Explanation:
This reaction is in equilibrium and would hence obey lechatelier's principle. This principle states that whenever a system at equilibrium undergoes a change, it would react in way so as to annul that change.
Since it is an endothermic reaction, increasing the temperature would cause the reaction to shift towards the right.
This means that it favours product formation and more of the product would be formed.
Diluting sulfuric acid with water is highly exothermic:
(Use data from the Appendix to find for diluting 1.00 mol of H2SO4(l) (d = 1.83 g/mL) to 1 L of 1.00 MH2SO4(aq) (d = 1.060 g/mL). )
Suppose you carry out the dilution in a calorimeter. The initial T is 25.2°C, and the specific heat capacity of the final solution is 3.458 J/gK. What is the final T in °C ?
Answer:
The correct answer is 51.2 degree C.
Explanation:
The standard enthalpy for H₂SO₄ (l) is -814 kJ/mole and the standard enthalpy for H₂SO₄ (aq) is -909.3 kJ/mole.
Now the dHreaction = dHf (product) - dHf (reactant)
= -909.3 - (-814)
dHreaction or q = -95.3 kJ of energy will be used for dissociating one mole of H₂SO₄.
The heat change in calorimetry can be determined by using the formula,
q = mass * specific heat capacity * change in temperature -----------(i)
Based on the given information, the density of H₂SO₄ is 1.060 g/ml
The volume of H₂SO₄ is 1 Liter
Therefore, the mass of H₂SO₄ will be, density/Volume = 1.060 g/ml / 1 × 10⁻³ ml = 1060 grams
The initial temperature given is 25.2 degrees C, or 273+25.2 = 298.2 K, let us consider the final temperature to be T₂.
ΔT = T₂ -T₁ = T₂ - 298.2 K
Now putting the values in equation (i) we get,
95.3 kJ = 1060 grams × 3.458 j/gK (T₂ - 298.2 K) (the specific heat capacity of the final solution is 3.458 J/gK)
(T₂ - 298.2 K) = 95300 J / 1060 × 3.458 = 26 K
T₂ = 298.2 K + 26 K
T₂ = 324.2 K or 324.2 - 273 = 51.2 degree C.
g The combustion of ethylene proceeds by the following reaction: C2H4(g) + 3 O2(g) → 2 CO2(g) + 2 H2O(g) If the rate of O2 is -0.23 M/s, then what is the rate (in M/s) of disappearance of C2H4?
Answer:
Explanation:
C₂H₄(g) + 3O₂(g) → 2CO₂(g) + 2H₂O(g)
In this reaction we see that 3 moles of O₂ reacts with one mole of C₂H₄ .
Hence rate of disappearance of O₂ is 3 times faster .
- d [O₂] / dt = - 3 d [ C₂ H₄ ] / dt
Putting the given value
.23 = 3 d [ C₂ H₄ ] / dt
d [ C₂ H₄ ] / dt = .23 / 3
= .077 M / s
Hence the rate of disappearance of C₂ H₄ is .077 moles / s .
Mrs. Wilson leaves her freshly-baked blueberry pie on the windowsill to cool. The delicious fragrance diffuses through the air with a diffusion coefficient of D = 0.2 cm2/s. How long does it take for Dennis to smell the pie in his treehouse 10 meters away? Give your answer in days, without entering the unit.
Answer:
Poop Buttt.
Explanation:
If a balloon takes up 625L at 273K, what will the new volume be when the balloon is heated to 353K.
Answer:
The new volume will be 808 L
Explanation:
Charles's law is a law that says that the volume of gas at constant pressure is directly proportional to its absolute temperature (in degrees Kelvin), that is, when the amount of gas and pressure are kept constant, the quotient between volume and temperature will always have the same value:
[tex]\frac{V}{T}=k[/tex]
Having a certain volume of gas V1 that is at a temperature T1 at the beginning of the experiment, by varying the volume of gas to a new value V2, the temperature will change to T2 and the following will be fulfilled:
[tex]\frac{V1}{T1} =\frac{V2}{T2}[/tex]
In this case:
V1= 625 LT1= 273 KV2= ?T2= 353 KReplacing:
[tex]\frac{625 L}{273 K} =\frac{V2}{353 K}[/tex]
Solving:
[tex]V2=353 K*\frac{625 L}{273 K}[/tex]
V2= 808 L
The new volume will be 808 L
Aluminum and oxygen react according to the following equation: 4Al + 3O2 -> 2Al2O3 In a certain experiment, 4.6g Al was reacted with excess oxygen and 6.8g of product was obtained. What was the percent yield of the reaction?
Answer:
Percent yield: 78.2%
Explanation:
Based on the reaction:
4Al + 3O₂ → 2Al₂O₃
4 moles of Al produce 2 moles of Al₂O₃
To find percent yield we need to find theoretical yield (Assuming a yield of 100%) and using:
(Actual yield (6.8g) / Theoretical yield) × 100
Moles of 4.6g of Al (Molar mass: 26.98g/mol) are:
4.6g Al × (1mol / 26.98g) = 0.1705 moles of Al.
