Answer:
-46.67 J.
Explanation:
We are given;
Initial Pressure = 15atm = 15 × 10^(3) J
Final pressure = 1atm = 1 × 10^(3) J
Temperature = 300k
The pressures were converted to Joules.
Formula for the entropy change is;
∆S_system = ∆S_surrounding = -(dQ)/T
-(dQ)/T = (-(15 × 10^(3)) - (1 × 10^(3))/300)
= -46.67 J.
calculate the molarity in a 0.550 m solution of NaCl in water. Assume that the solution density is 1.03g/mol
Answer:
M=0.549M
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to perform this calculation by firstly assuming we have 1 kg of water as the solvent so that we have 0.550 moles of NaCl as well. Moreover, we realize we have 1000 grams of water and the correct mass of the solution can be calculated by converting 0.550 moles of NaCl to grams by using its molar mass:
[tex]m_{solute}=0.550mol*\frac{58.44 g}{1mol}= 32.14g\\\\m_{solution}=1000g+32.14g=1032.14g[/tex]
And subsequently, the volume in liters by using the density and the correct conversion factor:
[tex]V_{solution}=1032.14g*\frac{1mL}{1.03g} *\frac{1L}{1000mL} =1.002L[/tex]
Finally, the molarity will be:
[tex]M=\frac{0.550mol}{1.002L} =0.549M[/tex]
Regards!
Chlorine can be prepared in the laboratory by the reaction of manganese dioxide with hydrochloric acid, HCl(aq) , as described by the chemical equation
MnO2(s)+4HCl(aq)⟶MnCl2(aq)+2H2O(l)+Cl2(g)
How much MnO2(s) should be added to excess HCl(aq) to obtain 185 mL Cl2(g) at 25 °C and 745 Torr ?
mass of MnO2:
Answer:
0.605 g
Explanation:
MnO₂(s) + 4HCl(aq) ⟶ MnCl₂(aq) + 2H₂O(l) + Cl₂(g)First we calculate how many Cl₂ moles need to be produced, using the PV=nRT formula:
P = 745 Torr ⇒ 745 / 760 = 0.980 atmV = 185 mL ⇒ 185 / 1000 = 0.185 Ln = ?R = 0.082 atm·L·mol⁻¹·K⁻¹T = 25 °C ⇒ 25 + 273.16 = 298.16 KInputting the data:
0.980 atm * 0.185 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 298.16 Kn = 0.00696 molThen we convert 0.00696 moles of Cl₂ to MnO₂ moles:
0.00696 mol Cl₂ * [tex]\frac{1molMnO_2}{1molCl_2}[/tex] = 0.00696 mol MnO₂Finally we convert 0.00696 moles of MnO₂ to grams, using its molar mass:
0.00696 mol MnO₂ * 86.94 g/mol = 0.605 gwhich straight-chain alkane would you predict to be the most viscous? all are liquids exhbiting the general bonding pattern ch3-(ch2)n-ch3
The question is incomplete, the complete question his;
Which straight chain alkane below would you predict to be the most viscous? Why? All are liquids exhibiting the general bonding pattern CH3-(CH2)n-CH3
C9H20
C10H22
C5H12
C6H14
C12H26
Answer:
C12H26
Explanation:
Generally, the viscosity of a liquid increases with increase in molecular mass of the substance.
Liquids of high molecular mass do not flow easily. This means that they posses high viscosity.
Thus, since C12H26 has the highest molecular mass among the options given in the question, C12H26 exhibits the greatest viscosity.
why do people who do a lot of physical work need more carbohydrate?
Answer:
A person doing physical work needs lots of good carbohydrates to keep their energy levels up and proteins to repair a muscle that might get wear and tear from overexertion. Carbohydrate will help the person work for more extended periods.
In the context of small molecules with similar molar masses, arrange the intermolecular forces by strength.
a. London dispersion forces
b. hydrogen bonding
c. dipole-dipole interactions
Answer:
Hydrogen bonding - London dispersion forces - dipole-dipole interactions
Strongest ----> Weakest
You are asked to prepare a buffer solution with a pH of 3.50. The following solutions, all 0.100 M, are available to you: HCOOH, CH3COOH, H3PO4 , NaCHOO, NaCH3COO, and NaH2PO4. What would be the best combination to make the required buffer solution? Select one:
a. NaH2PO4 and NaCHOO
b. H3PO4 and NaH2PO4
c. NaH2PO4 and HCOOH
d. CH3COOH and NaCH3COO e. HCOOH and NaCHOO
can someone helo me with this
Answer:
e. HCOOH and NaCHOO
Explanation:
For a buffer solution, both an acid and its conjugate base are required.
