Answer:
1.
Explanation:
The atomic number of sodium is 11 and there are 10 inner electrons,
So the effective nuclear charge required is 11 - 10 = 1.
The core charge of Na is +1. Sodium contains 11 protons and 10 electrons in the inner shell. Thus, the difference is 1 (positive value is taken for neutral atoms).
What are subatomic particles?An atom is composed of various subatomic particles. In the nucleus of an atom, various subatomic particles are present. Proton is one of them. Proton has a positive electric charge and it is stable in nature.
Neutrons are similar to protons in mass but they do not possess any charge and hence is the name. Neutrons are crucial for imparting stability to an atom.
Electrons possess negative charge and are found in the orbits. They revolve outside the nucleus and has an affinity towards the protons. Electrons are essential for various physical and chemical interactions in an atom.
Subatomic particles are therefore important in determining the structure of an atom. They act as building blocks for various matter present on earth.
Learn more about subatomic particles, here:
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10. The enthalpy of fusion for benzene (C6H6, 78.0 g/mol) is 127.40 kJ/kg, and its melting point is 5.5°C. What is the entropy change when 1 mole of benzene melts at 5.5°C?
Answer:
[tex]\Delta _fS=35.68\frac{J}{K}[/tex]
Explanation:
Hello,
In this case, the entropy of fusion is computed in terms of the enthalpy of fusion considering the fusion temperature in kelvins:
[tex]\Delta _fS=\frac{\Delta _fH}{T}[/tex]
Thus, since the enthalpy of fusion is given in kJ/kg we must compute the grams of benzene in mole of benzene via its molar mass:
[tex]m=1mol*\frac{78.0g}{1mol}=78g[/tex]
Next:
[tex]\Delta _fH=78g*\frac{127.4J}{g}=9937.2J[/tex]
Finally, the entropy:
[tex]\Delta _fS=\frac{9937.2J}{(5.5+273)K}\\\\\Delta _fS=35.68\frac{J}{K}[/tex]
Best regards.
why do we apply paint on an articles
Answer:
Answer: ... When iron is exposed to the environmental condition it terribly reacts with water and air and ends up with iron rusting. Iron articles are therefore painted off so that iron can not react with water and air.
I hope this will help you...
Butyl butyrate is an ester that is a naturally occurring oil used in the flavor industry for its fruity scent. If the steam distillation of butyl butyrate with water has partial pressures of 50 mm Hg and 710 mm Hg respectively, how much of the distillate is water
Answer:
The correct answer is 62.5 %.
Explanation:
Based on the given information, the partial pressure of butyl butyrate is 50 mmHg and the partial pressure of water is 710 mmHg.
Hence, the total pressure is 710+50 = 760 mmHg
According to Dalton's law of partial pressure,
Partial pressure = mole fraction * total pressure
Mole fraction of water is,
Partial pressure of water/Total pressure = 710/760 = 0.93
Similarly, the mole fraction of butyl butyrate is,
Partial pressure of butyl-butyrate/Total pressure = 50/760 = 0.07
Therefore, mole% of water is 0.93 * 100 = 93 %
For calculating mass%,
Mass of H2O = 0.93 * 18 = 16.8 grams (The molecular mass of water is 18 grams per mole)
The molecular mass of butyl-butyrate is 144 gram per mole
The mass of butyl-butyrate = 144 * 0.07 = 10.08 grams
The mass percent of water will be,
Mass % of water/Total mass % * 100 = 16.8 / 10.08 + 16.8 * 100 = 62.5%.
At 22 °C, an excess amount of a generic metal hydroxide, M(OH)2, is mixed with pure water. The resulting equilibrium solution has a pH of 10.22. What is the Ksp of the salt at 22 °C?
