Answer:
pH = 9.58
Explanation:
First of all, we need to determine the molarity of the solution.
We determine the molar mass of morphine:
12g/m . 17 + 1 g/m . 19 + 14 g/m + 16 g/m . 3 = 285.34 g/m
molar mass g/m, is the same as mg/mm
25 mg . 1 mmol / 285.34 mg = 0.0876 mmoles / 100 mL = 8.76×10⁻⁴ M
In diltuted solution, we must consider water.
Mass balance for morphine = [Morphine] + [Protonated Morphine]
8.76×10⁻⁴ M = [Morphine] + [Protonated Morphine]
As Kb is too small, I can skipped, the [Protonated Morphine]
8.76×10⁻⁴ M = [Morphine]
In the charge balance I will have:
[OH⁻] = [H⁺ morphine] + [H⁺]
Let's go to the Kb expression
Morphine + H₂O ⇄ MorphineH⁺ + OH⁻ Kb
Kb = [MorphineH⁺] [OH⁻] / [Morphine]
Kb = [MorphineH⁺] [OH⁻] / 8.76×10⁻⁴ M
So now, we need to clear [MorphineH⁺] to replace it in the charge balance
Kb . 8.76×10⁻⁴ M / [OH⁻] = [MorphineH⁺]
Now, the only unknown value is the [OH⁻]
[OH⁻] = Kb . 8.76×10⁻⁴ M / [OH⁻] + Kw/[OH⁻]
Remember that Kw = [H⁺] . [OH⁻]
[H⁺] = Kw/[OH⁻]
[OH⁻]² = 1.62×10⁻⁶ . 8.76×10⁻⁴ + 1×10⁻¹⁴
[OH⁻] = √(1.62×10⁻⁶ . 8.76×10⁻⁴ + 1×10⁻¹⁴)
[OH⁻] = 3.76×10⁻⁵ → - log [OH⁻] = pOH = 4.42
pH = 14 - pOH → 14 - 4.42 = 9.58
Balance the following equations: (c) H2(g)+I2(s)⟶HI(s)H2(g)+I2(s)⟶HI(s)
Answer: [tex]H_2(g)+I_2(g)\rightarrow 2HI(s)[/tex]
Explanation:
According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.
Thus in the reactants, there are 2 atoms of hydrogen and 2 atoms of iodine .Thus there has to be 2 atoms of hydrogen and 2 atoms of iodine in the product as well. Thus a coefficient of 2 is placed in front of HI.
The balanced chemical reaction is:
[tex]H_2(g)+I_2(g)\rightarrow 2HI(s)[/tex]
Magnesium oxide can be produced by heating magnesium metal in the presence of oxygen. When 10.1 g of Mg reacts with 10.5 g of O2, 11.9 g of MgO are collected.
a) Write the balanced equation for this reaction, including all states of matter.
b) What is the limiting reactant in this reaction?
c) What is the theoretical yield for this reaction?
d) What is the percent yield for this reaction?
Answer:
A. 2Mg(s) + O2(g) —> 2MgO(s)
B. Mg is the limiting reactant.
C. Theoretical yield of MgO is 16.83g.
D. The percentage yield is 70.7%
Explanation:
A. The balanced equation for the reaction. This is given below:
2Mg(s) + O2(g) —> 2MgO(s)
B. Determination of the limiting reactant.
First, we shall determine the mass of Mg and O2 that reacted and the mass of MgO produced from the balanced equation. This is illustrated below:
Molar mass of Mg = 24g/mol
Mass of Mg from the balanced equation = 2 x 24 = 48g.
Molar mass of O2 = 16x2 = 32g/mol.
Mass of O2 from the balanced equation = 1 x 32 = 32g
Molar mass of MgO = 24 + 16 = 40g/mol
Mass of MgO from the balanced equation = 2 x 40 = 80g
Summary:
From the balanced equation above,
48g of Mg reacted with 32g of O2 to produce 80g of MgO.
Now, we can obtain the limiting reactant as follow:
From the balanced equation above,
48g of Mg reacted with 32g of O2.
Therefore, 10.1g of Mg will react with = (10.1 x 32)/48 = 6.73g of O2.
From the calculations made above, we can see that only 6.73g out of 10.5g of O2 given is needed to react completely with 10.1g of Mg.