With the information above in mind, we can discard options a) and c), as those combinations are not of an acid and its conjugate base.
Now it is a matter of comparing the pKa (found in literature tables) of the acids of the remaining three acids:
H₃PO₄ pKa = 2.12CH₃COOH pKa = 2.8HCOOH pKa = 3.74The acid with the pKa closest to the desired pH is HCOOH, so the correct answer is e. HCOOH and NaCHOO
Under certain conditions, the substance mercury(II) oxide can be broken down to form mercury and oxygen. If 32.2 grams of mercury(II) oxide react to form 29.8 grams of mercury, how many grams of oxygen must simultaneously be formed
Explanation:
This is a decomposition reaction. Firstly, you will want to write the chemical equation out and balance it.
[tex]2Hg_2O->4Hg+O_2[/tex] (The -> is supposed to be an arrow, sorry!)
We see that there's only 1mol of Oxygen made in the products, we can do some simple math to solve for the amount of grams of Oxygen produced according to the amount of the reactant (Hg2O).
[tex]32.2gHg_2O*\frac{1molHg_2O}{417.18gHg_2O}*\frac{1molO_2}{2molHg_2O}*\frac{32gO_2}{1molO_2}[/tex]
I want to break this down, just in case:
The 417.18gHg2O is the molecular mass of the molecule (so I doubled Hg and added 16 to it to get this number).
As we can see in the chemical equation, 1mol Hg2O produces 2mol O because Oxygen is a diatomic molecule (so there will always be two of it when it's by itself).
And finally, in 1mol O2 there are 32g of O2.
** When you do math like this, always make sure that all of your units cancel out except for the units you're looking for. For example, here we're looking for the grams of Oxygen, so after everything else cancels out, we should only have grams O2.
So, 1.23gO2 should be your answer.
Choose the correct statement. A) The cathode is the electrode where the oxidation takes place. B) The cathode is the electrode where the reduction takes place. C) Both oxidation and reduction may take place at the cathode, depending on the cell. D) The cathode is always positive
If 0.250 L of a 5.90 M HNO₃ solution is diluted to 2.00 L, what is the molarity of the new solution?
Answer:
0.74 M
Explanation:
From the question given above, the following data were obtained:
Molarity of stock solution (M₁) = 5.90 M
Volume of stock solution (V₁) = 0.250 L
Volume of diluted solution (V₂) = 2 L
Molarity of diluted solution (M₂) =?
The molarity of the diluted solution can be obtained by using the dilution formula as illustrated below:
M₁V₁ = M₂V₂
5.90 × 0.250 = M₂ × 2
1.475 = M₂ × 2
Divide both side by 2
M₂ = 1.475 / 2
M₂ = 0.74 M
Thus, the molarity of the diluted solution is 0.74 M
Based on the equations below, which metal is the least active? Pb(NO3)2(aq) + Ni (s) --> Ni(NO3)2 (aq)+ Pb(s) Pb(NO3)2(aq) + Ag(s) --> No reaction Cu(
Answer:
Ni
Explanation:
An active metal is a highly reactive metal. Active metals are found high up in the activity series.
Active metals react with other metals that are lower than them in the activity thereby displacing the lower metals from a solution of their salts. This is what may have happened in the other two reactions.
Ni is the most active metal listed in the question since it can react a compounds with Pb(NO3)2(aq) to liberate Pb metal.
Calcium chloride and magnesium sulfate are common drying agents. What type of solvent should be dried with calcium chloride, and what type with magnesium sulfate
Answer: The type of solvent that should be dried with calcium chloride is esters while magnesium sulfate is diethyl ether
Explanation:
Drying agents are mainly hygroscopic substances that has the ability to absorb water on exposure to the atmosphere but not enough to form solutions. They are used in desiccators. Examples of drying agents include:
--> CALCIUM CHLORIDE: This is a compound of calcium that is found in soil water and sea water. It is prepared by the action of dilute hydrochloric acid on calcium trioxocarbonate(IV). The anhydrous salt is used in drying a wide variety of solvent including esters.
--> MAGNESIUM SULFATE: This is a slightly acidic drying agent. It works well in solvents like diethyl ether. It is a fast drying agent because it comes as a fine powder with a large surface area.
Suppose that you move from a Suppose that you move from a town near the ocean to a town in the mountains. To what atmospheric changes would your body need to adjust? town near the ocean to a town in the mountains. To what atmospheric changes would your body need to adjust?