Answer:
2.29x10⁻¹² is Ksp of the salt
Explanation:
The Ksp of the metal hydroxide is:
M(OH)₂(s) ⇄ M²⁺ + 2OH⁻
Ksp = [M²⁺] [OH⁻]²
As you can see in the reaction, 2 moles of OH⁻ are produced per mole of M²⁺. It is possible to find [OH⁻] with pH, thus:
pOH = 14- pH
pOH = 14 - 10.22
pOH = 3.78
pOH = -log[OH⁻]
1.66x10⁻⁴ = [OH⁻]
And [M²⁺] is the half of [OH⁻], [M²⁺] = 8.30x10⁻⁵
Replacing in Ksp formula:
Ksp = [8.30x10⁻⁵] [1.66x10⁻⁴]²
Ksp = 2.29x10⁻¹² is Ksp of the salt
What volume, in mL, of a 0.100 M KI solution contains enough KI to react exactly with 15.25 mL of a 0.200 M solution of Cu(NO3)2
Answer: Therefore, the volume of 0.001 M KI is 61 ml.
Explanation:
The balanced chemical reaction is :
[tex]2KI+Cu(NO_3)_2\rightarrrow CuI_2+2KNO_3[/tex]
According to the neutralization law,
[tex]n_1M_1V_1=n_2M_2V_2[/tex]
where,
[tex]M_1[/tex] = molarity of [tex]KI[/tex] solution = 0.100 M
[tex]V_1[/tex] = volume of [tex]KI[/tex] solution = ?
[tex]M_2[/tex] = molarity of [tex]Cu(NO_3)_2[/tex] solution = 0.200 M
[tex]V_2[/tex] volume of [tex]Cu(NO_3)_2[/tex] solution = 15.25 ml
[tex]n_1[/tex] = valency of [tex]KI[/tex] = 1
[tex]n_2[/tex] = valency of [tex]Cu(NO_3)_2[/tex] = 2
Putting in the values we get:
[tex]1\times 0.100\times V_1=2\times 0.200\times 15.25[/tex]
[tex]V_1=61ml[/tex]
Therefore, the volume of 0.001 M KI is 61 ml.
0.1 dm³ of argon contains x atoms .How many atoms are there in 0.2 dm³ of hydrogen gas under the same conditions?
A. )×/2
B.)2x
C.)4x
Answer:
B.) 2x
Explanation:
Hello,
In this case, we can apply the following rule of three, knowing that 0.1 dm³ equals x and 0.2 dm³ is the unknown:
[tex]0.1dm^3\longrightarrow x\\0.2dm^3 \longrightarrow ?[/tex]
Thus, solving for the unknown we find:
[tex]?=\frac{0.2dm^3*x}{0.1dm^3} \\\\?=2*x[/tex]
Therefore, the answer is B.) 2x.
Best regards.
The pressure in an automobile tire is 2.0 atm at 27°C. At the end of a journey on a hot summer day the pressure has risen to 2.2 atm. What is the temperature of the air in the tire? a. 272.72 K b. 330 K c. 0.014 K d. 175 K
Hey there!
For this we can use the combined gas law:
[tex]\frac{P_{1}V_{1} }{T_{1}} = \frac{P_{2}V_{2} }{T_{2}}[/tex]
We are only working with pressure and temperature so we can remove volume.
[tex]\frac{P_{1} }{T_{1}} = \frac{P_{2} }{T_{2}}[/tex]
P₁ = 2 atm
T₁ = 27 C
P₂ = 2.2 atm
Plug these values in:
[tex]\frac{2atm}{27C} = \frac{2.2atm}{T_{2}}[/tex]
Solve for T₂.
[tex]2atm = \frac{2.2atm}{T_{2}}*27C[/tex]
[tex]2atm * T_{2}={2.2atm}*27C[/tex]
[tex]T_{2}={2.2atm}\div2atm*27C[/tex]
[tex]T_{2}=1.1*27C[/tex]
[tex]T_{2}=29.7C[/tex]
Convert this to kelvin and get 302.85 K, which is closest to B. 330 K.
Hope this helps!
Select the correct answer
In an experiment, chemists heated a solid, red substance in a test tube. Vapors from the heated red substance condensed as a metallic liquid on
the side of the tube. The red substance eventually disappeared, leaving only the metallic liquid and producing a gas that escaped from the tube.
Which statement best describes what happened to the red substance in this experiment?