Therefore, Mg is the limiting reactant and O2 is the excess reactant.
C. Determination of the theoretical yield of MgO.
The limiting reactant is used in this case as it will produce the maximum yield of the reaction since all of I is used up in the reaction.
The theoretical yield can be obtain as illustrated below:
From the balanced equation above,
48g of Mg reacted to produce 80g of MgO.
Therefore, 10.1g of Mg will react to produce = (10.1 x 80)/48 = 16.83g of MgO.
Therefore, the theoretical yield of MgO is 16.83g.
D. Determination of the percentage yield.
This is illustrated below:
Actual yield of MgO = 11.9g
Theoretical yield of MgO = 16.83g
Percentage yield =..?
Percentage yield = Actual yield /Theoretical yield x 100
Percentage yield = 11.9/16.83 x 100
Percentage yield = 70.7%
Can a catalyst change an exothermic reaction into an endothermic reaction or vice versa? Please explain your answer.
Answer:
A catalyst cannot change an exothermic reaction into an endothermic reaction or vice versa.
Explanation:
Catalyst is basically a substance that enables a chemical reaction to occur at a faster rate as compared to the reaction without catalysis. It lowers the activation energy and temperature for a chemical reaction and a catalyst itself does not goes through any permanent chemical change. This means it does not get used in the process.
Exothermic and endothermic are the chemical reaction. Exothermic reactions absorb energy. This energy is absorbed in the form of heat. When the energy is released in the form of heat then this reaction is called endothermic. So one absorbs the heat and the other releases it.
As we know that the catalyst does not undergo change at the end of the reaction so the energy or heat whether is absorbed or emitted or you can say whether the reaction is exothermic or endothermic, the total energy stays unchanged during the reaction. So with and without a catalyst, if both have same reactants and products and the difference in enthalpy between products and reactants will be the same.
A pentavalent cation atom has 20 and 15 neutrons as protons. Find the electron quantity and mass number respectively. (40 pts.) a) 20 and 15 b) 15 and 20 c) 15 and 35 d) 35 and 15 e) 10 and 20
Answer:
C.
Explanation:
Since the mass number is the number of protons and neutrons added together, the answer is 35. Since the questions are respectively electron quantity and mass number, the only answer choice with 35 as the second choice is C, so that is the correct answer.
Description (with words) of water just above melting temperature. What intermolecular forces do you expect to find in water in liquid state
Answer:
intermolecular dipole-dipole hydrogen bonds
Explanation:
Water is a polar molecule. Recall that the central atom in water is oxygen. The molecule is bent, hence it has an overall dipole moment directed towards the oxygen atom. Since it has a permanent dipole moment, we expect that it will show dipole-dipole interactions in the liquid state.
Similarly, water contains hydrogen and oxygen. Recall that hydrogen bonds are formed when hydrogen is covalently bonded to highly electronegative elements. Hence, water in the liquid state exhibits strong hydrogen bonding. The unique type of dipole-dipole interaction in liquid water is actually hydrogen bonding, hence the answer.
Which best describes the act of using senses or tools to gather information? creating a hypothesis making an observation summarizing the results recording the measurements
Answer:
B - Making an Observation
Explanation:
Making an observation best describes the act of using senses or tools to gather information. Therefore, option B is correct.
What are senses in the scientific method?The five senses—sight, taste, touch, hearing, and smell—gather data about our surroundings that the brain interprets. Based on prior experience (and subsequent learning), as well as by combining the data from each sensor, we make sense of this information.
Information gleaned from your five senses is referred to as an observation. These are smell, taste, touch, hearing, and sight. When you see a bird or hear it sing, you notice it.
The term observation, which is also used to sense five aspects of the world including vision, taste, touch, smell, and hearing, is used to describe utilizing the senses to examine the world, employing tools to take measurements, and looking at prior research findings.
Thus, option B is correct.
To learn more about the scientific method, follow the link:
https://brainly.com/question/7508826
#SPJ7
Add distilled water to the beaker until the volume
totals 15 mL.
Record the amount of oil that dissolved.
Answer:
i guess oil never dissolve in water. As like dissolve like. water is polar so it dissolves only polar substances
Explanation:
Answer:
None
Explanation:
Answer on Edge 2022
The study of how humans impact the environment is
?