Answer:
all I can say is town near the ocean atmospheric changes will be cooler, warm, sea breeze, and fresh healthy air. Then when it comes to the mountain lot of change firstly there's a dry air
15. You are interested in separating 4-methylbenzoic acid from 1,4-dimethoxybenzene using a procedure similar to the extraction procedure we used in lab. You plan to use sodium bicarbonate instead of sodium hydroxide. a) Show the reaction between salicylic acid and sodium bicarbonate. Label the acid, base, conjugate acid, conjugate base. b) Give the pKa values of the acid and conjugate acid. c) Which base will work better, sodium hydroxide or sodium bicarbonate
Solution :
a). The separation of 4-methylbenzoic acid from 1,4-dimethoxybenzene will work but it will result in lower recovery.
In the reaction of acid-base to form a sodium 4 - methoxy benzoate, that is soluble in the water, 4-methoxy benzoic acid reacts with the sodium bicarbonate to give sodium 4-methoxybenzoate as well as carbonic acid.
b). The pKa for the 4-methoxybenzoic acid is [tex]4.46[/tex], and that of carbonic acid is [tex]6.37[/tex]
c). The Keq for the reaction is [tex]10(6.37 - 4.46) = 101.91[/tex]
Therefore, the equilibrium lies to the right and also the reaction favors the products and the separation works.
But the recovery will be low when compared to the extraction with Sodium hydroxide as the strong base will drive the equilibrium further to the right position, thus neutralizing all the acids virtually. And the weak base will partially neutralize the acid.
g You observed the formation of several precipitates in the Reactions in Solution lab exercise. Identify the precipitate in each of the following reactions: a. The yellow precipitate formed in the reaction between KI and Pb(NO3)2 is . b. The white precipitate formed in the reaction between BaCl2 and H2SO4 is . c. The brown precipitate formed in the reaction between NaOH and FeCl3 is . d. The blue precipitate formed in the reaction between CuSO4 and NaOH is .
Answer:
For a: Lead iodide is a yellow precipitate.
For b: Barium sulfate is a white precipitate.
For c: Ferric hydroxide is a brown precipitate.
For d: Copper (II) hydroxide is a blue precipitate.
Explanation:
Precipitation reaction is defined as the reaction where a solid precipitate (solid substance) is formed at the end of the reaction. It is insoluble in water.
For the given options:
For (a):The chemical reaction between KI and lead (II) nitrate follows:
[tex]2KI(aq)+Pb(NO_3)_2(aq)\rightarrow PbI_2(s)+2KNO_3(aq)[/tex]
The iodide of lead is generally insoluble in water. Thus, lead iodide is a yellow precipitate.
For b:The chemical reaction between barium chloride and sulfuric acid follows:
[tex]BaCl_2(aq)+H_2SO_4(aq)\rightarrow BaSO_4(s)+2HCl(aq)[/tex]
The sulfate of barium is insoluble in water. Thus, barium sulfate is a white precipitate.
For c:The chemical reaction between NaOH and ferric chloride follows:
[tex]3NaOH(aq)+FeCl_3(aq)\rightarrow Fe(OH)_3(s)+3NaCl(aq)[/tex]
The hydroxide of iron is insoluble in water. Thus, ferric hydroxide is a brown precipitate.
For d:The chemical reaction between NaOH and copper sulfate follows:
[tex]CuSO_4+2NaOH\rightarrow Cu(OH)_2+Na_2SO_4[/tex]
The hydroxide of copper is insoluble in water. Thus, copper (II) hydroxide is a blue precipitate.
(a) The yellow precipitate formed in the reaction between KI and Pb(NO3)2 would be PbI2 according to the equation:
[tex]Pb(NO_3)_2(aq) + 2KI(aq) ---> PbI2(s) + 2KNO_3(aq)[/tex]
(b) The white precipitate formed in the reaction between BaCl2 and H2SO4 would be BaSO4 according to the equation:
[tex]BaCl_2 (aq) + H_2SO_4 (aq) ---> BaSO_4 (s) + 2 HCl (aq)[/tex]
(c) The brown precipitate formed in the reaction between NaOH and FeCl3 would be Fe(OH)3 according to the equation:
[tex]FeCl_3 (aq) + NaOH (aq) ---> Fe(OH)_3 (s) + NaCl (aq)[/tex]
(d) The blue precipitate formed in the reaction between CuSO4 and NaOH would be Cu(OH)2 according to the equation:
[tex]CuSO_4(aq) + 2 NaOH (aq) ---> Cu(OH)_2 (s) + Na_2SO_4 (aq)[/tex]
More on precipitation reaction can be found here: https://brainly.com/question/24846690
5. The Rf of ibuprofen was found to be 0.32 when t-butyl methyl ether was used as the development solvent. What effect would there be on the Rf of ibuprofen if acetone had been used to develop the TLC plate?
Answer:
The Rf value of ibuprofen increases
Explanation:
TLC involves the elution of a solute using a mobile phase(solvent). The stationary phase is made of an adsorbent such as silica.
The extent of interaction between the solute and the mobile phase affects the Rf value. The greater the interaction between the solute and the solvent, the greater the Rf value.
On the other hand, the polarity of the solvent and the solute also affects the Rf value. If the solvent is changed from t-butyl methyl ether to acetone, the Rf value for ibuprofen increases because ibuprofen is polar and acetone is also polar hence there is greater interaction between the solvent and solute.
During a reaction, ΔH for reactants is −750 kJ/mol and ΔH for products is 920 kJ/mol. Which statement is correct about the reaction? (5 points)
Group of answer choices
It is endothermic because the energy required to break bonds in the reactants is less than the energy released when the products are formed.
It is endothermic because the energy required to break bonds in the reactants is greater than the energy released when the products are formed.
It is exothermic because the energy required to break bonds in the reactants is less than the energy released when the products are formed.
It is exothermic because the energy required to break bonds in the reactants is greater than the energy released when the products are formed.
Answer: The statement it is endothermic because the energy required to break bonds in the reactants is less than the energy released when the products are formed, is true.
Explanation:
A chemical reaction in which heat energy is released is called an exothermic reaction. For exothermic reactions, the value of [tex]\Delta H[/tex] is always negative.
A chemical reaction in which heat energy is absorbed is called an endothermic reaction. For endothermic reaction, the value of [tex]\Delta H[/tex] is always positive.
In endothermic reactions, energy required for breaking the bonds between reactants is less than the energy when products are formed due to which the value of [tex]\Delta H[/tex] remains positive.
Thus, we can conclude that the statement it is endothermic because the energy required to break bonds in the reactants is less than the energy released when the products are formed, is true.
It is endothermic because the energy required to break bonds in the reactants is greater than the energy released when the products are formed. The correct option is B.
The above reaction is endothermic because more energy is produced when new bonds form in the products (H = 920 kJ/mol) than is required to break bonds in the reactants (H = -750 kJ/mol).
In an endothermic process, more energy than is generated during bond creation is absorbed from the environment to dissolve existing bonds. This causes a net absorption of energy, which cools the system.
The reaction takes more energy than it releases, proving its endothermic nature, as seen by the positive difference between the energy needed to dissolve bonds and the energy released during bond formation.
Thus, the correct option is B.
For more details regarding endothermic process, visit:
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Your question seems incomplete, the probable complete question is:
During a reaction, ΔH for reactants is −750 kJ/mol and ΔH for products is 920 kJ/mol. Which statement is correct about the reaction? (5 points)
Group of answer choices
A. It is endothermic because the energy required to break bonds in the reactants is less than the energy released when the products are formed.
B. It is endothermic because the energy required to break bonds in the reactants is greater than the energy released when the products are formed.
C. It is exothermic because the energy required to break bonds in the reactants is less than the energy released when the products are formed.
D. It is exothermic because the energy required to break bonds in the reactants is greater than the energy released when the products are formed.
formula of
Al³⁺ and SO₄²⁻
Answer:
The formula of Al³⁺ and SO₄²⁻ is aluminum sulfate.
Explanation:
The formula for aluminum sulfate is Al₂(SO₄)₃. If we say in terms of ions. The ions are Al³⁺. It is a positive ion or the cation. Other ion is SO₄²⁻. It is sulfate ion. It is anion.
Aluminum sulphate is used in water purification and as a mordant in dyeing and printing textiles.
Hence, the formula of Al³⁺ and SO₄²⁻ is aluminum sulfate.
What intermolecular force or bond is primarily responsible for the solubility of chlorine (Cl2) in water
The question is incomplete, the common question is;
What intermolecular attractive force is primarily responsible for the solubility of chlorine, Cl2, in water?
a. dipole - dipole
b. hydrogen bonding
c. dipole-induced dipole
d. ion-dipole
e. ion-induced dipole
Answer:
dipole-induced dipole
Explanation:
We have to remember that water is a polar molecule hence it possesses a dipole moment. Its dipole moment is responsible for the ability of water to dissolve many substances.
On the other hand, Cl2 is a nonpolar molecule bound together by only weak dispersion forces.
Recall that dispersion forces involve transient appearance of a dipole in a molecule.
Water molecules can induce a dipole in Cl2 thereby causing the both molecules to interact and Cl2 dissolves in water.
Suppose an electron is transferred from a potassium atom to an unknown halogen atom. For which of the following halogen atoms would this process require the least amount of energy?
A. Cl
B. Br
C. I
Answer:
Cl
Explanation:
Electronegativity is the ability of an electron to attract electrons.
