Answer:
8.90 M
Explanation:
Step 1: Given data
Initial concentration (C₁): ?Initial volume (V₁): 635 mL = 0.635 LFinal concentration (C₂): 5.00 MFinal volume (V₂): 1.13 LStep 2: Calculate the initial concentration
We have a concentrated NaCl solution and we want to prepare a diluted one. We will use the dilution rule.
C₁ × V₁ = C₂ × V₂
C₁ = C₂ × V₂ / V₁
C₁ = 5.00 M × 1.13 L / 0.635 L
C₁ = 8.90 M
Answer:
[tex]\large \boxed{\text{8.90 mol/L}}[/tex]
Explanation:
We can use the dilution formula to calculate the concentration of the original solution.
[tex]\begin{array}{rcl}V_{1}c_{1} & = & V_{2}c_{2}\\\text{635 mL }\times c_{1} & = & \text{1130 mL} \times \text{5.00 mol/L}\\635 c_{1}&=& \text{5650 mol/L}\\c_{1}& = & \dfrac{5650}{635}\text{ mol/L}\\\\& = & \textbf{8.90 mol/L}\\\end{array}\\\text{The initial concentration was $\large \boxed{\textbf{8.90 mol/L }}$}[/tex]
0.22 L of HNO3 is titrated to equivalence using 0.18 L of 0.2 MNaOH. What is the concentration of the HNO3?
Answer:
0.16 M
Explanation:
Data provided as per the question is below:-
Volume of [tex]HNO_3[/tex] = 0.22 L
The Volume of NaOH = 0.18 L
Morality of NaOH = 0.2
According to the given situation, the calculation of the concentration of the [tex]HNO_3[/tex] is shown below:-
For equivalence,
Number of the equivalent of [tex]HNO_3[/tex] = Number of equivalents of NaOH
[tex]= \frac{0.18\times0.2}{0.22}[/tex]
[tex]= \frac{0.036}{0.22}[/tex]
= 0.16363 M
or
= 0.16 M
the pain reliever codeine is a weak base with a kb equal to 1.6 x 10^-6. what is the ph of a 0.05 m aqueous codeine solution
Answer:
[tex]pH=10.45[/tex]
Explanation:
Hello,
In this case, for the dissociation of the given base, we have:
[tex]base\rightleftharpoons OH^-+CA[/tex]
Whereas CA accounts for conjugated acid and OH⁻ for the conjugated base. In such a way, equilibrium expression is:
[tex]Kb=\frac{[OH^-][CA^+]}{[base]}[/tex]
And in terms of the reaction extent [tex]x[/tex] we can write:
[tex]1.6x10^{-6}=\frac{x*x}{0.05M-x}[/tex]
For which the roots are:
[tex]x_1=-0.000284M\\x_2=0.000282M[/tex]
For which clearly the result is the positive root which also equals the concentration of hydroxyl ions and we can compute the pOH:
[tex]pOH=-log([OH^-])=-log(0.000282)\\\\pOH=3.55[/tex]
And the pH:
[tex]pH=14-pOH=14-3.55\\\\pH=10.45[/tex]
Regards.
The pH of the solution is 10.45.
Let us represent codeine with the generic formula BH. We can set up the ICE table as follows;
:B(aq) + H2O(l) ⇄ BH(aq) + OH^-(aq)
I 0.05 0 0
C -x +x +x
E 0.05 - x x x
We know that the Kb of codeine is 1.6 x 10^-6, Hence;
1.6 x 10^-6 = x^2/0.05 - x
1.6 x 10^-6 (0.05 - x ) = x^2
8 x 10^-8 - 1.6 x 10^-6x = x^2
x^2 + 1.6 x 10^-6x - 8 x 10^-8 = 0
x = 0.00028 M
The concentration of hydroxide ions = 0.00028 M
Given that pOH = - log[0.00028 M]
pOH = 3.55
pH + pOH = 14
pH = 14 - 3.55
pH = 10.45
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A laboratory technician combines 35.9 mL of 0.258 M chromium(II) chloride with 35.8 mL 0.338 M potassium hydroxide. How many grams of chromium(II) hydroxide can precipitate
Answer:
0.52 g of chromium(II) hydroxide, Cr(OH)2.
Explanation:
We'll begin by calculating the number of mole of chromium (ii) chloride, CrCl2 in 35.9 mL of 0.258 M chromium(II) chloride solution.