As 4 moles of Al produce 2 moles of Al₂O₃, theoretical moles of Al₂O₃ obtained from 0.1705 moles of Al are:
0.17505 moles Al × (2 moles Al₂O₃ / 4 moles Al) = 0.0852 moles of Al₂O₃,
In grams (Molar mass Al₂O₃ = 101.96g/mol):
0.0852 moles of Al₂O₃ × (101.96g / mol) =
8.7g of Al₂O₃ can be produced (Theoretical yield)Thus, Percent yield is:
(6.8g / 8.7g) × 100 =
78.2%a. How many moles of copper equal 8.00 × 109 copper atoms?
b. How many moles of calcium equal 102.5 g calcium?
c. How many atoms of lead are present in 5.04 g lead?
d. How many aluminum atoms are present in 2.85 moles of aluminum?
e. What is the mass in grams of 1.08 × 103 moles of fluorine gas?
f. How many molecules of benzene (C6H6) are present in 0.584 g of benzene?
g. What is the mass of 5.09 × 109 atoms of hydrogen gas?
h. How many calcium atoms are present in 0.45 mol Ca3PO4?
Answer:
Explanation:
a ) one mole = 6.02 x 10²³ atoms
no of moles in given no of atoms
= 8 x 10⁹ / 6.02 x 10²³
= 1.329 x 10⁻¹⁴ moles .
b ) one mole of calcium = 40 gram
102 .5 g calcium
= 102 .5 / 40 moles
= 2.5625 moles
c )
no of moles in 5.04 g lead = 5.04 / 207
= 2.4347 x 10⁻² moles
= 2.4347 x 10⁻²x 6.02 x 10²³ no of atoms of lead
= 14.6568 x 10²¹ no of atoms .
d)
one mole = 6.02 x 10²³ atoms
2.85 mole = 17.157 x 10²³ atoms .
e )
moles of fluorine gas = 1.08 x 10³ / 6.02 x 10²³
= .1794 x 10⁻²⁰ moles
mass in grams = .1794 x 10⁻²⁰ x 38
= 6.8172 x 10⁻²⁰ grams
f )
no of moles in .584 g of benzene = .584 / 78
= 7.487 x 10⁻³ moles
no of molecules = 6.02 x 10²³ x 7.487 x 10⁻³
= 45.07 x 10²⁰ molecules .
g )
moles of atoms = 5.09 x 10⁹ / 6.02 x 10²³
= .8455 x 10⁻¹⁴ moles
mass in gram = .8455 x 10⁻¹⁴ x 1
= .8455 x 10⁻¹⁴ g
h )
.45 moles of Ca₃PO₄ = .45 x 6.02 x 10²³ molecules
= 2.709 x 10²³ molecules of Ca₃PO₄
no of atoms of Ca = 3 x 2.709 x 10²³
= 8.127 x 10²³ atoms of Ca .
Using the Ideal Gas Law, PV = nRT and the information below, solve the question. 5.01 moles of chlorine are held in a vessel with a fixed volume of 70L. What is the pressure of the gas in atm, if its temperature is 303K. REMEMBER TO USE R=0.08206L*atm / k*mol
Answer:
1.7795 atmExplanation:
[tex]n = 5.01\\V = 70L\\P = ?\\T = 303K\\R = 0.08206\\\\PV =nRT\\Make \:P \:Subject\:of\:the\:formula\\P = \frac{nRT}{V} \\\\P = \frac{5.01\times0.08206\times303}{70} \\\\P =\frac{124.5695}{70}\\ P = 1.7795[/tex]
If the concentration of Mg2+ in the solution were 0.039 M, what minimum [OH−] triggers precipitation of the Mg2+ ion? (Ksp=2.06×10−13.) Express your answer to two significant figures and include the appropriate units. nothing nothing
Answer:
2.30 × 10⁻⁶ M
Explanation:
Step 1: Given data
Concentration of Mg²⁺ ([Mg²⁺]): 0.039 M
Solubility product constant of Mg(OH)₂ (Ksp): 2.06 × 10⁻¹³
Step 2: Write the reaction for the solution of Mg(OH)₂
Mg(OH)₂(s) ⇄ Mg²⁺(aq) + 2 OH⁻(aq)
Step 3: Calculate the minimum [OH⁻] required to trigger the precipitation of Mg²⁺ as Mg(OH)₂
We will use the following expression.
Ksp = 2.06 × 10⁻¹³ = [Mg²⁺] × [OH⁻]²
[OH⁻] = 2.30 × 10⁻⁶ M
Which one of the following reactions must be carried out in an electrolytic cell rather than in an electrochemical cell?
a. Zn2+ + Ca → Zn + Ca2+
b. Al3+ + 3Br– → Al + (3/2)Br2
c. 2Al + 3Fe2+ → 2Al3+ + 3Fe
d. H2 + I2(s) → 2H+ + 2I–
Answer:
b. Al3+ + 3Br– → Al + (3/2)Br2
Explanation:
If we look at this reaction closely, aluminum was reduced while bromine was oxidized. The reduction potential of aluminum is -1.66 V while that of bromine is + 1.087. Recall that the more negative the redox potential of a chemical specie, the greater its tendency to function as a reducing agent donating electrons in an electrochemical reaction.