OA The experiment demonstrates the formation of plasma from the red substance.
B. The experiment demonstrates a physical property of the red substance.
OC The experiment shows that the red substance experienced a chemical change.
OD. The experiment shows that the red substance can be a solid, liquid, or gas.
CE. The experiment shows that the red substance is an element.
Answer:
C The experiment shows that the red substance experienced a chemical change.
Explanation:
Apparently, adding heat caused the red substance to decompose into a gas and a metallic liquid. If it were simply a phase change, the original red substance could be expected to return when the temperature cooled. Because the substance apparently decomposed, it is clearly not an element. At no point in the experiment is there any evidence of a plasma being formed.
The observed decomposition is a chemical change.
What is the value of ΔG at 25°C when the initial concentrations of A, B, and C are 1 M, 1 mM, and 1 μM, respectively?
Answer:
Explanation:
0,44
Rank the following in terms of increasing atomic radius Mg, Sc, Ca
Answer:
Mg, Sc, Ca
Explanation:
To figure out increasing atomic radii, we use Periodic Trends applied to the Elements of the Periodic Table to help us out. We know that the trend for atomic radii is increasing left and down. Since Ca is the furthest down and left of the 3, it has the largest atomic radius. Since Sc is next element to Ca, it would be the 2nd largest atomic radius of the 3. Since Mg is above Ca, it has the smallest atomic radius of the 3.
Calculate the vapor pressure of water above the solution. The vapor pressure of pure water at this temperature is 0.0313 atm. Assume complete dissociation of the solute. Express your answer in atmospheres to three significant figures. Pvap = atm
The question is incomplete, the solute was not given.
Let the solute be K₂CrO₄ and the solvent be water
Complete Question should be like this:
The density of a 0.438 M solution of potassium chromate (K₂CrO₄) at 298 K is 1.063 g/mL.
Calculate the vapor pressure of water above the solution. The vapor pressure of pure water at this temperature is 0.0313 atm. Assume complete dissociation.
Pvap = ________atm
Answer:
Pvap (of water above the solution) = 0.0306 atm
Dissolution of the solute
K₂CrO₄ => 2K⁺ + Cr₂O₄²⁻
Explanation:
Given
volume of solution = 1 Litre = 1000 mL of the solution
density of the solution = 1.063 g/mL
concentration of the solution= 0.438M
temperature of the solution= 298 K
vapour pressure of pure water = 0.0313atm
Recall: density = mass/volume
∴mass of solution = volume x density
m = 1000 x 1.063 = 1063 g
To calculate the moles of K₂CrO₄ = volume x concentration
= 1 x 0.438 = 0.438 mol
Mass of K₂CrO₄ = moles x molar mass = 0.438 x 194.19 = 85.055 g
Mass of water = mass of solution - mass of K₂CrO₄
= 1063 - 85.055 = 977.945 g
moles of water = mass/molar mass
∴ moles of water = 977.945/18.02 = 54.27 mol
Dissolution of the solute
K₂CrO₄ => 2K⁺ + Cr₂O₄²⁻
(dissolution is the process by which solute(K₂CrO₄) is passed into solvent(H₂O) to form a solution
moles of ions = 3 x moles of K₂CrO₄
= 3 x 0.438 = 1.314 mol
Vapor pressure of solution = mole fraction of water x vapor pressure of water
= 54.27/(54.27 + 1.314) x 0.0313 = 0.0306 atm
1.60 mL of a suspension of 320.0 mg/5.00 mL aluminum hydroxide is
added to 2.80 mL of hydrochloric acid. What is the molarity of the
hydrochloric acid?
Answer:
1.41 M
Explanation:
First we must use the information provided to determine the concentration of the aluminum hydroxide.