This can be obtained as follow:
Molarity of CrCl2 = 0.258 M
Volume = 35.9 mL = 35.9/1000 = 0.0359 L
Mole of CrCl2 =?
Molarity = mole /Volume
0.258 = mole of CrCl2 /0.0359
Cross multiply
Mole of CrCl2 = 0.258 x 0.0359
Mole of CrCl2 = 0.0093 mole
Next, we shall determine the number of mole of potassium hydroxide, KOH in 35.8 mL 0.338 M potassium hydroxide solution.
This can be obtained as follow:
Molarity of KOH = 0.338 M
Volume = 35.8 mL = 35.8/1000 = 0.0358 L
Mole of KOH =.?
Molarity = mole /Volume
0.338 = mole of KOH /0.0358
Cross multiply
Mole of KOH = 0.338 x 0.0358
Mole of KOH = 0.0121 mole.
Next, we shall write the balanced equation for the reaction. This is given below:
2KOH + CrCl2 → Cr(OH)2 + 2KCl
From the balanced equation above,
2 mole of KOH reacted with 1 mole of CrCl2 to produce 1 mole of Cr(OH)2.
Next, we shall determine the limiting reactant. This can be obtained as follow:
From the balanced equation above,
2 mole of KOH reacted with 1 mole of CrCl2.
Therefore, 0.0121 mole of KOH will react with = (0.0121 x 1)/2 = 0.00605 mole of CrCl2.
From the calculations made above, we can see that only 0.00605 mole out of 0.0093 mole of CrCl2 is needed to react completely with 0.0121 mole of KOH.
Therefore, KOH is the limiting reactant.
Next, we shall determine the number of mole of Cr(OH)2 produced from the reaction.
In this case, we shall be using the limiting reactant because it will give the maximum yield of Cr(OH)2.
The limiting reactant is KOH and the number of mole of Cr(OH)2 produced can be obtained as illustrated below:
From the balanced equation above,
2 mole of KOH reacted to produce 1 mole of Cr(OH)2.
Therefore, Therefore, 0.0121 mole of KOH will react with = (0.0121 x 1)/2 = 0.00605 mole of Cr(OH)2.
Finally, we shall convert 0.00605 mole of Cr(OH)2 to grams.
This is illustrated below:
Mole of Cr(OH)2 = 0.00605 mole
Molar mass of Cr(OH)2 = 52 + 2(16 + 1) = 52 + 2(17) = 86 g/mol
Mass of Cr(OH)2 =..?
Mole = mass /Molar mass
0.00605 = mass of Cr(OH)2/86
Cross multiply
Mass of Cr(OH)2 = 0.00605 x 86
Mass of Cr(OH)2 = 0.52 g
Therefore, 0.52 g of chromium(II) hydroxide, Cr(OH)2 was produced.
explain how the liquid in a thermometer changes so that it can be used to measure a temprature
Answer:
The liquid that is often used in thermometers is chrome.
It is khwon for raising its volule when the temperature raises and vice-versa. ● the temperature and the volume are proprtional to each other so using Mathematics, scientists have figured out a way to benefit from it to make a thermometer.
Name four types of salts
Answer:
Any ionic molecule formed of a base and an acid, which dissolves in water to produce ions is known as a salt. The four common types of salts are:
1. NaCl or sodium chloride is the most common kind of salt known. It is also known as table salt.
2. K2Cr2O7 or potassium dichromate refers to an orange-colored salt formed of chromium, potassium, and oxygen. It is toxic to humans and is also an oxidizer, which is a fire hazard.
3. CaCl2 or calcium chloride looks like table salt due to its white color. It is broadly used to withdraw ice from roads. It is hygroscopic.
4. NaHSO4 or sodium bisulfate produces from hydrogen, sodium, oxygen, and sulfur. It is also known as dry acid. It has commercial applications like reducing the pH of swimming pools and spas and others.
Compare and contrast an electrolytic cell with a voltaic cell. Provide one example where both are present in daily life.
Answer:
ELECTROLYTIC CELL
An electrochemical cell in which a non-spontaneous chemical reaction takes place when electric current is passed through the solution,is called an electrolytic cell.
EXAMPLE
Nelsons cell and Downs cell
VOLTAIC CELL OR GALVANIC CELL
The electrochemical cell in which a spontaneous chemical reaction takes place and generates electric current is called galvanic and voltaic cell.