However, in this reaction, aluminium is found to accept rather than donate electrons. Therefore, the process is non spontaneous and can only occur in an electrolytic cell, hence the answer.
b. [tex]Al^{3+} + 3 Br^{-}[/tex] → [tex]Al + \frac{3}{2} Br_{2}[/tex]
Electrolytic Cell v/s Electrochemical Cell:
Electrochemical cells convert chemical energy into electrical energy or vice versa while an electrolytic cell is a type of electrochemical cell in which electrical energy is converted into chemical energy.Electrochemical cells consist of a cathode (+) and an anode(-) while Electrolytic cells consist of a positively charged anode and a negatively charged cathode.Out of the given reactions, [tex]Al^{3+} + 3 Br^{-}[/tex] → [tex]Al + \frac{3}{2} Br_{2}[/tex] is carried out in an electrolytic cell rather than in an electrochemical cell.
As we know, In electrolytic cells, like galvanic cells, are composed of two half-cells
one is a reduction half-cell, the other is an oxidation half-cell.In the given reaction, aluminum is being reduced while bromine gets oxidized and the value of reduction potential of aluminum is -1.66 V while that of bromine is + 1.087. Therefore, the process is non spontaneous and can only occur in an electrolytic cell.
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The overall process that uptakes energy-poor molecules (CO2 and H2O) from their reservoirs in nature and converts them into energy-rich molecules is
Answer:
Photosynthesis
Explanation:
Photosynthesis is a process by which photosynthetic organisms use the energy of captured sunlight to convert energy-poor molecules such as carbon (iv), CO₂ and water (H₂O) into energy-rich organic molecules sch as carbohydrates e.g. glucose.
Photosynthesis occurs in a variety of bacteria and algae as well in vascular plants. The overall equation for the reaction of photosynthesis is as follows:
CO₂ + H₂O + light-------> (CH₂O) + O₂
It is a redox reaction in which water donates electrons (as hydrogen) for the reduction of CO₂ to carbohydrate, (CH₂O).
Carbohydrates are energy-rich molecules which serves as energy sources for many living organisms.
Which sample is most likely to experience the smallest temperature change upon observing 55KJ of heat? 
Answer:
100 g of water: specific heat of water 4.18 J/g°C
Explanation:
To know the correct answer to the question, we shall determine the temperature change in each case.
For 100 g of water:
Mass (M) = 100 g
Specific heat capacity (C) = 4.18 J/g°C
Heat absorbed (Q) = 55 KJ = 55000 J
Change in temperature (ΔT) =..?
Q = MCΔT
55000 = 100 x 4.18 x ΔT
Divide both side by 100 x 4.18
ΔT = 55000/ (100 x 4.18)
ΔT = 131.6 °C
Therefore the temperature change is 131.6 °C
For 50 g of water:
Mass (M) = 50 g
Specific heat capacity (C) = 4.18 J/g°C
Heat absorbed (Q) = 55 KJ = 55000 J
Change in temperature (ΔT) =..?
Q = MCΔT
55000 = 50 x 4.18 x ΔT
Divide both side by 50 x 4.18
ΔT = 55000/ (50 x 4.18)
ΔT = 263.2 °C
Therefore the temperature change is 263.2 °C
For 50 g of lead:
Mass (M) = 50 g
Specific heat capacity (C) = 0.128 J/g°C
Heat absorbed (Q) = 55 KJ = 55000 J
Change in temperature (ΔT) =..?
Q = MCΔT
55000 = 50 x 0.128 x ΔT
Divide both side by 50 x 0.128
ΔT = 55000/ (50 x 0.128)
ΔT = 8593.8 °C
Therefore the temperature change is 8593.8 °C.
For 100 g of iron:
Mass (M) = 100 g
Specific heat capacity (C) = 0.449 J/g°C
Heat absorbed (Q) = 55 KJ = 55000 J
Change in temperature (ΔT) =..?
Q = MCΔT
55000 = 100 x 0.449 x ΔT
Divide both side by 100 x 0.449
ΔT = 55000/ (100 x 0.449)
ΔT = 1224.9 °C
Therefore the temperature change is 1224.9 °C.
The table below gives the summary of the temperature change of each substance:
Mass >>> Substance >> Temp. Change
100 g >>> Water >>>>>> 131.6 °C
50 g >>>> Water >>>>>> 263.2 °C
50 g >>>> Lead >>>>>>> 8593.8 °C
100 g >>> Iron >>>>>>>> 1224.9 °C
From the table given above we can see that 100 g of water has the smallest temperature change.
