed inot water inside a coffee cup calorimeter resulting in a vigorous reaction. assume a total volume of 100 ml for the resulting solution. the temperature of the solution changes from 22 to 25.1 due to the reaction. how much heat in kj is generated per gram of potassium for this reaction? assume the density of the solution after the reaction is the same as the density of water
Answer:
5 kJ/g
Explanation:
There are two energy flows in this reaction.
q₁ = heat from reaction
q₂ = heat to warm the solution
q₁ + q₂ = 0
m₁ΔH + m₂CΔT = 0
Data:
m₁ = 0.258 g
V₂ = 100 mL
C = 4.184 J°C⁻¹g⁻¹
T_i = 22 °C
T_f = 25.1 °C
Calculations
(a) Mass of solution
[tex]\text{Mass} = \text{100 mL} \times \dfrac{\text{1.00 g}}{\text{1 mL}} = \text{100 g}[/tex]
(b) ΔT
ΔT = T_f - T_i = 25.1 °C - 22 °C = 3.1°C
(c) ΔH
[tex]\begin{array}{ccccl}m_{1}\Delta H & +& m_{2}C \Delta T& = &0\\\text{0.258 g}\times \Delta H& + & \text{100 g} \times 4.184 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$} \times 3.1 \, ^{\circ}\text{C} & = & 0\\0.258 \Delta H \text{ g} & + & \text{1300 J} & = & 0\\&&0.258 \Delta H \text{ g} & = & \text{-1300 J} & & \\& &\Delta H & = & \dfrac{\text{-1300 J}}{\text{0.258 g}}\\\\& & & = & \text{-5000 J/g}\\& & & = & \textbf{-5 kJ/g}\\\end{array}[/tex]
[tex]\text{The reaction produces $\large \boxed{\textbf{5 kJ}}$ per gram of potassium.}[/tex]
Note: The answer can have only one significant figure because you measured the initial temperature of the water only to the nearest degree.
From the calculation, the heat generated from the solution is -194.4 kJ/mol
What is a calorimeter?A calorimeter is an instrument that is used to measure heat.
Now we know that number of moles of the potassium = 0.258 g /39 g/mol = 0.0066 moles
Total mass present = 0.258 g + 100 g = 100.258 g
Temperature change = 25.1°C - 22°C = 3.1°C
Now;
H = -(100.258 * 4.128 * 3.1)/ 0.0066
= -194.4 kJ/mol
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1. 3A + 2B + C + 2D + 20 kJ
a) pressure is decreased_____
b) temperature is raised____
c) D is removed from the system_____
Final answer:
The reaction in question demonstrates the application of Le Chatelier's Principle, with shifts in equilibrium occurring in response to decreased pressure, increased temperature, and the removal of a reactant.
Explanation:
The reaction presented is dealing with changes in reaction conditions in a chemical equilibrium situation. This is directly related to Le Chatelier's Principle, which states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium moves to counteract the change. In this scenario:
Decreased pressure (arrow "b") will cause the equilibrium to shift towards the side with more gas molecules to increase the pressure.Increase in temperature (arrow "a") will cause the equilibrium to shift towards the endothermic direction, in this case to the left, absorbing the added heat.Removal of D from the system will cause the equilibrium to shift towards the right to produce more D, according to Le Chatelier's Principle.Which phrases identify the names of the main wind belts? Check all that apply.
polar easterlies
polar westerlies
prevailing westerlies
tropical easterlies
trade breezes
Answer:
polar easterliesprevailing westerliestropical easterliesExplanation:
Global winds are the winds that occur in the belts that are found all over the planet. Like local winds, global winds are caused by differences in heat in the atmosphere.
Polar Easterlies, from 60-90 degrees latitude.
Prevailing Westerlies, from 30-60 degrees latitude.
Tropical Easterlies, from 0-30 degrees latitude.
Answer:
A,C,D
Explanation:
I was very confused because I wasn't sure for this question so i went on brainly to find a answer and all of them said A,B,C and i was confused cause that was not right low and behold i put in this as a geuss and it was right hope this helped anyone wondering!! but correct answer is A,C,D
A gas sample occupies 8.77 L at 20 degrees Celsius and 3.98 atm. What is
the amount of mols of gas in the sample? *
Answer:
1.45 mol
Explanation:
Given data
Volume of the gas (V): 8.77 LTemperature of the gas (T): 20 °CPressure of the gas (P): 3.98 atmStep 1: Calculate the absolute temperature (Kelvin)
We will use the following expression.