EXAMPLE
Daniel cell
Explanation:
Electrolytic cell by the non-spontaneous reactions covert the chemical energy into electical and volataic cells are those in which spontaneous redox reaction takes place.
What is cell?Cell is a device which converts chemical energy into electrical energy.
Electrolytic Cell: An electrolytic cell is an electrochemical cell in which a non-spontaneous redox chemical reaction occurs when an electric current is conducted through the solution.
Example- Electrolysis of sodium chloride, by which formation of sodium metal and chlorine gas takes place.
Voltaic cell: Voltaic cells are electrochemical cells in which a spontaneous redox chemical reaction occurs and creates electric current. These cells are also known by the name of Galvanic cells.
Example- It is used in the form of batteries which can be portable easily.
Hence, in electrolytic cells non - spontaneous reaction occur and in voltaic cell spontaneous reaction is occured.
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21. What are the two main ways of working with clay?
Answer:
Diferentes tipos de arcilla
ARCILLA DE LADRILLOS. Contiene muchas impurezas. ...
ARCILLA DE ALFARERO. Llamada también barro rojo y utilizada en alfarería y para modelar. ...
ARCILLA DE GRES. Es una arcilla con gran contenido de feldespato. ...
ARCILLAS “BALL CLAY” O DE BOLA. ...
CAOLIN. ...
ARCILLA REFRACTARIA. ...
BENTONITA.
Explanation:
Answer:
Coil method and the slab method.
Explanation:
"What is the difference between the revertible and nonrevertible rII mutants that Benzer generated?"
Chlorine can be prepared in the laboratory by the reaction of manganese dioxide with hydrochloric acid, HCl(aq), as
described by the chemical equation
MnO,(s) + 4 HCl(aq)
MnCl(aq) + 2 H2O(l) + Cl (8)
How much MnO(s) should be added to excess HCl(aq) to obtain 175 mL C12(g) at 25 °C and 715 Torr?
mass of MnO2
Answer:
Explanation:
MnO₂(s) + 4 HCl(aq) = MnCl₂(aq) + 2 H₂O(l) + Cl₂
87 g 22.4 x 10³ mL
volume of given chlorine gas at NTP or at 760 Torr and 273 K
= 175 x ( 273 + 25 ) x 715 / (273 x 760 )
= 179.71 mL
22.4 x 10³ mL of chlorine requires 87 g of MnO₂
179.4 mL of chlorine will require 87 x 179.4 / 22.4 x 10³ g
= 696.77 x 10⁻³ g
= 696.77 mg .
To refine aluminum from its ore, aluminum oxide is electrolyzed to form aluminum and oxygen. At which electrode does oxygen form? options: A) Both the anode and the cathode B) Cathode C) Neither electrode D) Anode
Answer:
im pretty sure its the anode
Explanation:
To solve such, we must know the concept of electrolysis reaction. The correct option is option D among all given options. At anode electrode oxygen forms.
What is chemical reaction?Chemical reaction is a process in which two or more than two molecules collide in right orientation and energy to form a new chemical compound. The mass of the overall reaction should be conserved. There are so many types of chemical reaction reaction like combination reaction, double displacement reaction.
Electrolysis is the process of passing an electric current through a material to cause a chemical change. A chemical change occurs when a material loses or acquires the electron. To refine aluminum from its ore, aluminum oxide is electrolyzed to form aluminum and oxygen. At anode electrode oxygen forms.
Therefore, the correct option is option D among all given options. At anode electrode oxygen forms.
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A 0.753 g sample of a monoprotic acid is dissolved in water and titrated with 0.250 M NaOH. What is the molar mass of the acid if 21.5 mL of the NaOH solution is required to neutralize the sample?
Answer:
[tex]MM_{acid}=140.1g/mol[/tex]
Explanation:
Hello,
In this case, since the acid is monoprotic, we can notice a 1:1 molar ratio between, therefore, for the titration at the equivalence point, we have:
[tex]n_{acid}=n_{base} \\\\V_{acid}M_{acid}=V_{base}M_{base}\\\\n_{acid}=V_{base}M_{base}[/tex]
Thus, solving for the moles of the acid, we obtain:
[tex]n_{acid}=0.0215L*0.250\frac{mol}{L}=5.375x10^{-3}mol[/tex]
Then, by using the mass of the acid, we compute its molar mass:
[tex]MM_{acid}=\frac{0.753g}{5.375x10^{-5}mol} \\\\MM_{acid}=140.1g/mol[/tex]
Regards.