Fermentation is a complex chemical process of making wine by converting glucose C6H12O6 into ethanol C2H5OH and carbon dioxide: C6H12O6(s) ---> 2 C2H5OH (l) + 2 CO2(g) Calculate the mass of ethanol produced if 500.0 grams of glucose reacts completely.
Answer:
[tex]255.71~g~C_2H_5OH[/tex]
Explanation:
First, we have to check if the reaction is balanced:
[tex]C_6H_1_2O_6_(_s_)~->~2C_2H_5OH_(_l_)~+~2CO_2_(_g_)[/tex]
We have 6 carbon atoms on both sides, 12 hydrogens, and 6 oxygens. So, we can continue with the problem. If we want to calculate the mass of ethanol ([tex]C_2H_5OH[/tex]) we need to know:
1) Molar mass of glucose ([tex]C_6H_12O_6[/tex])
In this case, we have to know the atomic mass of each atom:
-) C 12 g/mol
-) O 16 g/mol
-) H 1 g/mol
With the formula we can calculate the molar mass:
(12*6) + (16*6) + (1*12) = 180 g/mol
2) Molar ratio between glucose and ethanol
In the balanced equation we have 1 mol of [tex]C_6H_12O_6[/tex] and 2 moles of [tex]C_2H_5OH[/tex]. So the molar mass is 1:2
3) Molar mass of ethanol ([tex]C_2H_5OH[/tex])
With the formula, we can calculate the molar mass:
(12*2) + (6*1) + (16*1) = 46 g/mol
Finally, we can to the calculation:
[tex]500.0g~C_6H_12O_6\frac{1mol~C_6H_12O_6}{180g~C_6H_12O_6}\frac{2mol~C_2H_5OH}{1mol~C_6H_12O_6}\frac{46g~C_2H_5OH}{1mol~C_2H_5OH}=255.71g~C_2H_5OH[/tex]
I hope it helps!
The branch of science which deals with chemicals and bonds is called chemistry.
The correct answer for the question is 255g of ethanol.
The process of digesting the glucose in absence of oxygen is called fermentation.
Before solving the question, we must check the atoms of compound and balanced it.
The molar mass of the glucose is as follows:
[tex]C_6H_12O_6 = (12*6) + (16*6) + (1*12) = 180 g/mol[/tex].After balancing the equation, the mole ratio of the glucose vs ethanol is 1:2
The molar mass of the ethanol is as follows:-
[tex]C_2H_5OH[/tex] =[tex]12*2) + (6*1) + (16*1) = 46 g/mol[/tex]
After diving each molar mass of each compound with respect to its mole ratio. The mass of the ethanol produced is 255g.
Hence, the correct answer is 255g.
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Give the electron configuration for the following atoms using appropriate noble gas inner core abbreviation: Bi Cr Sr P 2. Give a set of 4 possible quantum numbers for the most energetic electron(s) of: Bi Cr Sr P n = l = ml = ms = 3. What is the symbol and name of the element with the following electron configuration?
Answer:
See explanation
Explanation:
The noble gas core electron configuration involves writing the inert gas core of an atom followed by the valence electrons. This is shown for the following atoms;
Bismuth;
[Xe]4f14 5d10 6s2 6p3
Chromium;
[Ar]4s1 3d5
Strontium;
[Kr]5s2
Phosphorus;
[Ne]3s2 3p3
2.
Bi
6p- n=6, l= 1, ml= 1, ms= 1/2
Cr
3d- n=3, l=2, ml=2,ms=1/2
Sr
5s- n=5, l=0, ml=0, ms=1/2
P
3p- n=3, l= 1, ml= 1, ms=1/2
3.
a) Tin (Sn) - [Kr] 5s2 4d10 5p2
b) Caesium (Cs)- [Xe] 6s1
c) Copper (Cu)- [Ar] 4s1 3d10
Phosphofructokinase catalyzes the phosphorylation of fructose 6‑phosphate to fructose 1,6‑bisphosphate in glycolysis. Fructose 1,6‑bisphosphatase catalyzes the hydrolysis of fructose 1,6‑bisphosphate to fructose 6‑phosphate in gluconeogenesis.
fructose 6- phosphate
phosphofructokinase fructose 1 ,6-bisphosphatase
fructose 1,6 - bisphosphate
How does fructose-2,6-bisphosphate (F26BP) affect the activity of the enzymes phosphofructokinase-1 (PFK) and fructose I ,6-bisphosphatase (FBPase)?
a. increases PFK activity, increases FBPase activity
b. decreases PFK activity, increases FBPase activity
c. decreases PFK activity, decreases FBPase activity
d. increases PFK activity, decreases FBPase activity
Answer:
d. increases PFK activity, decreases FBPase activity
Explanation:
Fructose-2,6-bisphophate is formed by the phosphorylation of fructose-6-phosphate catalyzed by phosphofructokinase-2, PFK-2.
Fructose-2,6-bisphophate functions as an allosteric effector of the enzymes phosphofructokinase-1, PFK-1 and fructose-1,6-bisphosphatase, FBPase.