Mass of aluminum hydroxide= 320mg = 0.32 g
Molar mass of aluminum hydroxide= 78 g/mol
Volume of the solution= 5.00 ml
From;
m/M= CV
Where;
m= mass of aluminum hydroxide= 0.32 g
M= molar mass of aluminum hydroxide = 78 g/mol
C= concentration of aluminum hydroxide solution = the unknown
V= volume of aluminum hydroxide solution = 5.0 ml
0.32 g/78 g/mol = C × 5/1000
C = 4.1×10^-3/5×10^-3
C= 0.82 M
Reaction equation;
Al(OH)3(aq) + 3HCl(aq) -----> AlCl3(aq) + 3H2O(l)
Concentration of base CB= 0.82 M
Volume of base VB= 1.60 ml
Concentration of acid CA= the unknown
Volume of acid VA= 2.80 ml
Number of moles of acid NA = 3
Number of moles of base NB= 1
Using;
CA VA/CB VB = NA/NB
CAVANB = CBVBNA
CA= CB VB NA/VA NB
CA= 0.82 × 1.60 × 3/ 2.80 ×1
CA= 1.41 M
Therefore the concentration of HCl is 1.41 M
Why can tin (IV) sulfide and antimony (III) sulfide be separated from copper (II) sulfide and bismuth (III) sulfide by the addition of sodium hydroxide
Answer:
Tin (IV) sulfide and antimony (III) sulfide be separated from copper (II) sulfide and bismuth (III) sulfide by the addition of sodium hydroxide because they become soluble whereas copper (II) sulfide and bismuth (III) sulfide remain insoluble.
Explanation:
Sodium hydroxide is a basic solution which is used as a precipitating agent for metallic ions in the laboratory.
When a solution containing a mixture of the sulfides of the Group II cations, antimony (III), copper (II), tin (IV), and bismuth (III), is made basic by the addition of a base such as sodium hydroxide or ammonium hydroxide, the sulfide ion concentration will increase. The sulfides of antimony (III) and tin (IV) will then become soluble because antimony (III) and tin (IV) form stable complexes with sulfide, which are soluble in water, while the sulfides of copper (II) and bismuth (III) do not. The result is the dissolution of the antimony (III) sulfide and tin (IV) sulfide, separating them from the copper (II) sulfide and the bismuth (III) sulfide.
Sb₂S₃(s) + 3S²⁻(aq) ----> 2SbS₃³⁻(aq)
SnS₂(s) + S²⁻(aq) ----> SnS₃³⁻(aq)
A 9.38 gram sample of a compound is dissolved in 250.0 grams of benzene. The freezing point of this solution is 1.02 ˚C below that of pure benzene. What is the molar mass of this compound? Kf for benzene is 5.12 ˚C/m and you may assume i = 1.
Answer:
THE MOLAR MASS OF THE COMPOUND IS 45.31 G/MOL.
Explanation:
Using the formula;
Change in freezing point Δt = i Kf m
where;
i = Van Hoff constant = 1
Kf = freezing point dissociation constant = 5.12 °C /m
M = molarity = unknown
The freezing point of benzene = 5.444°C
Temperature of the final solution = 1.02°C
First is to calculate the change in the temperature
= Temperature of benzene - temperature of the solution
= 5.444 - 1.02
= 4.424 °C
Next is the use the above formula to solve for the molarity
Δt = i Kf m
4.244 = 1 * 5.12 * m / 0.250 kg of benzene
We have to divide the molarity by the mass of benzene used since it was not 1 kg if benzene that was used.
So therefore,
4.244 = 20.48 * m
m = 4.244 / 20.48
m = 0.207 mol.
Next is to calculate the molar mass of the compound:
Molarity = mass / molar mass
Molar mass = mass/ molarity
Molar mass = 9.38 g / 0.207 mol
Molar mass = 45.31 g /mol
The molar mass of the compound is 45.31 g/mol.
To what volume should you dilute 50.0 ml of 12 m hno3 solution to obtain a 0.100 m hno3 solution?
