Explanation:
The electron density number as well as the sublevel letter are used to describe valence electrons in an atom. The third total energy and subbasement p, for example, is denoted by 3p. The electron configuration of oxygen, for example, is 1s^2 2s^2 2p^4, which means the first two electrons will couple up in the 1s orbital, while the following two protons will pair up in the 2s orbital.
The sample atom is Carbon with electron configuration; 1s² 2s² 2p².
The principal energy level of an electron refers to the shellp in which the electron is located relative to the atom's nucleus. In this case only 2 energy levels exist in a carbon atom; which are energy level 1 and 2
The sublevels exist within a principal energy and the electron configuration of an atom is described with consideration of energy sublevels. The sublevels in a carbon atom are;
s and p energy sublevels.The orbitals in this configuration are: 1s 2s 2px 2py 2pz in which case; each orbital can accommodate 2 electrons each.
Ultimately, the location of an element on the periodic table with respect to group and period are used to determine the valency and no. of energy levels in the atom of that Element.
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A physical chemist measures the temperature inside a vacuum chamber. Here is the result.
T= - 35.903°C
Convert T to SI units.
This is an acronym of the word Systeme International in French. Its International System of Units (SI) is a metric system that would be universally acknowledged as a measurement standard, and the further discussion can be defined as follows:
SI units play an important function in research and innovation in science and technology. In this, the temperature will be measure in the kelvin.Conversion [tex]^{\circ} \ to \ K[/tex]:
let,
[tex]\to 0^{\circ}\ C + 273.15 = 273.15\ K\\\\[/tex]
So,
[tex]\to -35.903+273.15=237.247\ K[/tex]
Therefore, the final answer is "237.247".
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Which would most likely be reduced when combined with Zn(s)?
A. Fe²⁺
B. K⁺
C. Ni
D. Al³⁺
The reduction of the species defines the gain of electrons. The iron is most likely to be reduced when reacts with zinc. Thus option A is correct.
What is oxidizing agent?Oxidizing agents are the species that gain electrons and get reduced, their oxidation number gets reduced when the metal reacts.
In the reactivity, series zinc is placed before iron and hence is a reducing agent that gets oxidized. Down the series, the reducing ability decreases while the oxidizing increases.
Therefore, option A. iron will be reduced when reacts with zinc.
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7.23 One equivalent of sodium methanethiolate is added to an electrophile that has two leaving groups. Which product will be major
The question is incomplete, the complete question is shown in the image attached to this answer
Answer:
A
Explanation:
We can see from the conditions of the reaction that the incoming nucleophile is -SCH3 and there are two possible leaving groups in the substrate.
First of all, we have to look at the conditions of the reaction. We can see that the reaction is taking place in DMF, a polar aprotoc solvent. This condition favours the SN2 synchronous mechanism over the SN1 ionic mechanism.
Hence, the nucleophile at the 1-position is preferentially substituted owing to the conditions of the reaction.
Thus, option A is the major product of the reaction.
A
(c) 2 C(s) + MnO2(s)
Mn(s) + 2 CO(g)
O combination reaction
O decomposition reaction
O combustion reaction
O single-displacement reaction
Answer: The reaction, [tex]2C(s) + MnO_{2}(s) \rightarrow Mn(s) + 2CO(g)[/tex] is a single-displacement reaction.
Explanation:
A chemical reaction in which one element of a compound is replaced by another element participating in the reaction.
For example, [tex]2C(s) + MnO_{2}(s) \rightarrow Mn(s) + 2CO(g)[/tex]
Here, the element manganese is replaced by carbon atom. As only one element gets replaced so, it is a single-displacement reaction.
Thus, we can conclude that [tex]2C(s) + MnO_{2}(s) \rightarrow Mn(s) + 2CO(g)[/tex] is a single-displacement reaction.
At 298 K, AHO = -1652 kJ/mol and ASO = 0.097 kJ/(K•mol). What is the Gibbs
free energy of the reaction?
Based on the calculations, the Gibbs's free energy for this chemical reaction is equal to -1,680.906 kJ/mol.
Given the following data:
Enthalpy of reaction (ΔH°) = -1652 kJ/mol.Temperature = 298 K.Entropy of reaction (ΔS°) = 0.097 kJ/mol.What is Gibbs's free energy?Gibbs's free energy simply refers to the quantity of energy that is associated with a particular chemical reaction.