Fructose-2,6-bisphophate has opposite effects on the enzymes, PFK-1 and FBPase. When it binds to the allosteric site of the enzyme, PFK-1, it increases the enzymes's activity by increasing its affinity for its substrate fructose-6-phosphate and reduces its affinity for its allosteric inhibitors ATP and citrate. However, when it binds to FBPase, it reduces its activity by reducing its affinity for glucose, its substrate
What is the energy change associated with 1.5 mole of D being formed? 
Answer:
–36 KJ.
Explanation:
The equation for the reaction is given below:
2B + C —› D + E. ΔH = – 24 KJ
From the equation above,
1 mole of D required – 24 KJ of energy.
Now, we shall determine the energy change associated with 1.5 moles of D.
This can be obtained as illustrated below:
From the equation above,
1 mole of D required – 24 KJ of energy
Therefore,
1.5 moles of D will require = 1.5 × – 24 = –36 KJ.
Therefore, –36 KJ of energy is associated with 1.5 moles of D.
Some metal oxides, such as Sc2O3, do not react with pure water, but they do react when the solution becomes either acidic or basic. Do you expect Sc2O3 to react when the solution becomes acidic or when it becomes basic?
Write a balanced chemical equation to support your answer.
Answer:
[tex]Sc_2O_3[/tex] reacts with an acidic solution
Explanation:
Scandium Oxide [tex]Sc_2O_3[/tex] is a basic metal oxide which therefore reacts with acidic solution. An oxide is a compound that contains only two elements, one of which is oxygen .
The objective of this question is to Write a balanced chemical equation to support your answer.
The chemical equation to support the reaction of [tex]Sc_2O_3[/tex] with acidic solution is as follows:
Assuming the acidic solution to be HCl
[tex]\mathbf{Sc_2O_3_{(s)} + 6 HCl_{(aq)} ----> 2 ScCl_{3(aq)} + 3H_2O_{(l)}}[/tex]
The ionic equation :
[tex]\mathbf{Sc_2O_{3(s)} + 6H^+_{(aq)} ---> 2Sc^{3+}_{(aq)} + 3H_2O_{(l)}}[/tex]
In a 74.0-g aqueous solution of methanol, CH4O, the mole fraction of methanol is 0.140. What is the mass of each component?
Answer:
The correct answer is 16.61 grams methanol and 57.38 grams water.
Explanation:
The mole fraction (X) of methanol can be determined by using the formula,
X₁ = mole number of methanol (n₁) / Total mole number (n₁ + n₂)
X₁ = n₁/n₁ + n₂ = 0.14
n₁ / n₁ + n₂ = 0.14 ---------(i)
n₁ mole CH₃OH = n₁ mol × 32.042 gram/mol (The molecular mass of CH₃OH is 32.042 grams per mole)
n₁ mole CH₃OH = 32.042 n₁ g
n₂ mole H2O = n₂ mole × 18.015 g/mol
n₂ mole H2O = 18.015 n₂ g
Thus, total mole number is,
32.042 n₁ + 18.015 n₂ = 74 ------------(ii)
From equation (i)
n₁/n₁ + n₂ = 0.14
n₁ = 0.14 n₁ + 0.14 n₂
n₁ - 0.14 n₁ = 0.14 n₂
n₁ = 0.14 n₂ / 1-0.14
n₁ = 0.14 n₂/0.86 ----------(iii)
From eq (ii) and (iii) we get,
32.042 × 0.14/0.86 n₂ + 18.015 n₂ = 74
n₂ (32.042 × 0.14/0.86 + 18.015) = 74
n₂ = 74 / (32.042 × 0.14/0.86 + 18.0.15)
n₂ = 3.1854 mol
From equation (iii),
n₁ = 0.14/0.86 n₂
n₁ = 0.14/0.86 × 3.1854
n₁ = 0.5185 mol
Now, presence of water in the mixture is,
= 3.1854 mole × 18.015 gram per mole
= 57.38 grams
Methanol present in the mixture is,
= 0.5185 mol × 32.042 gram per mole
= 16.61 grams
p32p32 is a radioactive isotope with a half-life of 14.3 days. if you currently have 63.163.1 g of p32p32 , how much p32p32 was present 8.008.00 days ago
Answer:
92.93 g
Explanation:
Number of half lives that have elapsed in eight days =8/14.3 = 0.559
Fraction of the radioactive nuclide that remains after 0.559 half lives is given by
N/No=(1/2)^0.559
Where N= mass of radioactive nuclides remaining after a time t
No= mass of radioactive nuclides originally present
N/No=(1/2)^0.559= 0.679
Mass of nuclides present eight days before= 63.1g/0.679
Mass of nuclides present eight days before=92.93 g
Explain why o-vanillin does not fully protonate p-toluidine. Reference appropriate pKa values and include a balanced chemical reaction and an appropriate reaction arrow in your answer.
Answer:
Here's what I get
Explanation:
pKₐ of o-vanillin = 7.81; pKₐ of p-toluidine = 4.44
The higher the pKₐ, the weaker the acid.
Thus, o-vanillin is the weaker acid and has a stronger conjugate base.
The conjugate acid of p-toluidine is the stronger and has the weaker conjugate base.
The equation for the equilibrium is
H-OC₆H₃(OCH₃)CHO + CH₃C₆H₄NH₂ ⇌ ⁻OC₆H₃(OCH₃)CHO + CH₃C₆H₄NH₃⁺
weaker acid weaker base stronger base stronger acid
The reaction between the stronger acid and the stronger base pushes the position of equilibrium to the left.
Thus, o-vanillin does not fully protonate p-toluidine.
O-vanillin is a weaker acid than p-toluidine and has a more stable conjugate base; hence, o-vanillin does not fully protonate p-toluidine.
The pKa is defined as the negative logarithm of Ka. The dissociation constant of an acid Ka shows the extent of dissociation of an acid in solution. The higher the pKa, the lower the Ka and the weaker the acid.
The pKₐ of o-vanillin is 7.81 while the pKₐ of p-toluidine is 4.44. This means that o-vanillin is a weaker acid than p-toluidine and has a more stable conjugate base. Hence, o-vanillin does not fully protonate p-toluidine.
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A student obtained a clean flask. She weighed the flask and stopper on an analytical balance and found the total mass to be 34.232 g. She then filled the flask with water and found the new mass to be 60.167 g. The temperature of the water was measured to be
Answer:
25.99mL is the volume internal volume of the flask
Explanation:
To complete the question:
The temperature of the water was measured to be 21ºC. Use this data to find the internal volume of the stoppered flask
The flask was filled with water, that means the internal volume of the flask is equal to the volume that the water occupies.
To find the volume of the water you need to find the mass and by the use of density of water at 21ºC (0.997992g/mL), you can find the volume of the flask, thus:
Mass water = Mass filled flask - Mass of clean flask
Mass water = 60.167g - 34.232g
Mass water = 25.935g of water.
To convert this mass to volume:
25.935g × (1mL / 0.997992g) =
25.99mL is the volume internal volume of the flaskCalculate the pH of a buffer solution obtained by dissolving 25.025.0 g of KH2PO4(s)KH2PO4(s) and 38.038.0 g of Na2HPO4(s)Na2HPO4(s) in water and then diluting to 1.00 L.
Answer:
pH = 7.37
Explanation:
The buffer H₂PO₄⁻/HPO₄²⁻ has as pKa 7.21. To find the pH of a buffer you can use H-H equation:
pH = pKa + log₁₀ [HPO₄²⁻] / [H₂PO₄⁻]
Where [HPO₄²⁻] and [H₂PO₄⁻] are molar concentrations of each species. As volume is 1.00L, [HPO₄²⁻] and [H₂PO₄⁻] are MOLES.
Moles of 25.0g of KH₂PO₄ (Molar mass: 136.086g/mol):
25.0g KH₂PO₄ ₓ (1mol / 136.086g) = 0.1837 moles KH₂PO₄ = moles H₂PO₄⁻
Moles of 38.0g of Na₂HPO₄ (Molar mass: 141.96g/mol):
38.0g KH₂PO₄ ₓ (1mol / 141.96g) = 0.2677 moles Na₂HPO₄ = moles HPO₄²⁻
Replacing in H-H equation:
pH = pKa + log₁₀ [HPO₄²⁻] / [H₂PO₄⁻]
pH = 7.21 + log₁₀ [0.2677] / [0.1837]
pH = 7.37
What are the solutions to the quadratic equation 2x2 + 10x - 48 = 0?
Answer:
x = 3 , x= -8Explanation:
[tex]2x^2+10x-48\\=2\left(x^2+5x-24\right)\\x^2+5x-24\\=\left(x^2-3x\right)+\left(8x-24\right)\\=x\left(x-3\right)+8\left(x-3\right)\\=\left(x-3\right)\left(x+8\right)\\=2\left(x-3\right)\left(x+8\right)\\2\left(x-3\right)\left(x+8\right)=0\\x-3=0\\x = 0+3\\x = 3\\x+8=0\\x+8-8=0-8\\x=-8\\x=3,\:x=-8[/tex]
Review the reversible reactions given, along with the associated equilibrium constant Kat room temperature. In each case, determine whether the forward or reverse reaction is favored.
CH3COOH → CH3C00^- + H^+
Ka=1.8 x 10^-5
AgCl → Ag^+ + Cl^-
Ksp=1.6 x 10^-10
Al(OH)3 → Al^3+ + 3OH^-
Ksp=3.7 x 10^-15
A+B → C
K=4.9 x 10^3
Answer:
The answers to your questions are given below
Explanation:
The following data were obtained from the question:
CH3COOH → CH3C00^- + H^+
Equilibrium constant, Ka = 1.8 x 10^-5
AgCl → Ag^+ + Cl^-
Equilibrium constant, Ksp = 1.6 x 10^-10
Al(OH)3 → Al^3+ + 3OH^-
Equilibrium constant, Ksp = 3.7 x 10^-15
A+B → C
Equilibrium constant, K = 4.9 x 10^3
When the value of the equilibrium constant is grater than 1, it shows that the concentration of product is higher than that of the reactant and it implies that the forward reaction is favored.
When the value of the equilibrium constant is 1, it shows that the the concentration of the product and reactant are the same. Therefore neither the forward nor the reverse reaction is favored.
When the value of the equilibrium constant is lesser than 1, it shows that the concentration of the reactant is higher than the concentration of the product. Therefore, the reversed reaction is favored.
Now, we shall the question given above as follow:
A. CH3COOH → CH3C00^- + H^+
Equilibrium constant, Ka = 1.8 x 10^-5
Since the value of the equilibrium constant is lesser than 1, it means that the reverse reaction is favored.
B. AgCl → Ag^+ + Cl^-
Equilibrium constant, Ksp = 1.6 x 10^-10
Since the value of the equilibrium constant is lesser than 1, it means that the reverse reaction is favored.
C. Al(OH)3 → Al^3+ + 3OH^-
Equilibrium constant, Ksp = 3.7 x 10^-15
Since the value of the equilibrium constant is lesser than 1, it means that the reverse reaction is favored.
D. A+B → C
Equilibrium constant, K = 4.9 x 10^3
Since the value of the equilibrium constant is greater than 1, it means that the forward reaction is favored.
The reaction conditions are:
A. The reverse reaction is favored.
B. The reverse reaction is favored.
C. The reverse reaction is favored.
D. The forward reaction is favored.
Chemical reaction:
A. [tex]CH_3COOH[/tex] → [tex]CH_3COO^- + H^+[/tex]
Equilibrium constant, Ka = [tex]1.8 * 10^{-5}[/tex]
B. [tex]AgCl[/tex] → [tex]Ag^+ + Cl^-[/tex]
Equilibrium constant, Ksp = [tex]1.6 * 10^{-10}[/tex]
C. [tex]Al(OH)_3[/tex] → [tex]Al^{3+} + 3OH^-[/tex]
Equilibrium constant, Ksp = [tex]3.7 * 10^{-15}[/tex]
D. A+B → C
Equilibrium constant, K = [tex]4.9 * 10^3[/tex]
Conditions for Equilibrium constant:When the value of the equilibrium constant is greater than 1, it shows that the concentration of product is higher than that of the reactant and it implies that the forward reaction is favored.
When the value of the equilibrium constant is 1, it shows that the the concentration of the product and reactant are the same. Therefore neither the forward nor the reverse reaction is favored.
When the value of the equilibrium constant is lesser than 1, it shows that the concentration of the reactant is higher than the concentration of the product. Therefore, the reversed reaction is favored.
Thus, the reactions will be:
A. The reverse reaction is favored.
B. The reverse reaction is favored.
C. The reverse reaction is favored.
D. The forward reaction is favored.
Find more information about Equilibrium constant here:
brainly.com/question/12858312
Sometimes a nuclide is referenced by the name of the element followed by the:______
a. atomic number
b. mass number
c. electrical charge
d. none of the above
Answer:
The correct option is d
Explanation:
Nuclide is synonymous with groups of electrons or protons, that is, a nuclide is the grouping of nucleons.
which of the following compounds exhibits dipole -dipole forces as its strongest attraction between molecules? o2,ch3Br,CCl4,He,BrCH2CH2OH
Answer:
B. CH3Br
Explanation:
Dipole -Dipole interactions take place in polar molecules.
CH3Br exhibits dipole -dipole forces as its strongest attraction between molecules because it is a polar molecule due to the slightly negative dipole present on the Br molecule.
While O2 is a nonpolar molecule due to its linear structure, CCl4 has zero resultant dipole moment, Helium is non-polar and BrCH2CH2OH is a non polar compound having net dipole moment is zero.
Hence, the correct option is B. CH3Br.
The compound that exhibits dipole -dipole forces is CH3Br
Dipole -Dipole interactions:It should be taken place in polar molecules. Also, CH3Br should be the strongest attraction that lies between the molecules since it is treated as the polar molecule because of the slightly negative. While on the other hand, O2 should be non-polar molecule because of the linear structure.
Therefore, The compound that exhibits dipole -dipole forces is CH3Br
Learn more about force here: https://brainly.com/question/14379797
A four carbon chain; the second carbon is also single bonded to CH3. Spell out the full name of the compound
Answer:
This description shows a methyl group.
Explanation:
A student mixes wants to prepare 24.1 mmol of benzamide from benzoyl chloride and NH4OH. If the student uses excess 15 M NH4OH, how many mL of Benzoyl chloride must be used
Answer:
2.81mL
Explanation:
Based on the reaction:
C₆H₃COCl + 2NH₃ → C₆H₅CONH₂ + NH₄Cl
Benzoyl chloride + ammonia → Benzamide
1 mole of benzoyl chloride in excess of ammonia produce 1 mole of Benzamide.
Thus, assuming a theoretical yield, to produce 24.1mmoles of benzamide you require 24.1mmoles of benzoyl chloride.
As molar mass of benzoyl chloride is 141g/mol, mg you require are:
mg Benzoyl chloride: 24.1mmol × (141mg / 1mmol) = 3398.1mg = 3.3981g of benzoyl chloride.
to convert this mass to mL, you require density of Benzoyl chloride (1.21g/mL). Thus, mL you need are:
3.3981g × (1mL / 1.21g) =
2.81mLThe NMR spectrum of your final compound will contain extra peaks that were not present in your starting material. For what hydrogen nuclei do those peaks occur?
Answer:
The peaks are registered from tetramethyl silane (TMS)
Explanation:
Tetramethyl silane (TMS) is used as internal reference in proton nmr (H NMR) spectrometry.
Its peak is usually registered at about a 2.0 chemical shift means that the hydrogen atoms which caused that peak need a magnetic field two millionths less than the field needed by TMS to produce resonance. This is not affected by the chemical shift of the sample analysed.
I hope this helped.
Un globo lleno de helio tenia un volumen de 8.5 L en el suelo a 20°C y a una presión de 750 torr. Cuando se le soltó, el globo se elevo a una altitud donde la temperatura era de -20°C y la presión de 425 torr, ¿Cuál era el volumen del gas del globo en estas condiciones?
Answer:
El volumen del gas era 12.95 L
Explanation:
Se relaciona la presión y el volumen mediante la ley de Boyle, que dice:
“El volumen ocupado por una determinada masa gaseosa a temperatura constante, es inversamente proporcional a la presión”
La ley de Boyle se expresa matemáticamente como: P*V=k
Por otro lado, la Ley de Charles consiste en la relación que existe entre el volumen y la temperatura absoluta de una cierta cantidad de gas ideal, el cual se mantiene a una presión constante. Esta ley dice que cuando la cantidad de gas y de presión se mantienen constantes, el cociente que existe entre el volumen y la temperatura siempre tendrán el mismo valor:
[tex]\frac{V}{T}=k[/tex]
Por último, la Ley de Gay Lussac dice que la temperatura absoluta y la presión son directamente proporcionales. Es decir, cuando se mantiene todo lo demás constante, mientras suba la temperatura de un gas subirá también su presión. Y mientras la temperatura del gas baje, lo mismo ocurrirá con la presión:
[tex]\frac{P}{T}=k[/tex]
Combinado las mencionadas tres leyes se obtiene:
[tex]\frac{P*V}{T} =k[/tex]
Cuando se desean estudiar dos diferentes estados, uno inicial y una final de un gas, se puede aplicar:
[tex]\frac{P1*V1}{T1} =\frac{P2*V2}{T2}[/tex]
Recordando que la temperatura debe usarse en grados Kelvin, conoces los siguientes datos:
P1: 750 torrV1: 8.5 LT1: 20°C= 293°K (siendo 0°C=273°K)P2: 425 torrV2: ?T2: -20°C= 253 °KReemplazando:
[tex]\frac{750 torr*8.5 L}{293K} =\frac{425 torr*V2}{253 K}[/tex]
Resolviendo:
[tex]V2=\frac{750 torr*8.5 L}{293K} *\frac{253 K}{425 torr}[/tex]
V2= 12.95 L
El volumen del gas era 12.95 L
A volume of 30.0 mL of a 0.140 M HNO3 solution is titrated with 0.570 M KOH. Calculate the volume of KOH required to reach the equivalence point.
Answer:
7.37 mL of KOH
Explanation:
So here we have the following chemical formula ( already balanced ), as HNO3 reacts with KOH to form the products KNO3 and H2O. As you can tell, this is a double replacement reaction,
HNO3 + KOH → KNO3 + H2O
Step 1 : The moles of HNO3 here can be calculated through the given molar mass ( 0.140 M HNO3 ) and the mL of this nitric acid. Of course the molar mass is given by mol / L, so we would have to convert mL to L.
Mol of NHO3 = 0.140 M [tex]*[/tex] 30 / 1000 L = 0.140 M [tex]*[/tex] 0.03 L = .0042 mol
Step 2 : We can now convert the moles of HNO3 to moles of KOH through dimensional analysis,
0.0042 mol HNO2 [tex]*[/tex] ( 1 mol KOH / 1 mol HNO2 ) = 0.0042 mol KOH
From the formula we can see that there is 1 mole of KOH present per 1 moles of HNO2, in a 1 : 1 ratio. As expected the number of moles of each should be the same,
Step 3 : Now we can calculate the volume of KOH knowing it's moles, and molar mass ( 0.570 M ).
Volume of KOH = 0.0042 mol [tex]*[/tex] ( 1 L / 0.570 mol ) [tex]*[/tex] ( 1000 mL / 1 L ) = 7.37 mL of KOH