Answer:
The answer is "6L"
Explanation:
Formula:
[tex]\bold{C_1 \times V_1 = C_2 \times V_2 }\\\\V_2= \frac{C_1\times V_1}{C_2}[/tex]
Values:
[tex]\to C_1= 12 \ m\\\to V_1= 50 \ ml\\\to C_2= 0.100 \ m\\\\\\V_2= \frac{12 \times 50 }{0.100}[/tex]
[tex]= \frac{12 \times 50 }{0.100}\\\\= \frac{12 \times 50 \times 1000}{100}\\\\= \frac{600 \times 1000}{100}\\\\= 600 \times 10\\\\=6000 \ ml\\= 6 \ L[/tex]
which resonance form would be the most stable contributor to the intermediate arenium ion produced by electrophilic bromination of methoxybenzene
The question is incomplete as the options are missing, however, the correct complete question is attached.
Answer:
The correct answer is option A. ( check image)
Explanation:
The most stable contributor to the intermediate arenium ion produced by electrophilic bromination of methoxybenzene in given options is option a due to the fact that this resonating form follows the octet rule is satisfied for all atoms and additional π bond is present in between C-O that makes it more stable, while in other options there are positive charge which means they do not follows octet rule completely.
Thus, the correct answer is option A. ( check image)
The substance nitrogen has the following properties: normal melting point: 63.2 K normal boiling point: 77.4 K triple point: 0.127 atm, 63.1 K critical point: 33.5 atm, 126.0 K At temperatures above 126 K and pressures above 33.5 atm, N2 is a supercritical fluid . N2 does not exist as a liquid at pressures below atm. N2 is a _________ at 16.7 atm and 56.5 K. N2 is a _________ at 1.00 atm and 73.9 K. N2 is a _________ at 0.127 atm and 84.0 K.
Answer:
- N2 does not exist as a liquid at pressures below 0.127 atm.
- N2 is a solid at 16.7 atm and 56.5 K.
- N2 is a liquid at 1.00 atm and 73.9 K
- N2 is a gas at 0.127 atm and 84.0 K.
Explanation:
Hello,
At first, we organize the information:
- Normal melting point: 63.2 K.
- Normal boiling point: 77.4 K.
- Triple point: 0.127 atm and 63.1 K.
- Critical point: 33.5 atm and 126.0 K.
In such a way:
- N2 does not exist as a liquid at pressures below 0.127 atm: that is because below this point, solid N2 exists only (triple point).
- N2 is a solid at 16.7 atm and 56.5 K: that is because it is above the triple point, below the critical point and below the normal melting point.
- N2 is a liquid at 1.00 atm and 73.9 K: that is because it is above the triple point, below the critical point and below the normal boiling point.
- N2 is a gas at 0.127 atm and 84.0 K: that is because it is above the triple point temperature at the triple point pressure.
Best regards.
A 10.0 mL sample of a 2.5 M NaCl solution is diluted to 25.8 mL. What is the new concentration?
Answer:
[tex]C_2=0.97M[/tex]
Explanation:
Hello,
In this case, for dilution process, we can notice that the initial moles remain the same once the dilution is completed, therefore, both concentration and volume change considering:
[tex]n_1=n_2\\\\V_1C_1=V_2C_2[/tex]
In such a way for the given final volume, the resulting concentration is noticed to be:
[tex]C_2=\frac{V_1C_1}{V_2} =\frac{10mL*2.5M}{25.8mL}\\ \\C_2=0.97M[/tex]
This is supported by the fact that the higher the volume the lower the concentration.
Best regards.
Suppose you are using distillation to separate cyclohexane and toluene. The boiling point of cyclohexane is
Question
Suppose you are using distillation to separate cyclohexane and toluene. The boiling point of cyclohexane is _____, and the boiling point of toluene is ____
Answer:
The boiling point of cyclohexanol is 81°C and the boiling point of toluene is 111°C.
Explanation:
The boiling point of cyclohexanol is 81°C and the boiling point of toluene is 111°C.
Therefore, the liquid collected first should be cyclohexane. .
Cyclohexane is collected first in the distillation to separate cyclohexane and toluene.
In the distillation process the mixture of cyclohexane and toluene is heated therefore vapourizing the subtance. Under the boiling reflux, the vapour phase become richer in the low boilng component ,cyclohexane. The cyclohexane vapours continue to condense first as compare to high boiling component ,toluene. Hence, cyclohexane is collected first.
Calculate the pressure exerted by 0.5600 mole of in a 1.1000-L container at 25.0°C. (The gas constant is 0.08206 L·atm/mol·K. Take absolute zero to be –273.2°C.)
Use the ideal gas law.
(Enter your answer to four significant figures.)
Pressure =
atm
Use the van der Waals equation.
(For : a = 1.39 atm L2/mol2, and b = 0.0391 L/mol. Enter your answer to four significant figures.)
Pressure =
atm
Compare the results.
(Enter your answer to two significant figures.)
The
_________
is higher by
atm, or
%.
Answer:
using ideal gas equation =12.4576atm to 4.significant figure
using vander Waals equation = 12.3504
The differences is 0.10atm
The addition of 0.275 L of 1.62 M KCl to a solution containing Ag+ and Pb2+ ions is just enough to precipitate all of the ions
as AgCl and PbCl2. The total mass of the resulting precipitate is 62.37 g. Find the mass of PbCl, and AgCl in the precipitate.
Answer:
The mass of PbCl₂ is 45.88 grams and the mass of AgCl is 16.48 grams.
Explanation:
As mentioned in the given question, the addition of 0.275 L of 1.62 M KCl is done in a solution that comprise Ag⁺ and Pb²⁺ ions so that all the ions get precipitated. Therefore, the moles of KCl present is,
Moles of KCl = 0.275 L × 1.62 M = 0.445 moles
Now the reaction will be,
Ag⁺ + Pb²⁺ + KCl ⇒ AgCl + PbCl₂ + 3K⁺
Now let us assume that the formation of x moles of AgCl and y moles of PbCl₂ is taking place.
Therefore, mass of AgCl will be x × molecular mass, which will be equal to x × 143.32 grams = 143.32 x grams
Now the mass of PbCl2 formed will be,
y × molecular mass = y × 278.1 grams = 278.1 y grams
Now the total precipitate will be,
62.37 grams = 143.32 x + 278.1 y -----------(i)
Now as AgCl and PbCl₂ requires 1:2 ratio of KCl, this shows that x moles of AgCl will require x moles of KCl and y mol of PbCl₂ will require 2*y moles of PbCl₂. Therefore,
x + 2y = total mass of KCl
x + 2y = 0.445 moles ------ (ii)
On solving equation (i) and (ii) we get,
x as 0.115 and y as 0.165
Now the mass of AgCl will be,
143.32 × 0.115 = 16.48 grams
The mass of PbCl₂ will be,
278.1 × 0.165 = 45.88 grams.
Thulium-167 has a half-life of 9.25 days. If you begin with 48 grams of thulium-167, how much of the original isotope will remain after 37 days?
Answer:
2.3125g
Explanation:
Half-life referred to the time required for a quantity to reduce to half of its initial value, It used to calculate how unstable atoms undergo, or the period of time and atom can survive, radioactive decay.
Given:
t(1/2)= 9.25days
Initial mass of Thulium-167 = 48grams
We need to calculate the remaining amount after 37days.
Since we know that 1 half life = 9.25 days
Then 37 days means ( 37/9.25) half lives
37days means 4 half life
That means the 38grams of Thulium-167 will be halved by 4 times.
Then the ratio between the initial Amount and the amount remaining after 37 days can be calculated as. 0.5^(4)
= 37days × 0.5^(4)
= 2.3125g
the remaining amount of Thallium-167 after 37days is 2.3125g
In experimental Procedure, Part A.1. For preparing a set of standard solutions of FeNCS2+, the equilibrium molar concentration of FeNCS2+ is assumed to equal the initial molar concentration of the SCN- in the reaction mixture. Why is this assumption valid?A) The high concentration of the iron ion ensures that the reaction is favored to go to the rightB) The high concentration of the SCN- ion ensures that the reaction is favored to the rightC) The low concentration of the iron ion ensures that the reaction is favored to go to the rightD) This reaction will not reach equilibrium
Answer:
The high concentration of the iron ion ensures that the reaction is favored to go to the right
Explanation:
According to Le Chateliers principle, when a constraint such as a change in concentration, pressure or volume is imposed on a chemical system in equilibrium, the system will readjust itself in order to annul the constraint. This is done by shifting the position of equilibrium, either to the left hand side or the right hand side depending on the requirement of the system based on the imposed constraint. Equilibrium his now reestablished in the system as a result of this readjustment.
Since there is a high(excess) concentration of iron in the system, it can be safely assumed that the equilibrium concentration of the FeNCS2+ is equal to the initial concentration of SCN- ion present. This implies that the equilibrium position moves towards the right hand side according to Le Chateliers principle.
What is the maximum amount of silver (in grams) that can be plated out of 4.7 L of an AgNO3 solution containing 6.3 % Ag by mass
Answer:
296.1g of Ag is the maximum amount of silver
Explanation:
A solution of 6.3% Ag by mass contains 6.3g of Ag per 100g of solution. Thus, you need to calculate the mass of the solution and then, the mass of Ag present in solution, thus:
Mass of solution:
Assuming a density of 1g/mL:
[tex]4.7L \frac{1000mL}{1L} \frac{1g}{mL} = 4700g[/tex]
If the solution contains 6.3g of Ag per 100g of solution, the mass of Ag in 4700L is:
4700L × (6.3g Ag / 100g) =
296.1g of Ag is the maximum amount of silverA chemistry graduate student is given of a pyridine solution. Pyridine is a weak base with . What mass of should the student dissolve in the solution
Answer:
34.1g of C₅H₅NHCl the student need to dissolve to the solution.
Explanation:
Full question is:
A chemistry graduate student is given 500.mL of a 0.20M pyridine C5H5N solution. Pyridine is a weak base with =Kb×1.7x10−9 . What mass of C5H5NHCl should the student dissolve in the C5H5N solution to turn it into a buffer with pH =4.76 ?
Using H-H expression for weak bases, it is possible to find pH of a buffer thus:
pOH = pKb + log [BH⁺] / [B]
Where pKb is -log Kb = 8.77, [BH⁺] concentration of C₅H₅NHCl and [B] concentration of C₅H₅N (It is possible to take the moles of both compounds and not its concentration.
As pH the student wants is 4.76, pOH is:
pOH = 14 - pH = 14 - 4.76 = 9.24
Replacing:
9.24 = 8.77 + log [C₅H₅NHCl] / [C₅H₅N]
Moles of C₅H₅N are:
0.500L × (0.20mol / L) = 0.10mol C₅H₅N
Replacing again:
9.24 = 8.77 + log [C₅H₅NHCl] / [0.10mol]
2.9512 = [C₅H₅NHCl] / [0.10mol]
0.29512 moles = [C₅H₅NHCl].
As molar mass of C₅H₅NHCl is 115.56g/mol, mass of 0.29512 moles are:
0.29512 moles C₅H₅NHCl × (115.56g / mol) =
34.1g of C₅H₅NHCl the student need to dissolve to the solution.The rate constant for this first‑order reaction is 0.550 s−10.550 s−1 at 400 ∘C.400 ∘C. A⟶products A⟶products How long, in seconds, would it take for the concentration of AA to decrease from 0.690 M0.690 M to 0.220 M?
Answer:
[tex]t=2.08s[/tex]
Explanation:
Hello,
In this case, for first order reactions, we can use the following integrated rate law:
[tex]ln(\frac{[A]}{[A]_0} )=kt[/tex]
Thus, we compute the time as shown below:
[tex]t=-\frac{ln(\frac{[A]}{[A]_0} )}{k}=- \frac{ln(\frac{0.220M}{0.690M} )}{0.55s^{-1}} \\\\t=-\frac{-1.14}{0.550s^{-1}}\\ \\t=2.08s[/tex]
Best regards.
A sample of a compound is made up of 57.53 g C, 3.45 g H, and 39.01 g F. Determine the empirical formula of this compound.
Answer:
C7H5F3
Explanation:
The following data were obtained from the question:
Mass of Carbon (C) = 57.53g
Mass of Hydrogen (H) = 3.45g
Mass of Fluorine (F) = 39.01g
The empirical formula of the compound can be obtained as follow:
C = 57.53g
H = 3.45g
F= 39.01g
Divide each by their molar mass
C = 57.53/12 = 4.79
H = 3.45/1 = 3.45
F = 39.01/19 = 2.05
Divide each by the smallest
C = 4.79/2.05 = 2.3
H = 3.45/2.05 = 1.7
F = 2.05/2.05 = 1
Multiply through by 3 to express in whole number
C = 2.3 x 3 = 7
H = 1.7 x 3 = 5
F = 1 x 3 = 3
Therefore, the empirical formula for the compound is C7H5F3
The reaction rate is k[Ce4+][Mn2+] for the following reaction: 2Ce4+(aq) + Tl+(aq) + Mn2+(aq) → 2Ce3+(aq) + Tl3+(aq) + Mn2+(aq What is the catalyst?
Answer:
Manganese (II) ion, Mn²⁺
Explanation:
Hello,
In this case, given the overall reaction:
[tex]2Ce^{4+}(aq) + Tl^+(aq) + Mn^{2+}(aq) \rightarrow 2Ce^{3+}(aq) + Tl^{3+}(aq) + Mn^{2+}(aq)[/tex]
Thus, since manganese (II) ion, Mn²⁺ is both at the reactant and products, we infer it is catalyst, since catalysts are firstly consumed but finally regenerated once the reaction has gone to completion. Moreover, since inner steps are needed to obtain it, we can infer that the given rate law corresponds to the slowest step that is related with the initial collisions between Ce⁴⁺ and Mn²⁺
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The mass of an empty flask plus stopper is 44.232 g. When the flask is completely filled with water the new mass is 153.617 g. The flask is emptied and dried, and a piece of metal is added. The mass of the flask, stopper and metal is 143.557 g. Next, water is added to the flask containing the metal and the mass is found to be 226.196 g. What is the density of the metal (in g/cm3) to the proper number of significant figures
Answer:
3.7136g/cm³
Explanation:
Density is defined as the ratio between mass of a substance and its volume.
First, we will find the mass of the piece of the metal that is the difference between mass of metal + flask and mass of empty flask. That is:
Mass metal:
143.557g - 44.232g = 99.325g of the metal
Now, to find its volume you must know first the volume of the flask that can be obtained from the mass of water in the filled flask, that is:
153.617g - 44.232g = 109.385g of water = cm³ of water
In the second experiment, the mass of water = its volume is:
226.196g - 143.557g = 82.639g = 82.639cm³ of water
That means the volume the piece of metal is occupying is:
109.385cm³ - 82.639cm³ = 26.746cm³ of piece of metal
And its density is:
99.325g / 26.746cm³ =
3.7136g/cm³
If 40.0 g of molten iron(II) oxide reacts with 10.0 g of mag-nesium, what is the mass of iron produced
Answer:
[tex]m_{Fe}=23.0gFe[/tex]
Explanation:
Hello,
In this case, the undergoing chemical reaction is:
[tex]FeO+Mg\rightarrow Fe+MgO[/tex]
Thus, for the given masses of reactants we should compute the limiting reactant for which we first compute the available moles of iron (II) oxide:
[tex]n_{FeO}=40.0gFeO*\frac{1molFeO}{72gFeO} =0.556molFeO[/tex]
Next, we compute the consumed moles of iron (II) oxide by the 10.0 g of magnesium, considering their 1:1 molar ratio in the chemical reaction:
[tex]n_{FeO}^{consumed}=10.0Mg*\frac{1molMg}{24.3gMg}*\frac{1molFeO}{1molMg}=0.412molFeO[/tex]
Therefore, we can notice there is less consumed iron (II) oxide than available for which it is in excess whereas magnesium is the limiting reactant. In such a way, the produced mass of iron turns out:
[tex]m_{Fe}=0.412molFeO*\frac{1molFe}{1molFeO}*\frac{56gFe}{1molFe}\\ \\m_{Fe}=23.0gFe[/tex]
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