Suppose a student completes an experiment with an average value of 2.9 mL and a calculated standard deviation of 0.71 mL. What is the minimum value within a 1 SD range of the average
Answer:
The correct answer is 2.2 mL.
Explanation:
Given:
Average: 2.9 mL
SD: 0.71 mL
We can define a 1 SD range in which the value of volume (in mL) will be comprised:
Volume (mL) = Average ± SD = (2.9 ± 0.7) mL
Maximum value= Average + SD= 2.9 + 0.7 mL = 3.6 mL
Minimum value= Average - SD = 2.9 - 0.7 mL = 2.2 mL
Thus, the minimum value within a 1 SD range of the average is 2.2 mL
The minimum value within 1 SD is 2.19 mL
The z score is used to determine by how many standard deviations the raw score is above or below the mean. The z score is given by:
[tex]z=\frac{x-\mu}{\sigma} \\\\where\ x\ is\ raw\ score, \mu=mean,\sigma=standard\ deviation[/tex]
Given that μ = 2.9 mL, σ = 0.71 mL; hence:
The minimum value within 1 SD range = μ ± σ = 2.9 ± 0.71 = (2.19, 3.61)
Therefore the minimum value within 1 SD is 2.19 mL
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Methanol liquid burns readily in air. One way to represent this equilibrium is: 2 CO2(g) + 4 H2O(g)2 CH3OH(l) + 3 O2(g) We could also write this reaction three other ways, listed below. The equilibrium constants for all of the reactions are related. Write the equilibrium constant for each new reaction in terms of K, the equilibrium constant for the reaction above. 1) CH3OH(l) + 3/2 O2(g) CO2(g) + 2 H2O(g) K1 = 2) CO2(g) + 2 H2O(g) CH3OH(l) + 3/2 O2(g) K2 = 3) 2 CH3OH(l) + 3 O2(g) 2 CO2(g) + 4 H2O(g)
Answer:
Answers are in the explanation
Explanation:
It is possible to obtain K of equilibrium of related reactions knowing the laws:
A + B ⇄ C K₁
C ⇄ A + B K = 1 /K₁
The inverse reaction has the inverse K equilibrium
2A + 2B ⇄ 2C K = K₁²
The multiplication of the coefficients of reaction produce a k powered to the number you are multiplying the coefficients
For the reaction:
2 CO2(g) + 4 H2O(g) ⇄ 2 CH3OH(l) + 3 O2(g) K
1) CH3OH(l) + 3/2 O2(g) ⇄ CO2(g) + 2 H2O(g)
This is the inverse reaction but also the coefficients are dividing in the half, that means:
[tex]K_1 = \frac{1}{k^{1/2}} = (1/K)^{1/2}[/tex]
2) CO2(g) + 2 H2O(g) ⇄ CH3OH(l) + 3/2 O2(g)
Here,the only change is the coefficients are the half of the original reaction:
[tex]K_2 = K^{1/2}[/tex]
3) 2CH3OH(l) + 3 O2(g) ⇄ 2 CO2(g) + 4 H2O(g)
This is the inverse reaction. Thus, you have the inverse K of equilibrium:
[tex]K_3 = \frac{1}{K}[/tex]
According to the following reaction, how many grams of ammonia will be formed upon the complete reaction of 31.2 grams of hydrogen gas with excess nitrogen gas ? nitrogen(g) + hydrogen(g) ammonia(g)
Answer:
176.8 g of ammonia, NH3.
Explanation:
We'll begin by writing the balanced equation for the reaction. This is given below:
N2 + 3H2 —> 2NH3
Next, we shall determine the mass of H2 that reacted and the mass of NH3 produced from the balanced equation. This is illustrated below:
Molar mass of H2 = 2x1 = 2 g/mol
Mass of H2 from the balanced equation = 3 x 2 = 6 g
Molar mass of NH3 = 14 + (3x1) = 17 g/mol
Mass of NH3 from the balanced equation = 2 x 17 = 34 g.
From the balanced equation above,
6 g of H2 reacted to produce 34 g of NH3.
Finally, we shall determine the mass of ammonia, NH3 produced by reacting 31.2 g of H2.
This can be obtained as follow:
From the balanced equation above,
6 g of H2 reacted to produce 34 g of NH3.
Therefore, 31.2 g of H2 will react to produce = (31.2 x 34)/6 = 176.8 g of NH3.