Mathematically, the Gibbs's free energy for this chemical reaction can be calculated by using this formula:
ΔG° = ΔH° - ΔS°
Substituting the given parameters into the formula, we have;
ΔG° = -1652 × 10³ - (298 × 0.097)
ΔG° = -1652 × 10³ - 28.906
ΔG° = -1,680.906 kJ/mol.
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The chemical formula for hydrogen chloride is HCl A chemist determined by measurements that moles of hydrogen chloride participate in a chemical reaction. Calculate the mass of hydrogen chloride that participates. Be sure your answer has the correct number of significant digits.
The question is incomplete, the complete question is;
. A chemist determined by measurements that 0.015 moles of hydrogen chloride participate in a chemical reaction. Calculate the mass of hydrogen chloride that participates Round your answer to 2 significant digits. x S. ?
Answer:
0.54 g
Explanation:
Recall that;
Number of moles = mass/molar mass
Molar mass of HCl =36.5 g/mol
Mass= number of moles × molar mass
Mass= 0.015 moles × 36 g/mol
Mass= 0.54 g
The enthalpy of Sodium is 235 calories. The enthalpy Chlorine is 435 calories. The enthalpy of Sodium chloride 670 joules, what is the change in enthalpy for this reaction?
Answer:
ΔH = -2446J
Explanation:
Based on the reaction:
2 Na(s) + Cl2(g) → 2NaCl
We can find the enthalpy of this reaction using Hess's law:
The enthalpy of a reaction is equal to the sum of the enthalpy of products times their reaction quotient subtracting the enthalpy of reactants times their reaction quotient. For the reaction of the problem:
ΔH = 2ΔH(NaCl) - [2ΔH(Na) + ΔHCl2)]
ΔH(NaCl) = 670J
ΔH(Na) = 235cal * (4.184J/1cal) = 983J
ΔHCl2 = 435cal * (4.184J/1cal) = 1820J
ΔH = 2*670J - [2*983J + 1820J]
ΔH = 1340J - [3786J]
ΔH = -2446JAnswer:
the heat content of a system at constant pressure
Explanation:
In an exothermic chemical reaction:____.
A. The mass of the products is greater than the mass of the reactants.
B. The mass of the products is less then the mass of the reactants.
C. Heat is released as the reaction proceeds.
D. Heat is absorbed as the reaction proceeds.
Answer:
C
Explanation:
Think of 'exo' as exit and 'thermic' as relating to thermal energy/ heat. Thus, an exothermic release thermal energy as the reaction proceeds.
In an exothermic reaction, the total energy of the products is lesser than that of the reactants and ΔH (change in energy) is less than zero.
When heat is absorbed as the reaction proceeds, the chemical reaction is an endothermic reaction.
What volume is occupied by 0.104 mol of helium gas at a pressure of 0.91 atm and a temperature of 314 K ?
Answer:
The volume will be "2.95 L".
Explanation:
Given:
n = 0.104
p = 0.91 atm
T = 314 K
Now,
The Volume (V) will be:
= [tex]\frac{nRT}{P}[/tex]
By putting the values, we get
= [tex]\frac{0.104\times 0.0821\times 314}{0.91}[/tex]
= [tex]\frac{2.6810}{0.91}[/tex]
= [tex]2.95 \ L[/tex]
When you hold an object with a positive electric charge near an object with a
negative electric charge, it causes the negatively charged object to move. How
could you model this interaction with magnets? Describe the types of forces
involved and the energy transformations that take place.
Answer:
A positively charged object will exert a repulsive force upon a second positively charged. This repulsive force will push the two objects apart while a negatively charged object will exert a repulsive force upon a second negatively charged object. Objects with like charge repel each other
The interaction between objects with positive and negative electric charges can be analogously modeled using magnets. The Types of Forces Involved are; Attractive Magnetic Force, Repulsive Magnetic Force and the Energy Transformations are; Potential Energy Transformation, and Kinetic Energy Transformation.
In this analogy, magnets can represent the charges, and magnetic forces can represent the electric forces.
Interaction Between Magnets
Imagine we have two magnets: one with a north pole (N) and the other with a south pole (S). When you bring the north pole of one magnet close to the south pole of the other magnet, they are attracted to each other. Conversely, if you bring the north pole of one magnet near the north pole of the other magnet, they repel each other.
Types of Forces Involved:
Attractive Magnetic Force (Analogous to Electric Attraction):
When the north pole of one magnet is brought close to the south pole of another magnet, they experience an attractive magnetic force. Similarly, when objects with opposite electric charges were brought close together, then they will experience an attractive electric force.