[tex]K = \°C + 273.15\\K = 20\°C + 273.15 = 293K[/tex]
Step 2: Calculate the number of moles (n) of the gaseous sample
We will use the ideal gas equation.
[tex]P \times V = n \times R \times T\\n = \frac{P \times V}{R \times T} = \frac{3.98atm \times 8.77L}{\frac{0.0821atm.L}{mol.K} \times 293K} = 1.45 mol[/tex]
The diagram represents an energy pyramid. At each successive tophic level from
1 to 3, the amount of energy available to the next higher level
A)decrease
B)remains the same
C)increases
HELP PLZZZZ
this is my final test to to determine if I pass my grade
pls answer fast
Answer:
decrease
Explanation:
The flavor of beer can be tainted by a trace contaminant, called ortho-bromophenol. To reduce the incidents of contamination, beer manufacturers have used certified beer flavor standards to train professional beer tasters to recognize the flavor of ortho-bromophenol. Preparing these certified standards requires pure samples of ortho-bromophenol. Propose a synthesis of ortho-bromophenol starting from phenol. (Org. Synth. 1934, 14, 14.)
Choose from the reagents below.
A. dilute NaOH
B. conc. fuming, 1 mol H2SO4
C. HBr
D. Br2, FeBr3
E. Zn
F. Dilute H2SO4
G. HNO3
H. Hcl
I. Br2
Answer:
conc. fuming, 1 mol H2SO4
Dilute NaOH
Br2
Dilute H2SO4
Explanation:
The synthesis of ortho-bromophenol follows the reaction sequence shown in the image attached.
First of all, the phenol is sulphonated using concentrated sulphuric acid at 100°C. Carrying out the reaction at 100°C ensures that the para-isomer predominates. Lower temperatures favour the formation of the ortho isomer. Dilute sodium hydroxide is added before the addition of bromine.
Bromine molecule is then added. The incoming electrophile now attaches to the ortho position. Dilute acid is added at 100°C to remove the -SO3H thereby obtaining the Ortho-bromophenol
Answer:
Check the explanation
Explanation:
Kindly check the attached image below to see the step by step explanation to the question above.
The rate constant for this second‑order reaction is 0.190 M − 1 ⋅ s − 1 0.190 M−1⋅s−1 at 300 ∘ C. 300 ∘C. A ⟶ products A⟶products How long, in seconds, would it take for the concentration of A A to decrease from 0.820 M 0.820 M to 0.340 M?
Answer:
9.1 seconds
Explanation:
Given that for a second order reaction
1/[A]t = kt + 1/[A]o
Where [A]t= concentration at time = t= 0.340M
[A]o= initial concentration = 0.820M
k= rate constant for the reaction=0.190m-1s-1
t= time taken for the reaction (the unknown)
Hence;
(0.340)^-1 = 0.190×t + (0.820)^-1
t= (0.340)^-1 - (0.820)^-1/0.190
t= 9.1 seconds
Hence the time taken for the concentration to decrease from 0.840M to 0.340M is 9.1 seconds.
Final answer:
To calculate the time it takes for the concentration of compound A to decrease in a second-order reaction, the integrated rate law 1/[A] - 1/[A]0 = kt is used with the provided rate constant and initial and final concentrations.
Explanation:
Calculating Time for a Second-Order Reaction:
The question pertains to the time it takes for the concentration of compound A to decrease from 0.820 M to 0.340 M in a second-order reaction with a rate constant of 0.190 M-1·s-1 at 300 °C. The integrated rate law for a second-order reaction is given by:
1/[A] - 1/[A]0 = kt
Where [A]0 is the initial concentration, [A] is the concentration at time t, k is the rate constant, and t is the time. Using the provided concentrations, we can solve for t:
1/0.340 M - 1/0.820 M = (0.190 M-1·s-1)t
After calculating the left side, we can isolate t:
t = (1/0.340 M - 1/0.820 M) / (0.190 M-1·s-1)
Hence:
(0.340)^-1 = 0.190×t + (0.820)^-1
t= (0.340)^-1 - (0.820)^-1/0.190
t= 9.1 seconds
Hence the time taken for the concentration to decrease from 0.840M to 0.340M is 9.1 seconds.