Therefore, 176.8 g of ammonia, NH3 were obtained from the reaction.
Why don't siblings look exactly alike
Answer:
Your genes play a big role in making you who you are. ... But brothers and sisters don't look exactly alike because everyone (including parents) actually has two copies of most of their genes. And these copies can be different. Parents pass one of their two copies of each of their genes to their kids.
Divers often inflate heavy duty balloons attached to salvage items on the sea floor. If a balloon is filled to a volume of 1.20 L at a pressure of 6.25 atm, what is the volume of the balloon when it reaches the surface?
Answer:
7.50 L
Explanation:
The balloon has a volume of 1.20 L (V₁) when the pressure at the sea floor is 6.25 atm (P₁). When it reaches the surface, the pressure is that of the atmosphere, that is, 1.00 atm (P₂). If we consider the gas to behave as an ideal gas and the temperature to be constant, we can calculate the final volume (V₂) using Boyle's law.
P₁ × V₁ = P₂ × V₂
V₂ = P₁ × V₁ / P₂
V₂ = 6.25 atm × 1.20 L / 1.00 atm
V₂ = 7.50 L
A compound is found to contain 31.42 % sulfur, 31.35 % oxygen, and 37.23 % fluorine by weight. To answer the questions, enter the elements in the order presented above. 1. What is the empirical formula for this compound? 2. The molecular weight for this compound is 102.1 g/mol. What is the molecular formula for this compound?
Explanation:
To obtain the empirical and molecular formula of this compound from the percent composition of the elements, we follow the steps below;
Step 1: Divide the percentage composition by the atomic mass
Sulphur = 31.42 / 32 = 0.9819
Oxygen = 31.35 / 16 = 1.9594
Flourine = 37.23 / 19 = 1.9595
Step 2: Divide by the lowest number
Sulphur = 0.9819 / 0.9819 = 1
Oxygen = 1.9594 / 0.9819 ≈ 2
Flourine = 1.9595 / 0.9819 ≈ 2
This means the ratio of the elements is 1 : 2: 2
The empirical formular (simplest formular of a compound) of the compound is;
SO₂F₂
To obtain the molecular formular (Actual formular of a compound);
(SO₂F₂)n = 102.1
Inserting the atomic masses and solving for n;
(102)n = 102.1
n ≈ 1
The molecular formular is; (SO₂F₂)₁ = SO₂F₂
Which of the following do we need to know in order to calculate pH during an acid-base titration of a strong monoprotic acid with a strong monoprotic base? Select all that apply
a. the concentration of the acid
b. the concentration of the base titrant
c. the initial volume of the acid solution
d. the volume of the titrant used
Answer:
the volume of the titrant used
Explanation:
Acid-base titrations are usually depicted on special graphs referred to as titration curve. A titration curve is a graph that contains a plot of the volume of the titrant as the independent variable and the pH of the system as the dependent variable.
Hence, a titration curve is a graphical plot showing the pH of the analyte solution plotted against the volume of the titrant as the reaction is in progress. The titration curve is drawn by plotting data obtained during a titration, that is, volume of the titrant added (plotted on the x-axis) and pH of the system (plotted on the y-axis).
Draw a picture of what you imagine solid sodium chloride looks like at the atomic level. (Do NOT draw Lewis structures.) Make sure to include a key. Then describe what you've drawn and any assumptions you are making.
Answer:
Kindly check the explanation section.
Explanation:
PS: kindly check the attachment below for the required diagram that is the diagram showing solid sodium chloride looks like at the atomic level.
The chemical compound known as sodium chloride, NaCl has Molar mass: 58.44 g/mol, Melting point: 801 °C and
Boiling point: 1,465 °C. The structure of the solid sodium chloride is FACE CENTRED CUBIC STRUCTURE. Also, solid sodium chloride has a coordination number of 6: 6.
In the diagram below, the positive sign shows the sodium ion while the thick full stop sign represent the chlorine ion.
The NaCl has been the ionic structure with an equal number of sodium and chlorine ions bonded.
In the structure, there has been each Na ion bonded with the Cl ions. There has been the transfer of electrons between the structure in order to attain a stable configuration.
The expected structure of the NaCl would be the image attached below.
The image has been the cubic structure of NaCl. With the presence of Na ions at the vertex of the structure, there has been the presence of the Cl ion with every Na ion for the electron transfer.