Repulsive Magnetic Force (Analogous to Electric Repulsion):
When two magnets with the same pole (both north or both south) are brought close to each other, they experience a repulsive magnetic force. This is analogous to the repulsion between objects with like electric charges (both positive or both negative).
Energy Transformations;
When you bring the magnets closer together or move them apart, energy transformations occur:
Potential Energy Transformation;
As the magnets are moved closer together, the potential energy of the magnetic interaction decreases. This is because the magnets' magnetic fields interact more strongly, and they tend to move toward each other due to the attractive or repulsive forces.
Kinetic Energy Transformation;
If you let the magnets go after bringing them close together, they will move towards each other (in the case of attraction) or move apart (in the case of repulsion). This movement involves a transformation of potential energy into kinetic energy. The kinetic energy increases as the magnets move, and it's at its maximum when the magnets are farthest apart (in the case of repulsion) or when they collide (in the case of attraction).
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Someone can you please please help me
Answer:
FalseExplanation:
Target organ toxins are chemicals that can cause adverse effects or disease states manifested in specific organs of the body. Toxins do not affect all organs in the body to the same extent due to their different cell structures.
A sealed vessel initially contains 100 g of chlorine gas and 90 g of hydrogen gas. The two gases undergo reaction to form HCl. Which of the following statements is true?
a. 100 g HCl is produced
b. 190 g HCl is produced
c. less than 90 g HCl is produced
d. between 100 and 190 g of HCl is produced
Answer:
d.
Explanation:
H2 + Cl2 = 2HCL
From the equation 2g hydrogen combine with 71g of chlorine.
So 35.5 g Cl2 combines with 1g of H2
There are 100g of Cl2 so this will, by proportion, react with 100/35.5 g hydrogen.
This is 2.8 g hydrogen so the mass of HCl formed = 102.8 g.
The true statement is that d. between 100 and 190 g of HCl is produced.
How do find the mass of HCL?To find mass of HCL:
H2 + Cl2 = 2HCL
From the equation, 2g of hydrogen combines with 71g of chlorine.
So 35.5 g Cl2 combines with 1g of H2
There are 100g of Cl2 so this will, by proportion, react with 100/35.5 g of hydrogen.
This is 2.8 g hydrogen so the mass of HCl formed = 102.8 g.
Hydrogen chloride may be formed by the direct combination of chlorine (Cl2) gas and hydrogen (H2) gas.
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this was in my science test just want to see if anyone knows?? the wording is so confusing
the relative atomic mass of an element compares the mass of an atom of an element with the mass of an atom of.......
Answer:
the molar mass of the element
Consider the following reaction at equilibrium. What effect will decreasing the temperature have on the system? DH=+890kJCO2(g)+2H2O(l)<--> CH4(g)+2O2(g)a) the equilibrium constant will increaseb) the reaction will shift to the left in the direction of the reactantsc) the reaction will shift to the right in the direction of the productsd) no effect will be observed
Answer:
Option B. The reaction will shift to the left in the direction of the reactants.
Explanation:
The equation for the reaction is given below:
CO₂ + 2H₂O <=> CH₄ + O₂
Enthalpy change (ΔH) = +890 KJ
The reaction illustrated by the equation is endothermic reaction since the enthalpy change (ΔH) is positive.
Increasing the temperature of an endothermic reaction will shift the equilibrium position to the right and decrease the temperature will shift the equilibrium position to the left.
Therefore, decreasing the temperature of the system illustrated by the equation above, will shift the reaction to the left in the direction of the reactants.
Thus, option B gives the right answer to the question.
Which piece of equipment would be BEST to measure 5 mL of a liquid? A 50 mL beaker filled half way between 0 and 10 mL markings A 50 mL Erlenmeyer flask filled half way between 0 and 10 ml markings A 10 mL graduated cylinder A disposable dropper A 100 mL graduated cylinder
Answer:
Which piece of equipment would be BEST to measure 5 mL of a liquid?
A 50 mL beaker filled halfway between 0 and 10 mL markings
A 50 mL Erlenmeyer flask filled halfway between 0 and 10 ml marking
A 10 mL graduated cylinder
A disposable dropper
A 100 mL graduated cylinder
Explanation:
To measure 5 mL of a liquid, the best equipment is
a 10 mL graduated cylinder.
The remaining apparatus do not give an accurate measurement because readings are not marked on them.
I have 50.00 mL of 0.100 M ethyl amine (C2H5NH2). I gradually add a solution of 0.025 M nitric acid (HNO3) to the ethyl amine solution.
Required:
What is the pH after the addition of a total of 201 mL of the nitric acid?
Answer:
4.00 is the pH of the mixture
Explanation:
The ethyl amine reacts with HNO3 as follows:
C2H5NH2 + HNO3 → C2H5NH3⁺ + NO3⁻
To solve this question we need to find the moles of ethyl amine and the moles of HNO3:
Moles C2H5NH2:
0.0500L * (0.100mol/L) = 0.00500 moles ethyl amine
Moles HNO3:
0.201L * (0.025mol/L) = 0.005025 moles HNO3
That means HNO3 is in excess. The moles in excess are:
0.005025 moles HNO3 - 0.00500 moles ethyl amine =
2.5x10⁻⁵ moles HNO₃
In 50 + 201mL = 251mL = 0.251L:
2.5x10⁻⁵ moles HNO₃ / 0.251L = 9.96x10⁻⁵M = [H+]
As pH = -log [H+]
pH = -log 9.96x10⁻⁵M
pH = 4.00 is the pH of the mixture
A mixture is -
A. a pure substance made through chemical processes
B. a pure substance made through physical processes
C.
an impure substance made through chemical processes
D.
an impure substance made through physical processes
Answer:
C
Explanation:
an impure substance made through chemical process
In a solution, the solvent is
ANSWER:
A. always water
B. dissolved in the solute
C. present in larger amount than the solute is
D. always nonpolar
Answer:
dissolved in the solute
Explanation:
A solvent is the component that dissolves the solute and is present in larger amount. The type of solution is determined by the state of the solute and solvent. If you have NaCl, a solid, dissolved in water, a liquid, the type of solution is a solid/liquid solution.
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...How many moles are in the number of molecules below? I only need to know the 5th question.
Answer:
11
1. 6.02×10 23
this is the answer Hope it helps you
The N H 2 group makes the molecule an amine. Like alcohols, amines can participate in hydrogen bonding. If the drawn molecule encounters water molecules, there will be (select all that apply): a hydrogen bond between the hydrogen of water and the carbon of the amine. a hydrogen bond between the hydrogen of water and the nitrogen of the amine. a hydrogen bond between the oxygen of water and a hydrogen from the -NH2 group. a hydrogen bond between the oxygen of water and a hydrogen from the -CH3 group. no hydrogen bonds.
Answer:
a hydrogen bond between the hydrogen of water and the nitrogen of the amine
a hydrogen bond between the oxygen of water and a hydrogen from the -NH2 group
Explanation:
A hydrogen bond is formed between molecules in which hydrogen is bonded to a highly electronegative element.
In amines, hydrogen is bonded to nitrogen while in water, hydrogen is bonded to oxygen. Both are highly electronegative elements hence hydrogen bonding is possible between amines and water.
This hydrogen bond may involve;
The hydrogen of water and the nitrogen of the amine
Or
The oxygen of water and a hydrogen from the -NH2 group
A vessel is filled at a rate of 3.41 cubic micrometers per minute. If the vessel has a volume of 54 liters, how many
seconds will it take to fill the vessel? provide a step buy step explanation.
Answer:
9.5x10¹⁷ s
Explanation:
First we convert 3.41 cubic micrometers (um³) to liters (L), as such:
3.41 um³ * [tex](\frac{1m}{10^6um} )^3*\frac{1000L}{1m^3}[/tex] = 3.41x10⁻¹⁵ LWith the converted rate of 3.41x10⁻¹⁵ L/min, we can calculate how many minutes it would take to fill a 54 L vessel:
54 L ÷ 3.41x10⁻¹⁵ L/min = 1.58x10¹⁶ minFinally we convert 1.58x10¹⁶ minutes to seconds:
1.58x10¹⁶ * 60 = 9.5x10¹⁷ sAnswer True or False for each of the following statements. (a) The carburization surface was maintained at slightly less than 0.25 wt% carbon for each specimen. (b) Comparing the finished specimens at a depth of 0.20 mm, specimen A features the lowest carbon concentration. (c) Comparing the finished specimens as a whole, specimen D features the lowest overall amount of carbon.
