Answer:
pH = 1.62
Explanation:
We can solve this problem using Henderson-Hasselbach's equation:
pH = pKa + log[tex]\frac{[A^-]}{[HA]}[/tex]Where
pKa = -log(Ka) = 4.20[A⁻] = [NaC₇H₅O₂][HA] = [HC₇H₅O₂]Now we convert 132.8 g of NaC₇H₅O₂ into moles, using its molar mass:
132.8 ÷ 144.11 g/mol = 0.921 molesThen we calculate [NaC₇H₅O₂]:
0.921 moles / 300.0 mL = 0.003 MNow we can proceed to calculate the pH of the solution:
pH = 4.20 + log[tex]\frac{0.003}{1.17}[/tex]pH = 1.62Which product is often derived from the natural environment?
Answer:
coal, cotton, chinese tradictional medicine
D = 22.1 g/cm3, M = 523.1 g, V = ? mL
Answer:
The volume is 22.66 mL
Explanation:
D = 22.1 g/cm³, M = 523.1 g, V = ? mL
For Volume
Density = Mass ÷ Volume
Volume = Mass ÷ Density
V = M ÷ D
V = 523.1 g ÷ 22.1 g/cm³
V = 23.66 cm³
Now, 1 cm³ = 1 mL
So,
V = 22.66 mL
Thus, The volume is 22.66 mL
-TheUnknownScientist
Which is a correct comparison between the modern quantum model and John Dalton’s model of the atom?
Answer:
No comparison => John Dalton did NOT postulate an atomic structure.
Explanation:
Dalton's postulates are based upon generalizations of observable phenomena. Micro structure was not proposed. The atomic structure that's accepted in the scientific community evolved from conclusions of ...
Thompson's raisen pudding model => electrons suspended in a positive matrix
Rutherford's shell model => dense positive nucleus surrounded by an electron cloud. (No fine structure postulated)
Bohr Concentric Ring Model => electrons orbiting a positive nucleus in discrete energy levels much like planets orbiting the sun.
Schrodinger - Dirac Quantum Models => electrons occupying specific orbital energy levels as based upon statistical wave mechanics.
When a lead acid car battery is recharged by the alternator, it acts essentially as an electrolytic cell in which solid lead(II) sulfate is reduced to lead at the cathode and oxidized to solid lead(II) oxide at the anode. Suppose a current of is fed into a car battery for seconds. Calculate the mass of lead deposited on the cathode of the battery. Be sure your answer has a unit symbol and the correct number of significant digits.
The question is incomplete, the complete question is;
When a lead acid car battery is recharged by the alternator, it acts essentially as an electrolytic cell in which solid lead(II) sulfate PbSO₄ is reduced to lead at the cathode and oxidized to solid lead(II) oxide PbO at the anode.
Suppose a current of 96.0 A is fed into a car battery for 37.0 seconds. Calculate the mass of lead deposited on the cathode of the battery. Round your answer to 3 significant digits. Also, be sure your answer contains a unit symbol.
Answer:
3.81 g of lead
Explanation:
The equation of the reaction is;
Pb^2+(aq) + 2e ---->Pb(s)
Quantity of charge = 96.0 A * 37.0 seconds = 3552 C
Now we have that 1F = 96500 C so;
207 g of lead is deposited by 2 * 96500 C
x g of lead is deposited by 3552 C
x = 207 * 3552/2 * 96500
x = 735264/193000
x = 3.81 g of lead
Which of the statements below about an acid-base buffer solution is/are true?
I. It can be prepared by combining a strong acid with a salt of its conjugate base.
II. It can be prepared by combining a weak acid with a salt of its conjugate base.
III. It can be prepared by combining a weak base with its conjugate acid.
IV. The pH of a buffer solution does not change when the solution is diluted.
V. A buffer solution resists changes in its pH when an acid or base is added to it.
A. I, II, and IV.
B. II, III, and V.
C. II, III, IV, and V.
D. I, II, IV, and V.
E. II, III, and IV.
Answer:
C. II, III, IV, and V.
Explanation:
Acid buffer is generally formed by the combination of a weak acid as well as the salt of the conjugate base.
Basic buffer is formed the combination of a weak base and also the salt of the conjugate acid.
On dilution the ration of the concentration terms of the salt and weak acid/base does not change. Hence the pH of the buffer solution does not change.
When acid or base is added to buffer, it resists changes in the pH.
Therefore, option (C) is correct.
Calculate the density, in grams per liter, of a gas at STP if 3.56 L of the gas at 36.7 °C and 758.5 mmHg weighs 0.433 g.
density:? g/L
Answer:
the density of the Gas at STP is 0.227 g/L .
Explanation:
This question involves the combined gas law . The equation for the combined gas law
A sample of Ne(g) has a volume 250 mL at 752 mm Hg. What is the
new volume if the temperature and amount of gas held constant, the
pressure is;
a) lowered to 385 mm Hg.
b) Increased to 3.68 atm.
Answer: a) 525 ml
b) 67.2 ml
Explanation:
Boyle's Law: This law states that pressure is inversely proportional to the volume of the gas at constant temperature and number of moles.
[tex]P\propto \frac{1}{V}[/tex] (At constant temperature and number of moles)
[tex]P_1V_1=P_2V_2[/tex]
where,
a) [tex]P_1[/tex] = initial pressure of gas = 752 mm Hg
[tex]P_2[/tex] = final pressure of gas = 385 mm Hg
[tex]V_1[/tex] = initial volume of gas = 250 ml
[tex]V_2[/tex] = final volume of gas = ?
[tex]752\times 250=385\times V_2[/tex]
[tex]V_2=525ml[/tex]
Therefore, the volume at 385 mm Hg is 525 ml.
b) [tex]P_1[/tex] = initial pressure of gas = 752 mm Hg
[tex]P_2[/tex] = final pressure of gas = 3.68 atm = 2796.8 mm Hg (760mmHg=1atm)
[tex]V_1[/tex] = initial volume of gas = 250 ml
[tex]V_2[/tex] = final volume of gas = ?
[tex]752\times 250=2796.8\times V_2[/tex]
[tex]V_2=67.2ml[/tex]
Therefore, the volume at 3.68 atm is 67.2 ml.
what are super fuels? Describe both types with examples. In your view, what makes super fuels?
[tex]\bcancel{\huge\pink{\fbox{\tt{࿐αɴѕωєя࿐}}}}[/tex]
Super fuel or high performance fuel is petrol with a higher octane rating. Most standard brands of fuel have a 95 octane rating. But super fuel typically has a higher rating - around 98 - which can make the engine work more efficiently and improve performance.What was the purpose of letting the transformed cells sit in LB for a few minutes before spreading them onto the plates?
A. This allows time for the cells to express the antibiotic resistance gene
B. This allows the cells to take up the plasmid after the heat shock procedure
C. This allows time for the cells to warm up before plating
D. This allows cells time to start glowing green
Answer: D
Explanation:
If you dilute 18.8 mL of a 3.5 M solution to make 296.6 mL of solution, what is the molarity of the dilute solution?
Answer:
0.22M
Explanation:
We will be using the law of dilutions. We are simply increasing the amount of solvent to create a larger volume of solution.
So: moles before dilution = moles after dilution & [tex]moles_{concentrated} = moles_{dilute}[/tex]. And M = moles/liter of solution, so if we express this as moles = M x [tex]L_{soln}[/tex].
That is how we derive the formula we will be using: [tex]M_{concentrated} * Vol_{conc} = M_{dilute} * Vol_{dilute}[/tex]
or
[tex]M_{1} * Vol_{1} = M_{2} * Vol_{2}[/tex]
Applying this formula to our problem, we can substitute the variables with the given values to find the molarity of the dilute solution.
M1 = 3.5M
V1 = 18.8mL
M2 = ?
V2 = 296.6mL
Equation: (3.5M)(18.8mL) = (296.6mL)(M2)
==> 65.8M*mL = 296.6mL * M2
==> M2 = (65.8 M*mL)/296.6mL
==> M2 = 0.22M
Need help with this question please.
Answer:
12.8
Explanation:
14 = pOH + pH
pH = 14 - pOH
pH = 14 - 1.2
pH = 12.8
If aluminum has a mass of 22.3 g, how many liters of oxygen gas are required at STP?
Answer:
27.8
Explanation:
How many moles of argon, Ar, are in 1.31×1024 Ar atoms?
Answer:
2.18 mol Ar.
Explanation:
Hello there!
In this case, according to the definition of mole as the amount of particles of a given substance, it is possible to introduce the Avogadro's number to assert that 1 mole of any element contains 6.022x10²³ atoms; thus, the moles of Ar in the given amount of atoms turns out to be:
[tex]1.31x10^{24}Ar atoms*\frac{1molAr}{6.022x10^{23}atoms}\\\\=2.18molAr[/tex]
Best regards!
Express your answer as a balanced half-reaction. Identify all of the phases in your answer.
(acidic) Cr2O7 2−(aq)⟶Cr 3+(aq)
(acidic) CrO4 2−(aq)⟶Cr(OH)4 −(aq)
(acidic) Bi 3+(aq)⟶BiO3 −(aq)
(acidic) CIO −(aq)⟶Cl −(aq)
(^for CIO - that is an i not an L)
Answer:
1. Cr₂O₇²−(aq) + 14H+ (aq) + 6e- ---->⟶2 Cr³+(aq) + 7H₂O (l)
2. CrO₄²− (aq)⟶+ 4H+ (aq) + 3e- ---> Cr(OH)₄ −(aq)
3. Bi³+ (aq) + 3H₂O (l) ---> BiO₃− (aq) + 6 H+ (aq) + 2 e-
4. CIO −(aq)⟶+ 2H+ (aq) + 2e- ---> Cl −(aq) + H₂O
Explanation:
The given equations are redox reaction equations expressed as as half reactions.
The first step is to identify whether the half-reaction is oxidation reduction.
Then the number of electrons gained or lost are added on the right side of the equation.
Appropriate H+ ions and water molecules are added where necessary since the reaction takes place in acidic environment
The atoms of elements involved in the reaction are balanced by adding the correct coefficients.
1. (acidic) Cr2O7 2−(aq)⟶Cr 3+(aq)
The half-reaction is reduction as the oxidation number of chromium changes from +6 to +3. Two Cr⁶+ ions accepts 3 electrons each to form Cr³+ ions
Cr₂O₇²−(aq) + 6e- ---->⟶2 Cr³+(aq)
Cr₂O₇²−(aq) + 14H+ (aq) + 6e- ---->⟶2 Cr³+(aq) + 7H₂O (l)
2. (acidic) CrO₄²− (aq)⟶---> Cr(OH)₄ −(aq)
The half-reaction is a reduction. One Cr⁶+ accepts 3 electrons to become Cr³+
CrO₄²− (aq)⟶+ 3e- ---> Cr(OH)₄ −(aq)
CrO₄²− (aq)⟶+ 4H+ (aq) + 3e- ---> Cr(OH)₄ −(aq)
3, (acidic) Bi³+ (aq)⟶---> BiO₃− (aq)
The half-reaction is an oxidation. One Bi³+ ion gives up two electrons to become Bi⁵+
Bi³+ (aq)⟶---> BiO₃− (aq) + 2e-
Bi³+ (aq) + 3H₂O (l) ---> BiO₃− (aq) + 6 H+ (aq) + 2 e-
4. (acidic) CIO −(aq)⟶---> Cl −(aq)
The half-reaction is a reduction. One Cl+ ion accepts two electrons to become Cl- ion.
CIO −(aq) + 2e-⟶---> Cl −(aq)
CIO −(aq)⟶+ 2H+ (aq) + 2e- ---> Cl −(aq) + H₂O
Fish need about 5 ppm oxygen dissolved in water to survive. Will water with 5 mg oxygen per
liter sustain the fish?
5mg in liter is 5 ppm
Explanation: 1 ppm is one part per million.
1 ppm is 1 mg is one part of million from 1 kg = 1000 000 mg
1 litre water is 1. Kg.
Hi can someone please help me with this. I would really appreciate it. It’s kind of urgent. Thank u so much!
Answer:
a) The system is in equilibrium
b) No other force is acting on the system
Explanation:
a) The forces in a system is balanced until and unless there is equal and opposite force acting on the system. In this image the upward force is equal to the downward force on the book and the table and hence it is in equilibrium
b) An object which is in static mode, produces downward force because of weight and in response to that a normal reaction is produced in upward direction. Apart from these two forces, no other force is acting on this book and table system.
Caffeine is a compound found in some natural coffees and teas and in some colas. a. Determine the empirical formula for caffeine, using the following composition of a 100.00-g sample. 49.47 grams of carbon, 28.85 grams of nitrogen, 16.48 grams of oxygen, and 5.20 grams of hydrogen b. If the molar mass of caffeine is 194.19 g/mol, calculate its molecular formula.
Answer: The molecular formula will be [tex]C_8N_4H_{10}O_2[/tex]
Explanation:
Mass of C= 49.47 g
Mass of N = 28.85 g
Mass of O = 16.48 g
Mass of H = 5.20 g
Step 1 : convert given masses into moles.
Moles of C =[tex]\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{49.47g}{12g/mole}=4.12moles[/tex]
Moles of N =[tex]\frac{\text{ given mass of N}}{\text{ molar mass of N}}= \frac{28.85g}{14g/mole}=2.06moles[/tex]
Moles of O =[tex]\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{16.48g}{16g/mole}=1.03moles[/tex]
Moles of H =[tex]\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{5.20g}{1g/mole}=5.20moles[/tex]
Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.
For C = [tex]\frac{4.12}{1.03}=4[/tex]
For N = [tex]\frac{2.06}{1.03}=2[/tex]
For O =[tex]\frac{1.03}{1.03}=1[/tex]
For H = [tex]\frac{5.20}{1.03}=5[/tex]
The ratio of C : N: O: H = 4: 2: 1: 5
Hence the empirical formula is [tex]C_4N_2OH_5[/tex]
The empirical weight of [tex]C_4N_2OH_5[/tex] = 4(12)+2(14)+1(16)+5(1)= 97 g.
The molecular weight = 194.19 g/mole
Now we have to calculate the molecular formula.
[tex]n=\frac{\text{Molecular weight}}{\text{Equivalent weight}}=\frac{194.19}{97}=2[/tex]
The molecular formula will be = [tex]2\times C_4N_2H_5O=C_8N_4H_{10}O_2[/tex]
13. According to Arrhenius definition which of the following is an acid *
1 point
NaCl
КСІ
Ο Ο Ο
HCI
Al(OH)3
Answer: hcl
Explanation:
A que se denomina función química
En química, el grupo de algunas sustancias compuestas que poseen propiedades químicas semejantes, denominadas propiedades funcionales, recibe el nombre de función química. ... Además están divididas en ácidos, bases, sales y óxidos; y funciones orgánicas que son las relativas a los compuestos orgánicos.
BRAINLIST
What is the formula mass of ZrF4?
Answer:
167.217g/mol
Explanation:
Formula mass is defined as the mass in grams that a mole of a molecule weighs. To solve the formula mass of ZrF₄ we require the molar mass of Zr and of F (Molar mass Zr: 91.225g/mol; F: 18.998g/mol)
In this molecule, there is 1 mole of Zr and 4 moles of F. The formula mass is:
Zr = 1*91.225g/mol = 91.225g/mol
F = 4*18.998g/mol = 75.992g/mol
Formula mass: 91.225g/mol + 75.992g/mol
167.217g/molChemical formula for Aluminum Oxide
Answer: Al₂O₃
Explanation:
2 CO + O2 → 2 CO2, how many grams of oxygen is required to produce 1.0 mole of CO2?
Answer:
16 g
Explanation:
Step 1: Write the balanced equation
2 CO + O₂ → 2 CO₂
Step 2: Calculate the moles of O₂ required to produce 1.0 moles of CO₂
The molar ratio of O₂ to CO₂ is 1:2.
1.0 mol CO₂ × 1 mol O₂/2 mol CO₂ = 0.50 mol O₂
Step 3: Calculate the mass corresponding to 0.50 moles of O₂
The molar mass of O₂ is 32.00 g/mol.
0.50 mol × 32.00 g/mol = 16 g
An analytical chemist is titrating 111.0 mL of a 0.3700 M solution of aniline (C6H5NH2) with a 0.3500 M solution of HNO3. The pK_b of aniline is 9.37. Calculate the pH of the base solution after the chemist has added 79.1 mL of the HNO_3 solution to it.
Answer:
The answer is "4.31"
Explanation:
aniline millimoles [tex]= 111 \times 0.37 = 41.07[/tex]
added [tex]HNO_3[/tex] millimoles [tex]= 79.1 \times 0.35 = 27.685[/tex]
[tex]\to 41.07 - 27.685 = 13.385[/tex] millimoles aniline left
[tex]\to 27.685[/tex] millimoles salt formed
total volume[tex]= 111 + 79.1 = 190.1\ mL\\\\[/tex]
[tex]\to [aniline] = \frac{13.385}{190.1} = 0.07 \ M\\\\\to [salt] =\frac{ 27.685}{ 190.1} = 0.146\ M\\\\\to pOH = pKb + \frac{\log [salt]}{ [base]}\\\\\to pOH = 9.37 + \frac{\log [0.146]}{[0.07]}\\\\\to pOH = 9.69\\\\\to pH = 14 - 9.69\\\\\to pH = 4.31\\[/tex]
Zn(s), Zn(NO3)2 (0.3 M) || Cu(s), CuCl2 (0.5 M)
(Need help solving these questions, so I can solve the rest of the questions like this in my lab. It is greatly appreciated. Thank you)
2. Write the half reactions that occur at the anode and cathode in this electrochemical cell. Annotate which half reaction occurs at the anode and the cathode.
3. Write the overall balance reaction of this electrochemical cell.
4. Calculate E°cell of this electrochemical cell. (include units)
5. Calculate the reaction quotient (Q) of this reaction.
6. Calculate the expected Ecell for this reaction
Answer:
See Explanation
Explanation:
At the anode;
Zn(s) -----> Zn^2+(aq) + 2e
At the cathode;
Cu^2+(aq) + 2e ------> Cu(s)
Overall electrochemical reaction;
Zn(s) + Cu^2+(aq) ------> Zn^2+(aq) + Cu(s)
E°cell = E°cathode - E°anode
E°cell = 0.34 - (-0.76)
E°cell = 1.1 V
Q = [0.3 M]/[0.5 M]
Q = 0.6
From Nernst equation;
Ecell = E°cell - 0.0592/n log Q
Ecell = 1.1 - 0.0592/2 log (0.6)
Ecell = 1.1 - 0.0296 log (0.6)
Ecell = 1.11 V
i need help can someone help me
Answer:
Option D. The number of oxygen atom is the same before and after the reaction.
Explanation:
From the question given above, the following were obtained:
Robin's equation:
H₂ + O₂ —> H₂O
Alex's equation
2H₂ + O₂ —> 2H₂O
To know which equation better represents the reaction, we shall determine which of the equation is balanced.
For Robin:
H₂ + O₂ —> H₂O
Element >>> Reactant >>> Product
H >>>>>>>>> 2 >>>>>>>>>> 2
O >>>>>>>>> 2 >>>>>>>>>> 1
Robin's equation is not balanced because the number of atoms of each element in the reactant and product are not equal.
For Alex:
2H₂ + O₂ —> 2H₂O
Element >>> Reactant >>> Product
H >>>>>>>>> 4 >>>>>>>>>> 4
O >>>>>>>>> 2 >>>>>>>>>> 2
Alex' equation is balanced because the number of atoms of each element in the reactant and product are equal.
Thus, option D gives the right answer to the question.
If 5.25 mL of HCl requires 4.96 mL of 0.9845 M NaOH to reach the equivalence point,
what is the concentration of the HCI?
[tex]M_{A}V_{A}=M_{B}V_{B}\\(5.25)M_{A}=(4.96)(0.9845)\\M_{A}=\frac{(4.96)(0.9845)}{5.25} \approx \boxed{0.93 \text{ M}}[/tex]
Draw a structural formula for the organic product formed by treating butanal with the following reagent: NaBH4 in CH3OH/H2O You do not have to consider stereochemistry. You do not have to explicitly draw H atoms. Do not include lone pairs in your answer. They will not be considered in the grading. Include counter-ions, e.g., Na , I-, in your submission, but draw them in their own separate sketcher. Do not draw organic or inorganic by-products.
Answer:
Please find the solution in the attachment file.
Explanation:
In the following reaction, if you wanted to produce more hydrochloric acid (HCl), what should you do? (2 points)
4HCl + O2 ⇄ 2H2O + Cl2
a. add more H2O
b. add more O2
c. remove H2O
d. remove Cl2
Answer:
Add more H2O
Explanation:
Took the test
If you wanted to produce more hydrochloric acid (HCl), you should add more [tex]H_2O[/tex] to 4HCl + [tex]O_2[/tex] ⇄ 2[tex]H_2O[/tex] + [tex]Cl_2[/tex]. Hence, option A is correct.
What is hydrochloric acid?Hydrochloric acid [[tex]H^+[/tex]](aq) [tex]Cl^-[/tex](aq) or [tex]H_3O^+ Cl^-[/tex]], also known as muriatic acid, is an aqueous solution of hydrogen chloride.
For instance, adding a strong acid such as HCl to water results in the reaction HCl + [tex]H_2O[/tex] → [tex]H_3O^+[/tex] + [tex]Cl^-[/tex].
In other words, the proton ([tex]H^+[/tex]) from the acid binds to neutral water molecules to form [tex]H_3O^+[/tex] raising the concentration of [tex]H^+[/tex].
Hence, option A is correct.
Learn more about hydrochloric acid here:
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Can someone help me please!!
Answer:
1. AgNO₃ (aq) + NaCl (aq) ----> NaNO₃ (aq) + AgCl (s)
2. Li₂SO₄ (aq) + BaCl₂ (aq) ----> 2 LiCl (aq) + BaSO₄ (s)
3. 2 NaOH (aq) + MgCl₂ (aq) ----> 2 NaCl (aq) + Mg(OH)₂ (s)
Explanation:
The reaction involving the mixing of two soluble solutions to produce a precipitate is known as a precipitation reaction.
A precipitation reaction is double-replacement reaction (a reaction that exchanges the cations or the anions of two ionic compounds) in which one product is a solid precipitate.
Precipitation reactions at useful in the identification of various ions present in a solution. In order to predict the reactions that will produce a precipitate, solubility rules as given in the solubility table below can be used.
From the tables, the reactions that will produce a precipitate, as well as their balanced molecular equations are as follows:
1. AgNO₃ (aq) + NaCl (aq) ----> NaNO₃ (aq) + AgCl (s)
2. Li₂SO₄ (aq) + BaCl₂ (aq) ----> 2 LiCl (aq) + BaSO₄ (s)
3. 2 NaOH (aq) + MgCl₂ (aq) ----> 2 NaCl (aq) + Mg(OH)₂ (s)
ASAP PLEASE AND THANK YOU
What is the molar mass of a pure gas that has the density of 1.40 g/L at STP?
Answer:
bro what Is this like I dont even kno
Answer:
O2 is the answer I believe