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o prepare vegetables for finishing by grilling, sautéing, pan frying, deep frying, or stewing, you should _______ them to cook them to partial doneness
Answer:
To prepare vegetables for finishing by grilling, sautéing, pan frying, deep frying, or stewing, you should parboil them to cook them to partial doneness.
If each NADHNADH generates 3 ATPATP molecules and each FADH2FADH2 generates 2 ATPATP molecules, calculate the number of ATPATP molecules generated from one saturated 18 ‑carbon fatty acid.
Answer:
[tex]128~ATP[/tex]
Explanation:
The metabolic pathway by which energy can be obtained from a fatty acid is called "beta-oxidation". In this route, acetyl-Coa is produced by removing 2 carbons from the fatty acid for each acetyl-Coa produced. In other words, for each round, 1 acetyl Coa is produced and for each round 2 carbons are removed from the initial fatty acid. Therefore, the first step is to calculate the number of rounds that will take place for an 18-carbon fatty acid using the following equation:
[tex]Number~of~Rounds=\frac{n}{2}-1[/tex]
Where "n" is the number of carbons, in this case "18", so:
[tex]Number~of~Rounds=\frac{18}{2}-1~=~8[/tex]
We also have to calculate the amount of Acetyl-Coa produced:
[tex]Number~of~Acetyl-Coa=\frac{18}{2}~=~9[/tex]
Now, we have to keep in mind that in each round in the beta-oxidation we will have the production of 1 [tex]FADH_2[/tex] and 1 [tex]NADH[/tex]. So, if we have 8 rounds we will have 8 [tex]FADH_2[/tex] and 8 [tex]NADH[/tex].
Finally, for the total calculation of ATP. We have to remember the yield for each compound:
-) [tex]1~FADH_2~=~2~ATP[/tex]
-) [tex]1~NADH~=~3~ATP[/tex]
-) [tex]Acetyl~CoA~=~10~ATP[/tex]
Now we can do the total calculation:
[tex](8*2)~+~(8*3)~+~(9*10)=130~ATP[/tex]
We have to subtract "2 ATP" molecules that correspond to the activation of the fatty acid, so:
[tex]130-2=128~ATP[/tex]
In total, we will have 128 ATP.
I hope it helps!
Draw the structure for the organic radical species produced by reaction of the compound with a chlorine atom. Assume reaction occurs at the weakest C-H bond.
Answer:
See explanation
Explanation:
The reaction of chlorine with the pictured compound will occur via free radical mechanism. The stability of the free radical formed will depend on its structure.
The order of stability of free radicals is methyl < primary < secondary < tertiary. Hence a tertiary carbon free radical is the most stable.
Looking at the compound, the radical will form at the position shown in the image attached since it will lead to a secondary free radical which is more stable.
The structure that should be drawn is shown below.
The reaction of chlorine:It should be within the pictured compound that will arise via a free radical mechanism. The stability should be based on the structure. The stability of the order of free radicals should be methyl < primary < secondary < tertiary. Thus, a tertiary carbon free radical should be most stable.
Here look at the compound, the radical should form at the position that should be shown in the image that resulted in the secondary free radical i.e. more stable.
Why are cells important to an organisms survival
Answer:
Cells are the basic structures of all living organisms. Cells provide structure for the body, take in nutrients from food and carry out important functions. ... These organelles carry out tasks such as making proteins?, processing chemicals and generating energy for the cell
Answer: I absolutely love this question! Biology is so interesting, so I always love to answer the curiosity of others regarding biology, such as that!
Cells are simply the basic structures of all organisms, that are living, of course! Cells provide structure for the body, and they also take in nutrients that your body needs from food and they carry out important functions. These organelles carry out tasks such as making proteins, processing chemicals, and generating energy for the cell. Isn’t that cool?
Hope this helped! <3
When salt is added to water, all of the following happens except? A. The salt breaks into positive chlorine ions and negative sodium icons B. the positive part of the water molecule is attracted to the negative ions C. The negative part of the water molecule is attracted to the positive ions D. The water molecules surround the dissociated ions
Answer:
The salt breaks into positive chlorine ions and negative sodium icons
Explanation:
The question requested for the wrong option in the list. If we look at the option selected, we will notice that sodium ions are positively charged ions since sodium is a metal. Metals produce cations (positive ions) because they loose electrons. Therefore, a sodium ion can never be negatively charged.
Similarly, chlorine is a highly electronegative nonmetal. It gains electrons in an ionic bond. Hence chlorine ions can not be positive.