Now, due to the fact that halogens need just one more electron to become stable in their outermost shell, it means all halogens are electronegative.
However, the smaller the atomic number, the bigger the charge density and thus the more electronegative.
Thus, it is the halogen element with the highest atomic number further down the periodic table that will have the least electro negativity and thus require highest amount of energy to attract other electrons.
Thus, since chlorine (Cl) has the least atomic number of 17, then it means that it will be the one that will easily accept the electrons the most from other elements. Therefore the process of transferring electrons from potassium to chlorine will take the least amount of energy.
The Bohr model of the atom explains why emission spectra are discrete. It could also be used to explain the photoelectric effect. Which is a correct explaination of the photoelectric effect according to the model
Answer:
photoelectric effect, phenomenon in which electrically charged particles are released from or within a material when it absorbs electromagnetic radiation. The effect is often defined as the ejection of electrons from a metal plate when light falls on it.
Calculate the osmotic pressure of 5.0g of sucrose ssolution in 1L. Answer should be in Torr
Answer: The osmotic pressure of 5.0g of sucrose solution in 1 L is 271.32 torr.
Explanation:
Given: Mass = 5.0 g
Volume = 1 L
Molar mass of sucrose = 342.3 g/mol
Moles are the mass of a substance divided by its molar mass. So, moles of sucrose are calculated as follows.
[tex]Moles = \frac{mass}{molarmass}\\= \frac{5.0 g}{342.3 g/mol}\\= 0.0146 mol[/tex]
Hence, concentration of sucrose is calculated as follows.
[tex]Concentration = \frac{moles}{Volume (in L)}\\= \frac{0.0146 mol}{1 L}\\= 0.0146 M[/tex]
Formula used to calculate osmotic pressure is as follows.
[tex]\pi = CRT[/tex]
where,
[tex]\pi[/tex] = osmotic pressure
C = concentration
R = gas constant = 0.0821 L atm/mol K
T = temperature
Substitute the values into above formula as follows.
[tex]\pi = CRT\\= 0.0146 \times 0.0821 L atm/mol K \times 298 K\\= 0.357 atm (1 atm = 760 torr)\\= 271.32 torr[/tex]
Thus, we can conclude that the osmotic pressure of 5.0g of sucrose solution in 1 L is 271.32 torr.
Consider the titration of 30 mL of 0.030 M NH3 with 0.025 M HCl. Calculate the pH after the following volumes of titrant have been added: a) 0 mL; b) 10 mL; c) 20 mL; d)35 mL; e) 36 mL; f) 37 mL.
Answer:
a)10.87
b)9.66
c)9.15
d)7.71
e) 5.56
f) 3.43
Explanation:
tep 1: Data given
Volume of 0.030 M NH3 solution = 30 mL = 0.030 L
Molarity of the HCl solution = 0.025 M
Step 2: Adding 0 mL of HCl
The reaction: NH3 + H2O ⇔ NH4+ + OH-
The initial concentration:
[NH3] = 0.030M [NH4+] = 0M [OH-] = OM
The concentration at the equilibrium:
[NH3] = 0.030 - XM
[NH4+] = [OH-] = XM
Kb = ([NH4+][OH-])/[NH3]
1.8*10^-5 = x² / 0.030-x
1.8*10^-5 = x² / 0.030
x = 7.35 * 10^-4 = [OH-]
pOH = -log [7.35 * 10^-4]
pOH = 3.13
pH = 14-3.13 = 10.87
Step 3: After adding 10 mL of HCl
The reaction:
NH3 + HCl ⇔ NH4+ + Cl-
NH3 + H3O+ ⇔ NH4+ + H2O
Calculate numbers of moles:
Moles of NH3 = 0.030 M * 0.030 L = 0.0009 moles
Moles HCl = 0.025 M * 0.010 L = 0.00025 moles
Moles NH4+ = 0 moles
Number of moles at the equilibrium:
Moles NH3 = 0.0009 -0.00025 =0.00065 moles
Moles HCl = 0
Moles NH4+ = 0.00025 moles
Concentration at the equilibrium:
[NH3]= 0.00065 moles / 0.040 L = 0.01625M
[NH4+] = 0.00625 M
pOH = pKb + log [NH4+]/[NH3]
pOH = 4.75 + log (0.00625/0.01625)
pOH = 4.34
pH = 9.66
Step 3: Adding 20 mL of HCl
Calculate numbers of moles:
Moles of NH3 = 0.030 M * 0.030 L = 0.0009 moles
Moles HCl = 0.025 M * 0.020 L = 0.00050 moles
Moles NH4+ = 0 moles
Number of moles at the equilibrium:
Moles NH3 = 0.0009 -0.00050 =0.00040 moles
Moles HCl = 0
Moles NH4+ = 0.00050 moles
Concentration at the equilibrium:
[NH3]= 0.00040 moles / 0.050 L = 0.008M
[NH4+] = 0.01 M
pOH = pKb + log [NH4+]/[NH3]
pOH = 4.75 + log (0.01/0.008)
pOH = 4.85
pH = 14 - 4.85 = 9.15
Step 4: Adding 35 mL of HCl
Calculate numbers of moles:
Moles of NH3 = 0.030 M * 0.030 L = 0.0009 moles
Moles HCl = 0.025 M * 0.035 L = 0.000875 moles
Moles NH4+ = 0 moles
Number of moles at the equilibrium:
Moles NH3 = 0.0009 -0.000875 =0.000025 moles
Moles HCl = 0
Moles NH4+ = 0.000875 moles
Concentration at the equilibrium:
[NH3]= 0.000025 moles / 0.065 L = 3.85*10^-4M
[NH4+] = 0.000875 M / 0.065 L = 0.0135 M
pOH = pKb + log [NH4+]/[NH3]
pOH = 4.75 + log (0.0135/3.85*10^-4)
pOH = 6.29
pH = 14 - 6.29 = 7.71
Step 5: adding 36 mL HCl
Calculate numbers of moles:
Moles of NH3 = 0.030 M * 0.030 L = 0.0009 moles
Moles HCl = 0.025 M * 0.036 L = 0.0009 moles
Moles NH4+ = 0 moles
Number of moles at the equilibrium:
Moles NH3 = 0.0009 -0.0009 =0 moles
Moles HCl = 0
Moles NH4+ = 0.0009 moles
[NH4+] = 0.0009 moles / 0.066 L = 0.0136 M
Kw = Ka * Kb
Ka = 10^-14 / 1.8*10^-5
Ka = 5.6 * 10^-10
Ka = [NH3][H3O+] / [NH4+]
Ka =5.6 * 10^-10 = x² / 0.0136
x = 2.76 * 10^-6 = [H3O+]
pH = -log(2.76 * 10^-6)
pH = 5.56
Step 6: Adding 37 mL of HCl
Calculate numbers of moles:
Moles of NH3 = 0.030 M * 0.030 L = 0.0009 moles
Moles HCl = 0.025 M * 0.037 L = 0.000925 moles
Moles NH4+ = 0 moles
Number of moles at the equilibrium:
Moles NH3 = 0.0009 -0.000925 =0 moles
Moles HCl = 0.000025 moles
Concentration of HCl = 0.000025 moles / 0.067 L = 3.73 * 10^-4 M
pH = -log 3.73*10^-4= 3.43
The pH of the solution in the titration of 30 mL of 0.030 M NH₃ with 0.025 M HCl, is:
a) pH = 10.86
b) pH = 9.66
c) pH = 9.15
d) pH = 7.70
e) pH = 5.56
f) pH = 3.43
Calculating the pH a) 0 mL
Initially, the pH of the solution is given by the dissociation of NH₃ in water.
NH₃ + H₂O ⇄ NH₄⁺ + OH⁻ (1)
The constant of the above reaction is:
[tex] Kb = \frac{[NH_{4}^{+}][OH^{-}]}{[NH_{3}]} = 1.76\cdot 10^{-5} [/tex] (2)
At the equilibrium, we have:
NH₃ + H₂O ⇄ NH₄⁺ + OH⁻ (3)
0.030 M - x x x
[tex] 1.76\cdot 10^{-5}*(0.030 - x) - x^{2} = 0 [/tex]
After solving for x and taking the positive value:
x = 7.18x10⁻⁴ = [OH⁻]
Now, we can calculate the pH of the solution as follows:
[tex] pH = 14 - pOH = 14 + log(7.18\cdot 10^{-4}) = 10.86 [/tex]
Hence, the initial pH is 10.86.
b) 10 mL
After the addition of HCl, the following reaction takes place:
NH₃ + HCl ⇄ NH₄⁺ + Cl⁻ (4)
We can calculate the pH of the solution from the equilibrium reaction (3).
[tex] 1.76\cdot 10^{-5}(Cb - x) - (Ca + x)*x = 0 [/tex] (5)
Finding the number of moles of NH₃ and NH₄⁺
The number of moles of NH₃ (nb) and NH₄⁺ (na) are given by:
[tex] n_{b} = n_{i} - n_{HCl} [/tex] (6)
[tex] n_{b} = 0.030 mol/L*0.030 L - 0.025 mol/L*0.010 L = 6.5\cdot 10^{-4} moles [/tex]
[tex] n_{a} = n_{HCl} [/tex] (7)
[tex] n_{a} = 0.025 mol/L*0.010 L = 2.5 \cdot 10^{-4} moles [/tex]
Calculating the concentrations of NH₃ and NH₄⁺The concentrations are given by:
[tex] Cb = \frac{6.5\cdot 10^{-4} moles}{(0.030 L + 0.010 L)} = 0.0163 M [/tex] (8)
[tex] Ca = \frac{2.5 \cdot 10^{-4} mole}{(0.030 L + 0.010 L)} = 6.25 \cdot 10^{-3} M [/tex] (9)
Calculating the pHAfter entering the values of Ca and Cb into equation (5) and solving for x, we have:
[tex] 1.76\cdot 10^{-5}(0.0163 - x) - (6.25 \cdot 10^{-3} + x)*x = 0 [/tex]
x = 4.54x10⁻⁵ = [OH⁻]
Then, the pH is:
[tex] pH = 14 + log(4.54\cdot 10^{-5}) = 9.66 [/tex]
Hence, the pH is 9.66.
c) 20 mLWe can find the pH of the solution from the reaction of equilibrium (3).
Calculating the concentrations of NH₃ and NH₄⁺The concentrations are (eq 8 and 9):
[tex] Cb = \frac{0.030 mol/L*0.030 L - 0.025 mol/L*0.020 L}{(0.030 L + 0.020 L)} = 8.0\cdot 10^{-3} M [/tex]
[tex] Ca = \frac{0.025 mol/L*0.020 L}{(0.030 L + 0.020 L)} = 0.01 M [/tex]
Calculating the pHAfter solving the equation (5) for x, we have:
[tex] 1.76\cdot 10^{-5}(8.0\cdot 10^{-3} - x) - (0.01 + x)*x = 0 [/tex]
x = 1.40x10⁻⁵ = [OH⁻]
Then, the pH is:
[tex] pH = 14 + log(1.40\cdot 10^{-5}) = 9.15 [/tex]
So, the pH is 9.15.
d) 35 mLWe can find the pH of the solution from reaction (3).
Calculating the concentrations of NH₃ and NH₄⁺[tex] Cb = \frac{0.030 mol/L*0.030 L - 0.025 mol/L*0.035 L}{(0.030 L + 0.035 L)} = 3.85\cdot 10^{-4} M [/tex]
[tex] Ca = \frac{0.025 mol/L*0.035 L}{(0.030 L + 0.035 L)} = 0.0135 M [/tex]
Calculating the pHAfter solving the equation (5) for x, we have:
[tex] 1.76\cdot 10^{-5}(3.85\cdot 10^{-4} - x) - (0.0135 + x)*x = 0 [/tex]
x = 5.013x10⁻⁷ = [OH⁻]
Then, the pH is:
[tex] pH = 14 + log(5.013\cdot 10^{-7}) = 7.70 [/tex]
So, the pH is 7.70.
e) 36 mL Finding the number of moles of NH₃ and NH₄⁺[tex] n_{b} = 0.030 mol/L*0.030 L - 0.025 mol/L*0.036 L = 0 [/tex]
[tex] n_{a} = 0.025 mol/L*0.036 L = 9.0 \cdot 10^{-4} moles [/tex]
Since all the NH₃ reacts with the HCl added, the pH of the solution is given by the dissociation reaction of the NH₄⁺ produced in water.
At the equilibrium, we have:
NH₄⁺ + H₂O ⇄ NH₃ + H₃O⁺
Ca - x x x
[tex] Ka = \frac{x^{2}}{Ca - x} [/tex]
[tex] Ka(Ca - x) - x^{2} = 0 [/tex] (10)
Calculating the acid constant of NH₄⁺
We can find the acid constant as follows:
[tex] Kw = Ka*Kb [/tex]
Where Kw is the constant of water = 10⁻¹⁴
[tex] Ka = \frac{1\cdot 10^{-14}}{1.76 \cdot 10^{-5}} = 5.68 \cdot 10^{-10} [/tex]
Calculating the pH
The concentration of NH₄⁺ is:
[tex] Ca = \frac{9.0 \cdot 10^{-4} moles}{(0.030 L + 0.036 L)} = 0.0136 M [/tex]
After solving the equation (10) for x, we have:
x = 2.78x10⁻⁶ = [H₃O⁺]
Then, the pH is:
[tex] pH = -log(H_{3}O^{+}) = -log(2.78\cdot 10^{-6}) = 5.56 [/tex]
Hence, the pH is 5.56.
f) 37 mLNow, the pH is given by the concentration of HCl that remain in solution after reacting with NH₃ (HCl is in excess).