how is the periodic table of the elements arranged
a) alphabetically
b)in a neat grid with 11 rows and 10 columns
c) by aligning groups with common properties
d) in a neat grid with 10 rows and 11 columns
Answer:
I am pretty sure the answer is c)
Explanation:
Alkali Metals, Alkaline Earth Metals, Transition Metals, Metalloids or Semimetals, Nonmetals, Halogens, and Noble Gases are the groups.
HOPE I HELPED :)
Final answer:
The periodic table is arranged by aligning groups with common properties, based on atomic number, into periods and groups. Elements in the same group share similar chemical properties, aiding in the study of chemical behavior.
Explanation:
The question asks how the periodic table of the elements is arranged. The correct answer is that the elements in the periodic table are arranged by aligning groups with common properties. In more detail, the periodic table is organized based on the atomic number of elements, and they are placed in a layout where they form rows called periods and columns known as groups. These groups are numbered 1-18, and elements within the same group share many chemical properties. This organization allows for the grouping of elements with similar properties and helps in the study of their chemical behavior.
For instance, the alignment is such that the elements are ordered in terms of increasing atomic number from left to right across the table. This arrangement makes it easy for us to identify elements with similar properties because they are positioned in the same column. The periodic table includes seven horizontal rows (periods) and 18 vertical columns (groups), making it a powerful tool for understanding chemical reactions and properties of elements. This systematic arrangement reflects the periodic law, which states that the properties of elements are periodic functions of their atomic numbers.
A chemist titrates 220.0 mL of a 0.1917M propionic acid (HC2H,CO2) solution with 0.1787 M KOIH solution at 25 °C. Calculate the pH at equivalence. The pK of propionic acid is 4.89.
Answer:
pH = 8.93
Explanation:
In this case, this titration is the case of a weak acid and a strong base. Now, at the equivalence point, it's supposed that we have the same moles of each reactant in solution, and we will expect that the pH would have to be 7. However, as the acid is pretty weak, there's a little difference in the solution because of the grade of dissociation of the acid, and the pH will be higher than 7. To know this, we first need to calculate the volume of added base:
M₁V₁ = M₂V₂
With this expression, let's calculate the volume of the base:
V₂ = M₁V₁ / M₂
V₂ = 0.1917 * 220 / 0.1787 = 236 mL
So, at the equivalence point, 236 mL are needed to neutralize this reaction. As the moles are the same for each reactant, we just need to calculate the concentration of the acid in this part. This will be the sum between the initial volume of acid and the calculated volume of base:
V of solution = 236 + 220 = 456 mL or 0.456 L
Then, the new concentration of the acid is:
[C₂H₅COOH] = 0.1917 * 0.220 / 0.456 = 0.0924 M
Now, the reaction with the base is the following:
C₂H₅COOH + KOH --------> C₂H₅COOK + H₂O
This means that in the equivalence point we have the propionic potassium and water, so, if take this and dissociates into it's ions we can calculate the pH of the solution:
C₂H₅COO⁻ + H₂O <-------> C₂H₅COOH + OH⁻
With this reaction in solution in the equivalence point, we just need the Kb of propionate ion, and this can be calculated with the value of the pKa of the acid:
Ka = 10^(-pKa)
Ka = 1.29x10⁻⁵
Now the value of Kb can calculated using the following expression:;
Kb = Kw / Ka ---> replacing we have
Kb = 1x10⁻¹⁴ / 1.29x10⁻⁵
Kb = 7.75x10⁻¹⁰
Now, with this value and the above reaction we can write an ICE chart to calculate the [OH⁻] and then, the pH of solution:
C₂H₅COO⁻ + H₂O --------> C₂H₅COOH + OH⁻ Kb = 7.75x10⁻¹⁰
i) 0.0924 0 0
e) 0.0924-x x x
The Kb expression:
Kb = [C₂H₅COOH] [OH⁻] / [C₂H₅COO⁻]
7.75x10⁻¹⁰ = x² / 0.0924-x ---> Kb is very small, so this substraction can be neglected to just 0.0924 assuming x will be very small too.
7.75x10⁻¹⁰ = x² / 0.0924
7.75x10⁻¹⁰ * 0.0924 = x²
x = [OH⁻] = 8.46x10⁻⁶ M
With this value, we can calculate pOH and then the pH:
pOH = -log(8.46x10⁻¹⁰) = 5.07
Finally the pH:
pH = 14 - pOH
pH = 14 - 5.07
pH = 8.93The nonvolatile, nonelectrolyte estrogen (estradiol), C18H24O2 (272.4 g/mol), is soluble in benzene C6H6.
How many grams of estrogen are needed to generate an osmotic pressure of 4.45 atm when dissolved in 234 ml of a benzene solution at 298 K.
_________grams estrogen
Answer: 11.6g of estrogen are needed to generate an osmotic pressure of 4.45 atm when dissolved in 234 ml of a benzene solution at 298 K.
Explanation:
To calculate the amount of solute, we use the equation for osmotic pressure, which is:
[tex]\pi=iMRT[/tex]
Or,
[tex]\pi=i\times \frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}\times RT[/tex]
where,
[tex]\pi[/tex] = osmotic pressure of the solution = 4.45 atm
i = Van't hoff factor = 1 (for non-electrolytes)
Let Mass of solute (estrogen) = x g
Volume of solution = 234 mL
R = Gas constant = [tex]0.0821Latmmol^{-1}K^{-1}[/tex]
T = temperature of the solution = [tex]298K[/tex]
Putting values in above equation, we get:
[tex]4.45=1\times \frac{x\times 1000}{272.4\times234}\times 0.0821Latmmol^{-1}K^{-1}\times 298K[/tex]
[tex]x=11.6g[/tex]
Hence, 11.6g of estrogen are needed to generate an osmotic pressure of 4.45 atm when dissolved in 234 ml of a benzene solution at 298 K.
The question is about calculating the quantity of estrogen needed to generate a specific osmotic pressure in a benzene solution using the Van 't Hoff equation. The moles of estrogen needed is calculated first, then converted into grams using the molar mass of estrogen.
Explanation:This question is based on the concept of osmotic pressure and solution chemistry, for a nonvolatile, nonelectrolyte compound in solution. The Van 't Hoff equation (π= nRT/V) can be used to solve the problem, where π refers to the osmotic pressure, n is the amount of solute in moles, R is the gas constant (0.0821 L·atm/K·mol for this problem), T is the temperature in Kelvin, and V is the volume in liters. Given the osmotic pressure (4.45 atm), the temperature (298 K), and the volume (0.234 L), you can find the number of moles of estrogen needed. After calculating the amount in moles, use the molar mass of the estrogen (272.4 g/mol) to find the mass in grams. Hence, the quantity of estrogen needed can be calculated.
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Determine the electron geometry (eg), molecular geometry (mg), and polarity of SO2. Determine the electron geometry (eg), molecular geometry (mg), and polarity of SO2. eg=trigonal planar, mg=bent, polar eg=tetrahedral, mg=bent, polar eg=linear, mg=linear, nonpolar eg=trigonal pyramidal, mg=trigonal pyramidal, polar eg=tetrahedral, mg=tetrahedral, nonpolar
Answer:
See explaination
Explanation:
The electrons geometry shows the special distribution of the electrons around of the central atom of the molecule.
The molecular geometry shows the special distribution of the atoms that form the molecule.
Please kindly check attachment for further solution.
Calculate the concentration in M) of hydroxide ions in a solution with a pOH of 2.468
A) 2.94 x 102
B) 3.40 x 1011
OC) 3.40 x 10-3
OD) 2.94 x 10-12
Answer:
C) 3.40 x 10-3
Explanation:
Hello,
In this case, for the given pOH, we are able to compute the concentration of hydrixide ions by applying the following formula:
[tex]pOH=-log([OH^-])[/tex]
[tex][OH^-]=10^{-pOH}=10^{-2.468}\\[/tex]
[tex][OH^-]=0.0034M=3.4x10^{-3}M[/tex]
Therefore, answer is C).
Best regards.
Answer:
3.40 × 10⁻³ M
Explanation:
The pOH scale is used to express the acidity or basicity of a solution.
If pOH < 7, the solution is basic.If pOH = 7, the solution is neutral.If pOH > 7, the solution is acid.The pOH of this solution is 2.468, so it is basic. We can calculate the concentration of hydroxide ions in the solution using the following expression.
[tex]pOH = -log [OH^{-} ]\\\[[OH^{-}] = antilog-pOH = antilog-2.468 = 3.40 \times 10^{-3} M[/tex]
The [OH-] of an aqueous solution is 6.4 x 10-5 M. What is the [H3O+]?
Answer: The [tex][H_3O^+][/tex] is [tex]1.55\times 10^{-10}M[/tex]
Explanation:
pH or pOH is the measure of acidity or alkalinity of a solution.
pOH is calculated by taking negative logarithm of hydroxide ion concentration.
[tex]pOH=-\log [OH^-][/tex]
Putting in the values:
[tex]pOH=-\log[6.4\times 10^{-5}][/tex]
[tex]pOH=4.19[/tex]
[tex]pH+pOH=14[/tex]
[tex]pH=9.81[/tex]
[tex]9.81=-\log [H_3O^+][/tex]
[tex][H_3O^+]=1.55\times 10^{-10}[/tex]
Thus the [tex][H_3O^+][/tex] is [tex]1.55\times 10^{-10}M[/tex]
The value of [H₃O⁺] will be "1.55 × 10⁻¹⁰ M". To understand the calculation, check below.
Aqueous solutionAccording to the question,
The value of [OH⁻] = 6.4 × 10⁻⁵
By taking hydroxide ion concentration log,
→ pOH = log[OH⁻]
By putting the value of [OH⁻],
pOH = log[6.4 × 10⁻⁵]
= 4.19
hence,
→ pH + pOH = 14
9.81 = log [H₃O⁺]
[H₃O⁺] = 1.55 × 10⁻¹⁰ M
Thus the above answer is correct.
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The Earth's biosphere is consists of
Answer:
The Earth's bioshere consist of the parts of Earth where life exists. Ecosystems.
Explanation:
What mass of NaOH must be used to prepare 2.5 L of a 0.010 M solution
Answer:
[tex]m = 1\,g[/tex]
Explanation:
The molarity is the ratio of the amount of moles solvent to volume of the solute.
[tex]0.010\,M = \frac{0.010\,moles\,NaOH}{1\,L}[/tex]
The quantity of solute is determined by simple rule of three:
[tex]n = \left(\frac{2.5\,L}{1\,L} \right)\cdot (0.010\,mole)[/tex]
[tex]n = 0.025\,moles[/tex]
The molecular weight of NaOH is [tex]39.997\,\frac{g}{mole}[/tex], the mass of solute is:
[tex]m = 1\,g[/tex]
The electron configurations of two unknown elements are x and y. X: 1s2 2s2 2p6 Y: 1s2 2s2 2p6 3s2 3p5 which statement is most likely correct about the two elements
Answer:
X: 2,8 = Neon
Y: 2,8,7= Chlorine
Neon has an atomic number of 10. It is referred to as a noble gas and it’s a mono atomic element.
Chlorine has an atomic number of 17. It is referred to as an Halogen. Chlorine exists in nature as a diatomic molecule.
However the question is incomplete and will need more information.
I need help pls.!!!
Placing a slightly soluble ionic solid in a solution containing common ions will Select the correct answer below:
increase the molar solubility of the solid relative to its molar solubility in pure water.
decrease the molar solubility of the solid relative to its molar solubility in pure water.
have no effect on the molar solubility of the solid.
depends on the solid
Answer:
decrease the molar solubility of the solid relative to its molar solubility in pure water.
Explanation:
The common ion effect is defined as a decrease in the solubility of a solute because of the addition of a second solute with which it has a common ion. If a solution already contains a solute MX and another ionic solid containing BX is added to the solution, the X^- anion is common to the both species. Hence the presence of MX will decrease the solubility of BX compared to the solubility of BX in pure water.
Generally, when a soluble ionic solid is placed in a solution containing a common ion, the molar solubility of the solid decreases compared to its molar solubility in pure water.
under a pressure of 1.0 atm , a confined gas has a volume of 750mL. at constant temperature, the pressure is increased until the gas has a volume of 610 mL. what is the new pressure
Using Boyle's Law which states that P1V1 = P2V2 for a gas at constant temperature, and given the initial conditions of 1.0 atm and 750 mL, when the volume decreases to 610 mL, the new pressure is calculated to be 1.23 atm.
Explanation:The question relates to the behavior of gases under different pressures and volumes, which is generally described by Boyle's Law. Boyle's Law states that for a given mass of gas at constant temperature, the volume of the gas varies inversely with the pressure. We use the formula P1V1 = P2V2, where P1 and V1 are the initial pressure and volume, and P2 and V2 are the new pressure and volume.
Given that the initial pressure (P1) is 1.0 atm and the initial volume (V1) is 750 mL, and the final volume (V2) is 610 mL, we want to find the new pressure (P2). Rearranging the formula to solve for P2 gives us P2 = P1V1/V2. Substituting our known values in we get:
P2 = (1.0 atm * 750 mL) / 610 mL = 1.23 atm (rounded to two decimal places).
Therefore, the new pressure when the volume is decreased to 610 mL while maintaining a constant temperature is 1.23 atm.
The pKa of propanoic acid (propionic acid), CH3CH2COOH, is 4.87. Consider an equilibrium mixture of propanoic acid and its conjugate base with a pH of 4.87. Draw the structure of the form (acid or conjugate base) that predominates after a decrease in [H3O ]. Include all hydrogen atoms and any appropriate formal charges.
Answer:
Explanation:
check below for explicit explanation.
When the number of hydronium ions decreases in a solution containing propanoic acid and its conjugate base, the solution shifts towards the acidic side. Therefore, the structure that would predominate in this circumstance is the acidic form, that is, propanoic acid itself.
Explanation:The equilibrium pH of a solution is equal to the pKa of the acid when the concentrations of the acid and its conjugate base are equal. In this case, where the propanoic acid has a pKa of 4.87 and the pH of the mixture is also 4.87, we have an equal concentration of the acid and its conjugate base. When a decrease in [H₃O⁺] occurs, the equilibrium of the solution shifts to compensate, and shifts towards the acidic side because the concentration of [H₃O⁺] (hydronium ions) has decreased. This means that the form of the propanoic acid that would predominate after a decrease in [H₃O⁺], would be the acidic form. Therefore, the structure that predominates is that of propanoic acid, CH₃CH₂COOH, not its conjugate base.
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Ammonia gas combines with excess oxygen gas to produce nitric oxide and water. How many grams of ammonia gas would have to react in order to release 154 kJ of energy
Answer:
11.6g of NH₃(g) have to react
Explanation:
For the reaction:
4 NH₃(g) + 5 O₂(g) → 4 NO(g) + 6 H₂O(g) ΔH = -905kJ
4 moles of ammonia produce 905kJ
Thus, if you want to produce 154kJ of energy you need:
154kJ × (4 mol NH₃ / 905kJ) = 0.681moles of NH₃. In mass -Molar mass ammonia is 17.031g/mol-
0.681mol NH₃ × (17.031g / mol) = 11.6g of NH₃(g) have to react
How many joules of heat are lost by 1000g of granite as it cools from 41.2 Celsius to -12.9 Celsius?
Answer: The amount of heat released is 42739 Joules
Explanation:
The quantity of heat required to raise the temperature of a substance by one degree Celsius is called the specific heat capacity.
[tex]Q=m\times c\times \Delta T[/tex]
Q = Heat released =?
c = heat capacity of granite = [tex]0.790J/g^0C[/tex]
Initial temperature = [tex]T_i[/tex] = [tex]41.2^0C[/tex]
Final temperature of the calorimeter = [tex]T_f[/tex] = [tex]-12.9^0C[/tex]
Change in temperature ,[tex]\Delta T=T_f-T_i=(-12.9-41.2)^0C=-54.1^0C[/tex]
Putting in the values, we get:
[tex]Q=1000g\times 0.790J/g^0C\times -54.1^0C=-42739J[/tex]
As heat comes out to be negative, that means the heat has been released and the amount of heat released is 42739 Joules
An isothermal chromatogram at 90 °C shows an elution order of 1-pentanol followed by ethylene diamine followed by diethylene glycol. But in order to reduce the overall time of the experiment, a temperature programmed chromatogram is tried, beginning at 60 °C and increasing to 190 °C over 5 minutes then holding at 190 °C for 4 minutes. Predict the elution order of the compounds using the temperature programmed conditions.
Answer:
The new elution order expected will be the following:
ethylene diamine1-pentanoldiethylene glycolPlease see below for details and explanation.
Explanation:
Which compound will elute first depends on a number of factors. The compound with the lowest boiling point will elute before another compound with a higher boiling point and so on. By extension, the volatility of the compound will also be considered when predicting elution order. Thirdly, how the solutes interact with each other during the stationary phase. I've listed the boiling points below:
1-pentanol 138 °C
ethylene diamine 116 °C
diethylene glycol 245 °C
The advantage of using temperature programmed chromatogram is that it changes retention times (time needed for the solute to pass through the column). And it will be according to the respective boiling points.
Hope that answers the question, have a great day!
For the reaction: N2(g) + 2 O2(g) ⇌ 2 NO2(g), Kc = 8.3 × 10-10 at 25°C. What is the concentration of N2 gas at equilibrium when the concentration of NO2 is twice the concentration of O2 gas? For the reaction: N2(g) + 2 O2(g) ⇌ 2 NO2(g), Kc = 8.3 × 10-10 at 25°C. What is the concentration of N2 gas at equilibrium when the concentration of NO2 is twice the concentration of O2 gas? 4.2 × 10-10 M 2.1 × 10-10 M 2.4 × 109 M 4.8 × 109 M 1.7 x 10 -9 M
Answer: Concentration of N₂ is 4.8.[tex]10^{9}[/tex] M.
Explanation: [tex]K_{c}[/tex] is a constant of equilibrium and it is dependent of the concentrations of the reactants and the products of a balanced reaction. For
N2(g) + 2 O2(g) ⇄ 2 NO2(g)
[tex]K_{c}[/tex] = [tex]\frac{[NO2]^{2} }{[N2][O2]^{2} }[/tex]
From the question concentration of NO2 is twice of O2:
[NO2] = 2[O2]
Substituting this into [tex]K_{c}[/tex]:
[tex]K_{c}[/tex] = [tex]\frac{[2O2]^{2} }{[N2][O2]^{2} }[/tex]
8.3.[tex]10^{-10}[/tex] = [tex]\frac{4O2^{2} }{[N2].O2^{2} }[/tex]
[N2] = [tex]\frac{4O2^{2} }{8.3.10^{-10}.O2^{2} }[/tex]
[N2] = [tex]\frac{4}{8.3.10^{-10} }[/tex]
[N2] = 4.8.[tex]10^{9}[/tex]
The concentration of N2 in the equilibrium is [N2] = 4.8.[tex]10^{9}[/tex]M.
The concentration of [tex]N_2[/tex] is [tex]4.8.10^9M[/tex]
Calculation of concentration:Since k_e represent the equilibrium constant and based on the concentrations of the reactants and the products of a balanced reaction.
Also, the reaction: N2(g) + 2 O2(g) ⇌ 2 NO2(g), And, the concentration of NO2 is twice the concentration of O2 gas
Kc = 8.3 × 10-10 at 25°C.
NO2 is twice of O2.
Now
[tex]8.310^{-10} = \frac{4O_2^2}{N_2O_2^2} \\\\N_2 = \frac{4O_2^2}{8.3.10^{-10}O_2^2}\\\\ = 4\div 8.3.10^{-10}\\\\= 4.8.10^9[/tex]
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What type of decay occurs during the fission of 240pu? 240 94 Pu 236 92 U+\ \ Select one Y, B-, A, or B +
Answer:
Alpha decay
Explanation:
Observing the question given, we discovered that the mass number of 240 94 Pu reduced by 4 and the atomic number reduce by 2 in the product obtained from the reaction. This clearly indicates that 240 94 Pu is undergoing alpha decay because atoms that ungoes alpha decay has a decrease of 4 in the mass number and a decrease 2 in the atomic number.
The attached photo gives further explanation...