Humans impact the physical environment in many ways: overpopulation, pollution, burning fossil fuels, and deforestation.
The study of how humans impact the environment is known as "environmental science," and it describes both the positive and negative effects of humans on the climate.
What is the significance of environmental science?Environmental science is the study of the environment, and it explains how the environment changes through such factors as human interaction, the biotic and abiotic parts of the environment, the temperature, humidity, etc. This branch of science explains the different interactions of humans and their adverse effects, such as pollution caused by human interference, global warming, climate change, etc.
Hence, the study of how humans impact the environment is known as "environmental science," and it describes both the positive and negative effects of humans on the climate.
Learn more about environmental science here.
https://brainly.com/question/4593618
#SPJ2
Determine the limiting reactant (LR) and the mass (in g) of nitrogen that can be formed from 50.0 g N 2O 4 and 45.0 g N 2H 4. Some possibly useful molar masses are as follows: N 2O 4 = 92.02 g/mol, N 2H 4 = 32.05 g/mol.
N 2O 4( l) + 2 N 2H 4( l) → 3 N 2( g) + 4 H 2O( g)
a) LR = N2O4, 45.7 g N2 formed
b) LR = N2O4, 105 g N2 formed
c) LR = N2H4, 13.3 g N2 formed
d) LR = N2H4, 59.0 g N2 formed
e) No LR, 45.0 g N2 formed
Answer:
Option A. LR = N2O4, 45.7g N2 formed
Explanation:
The balanced equation for the reaction is given below:
N2O4(l) + 2N2H4(l) → 3N2(g) + 4H2O(g)
Next, we shall determine the masses of N2O4 and N2H4 that reacted and mass of N2 produced from the balanced equation. This is illustrated below:
Molar mass of N2O4 = 92.02 g/mol
Mass of N2O4 from the balanced equation = 1 x 92.02 = 92.02 g
Molar mass of N2H4 = 32.05 g/mol
Mass of N2H4 from the balanced equation = 2 x 32.05 = 64.1g
Molar mass of N2 = 2x14.01 = 28.02g/mol
Mass of N2 from the balanced equation = 3 x 28.02 = 84.06g
Summary:
From the balanced equation above,
92.02g of N2O4 reacted with 64.1g of N2H4 to produce 84.06g of N2.
Next, we shall determine the limiting reactant. This can be obtained as follow:
From the balanced equation above,
92.02g of N2O4 reacted with 64.1g of N2H4.
Therefore, 50g of N2O4 will react with = (50 x 64.1)/92.02 = 34.83g of N2H4.
From the calculations made above, we can see that only 34.83g out 45g of N2H4 is required to react completely with 50g of N2O4.
Therefore, N2O4 is the limiting reactant and N2H4 is the excess reactant.
Finally, we shall determine the mass of N2 produced from the reaction.
In this case the limiting reactant will be used as it will produce the maximum yield of N2 since all of it is used up in the reaction.
The limiting reactant is N2O4 and the mass N2 produced can be obtained as illustrated below:
From the balanced equation above,
92.02g of N2O4 reacted to produce 84.06g of N2.
Therefore 50g of N2O4 will react to produce = (50 x 84.06)/92.02 = 45.7g of N2.
Therefore, 45.7g of N2 were produced from the reaction.
At the end of the day,
The limiting reactant is N2O4 and 45.7g of N2 were produced from the reaction.
Which of the following would have a fixed shape and volume? (3 points) nitrogen gas solid wood liquid water neon gas
Answer:
Solid Wood
Explanation:
Wood is like a solid block, whereas gases flow freely and liquids spread to fill the shape of their container.
Please let me know if I misunderstood the question, by the way.
Each unknown mixture contains 5 metal constituents. Select the 5 metal ions that you have identified as being present in your mixture. Please double check your selections before you hit the submit button. a. Ca b. Co c. Cr d. Fe e. K f. Mn g. Zn
Explanation:
A metal ion is a type of atom compound that has an electric charge.
Such atoms willingly lose electrons in order to build positive ions called cations. The selected Ions are :
[tex]1. Mn^2^+\\\ 2. Ca^2^+\\\ 3. Co^2^+\\\ 4. Fe^2^-\\\ 5. K^+[/tex]
a ______ consumer is a heterotroph that directly eats an autotroph. A primary B. Quaterany C tertiary D secondary
A 5.22 × 10−3−mol sample of HY is dissolved in enough H2O to form 0.088 L of solution. If the pH of the solution is 2.37, what is the Ka of HY?
Answer:
3.07 × 10⁻⁴
Explanation:
Step 1: Calculate the concentration of H⁺
We will use the definition of pH.
[tex]pH = -log [H^{+} ]\\\[ [H^{+} ] = antilog -pH = antilog -2.37 = 4.27 \times 10^{-3} M[/tex]
Step 2: Calculate the concentration of HY
5.22 × 10⁻³ mol of HY are dissolved in 0.088 L. The concentration of the acid (Ca) is:
[tex]Ca = \frac{5.22 \times 10^{-3} mol }{0.088L} = 0.0593M[/tex]
Step 3: Calculate the acid dissociation constant (Ka)
We will use the following expression.
[tex]Ka = \frac{[H^{+}]^{2} }{Ca} = \frac{(4.27 \times 10^{-3} )^{2} }{0.0593} = 3.07 \times 10^{-4}[/tex]
Identify the elements that correspond to the following generalized electron configuration: (noble gas]ns2(n − 2) f6
Express your answer as the element symbol. If there is more than one answer, separate them by a comma.
Answer:
Samarium:
Electron configuration:
Samarium
Explanation:
Samarium is a chemical element that belongs to the lanthanoid series. The lanthanoids are the chemical elements that follow lanthanum. They are all known to possess 4f orbitals. The 4f electrons are found in the antepenultimate shell of the elements of the lanthanoid series and they do not take part in chemical bonding. They are neither removed in bonding nor do they take part in crystal field stabilization of lanthanoid complexes.
The electronic configuration of samarium is; 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s2 4f6 while the condensed, short hand electronic configuration is [Xe] 4f6 6s2. This corresponds to (noble gas]ns2(n − 2) f6 as required by the question, hence the answer provided above.
Turn on Write equation. What you see is an equation that shows the original uranium atom on the left. The boxes on the right represent the daughter product—the atom produced by radioactive decay—and the emitted alpha particle.
Answer:
Uranium-238 undergoes alpha decay to form Thorium-234 as daughter product.
Explanation:
Alpha decay is indicative of loss of the equivalents of a helium particle emission. The reaction equation for this reaction is shown below:
[tex]_{92} ^{238} U_{}[/tex]→ [tex]_{90} ^{234} Th_{} + _{2} ^{4} He_{}[/tex]
I hope this explanation is clear and explanatory.
Pb(OH)Cl, one of the lead compounds used in ancient Egyptian cosmetics, was prepared from PbO according to the following recipe: PbO(s) NaCl(aq) H2O(l) --> Pb(OH)Cl(s) NaOH(aq) How many grams of PbO and how many grams of NaCl would be required to produce 10.0 g of Pb(OH)Cl
Answer:
8.59 g
2.25 g
Explanation:
According to the given situation the calculation of grams of PbO and grams of NaCL is shown below:-
Moles of Pb(OH)CL is
[tex]= \frac{Mass}{Molar\ mass}[/tex]
[tex]= \frac{10.0 g}{259.65g / mol}[/tex]
= 0.0385 mol
Mass of PbO needed is
[tex]= 0.385mol Pb(OH) Cl\times \frac{1 mol PbO}{1molpb (OH) cl} \times \frac{223.2g PbO}{1mol PbO}[/tex]
After solving the above equation we will get
= 8.59 g
Mass of NaCL needed is
[tex]= \frac{1mol\ NaCl}{1molPb\ (OH)Cl} \times \frac{58.45NaCl}{1mol NaCl}[/tex]
After solving the above equation we will get
= 2.25 g
Therefore we have applied the above formula.
Without doing any calculations, match the following thermodynamic properties with their appropriate numerical sign for the following endothermic reaction. 2N2(g) + O2(g)2N2O(g) Clear All > 0 Hrxn < 0 Srxn = 0 Grxn > 0 low T, < 0 high T Suniverse < 0 low T, > 0 high T
Answer:
∆H > 0
∆Srxn <0
∆G >0
∆Suniverse <0
Explanation:
We are informed that the reaction is endothermic. An endothermic reaction is one in which energy is absorbed hence ∆H is positive at all temperatures.
Similarly, absorption of energy leads to a decrease in entropy of the reaction system. Hence the change in entropy of the reaction ∆Sreaction is negative at all temperatures.
The change in free energy for the reaction is positive at all temperatures since ∆S reaction is negative then from ∆G= ∆H - T∆S, we see that given the positive value of ∆H, ∆G must always return a positive value at all temperatures.
Since entropy of the surrounding= - ∆H/T, given that ∆H is positive, ∆S surrounding will be negative at all temperatures. This is so because an endothermic reaction causes the surrounding to cool down.
When a strip of magnesium metal is placed in an aqueous solution of copper(II) nitrate, elemental copper coats the surface of the magnesium strip and aqueous magnesium nitrate forms.
1. As a reactant, what is the charge on copper?
2. As a product, what is the charge on copper?
Answer:
1. 2+ ([tex]Cu^{2+}[/tex]).
2. 0 ([tex]Cu^0[/tex]).
Explanation:
Hello,
In this case, the described chemical reaction is a redox reaction in fact, since the oxidation states of both magnesium and copper change as shown due to the displacement:
[tex]Mg(s)+Cu(NO_3)_2(aq)\rightarrow Mg(NO_3)_2(aq)+Cu(s)[/tex]
Therefore:
1. Since copper is the cation in the copper (II) nitrate, the (II) means that its charge is 2+ ([tex]Cu^{2+}[/tex]).
2. Since copper is alone, it means no electrons are being neither shared not given, its charge is 0 ([tex]Cu^0[/tex]).
Best regards.
Question 11: How does the energy of a photon emitted when the electron moves from the 3rd orbital to the 2nd orbital compare to the energy of a photon absorbed when the electron moves from the 2nd orbital to the 3rd orbital?
Answer:
Explanation:
The energy of a photon emitted when the electron moves from the 3rd orbital to the 2nd orbital is exactly same as the energy of a photon absorbed when the electron moves from the 2nd orbital to the 3rd orbital
Calculate the pH for the following 1.0M weak acid solutions:a. HCOOH Ka = 1.8 x 10-4 [
Answer: pH=2.38
Explanation:
To calculate the pH, let's first write out the equation. Then, we will make an ICE chart. The I in ICE is initial quantity. In this case, it is the initial concentration. The C in ICE is change in each quantity. The E is equilibrium.
HCOOH ⇄ H⁺ + HCOO⁻
I 1.0M 0 0
C -x +x +x
E 1.0-x x x
For the steps below, refer to the ICE chart above.
1. Since we were given the initial of HCOOH, we can fill this into the chart.
2. Since we were not given the initial for H⁺ and HCOO⁻, we will put 0 in their place.
3. For the change, we need to add concentration to the products to make the reaction reach equilibrium. We would add on the products and subtract from the reactants to equalize the reaction. Since we don't know how much the change in, we can use variable x.
4. We were given the Kₐ of the solution. We know [tex]K_{a} =\frac{product}{reactant}=\frac{[H^+][HCOO^-]}{[HCOOH]}[/tex].
5. The problem states that the Kₐ=1.8×10⁻⁴. All we have to so is to plug it in and to solve for x.
[tex]1.8*10^-^4 =\frac{x^2}{0.1-x}[/tex]
6. Once we plug this into the quadratic equation, we get x=0.00415.
7. The equilibrium concentration of [H⁺]=0.00415. pH is -log(H⁺).
-log(0.00415)=2.38
Our pH for the weak acid solution is 2.38.
What volume of 6.00 M hydrochloric acid is needed to prepare 500 mL of 0.100 M solution?
Answer:
8.33mL or .0083L
Explanation:
Use m1 * V1 = m2 * V2
6.00M(x) = 0.100M(500mL)
solve for x
x= (.1 * 500) / 6
x=8.333 mL
Calculate the free energy of formation of NaBr(s) given the following information: NaBr(s) → Na(s) + 1/2Br2(l), ∆G° = 349 kJ/mol
The given question is incomplete, the complete question is:
Calculate the free energy of formation of NaBr(s) given the following information: NaBr(s) → Na(s) + 1/2Br2(l), ΔG° = 349 kJ/mol
A) –309 kJ/mol
B) –329 kJ/mol
C) None of the above
D) –349 kJ/mol
E) –369 kJ/mol
Answer:
The correct answer is option D, that is, -349 kJ/mol.
Explanation:
Based on the given information, the reaction is:
NaBr (s) ⇔ Na (s) + 1/2 Br₂ (l), the ΔG° of the reaction given is 349 kJ per mole. In the given question, it is clearly mentioned that there is a need to determine the free energy of the formation of NaBr. Thus, there is a need to keep Na (s) and Br₂ (l) at the reactant side and NaBr (s) at the product side.
Therefore, there is a need to reverse the reaction and change the sign on ΔG.
Now the reaction will become,
Na (s) + 1/2 Br₂ (l) ⇔ NaBr (s), and the ΔG° will now become -349 kJ per mole. Hence, -349 kJ per mole is the free energy of the formation of NaBr (s).
if hybridization patterns of NH and SF gave sp3 and sp3d2 respecticely what is the value of 2 x + 3 y
Answer:
The value of 2x+3y is 26
Explanation:
Because NH pattern gave hybridization Sp3
Just like NH3 Or NH2-
And SF pattern gave hybridization Sp3d2
Also just like SF6
X is the static no. of NH pattern
So X=4
Y is the static no. Of SF pattern
So Y = 6
Value of 2X+3Y
2*4 + 3*6
26
An unknown element, X, reacts with oxygen to form two oxides with the general formula X_2O_3 and X_2O_5. The pure element is strongly paramagnetic, and the oxides are amphoteric. Which of the following elements could be represented by X?a) Scb) Bc) Nd) Sbe) S
Answer:
The unknown element is Sb
Explanation:
The first thing we must note is that the unknown element must be a member of group 15 in the periodic table. This is clear from the fact that the two oxides formed are X2O3 and X2O5. This implies that the unknown element X must have a valency of 3 or 5. This corresponds to our knowledge that the outermost electron configuration of group 15 elements is ns2np3. Hence, group fifteen elements can have a valency of 3 or 5.
The electronic configuration of antimony is; [Kr]4d10 5s2 5p3. This implies that the atom is paramagnetic since there are three unpaired 5p electrons. The oxides of antimony are known to be amphoteric. An ampohoteric oxide reacts with both acid and base, hence the answer.
Dissolving NaOH(s) in water is exothermic. Two calorimetry experiments are set up. Experiment 1: 2 g of NaOH are dissolved in 100 mL of water Experiment 2: 4 g of NaOH are dissolved in 200 mL of water Which of the following statements is true?a. both temperature changes will be the sameb. the second temeprature change will be approximately twice the firstc. the second temperature change will be approximately four times the firstd. the second temperature change will be approximately one-half of the firste. the second temperature change will be approximately one-fourth the first
Answer:
a. both temperature changes will be the same
Explanation:
When sodium hydroxide (NaOH) is dissolved in water, a determined amount is released to the solution following the equation:
Q = m×C×ΔT
Where Q is the heat released, m is the mass of the solution, C is the specific heat and ΔH is change in temperature.
Specific heat of both solutions is the same (Because the solutions are in fact the same). Specific heat = C.
m is mass of solutions: 102g for experiment 1 and 204g for experiment 2.
And Q is the heat released: If 2g release X heat, 4g release 2X.
Thus, ΔT in the experiments is:
Experiment 1:
X / 102C = ΔT
Experiment 2:
2X / 204C = ΔT
X / 102C = ΔT
That means,
a. both temperature changes will be the same
A package contains 1.33 lb of ground round. If it contains 29% fat, how many grams of fat are in the ground round? The book is saying 91g I keep getting 175g. Can someone please explain?
Answer:
To obtain the grams of fat that the ground round has, knowing that it weighs 1.33 pounds we must pass this value to grams. Since 1 pound equals 453.59 grams, 1.33 pounds equals 603.27 (453.59 x 1.33).
Now, to obtain 29 percent of 603.27, we must make the following calculation: 603.27 / 100 x 29, which gives a total of 174.94 grams.
In this way, your reasoning is correct and it is probably a mistake in the book.
"Calculate the pH during the titration of 20.00 mL of 0.1000 M HF(aq) with 0.2000 M NaOH(aq) after 9.4 mL of the base have been added. Ka of hydrofluoric acid
Answer:
The answer is " 10.39"
Explanation:
Calculating acid moles:
[tex]= 0.02000 \ L \times 0.1000 \ M \\\\= 0.002000[/tex]
Calculating NaOH moles:
[tex]= 0.02012 \ L \times 0.1000 \ M \\\\= 0.002012[/tex]
calculating excess in OH- Moles:
[tex]= 0.002012 - 0.002000\\\\=0.000012[/tex]
calculating total volume:
[tex]= 20.00 + 20.12\\\\ = 40.12 mL \\\\= 0.04012 L[/tex]
[tex][OH-]= \frac{0.000012} { 0.0472}[/tex]
[tex]=0.00025 M[/tex]
[tex]pOH = - \log 0.00025[/tex]
= 3.6
[tex]pH = 14 - pOH[/tex]
= 10.39
how do you fight off ADHD medication
Answer:A medication break can ease side effects. A lack of appetite, weight loss, sleep troubles, headaches, and stomach pain are common side effects of ADHD medication.
Explanation: It may boost your child’s growth. Some ADHD medications can slow a child’s growth in height, especially during the first 2 years of taking it. While height delays are temporary and kids typically catch up later, going off medication may lead to fewer growth delays.
It won’t hurt your child. Taking a child off ADHD medication may cause their ADHD symptoms to reappear. But it won’t make them sick or cause other side effects.
If the pKaof HCHO2is 3.74 and the pH of an HCHO2/NaCHO2solution is 3.11, which of the following is true?
A. [HCHO2] < [NaCHO2]
B. [HCHO2] = [NaCHO2]
C. [HCHO2] << [NaCHO2]
D. [HCHO2] > [NaCHO2]
E. It is not possible to make a buffer of this pH from HCHO2 and NaCHO2
Answer:
D. [HCHO₂] > [NaCHO₂]
Explanation:
Formic acid, HCHO₂, is a weak acid that, in presence of its conjugate base, NaCHO₂ (CHO₂⁻), produce a buffer following H-H equation:
pH = pKa + log [CHO₂⁻] / [HCHO₂]
As pKa of the acid is 3.74 and pH of the solution is 3.11:
3.11 = 3.74 + log [CHO₂⁻] / [HCHO₂]
-0.63 = log [CHO₂⁻] / [HCHO₂]
0.2344 = [CHO₂⁻] / [HCHO₂]
A ratio [CHO₂⁻] / [HCHO₂] < 1, means:
[HCHO₂] > [CHO₂⁻]When hydrocarbons are burned in a limited amount of air, both CO and CO2 form. When 0.430 g of a particular hydrocarbon was burned in air, 0.446 g of CO, 0.700 g of CO2, and 0.430 g of H2O were formed.
Required:
a. What is the empirical formula of the compound?
b. How many grams of O2 were used in the reaction?
c. How many grams would have been required for complete combustion?
Answer:
(a) The empirical formula of the compound is
m(CxHy) + m(O2) = m(CO) + m(CO2) + m(H2O).
(b) The grams of O2 that were used in the reaction is 1.146 g
(c) The amount of O2 that would have been required for complete combustion is 1.401 g.
Explanation:
a. m(CxHy) + m(O2) = m(CO) + m(CO2) + m(H2O)
(b) Using law of conservation of mass from above
m(O2) = m(CO) + m(CO2) + m(H2O) - m(CxHy)
m(O2) = 0.446 + 0.700 + 0.430 - 0.430
m(O2) = 1.146 g
The grams of O2 that were used in the reaction is 1.146 g
(c) for complete combustion, we need to oxidized CO to CO2
Then, 2CO +O2 = 2CO2
m(add)(O2) = M(O2)*¢(O2)/2 = M(O2) * {(m(CO))/(2M(CO))}
m(add)(O2) = 32 * {(0.446)/(2*28)} = 0.255 g
Note; Molar mass of O2 = 32, CO = 28
m(total)(O2) = m(O2) + m(add)(O2)
m(total)(O2) = 1.146 + 0.255 = 1.401 g
The amount of that grams would have been required for complete combustion is 1.401 g.
Note (add) and (total) were used subscript to "m"