Calculating the concentration of HCl
[tex] C_{HCl} = \frac{0.025 mol/L*0.037 L - 0.030 mol/L*0.030 L}{(0.030 L + 0.037 L)} = 3.73 \cdot 10^{-4} M = [H_{3}O^{+}] [/tex]
Calculating the pH
[tex] pH = -log(H_{3}O^{+}) = -log(3.73 \cdot 10^{-4}) = 3.43 [/tex]
Therefore, the pH is 3.43.
Find more about pH here:
brainly.com/question/491373
I hope it helps you!
In an endothermic reaction, reactants are __ products.
equal to
less stable than
more stable than
equally stable than
Give the symbol for an element that is:__________
a. a halogen: _______________
b. an alkali metal: _______________
c. a noble gas: _______________
d. an alkaline earth metal : ____________
a. Br, Cl, F
b. Na, K, Ba
c. He, Ar, Ne
d. Ca, Ba, Mg
Answer:
a. halogen : F ,Cl ,Br l ,At
b an alkali metal: Na,Li, Rb, Cs
c. a noble gas: He, Ne, Kr, Ar
d. an alkaline earth metal: Be,Mg,Ca, Sr
hope it helps
stay safe healthy and happy...Which is the primary type of radiation from the sun that is absorbed by the ozone layer?
A. infrared radiatin
B. UV-B
C. X-rays
D. UV-C
E. UV-A
the answer to the question is B.UV-B
A sample of hydrogen nitrate or nitric acid, HNO 3 contains 18.8 x 1022 molecules.
How much mass of nitric acid are in the sample?
Answer:
19.7 g.
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to realize this problem can be solved by using a molecules-moles-mass relationship, starting with the given molecules, using the Avogadro's number and the molar mass of nitric acid (63.01 g/mol):
[tex]18.8x10^{22}molec*\frac{1mol}{6.022x10^{23}molec}* \frac{63.01g}{1mol} \\\\=19.7g[/tex]
Regards!
What is the empirical formula for a compound if 300.00 g of it is known to contain 82.46224 g of molybdenum, 45.741 g of chlorine and the rest is bromine
Answer:
MoClBr₂
Explanation:
First we calculate the mass of bromine in the compound:
300.00 g - (82.46224 g + 45.741 g) = 171.79676 gThen we calculate the number of moles of each element, using their respective molar masses:
82.46224 g Mo ÷ 95.95 g/mol = 0.9594 mol Mo45.741 g Cl ÷ 35.45 g/mol = 1.290 mol Cl171.79676 g Br ÷79.9 g/mol = 2.150 mol BrNow we divide those numbers of moles by the lowest number among them:
0.9594 mol Mo / 0.9594 = 11.290 mol Cl / 0.9594 = 1.34 ≅ 12.150 mol Br / 0.9594 = 2.24 ≅ 2Meaning the empirical formula is MoClBr₂.
Which statement is about population density
Explanation:
Population density is defined as the number of people present per square kilometre. Population density of India according to 2011 census is 382 persons per square kilometres.
Indicate if the following are the correct ground state electron configurations
for the atom listed by choosing correct or incorrect from the drop down menu.
1. Cr: [Ar]4s03d6
2. Zr: [Kr]5s23f144d2
3. Fe: [Ar]4s23d6
4. Co3+: [Ar]4s03d6
5. Ti2+: [Ar]4s03d2
6. Cu+: [Ar]4s23d8
Answer:
1) incorrect
2) incorrect
3) correct
4) correct
5) correct
6) incorrect
Explanation:
The correct electronic configuration of chromium is; [Ar] 3d⁵ 4s¹
The correct electronic configuration for Zr is; [Kr] 4d² 5s²
The correct electronic configuration of Cu^+ is; [Ar] 3d¹⁰
The electronic configuration of an atom refers to the arrangement of electrons in the atoms of such element.
The appropriate number of electrons and its properly written electronic configuration is clearly shown in this answer.
what class of organic compound is formed when cyclopentanone reacts with ethylamine in the presence of trace acid
The question is incomplete, the complete question is;
What functional group results when cyclopentanone reacts with ethylamine in the presence of trace acid? A) cyanohydrin B) semicarbazone C) imine D) enamine E) oxime
Answer:
imine
Explanation:
An imine is an unsaturated amine. An imine contains the carbon- nitrogen double bond.
Imines are obtained when a carbonyl compound is condensed with NH3 or an amine. The reaction involves several steps in its mechanism.
Since cyclopentanone is a ketone (carbonyl compound) and ethylamine is an amine,in the presence of trace acid, condensation of the two compounds occur to yield an imine