Solid sodium iodide is slowly added to a solution that is 0.0050 M Pb 2+ and 0.0050 M Ag +. [K sp (PbI 2) = 1.4 × 10 –8; K sp (AgI) = 8.3 × 10 –17] Calculate the Ag + concentration when PbI 2 just begins to precipitate.
Answer:
[Ag⁺] = 5.0x10⁻¹⁴M
Explanation:
The product solubility constant, Ksp, of the insoluble salts PbI₂ and AgI is defined as follows:
Ksp(PbI₂) = [Pb²⁺] [I⁻]² = 1.4x10⁻⁸
Ksp(AgI) = [Ag⁺] [I⁻] = 8.3x10⁻¹⁷
The PbI₂ just begin to precipitate when the product [Pb²⁺] [I⁻]² = 1.4x10⁻⁸
As the initial [Pb²⁺] = 0.0050M:
[Pb²⁺] [I⁻]² = 1.4x10⁻⁸
[0.0050] [I⁻]² = 1.4x10⁻⁸
[I⁻]² = 1.4x10⁻⁸ / 0.0050
[I⁻]² = 2.8x10⁻⁶
[I⁻] = 1.67x10⁻³So, as the [I⁻] concentration is also in the expression of Ksp of AgI and you know concentration in solution of I⁻ = 1.67x10⁻³M:
[Ag⁺] [I⁻] = 8.3x10⁻¹⁷
[Ag⁺] [1.67x10⁻³] = 8.3x10⁻¹⁷
[Ag⁺] = 5.0x10⁻¹⁴MA solution of HCOOH has 0.16M HCOOH at equilibrium. The Ka for HCOOH is 1.8×10−4. What is the pH of this solution at equilibrium? Express the pH numerically.
Answer:
[tex]pH=2.28[/tex]
Explanation:
Hello,
In this case, for the acid dissociation of formic acid (HCOOH) we have:
[tex]HCOOH(aq)\rightarrow H^+(aq)+HCOO^-(aq)[/tex]
Whose equilibrium expression is:
[tex]Ka=\frac{[H^+][HCOO^-]}{[HCOOH]}[/tex]
That in terms of the reaction extent is:
[tex]1.8x10^{-4}=\frac{x*x}{0.16-x}[/tex]
Thus, solving for [tex]x[/tex] which is also equal to the concentration of hydrogen ions we obtain:
[tex]x=0.00528M[/tex]
[tex][H^+]=0.00528M[/tex]
Then, as the pH is computed as:
[tex]pH=-log([H^+])[/tex]
The pH turns out:
[tex]pH=-log(0.00528M)\\\\pH=2.28[/tex]
Regards.
For the reaction 3H 2(g) + N 2(g) 2NH 3(g), K c = 9.0 at 350°C. What is the value of ΔG at this temperature when 1.0 mol NH 3, 5.0 mol N 2, and 5.0 mol H 2 are mixed in a 2.5 L reactor?
Answer:
ΔG = - 31.7kJ/mol
Explanation:
It is possible to find ΔG of a reaction at certain temperature knowing Kc following the equation:
ΔG = ΔG° + RT ln Q
ΔG° = -RT lnKc
ΔG = -RT lnKc + RT ln Q (1)
Where R is gas constant (8.314J/molK), T absolute temperature (350°C + 273.15 = 623.15K) and Q reaction quotient
For the reaction,
3H₂(g) + N₂(g) ⇄ 2NH₃(g)
Q = [NH₃]² / [H₂]³[N₂]
Where the concentrations of each chemical are:
[NH₃] = 1.0mol / 2.5L = 0.4M
[H₂] = 5.0mol / 2.5L = 2M
[N₂] = 2.5mol / 2.5L = 1M}
Q = [0.4M]² / [2M]³[1M]
Q = 0.02
And replacing in (1):
ΔG = -RT lnKc + RT ln Q
ΔG = -8.314J/molK*623.15K ln 9 + 8.314J/molK*623.15K ln 0.02
ΔG = - 31651J/mol
ΔG = - 31.7kJ/molA laboratory assistant needs to prepare 217 mL of 0.246 M solution. How many grams of calcium chloride will she need
Answer:
5.92 g
Explanation:
Convert milliliters to liters.
217 mL = 0.217 L
Since molarity (M) is moles per liter(mol/L), multiply the molarity by the volume to find out how many moles you will need.