Answer:
verdadero/a
falso/b
verdadero/c
Explanation:
Please helpplease please..help
Answer:
vague symptoms are characteristic of an acute toxin, because of the of the lack of well defined consistency that these symptoms have in relation to the course of the disease progress.
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Por favor, me ajudem.
Please, help me.
Someone can teach me English, because I'm from Brazil and if you know how to speak Spanish, it's better for me.
1) Write a balanced equation to show the reaction of gaseous ethane with gaseous oxygen to form carbon monoxide gas and water vapor.
Answer:
C₂H₆(g) + 2.5 O₂(g) ⇒ 2 CO(g) + 3 H₂O(g)
Explanation:
Let's consider the unbalanced equation in which gaseous ethane reacts with gaseous oxygen to form carbon monoxide gas and water vapor. This is an incomplete combustion reaction.
C₂H₆(g) + O₂(g) ⇒ CO(g) + H₂O(g)
We will balance it using the trial and error method.
First, we will balance C atoms by multiplying CO by 2 and H atoms by multiplying H₂O by 3.
C₂H₆(g) + O₂(g) ⇒ 2 CO(g) + 3 H₂O(g)
Then, we get the balanced equation by multiplying O₂ by 2.5.
C₂H₆(g) + 2.5 O₂(g) ⇒ 2 CO(g) + 3 H₂O(g)
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.
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The table shows the nature of reactants and products formed in a certain type of chemical reaction.
Nature of Reactants and Products
Reactants
Metal + lonic compound
Products
Metal + lonic compound
Which of the following is true about the type of chemical reaction?
it is a single replacement reaction, and the anions in the two inic compounds are different.
It is a single replacement reaction, and the cations in the two ionic compounds are different.
It is a double replacement reaction, and the anions in the two ionic compounds are different.
It is a double replacement reaction, and the cations in the two ionic compounds are different.
Answer: It is true about the type of reaction that it is a single replacement reaction, and the cations in the two ionic compounds are different.
Explanation:
When one element in a compound is replaced by another element in a chemical reaction then it is called a single replacement reaction.
For example, [tex]K + NaCl \rightarrow KCl + Na[/tex]
Here, potassium metal is replacing the sodium metal in the sodium chloride compound.
As metals become cation by losing an electron in a chemical reaction.
Thus, we can conclude that it is true about the type of reaction that it is a single replacement reaction, and the cations in the two ionic compounds are different.
Answer: Its A
Explanation:
a single replacement reactions, and the ANIONS in the two ionic compounds are different
a 150 j of energy is added to a system that does 50 j of work is done.by how m uch will tyhe internal energy of system be raised?
The internal energy of system is raised by 3 times
Calculate the vapor pressure (in torr) at 298 K in a solution prepared by dissolving 46.8 g of the non-volatile non-electrolye glucose in 117 g of methanol. The vapor pressure of methanol at 298 K is 122.7 torr. Enter your answer to 2 decimal places.
Answer: The total partial pressure of the solution is 131.37 torr.
Explanation:
The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] ......(1)
For glucose:Given mass of glucose = 46.8 g
Molar mass of glucose = 180 g/mol
Plugging values in equation 1:
[tex]\text{Moles of glucose}=\frac{46.8g}{180g/mol}=0.26 mol[/tex]
For methanol:Given mass of methanol = 117 g
Molar mass of methanol = 32 g/mol
Plugging values in equation 1:
[tex]\text{Moles of methanol}=\frac{117g}{32g/mol}=3.66 mol[/tex]
Mole fraction is defined as the moles of a component present in the total moles of a solution. It is given by the equation:
[tex]\chi_A=\frac{n_A}{n_A+n_B}[/tex] .....(2)
where n is the number of moles
Putting values in equation 2:
[tex]\chi_{methanol}=\frac{3.66}{0.26+3.66}=0.934[/tex]
Raoult's law is the law used to calculate the partial pressure of the individual gases present in the mixture. The equation for Raoult's law follows:
[tex]p_A=\chi_A\times p_T[/tex] .....(3)
where [tex]p_A[/tex] is the partial pressure of component A in the mixture and [tex]p_T[/tex] is the total partial pressure of the mixture
We are given:
[tex]p_{methanol}=122.7torr\\\chi_{methanol}=0.934[/tex]
Putting values in equation 3, we get:
[tex]122.7torr=0.066\times p_T\\\\p_T=\frac{122.7torr}{0.934}=131.37torr[/tex]
Hence, the total partial pressure of the solution is 131.37 torr.