0.217 L × 0.246 M = 0.05338 mol
Now, convert the moles to grams using the molar mass. The molar mass of calcium chloride is 110.98 g/mol.
0.05338 mol × 110.98 g/mol = 5.924 g ≈ 5.92 g
You will need 5.92 g of calcium chloride.
Using these metal ion/metal standard reduction potentials Cd2+(aq)|Cd(s) Zn2+(aq)|Zn(s) Ni2+(aq)|Ni(s) Cu2+(aq)/Cu(s) -0.40 V -0.76 V ‑0.25 V +0.34 V Calculate the standard cell potential for the cell whose reaction is Ni2+(aq) + Zn(s) →Zn2+(aq)+ Ni(s)
Answer: The standard cell potential for the cell is +0.51 V
Explanation:
Given : [tex]E^0_{Ni^{2+}/Ni}=-0.25V[/tex]
[tex]E^0_{Zn^{2+}/Zn}=-0.76V[/tex]
The given reaction is:
[tex]Ni^{2+}(aq)+Zn(s)\rightarrow Zn^{2+}(aq)+Ni(s)[/tex]
As nickel is undergoing reduction, it acts as cathode and Zinc is undergoing oxidation, so it acts as anode.
[tex]E^0_{cell}=E^0_{cathode}-E^0_{anode}[/tex]
where both [tex]E^0[/tex] are standard reduction potentials.
Thus putting the values we get:
[tex]E^0_{cell}=-0.25-(-0.76)[/tex]
[tex]E^0_{cell}=0.51V[/tex]
Thus the standard cell potential for the cell is +0.51 V
Calculate the enthalpy change (∆H) for the reaction- N2(g) + 3 F2(g) –––> 2 NF3(g) given the following bond enthalpies: N≡N 945 kJ/mol F–F 155 kJ/mol N–F 283 kJ/mol
Answer:
– 844 kJ/mol.
Explanation:
The following data were obtained from the question:
N2(g) + 3 F2(g) –––> 2 NF3(g)
Enthalpy of N≡N (N2) = 945 kJ/mol
Enthalpy of F–F (F2) = 155 kJ/mol
Enthalpy of N–F3 (NF3) = 283 kJ/mol
Enthalpy change (∆H) =?
Next, we shall determine the enthalpy of reactant.
This is illustrated below:
Enthalpy of reactant (Hr) = 945 + 3(155)
Enthalpy of reactant (Hr) = 945 + 465
Enthalpy of reactant (Hr) = 1410 kJ/mol
Next, we shall determine the enthalpy of the product.
This is illustrated below:
Enthalpy of product (Hp) = 2 x 283
Enthalpy of product (Hp) = 566 kJ/mol
Finally, we shall determine the enthalpy change (∆H) for the reaction as follow:
Enthalpy of reactant (Hr) = 1410 kJ/mol
Enthalpy of product (Hp) = 566 kJ/mol
Enthalpy change (∆H) =?
Enthalpy change (∆H) = Enthalpy of product (Hp) – Enthalpy of reactant (Hr)
Enthalpy change (∆H) = 566 – 1410
Enthalpy change (∆H) = – 844 kJ/mol
Answer:
– 844 kJ/mol.
Explanation:
The following data were obtained from the question:
N2(g) + 3 F2(g) –––> 2 NF3(g)
Enthalpy of N≡N (N2) = 945 kJ/mol
Enthalpy of F–F (F2) = 155 kJ/mol
Enthalpy of N–F3 (NF3) = 283 kJ/mol
Enthalpy change (∆H) =?
Next, we shall determine the enthalpy of reactant.
This is illustrated below:
Enthalpy of reactant (Hr) = 945 + 3(155)
Enthalpy of reactant (Hr) = 945 + 465
Enthalpy of reactant (Hr) = 1410 kJ/mol
Next, we shall determine the enthalpy of the product.
This is illustrated below:
Enthalpy of product (Hp) = 2 x 283
Enthalpy of product (Hp) = 566 kJ/mol
Finally, we shall determine the enthalpy change (∆H) for the reaction as follow:
Enthalpy of reactant (Hr) = 1410 kJ/mol
Enthalpy of product (Hp) = 566 kJ/mol
Enthalpy change (∆H) =?
Enthalpy change (∆H) = Enthalpy of product (Hp) – Enthalpy of reactant (Hr)
Enthalpy change (∆H) = 566 – 1410
Enthalpy change (∆H) = – 844 kJ/mol
Explanation: