Which of the following is a salt that will form from the combination of a strong base with a weak acid?

Select the correct answer below:
A. NaHCO3
B. H2O
C. CH3CO2H
D. NH4Cl

Answers

Answer 1

Answer:

A. NaHCO₃

Explanation:

NaHCO₃ ⇒ NaOH + H₂CO₃

NaOH is a strong base and H₂CO₃ is a weak acid. Therefore, NaHCO₃ is a salt of a strong base-weak acid reaction. The salt is basic because carbonic acid (H₂CO₃) is a weak acid so it remains undissociated. So, there is a presence of additional OH⁻ ions that makes the solution basic.

Hope that helps.


Related Questions

15.Vicinal coupling is:A)coupling between 1H nuclei attached to adjacent C atoms.B)coupling between 1H nuclei in an alkene.C)coupling between 1H nuclei attached to the same C atom.D)coupling between 1H nuclei in an alkane.

Answers

Answer:

A)coupling between 1H nuclei attached to adjacent C atoms.

Explanation:

The word ‘vicinal’ in chemistry means three bonds from the functional groups. The two functional groups are in a relationship with the atoms in adjacent position to them.

The 1H nuclei consists of two Hydrogen nucleus which acts as the functional groups. They are however attached and in a relationship with the adjacent C atoms. This makes option A the right choice.

What is the pH of 10.0 mL solution of 0.75 M acetate after adding 5.0 mL of 0.10 M HCl (assume a Ka of acetic acid of 1.78x10-5)

Answers

Answer:

5.90

Explanation:

Initial moles of CH3COO- = 10.0/1000 x 0.75 = 0.0075 mol

Moles of HCl added = 5.0/1000 x 0.10 = 0.0005 mol

CH3COO- + HCl => CH3COOH + Cl-

Moles of CH3COO- left = 0.0075 - 0.0005 = 0.007 mol

Moles of CH3COOH formed = moles of HCl added = 0.0005 mol

pH = pKa + log([CH3COO-]/[CH3COOH])

= -log Ka + log(moles of CH3COO-/moles of CH3COOH)

= -log(1.78 x 10^(-5)) + log(0.007/0.0005)

= 5.90

Answer:

The correct answer is 5.895.

Explanation:

The reaction will be,

CHCOO⁻ + H+ ⇔ CH₃COOH

Both the HCl and the acetate are having one n factor.

The millimoles of CH₃COO⁻ is,

= Volume in ml × molarity = 10 × 0.75 = 7.5

The millimoles of HCl = Volume in ml × molarity = 5 × 0.1 = 0.5

Therefore, 0.5 will be the millimoles of CH₃COOH formed, now the millimoles of the CH₃COO⁻ left will be, 7.5-0.5 = 7.0

The volume of the solution is, 10+5 = 15 ml

The molarity of CH₃COO⁻ is, millimoles / volume in ml = 7/15

The molarity of CH₃COOH is 0.5/15

pH = pKa + log[CH₃COO⁻]/[CH₃COOH]

= 4.74957 + 1.146

= 5.895


A mixture of gases includes helium, neon, argon and krypton. The partial pressures are as follows:
Partial pressure, mm Hg
He 152
Ne 164
Ar 126
Kr 169
What is the total pressure of the mixture after removing the argon component? Report your answer rounded
to the nearest mm Hg but without units

Answers

Answer:

the answer to your question is ar 126

A solution contains 2.2 × 10-3 M in Cu2+ and 0.33 M in LiCN. If the Kf for Cu(CN)42- is 1.0 × 1025, how much copper ion remains at equilibrium?

Answers

Answer:

[Cu²⁺] = 2.01x10⁻²⁶

Explanation:

The equilibrium of Cu(CN)₄²⁻ is:

Cu²⁺ + 4CN⁻ ⇄ Cu(CN)₄²⁻

And Kf is defined as:

Kf = 1.0x10²⁵ = [Cu(CN)₄²⁻] / [Cu²⁺] [CN⁻]⁴

As Kf is too high you can assume all Cu²⁺ is converted in Cu(CN)₄²⁻ -Cu²⁺ is limiting reactant-, the new concentrations will be:

[Cu²⁺] = 0

[CN⁻] = 0.33M - 4×2.2x10⁻³ = 0.3212M

[Cu(CN)₄²⁻] = 2.2x10⁻³

Some [Cu²⁺] will be formed and equilibrium concentrations will be:

[Cu²⁺] = X

[CN⁻] = 0.3212M + 4X

[Cu(CN)₄²⁻] = 2.2x10⁻³ - X

Where X is reaction coordinate

Replacing in Kf equation:

1.0x10²⁵ = [2.2x10⁻³ - X] / [X] [0.3212M +4X]⁴

1.0x10²⁵ = [2.2x10⁻³ - X] / 0.0104858X + 0.524288 X² + 9.8304 X³ + 81.92 X⁴ + 256 X⁵

1.04858x10²³X + 5.24288x10²⁴ X² + 9.8304x10²⁵ X³ + 8.192x10²⁶ X⁴ + 2.56x10²⁷ X⁵ = 2.2x10⁻³ - X

1.04858x10²³X + 5.24288x10²⁴ X² + 9.8304x10²⁵ X³ + 8.192x10²⁶ X⁴ + 2.56x10²⁷ X⁵ - 2.2x10⁻³ = 0

Solving for X:

X = 2.01x10⁻²⁶

As

[Cu²⁺] = X

[Cu²⁺] = 2.01x10⁻²⁶

Write chemical equations for the reactions that occur when solutions of the following substances are mixed:

c. OH- and H₂PO₄-
d. C₅H₅N pyridine, a base) and HC₂H₃O₂ (acetic acid)

Answers

#c

[tex]\\ \rm\hookrightarrow H_2PO_4^-+OH-\longrightarrow HPO_4^{2-}+H_2O[/tex]

H_2PO_4 is a base according to bronsted Lowry theory.So it loses a H+ and OH- takes that to form water

#2

[tex]\\ \rm\hookrightarrow C_5H_5N+C_2H_4O_2\longrightarrow C_6H_6N^++C_2H_3O_2^-[/tex]

Pyridine is a bronsted base .So it forms pyridine cation by taking one H^+ from acetic acid

Beginning with Na, record the number of energy levels, number of protons, and atomic radius for each element in period 3.

Answers

Answer:

Sodium, magnesium, aluminium, silicon, phosphorus, sulfur, chlorine, and argon are the elements of third period.

Explanation:

There are three energy levels in sodium atom. It has 11 electrons revolving around the nucleus. the atomic radius of sodium atom is 227 ppm. Magnesium, aluminium, silicon, phosphorus, sulfur, chlorine, and argon has also three energy levels like sodium because all these elements belongs to third period. There are 12 electrons present in magnesium, 13 in aluminium, 14 in silicon, 15 in phosphorus, 16 in sulfur, 17 in chlorine, and 18 electrons in argon. The atomic radius of magnesium atom is 173 ppm.  The atomic radius of aluminium atom is 143 ppm.  The atomic radius of silicon atom is 111 ppm.  The atomic radius of phosphorus atom is 98 ppm.  The atomic radius of sulfur atom is 87 ppm. The atomic radius of chlorine atom is 79 ppm and the atomic radius of argon atom is 71 ppm.

Which substance has the highest boiling point?
Select the correct answer below:

A) ethyl ether
B) ethyl alcohol
C) water
D) these all have the same boiling point

Answers

Answer:

D) these all have the same boiling point

What class of organic product results when 1-heptyne is treated with a mixture of mercuric acetate in aqueous sulfuric acid, and then HOCH2CH2OH with catalytic sulfuric acid

Answers

Answer:

2-methyl-2-pentyl-1,3-dioxolane

Explanation:

In this case, we have two reactions:

First reaction:

1-heptyne + mercuric acetate -------> Compound A

Second reaction:

Compound A + HOCH2CH2OH -------> Compound C

First reaction

In the first reaction, we have as a main functional group a triple bond. We have to remember that mercuric acetate in sulfuric acid will produce a ketone. The carbonyl group (C=O) would be placed in the most substituted carbon of the triplet bond (in this case, carbon 2). With this in mind, we will have as a product: heptan-2-one. (See figure 1).

Second reaction

In this reaction, we have as reagents:

-) Heptan-2-one

-) Ethylene-glycol [tex]HOCH_2CH_2OH[/tex]

-) Sulfuric acid [tex]H_2SO_4[/tex]

When we put ethylene-glycol with a ketone or an aldehyde we will form a cyclic acetal. In this case, this structure would be formed on carbon 2 forming 2-methyl-2-pentyl-1,3-dioxolane. (See figure 2).

I hope it helps!

Based on relative bond strengths, classify these reactions as endothermic (energy absorbed) or exothermic (energy released).
Strongest Bond
A-B
A-A
B-B
C-C
B-C
A-C
1. A2 + C2 rightarrow 2AC
2. B2 + C2 rightarrow 2BC
3. A + BC rightarrow AB + C
4. A2 + B2 rightaarrow 2AB
5. AB + C rightarrow AC + B
a. endothermic
b. exothermic

Answers

Answer:

1. Exothermic 2.  Exothermic 3. Endothermic 4. Endothermic 5. Exothermic.

Explanation:

1. An A-A and a C-C bond results in 2 A-C bonds which are lower than the A-A and C-C bonds so this reaction is exothermic.

2. A B-B bond and a C-C bond results in 2 B-C bonds which are lower than the first 2 bonds so this reaction is also exothermic.

3. There is no bond for single A, a single B-C bond results in a A-B bond and a C molecule. A-B bond is stronger than the B-C bond so the reaction absorbed energy along the way. This shows that it is endothermic.

4. An A-A bond and a B-B bond results in 2 A-B bonds which are stronger than the first two bonds so this reaction is also endothermic.

5. An A-B bond and a C molecule result in an A-C bond and a B molecule. A-C bond is weaker than the A-B bond so there is energy released. This reaction is exothermic.

I hope this answer helps.

At a given temperature the vapor pressures of benzene and toluene are 183 mm Hg and 59.2 mm Hg, respectively. Calculate the total vapor pressure over a solution of benzene and toluene with X Benzene

Answers

CHECK COMPLETE QUESTION BELOW

At a given temperature the vapor pressures of benzene and toluene are 183 mm Hg and 59.2 mm Hg, respectively. Calculate the total vapor pressure over a solution of benzene and toluene with Xbenzene = 0.580.

Answer:

The total vapor pressure is [tex]81.3 mmHg[/tex]

Explanation:

We will be making use of Dalton and Raoults equation in order to calculate the total pressure,

Which is [tex]PT= (PA × XA) +(PB ×XB)[/tex]

PT= total vapor pressure

From the question

Benene's Mole fraction = 0.580

then to get Mole fraction of toluene we will substract the one of benzene from 1. because total mole fraction is always 1.

= (1 - 0.580) = 0.420

Vapor pressure of benzene given = 183 mmHg

Vapor pressure of toluene given= 59.2 mmHg

If we substitute those value into above equation, we have

PT=(183×0.580)+(59.2×0.420)

=81.3mmHg

Therefore,, the total vapor pressure of the solution is 81.3 mmHg

The number of moles of H2O which contains 4g of oxygen?

Answers

Answer:

16G = 1 mole ; then 4G = how many moles? 4/16 = 0.25 mole; That means 4 grams of oxygen is 0.25 moles.

Explanation:

A mole of water molecules contains 2 moles of hydrogen atoms and 1 mole of oxygen atoms.

Methyl iodide reacts irreversibly with azide ion with rate = k[CH3I][N3–]. CH3I(aq) + N3–(aq) → CH3N3(aq) + I–(aq) The reaction is carried out with an initial concentration of CH3I of 0.01 M. Which statement about the reaction is correct?

Answers

Answer:

(D) The reaction cannot take place in a single elementary step

Explanation:

Statements are:

(A) The time it takes for [CH3I] to decrease to 0.005 M is independent of [N3-], as long as [N3] >> [CH3I].

B) If the initial concentrations of azide and CH3I are equal, then it takes half as long for [CH3I] to decrease to 0.005 M as it does for it to decrease from 0.005 M to 0.0025 M.

(C) The reaction rate is significantly smaller if excess I- is added to the solution.

(D) The reaction cannot take place in a single elementary step.

The rate of the reaction is:

rate = k[CH3I][N3–].

That means rate depends of concentration of CH₃I as much as N₃⁻ concentration

(A) The time it takes for [CH3I] to decrease to 0.005 M is independent of [N3-], as long as [N3] >> [CH3I]. FALSE. The reaction rate depends of N₃⁻ as much as CH₃I

B) If the initial concentrations of azide and CH3I are equal, then it takes half as long for [CH3I] to decrease to 0.005 M as it does for it to decrease from 0.005 M to 0.0025 M. FALSE. Reaction is second-order. Half-life is 1/K[A]₀. If initial concentration is 0.1M, to a concentration of 0.005M it takes:

1/K*0.1. If initial concentration is 0.005M it takes 1/K*0.005. That means it takes half to decrease from 0.005M to 0.0025 as it does for it to decrease from 0.01M to 0.005M.

(C) The reaction rate is significantly smaller if excess I- is added to the solution. FALSE. Reaction rate is independent of I⁻

(D) The reaction cannot take place in a single elementary step. TRUE. As this reaction is a single-replacement reaction implies the formation  of 1 C-N bond. But also the rupture of the C-I bond is impossible to explain this kind of reaction in a single elementary step.

2.Which of the alcohols listed below would you expect to react most rapidly with PBr3?A)CH3CH2CH2CH2CH2CH2OHB)(CH3CH2)2CH(OH)CH2CH3C)(CH3CH2)2CHOHCH3D)(CH3CH2)3COHE)(CH3CH2)2C(CH3)OH

Answers

Answer:

A) CH3CH2CH2CH2CH2CH2OH

Explanation:

For this question, we have the following answer options:

A) CH3CH2CH2CH2CH2CH2OH

B) (CH3CH2)2CH(OH)CH2CH3

C) (CH3CH2)2CHOHCH3

D) (CH3CH2)3COH

E) (CH3CH2)2C(CH3)OH

We have to remember the reaction mechanism of the substitution reaction with [tex]PBr_3[/tex]. The idea is to generate a better leaving group in order to add a "Br" atom.

The [tex]PBr_3[/tex] attacks the "OH" generation new a bond to P (O-P bonds are very strong), due to this new bond we will have a better leaving group that can remove the oxygen an allow the attack of the Br atom to generating a new C-Br bond. This is made by an Sn2 reaction. Therefore we will have a faster reaction with primary substrates. In this case, the only primary substrate is molecule A. So, "CH3CH2CH2CH2CH2CH2OH" will react faster.

See figure 1

I hope it helps!

When HCl is added to water, the [H3O+] = 0.6 M. What is the [OH-]?

What is the pH of the solution?

Answers

Answer:

pH=0.22.

[tex][OH^-]=1.66x10^{-14}M[/tex]

Explanation:

Hello,

In this case, since the pH is directly computed from the given concentration of hydronium ions:

[tex]pH=-log([H_3O^+])=-log(0.6M)=0.22[/tex]

It is widely known that the pH and POH are directly related via:

[tex]pH+pOH=14[/tex]

Therefore, the pOH is:

[tex]pOH=14-pH\\\\pOH=14-0.22=13.78[/tex]

Thus, the concentration of hydroxyl ions are computed from the pOH:

[tex]pOH=-log([OH^-]}\\\\[/tex]

[tex][OH^-]=10^{-pOH]=10^{-13.78}[/tex]

[tex][OH^-]=1.66x10^{-14}M[/tex]

Regards.

Because of movements at the Mid-Atlantic Ridge, the Atlantic Ocean widens by about 2.5 centimeters each year. Explain which type of plate boundary causes this motion.

Answers

Answer:

A divergent plate boundary  

Explanation:

At a divergent boundary, the plates pull away from each other and generate new crust.

 

Answer:

Because the ocean becomes larger, this is a divergent plate boundary. Divergent plates cause the ocean floor to expand, making the ocean larger.

Explanation:

PLATO ANSWER

What would happen to the rate of a reaction with rate law rate = k [NO]2[Hz] if
the concentration of NO were doubled?

Answers

The rate of a reaction with this rate law would increase by a factor of 4 if NO concentration were doubled.

Answer:

The rate would have doubled

Explanation:

How many moles of barium sulfate are produced from 0.100 mole of barium chloride?

Answers

Answer:

0.100 moles of barium sulfate are produced from 0.100 moles of barium chloride.

Explanation:

Barium chloride and sodium sulfate react according to the following balanced reaction:

BaCl₂ + Na₂SO₄ → BaSO₄ + 2 NaCl

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of reagent and products participate in the reaction:

BaCl₂: 1 moleNa₂SO₄: 1 moleBaSO₄: 1 moleNaCl : 2 moles

Then you can apply the following rule of three: if 1 mole of BaCl₂ produces 1 mole of BaSO₄, 0.100 mole of BaCl₂ how many moles of BaSO₄ does it produce?

[tex]amount of moles of BaSO_{4} =\frac{0.100 mole of BaCl_{2}* 1 mole of BaSO_{4} }{1 mole of BaCl_{2}}[/tex]

amount of moles of BaSO₄= 0.100

0.100 moles of barium sulfate are produced from 0.100 moles of barium chloride.

Wine goes bad soon after opening because the ethanol dissolved in it reacts with oxygen gas to form water and aqueous acetic acid , the main ingredient in vinegar. Calculate the moles of oxygen needed to produce of acetic acid. Be sure your answer has a unit symbol, if necessary, and round it to significant digits.

Answers

Answer:

0.500 moles of oxygen

Explanation:

Full question says: "Calculate the moles of oxygen needed to produce 0.500 moles of acetic acid.

The reaction of ethanol (C₂H₅OH) with oxygen (O₂) is:

C₂H₅OH + O₂ → CH₃COOH + H₂O

Where 1 mole of ethanol reacts per mole of oxygen to produce 1 mole of acetic acid (CH₃COOH) and 1 mole of water

Based on the chemical equation (1 mole of oxygen produce 1 mole of acetic acid; Ratio 1:1). Thus, if you want to produce 0.500 moles of acetic acid you will need:

0.500 moles of oxygen

Classify the following oxides as acidic, basic, amphoteric, or neutral:
(a) Select the acidic oxides:
A. CO₂
B. N₂O₅
C. Al₂O₃
D. NO
E. SO₃
(b) Select the basic oxides:
A. CaO
B. CO
C. SO₃
D. K₂O
E. BaO
(c ) Select the amphoteric oxides:
A. K₂O
B. Al₂O₃
C. CaO
D. CO₂
E. SnO₂
d) Select the neutral oxides:
A. CO
B. NO
C. SNO₂
D. N₂O₅
E. BaO

Answers

Answer:

(a).

» E. SO₃, sulphur trioxide.

(b).

» A. CaO, Calcium oxide.

» D. K₂O, potassium oxide.

» E. BaO, barium oxide.

(c).

» B. Al₂O₃, Aluminium oxide.

» E. SnO₂, tin (IV) oxide.

(d).

» A. CO, carbon monoxide.

» B. NO, nitrogen monoxide.

Explanation:

[tex]{ \underline{ \sf{ \blue{christ \:† \: alone }}}}[/tex]

Explain your reasoning. Match each explanation to the appropriate blanks in the sentences on the right.
1. the atomic radius decreases
2. the number of gas molecules decreases
3. molar mass and structure complexity decreases
4. structure complexity decreases
5. molar mass decreases
6. each phase (gas, liquid, solid) becomes more ordered
A (I_2(g), Br_2 (g), Cl_2 (g), F_2 (B): The ranking can best be explained by the trend entropy decreases as______.
B (H_2O_2 (g), H_2S(g), H_2O(g): The ranking can best be explained by the decreases a trend entropy decreases as_______.
C. (C(s, amorphous), C(s, graphite), C(s, diamond): The ranking can best be explained by the trend entropy decreases as_______.

Answers

Answer:

A (I_2(g), Br_2 (g), Cl_2 (g), F_2 (B): The ranking can best be explained by the trend entropy decreases as 5. molar mass decreases.

B (H_2O_2 (g), H_2S(g), H_2O(g): The ranking can best be explained by the decreases a trend entropy decreases as 3. molar mass and structure complexity decreases.

C. (C(s, amorphous), C(s, graphite), C(s, diamond): The ranking can best be explained by the trend entropy decreases as 4. structure complexity decreases.

Explanation:

Hello.

In this case, we can understand a higher entropy when more disorder is present and a lower entropy when less disorder is present, thus:

A (I_2(g), Br_2 (g), Cl_2 (g), F_2 (B): The ranking can best be explained by the trend entropy decreases as 5. molar mass decreases since iodine has the greatest molar mass (254 g/mol) and fluorine the least molar mass (38 g/mol).

B (H_2O_2 (g), H_2S(g), H_2O(g): The ranking can best be explained by the decreases a trend entropy decreases as 3. molar mass and structure complexity decreases since hydrogen peroxide weights 34 g/mol as well as hydrogen sulfide but the peroxide has more bonds (more complex, higher entropy).

C. (C(s, amorphous), C(s, graphite), C(s, diamond): The ranking can best be explained by the trend entropy decreases as 4. structure complexity decreases since diamond has a well-ordered structure and amorphous carbon has a very disordered one.

Best regards.

Consider the following reaction at 298K.
I2 (s) + Pb (s) = 2 I- (aq) + Pb2+ (aq)
Which of the following statements are correct?
Choose all that apply.
ΔGo > 0
The reaction is product-favored.
K < 1
Eocell > 0
n = 2 mol electrons
B-

Answers

Answer:

Eªcell > 0; n = 2

Explanation:

The reaction:

I2 (s) + Pb (s) → 2 I- (aq) + Pb2+ (aq)

Is product favored.

A reaction that is product favored has ΔG < 0 (Spontaneous)

K > 1 (Because concentration of products is >>>> concentration reactants).

Eªcell > 0 Because reaction is spontaneous.

And n = 2 electrons because Pb(s) is oxidizing to Pb2+ and I₂ is reducing to I⁻ (2 electrons). Statements that are true are:

Eªcell > 0; n = 2

A man weighs 185 lb. What is his mass in grams?
Please show work.
Thank you

Answers

Answer:

83914.52 grams

Explanation:

Given that,

Weight of a man is 185 lb

We need to find his weight in grams

For this, we must know the relation between lb and grams.

We know that,

1 lb = 453.592 grams

To find the mass of man in grams, the step is :

185 lb = (453.592 × 185) grams

= 83914.52 grams

So, the mass of a man is 83914.52 grams.

Sighting along the C2-C3 bond of 2-methylbutane, the least stable conformation (Newman projection) has a total energy strain of ______kJ/mol

Answers

Answer:

21 KJ/mol

Explanation:

For this question, we have to start with the linear structure of 2-methylbutane. With the linear structure, we can start to propose all the Newman projections keep it in mind that the point of view is between carbons 2 and 3 (see figure 1).

Additionally, we have several energy values for each interaction present in the Newman structures:

-) Methyl-methyl gauche: 3.8 KJ/mol

-) Methyl-H eclipse: 6.0 KJ/mol

-) Methyl-methyl eclipse: 11.0 KJ/mol

-) H-H eclipse: 4.0 KJ/mol

Now, we can calculate the energy for each molecule.

Molecule A

In this molecule, we have 2 Methyl-methyl gauche interactions only, so:

(3.8x2) = 7.6 KJ/mol

Molecule B

In this molecule, we have a Methyl-methyl eclipse interaction a Methyl-H eclipse interaction and an H-H eclipse interaction, so:

(11)+(6)+(4) = 21 KJ/mol

Molecule C

In this molecule, we have 1 Methyl-methyl gauche interaction only, so:

3.8 KJ/mol

Molecule D

In this molecule, we have three Methyl-H eclipse interaction, so:

(6*3) = 18 KJ/mol

Molecule E

In this molecule, we have 1 Methyl-methyl gauche interaction only, so:

3.8 KJ/mol

Molecule F

In this molecule, we have a Methyl-methyl eclipse interaction a Methyl-H eclipse interaction and an H-H eclipse interaction, so:

(11)+(6)+(4) = 21 KJ/mol

The structures with higher energies would be less stable. In this case, structures B and F with an energy value of 21 KJ/mol (see figure 2).

I hope it helps!

What does the third quantum number (m) describe?
A. Which energy level the electron is in
B. What type of orbital the electron is in
C. What direction the electron is spinning
D. The specific orbital within a sublevel

Answers

I think c the best answer

The third quantum number (m) describes the specific orbital within a sublevel. The correct answer is option D.

Quantum numbers are a set of four numbers that describe the properties of an electron in an atom. They specify the energy, position, and orientation of an electron in an atom.

The third quantum number (m) is also called the magnetic quantum number. It describes the orientation of the orbital in space. Each orbital within a sublevel has a different orientation in space, and the magnetic quantum number tells us which orbital we are talking about.

The value of m can range from -l to +l, where l is the second quantum number (the angular momentum quantum number). The value of l determines the number of orbitals in a sublevel, and the value of m determines the specific orbital within that sublevel.

In conclusion, the third quantum number (m) describes the specific orbital within a sublevel. It tells us how the orbital is oriented in space, and it can have values ranging from[tex]\rm -l \ to +l[/tex], where l is the second quantum number.

Option D is the correct answer.

Learn more about quantum number here:

https://brainly.com/question/32116992

#SPJ4

A person has a standing order for 6 doses of acetaminophen in a 24 hour period for fevers
acetaminophen requires what type of reporting?
ut of
Select one:
a. Routine
estion
b. Emergent
C. Certain Time
d. Urgent
4
A change in the type of seizure the person usually experiences that causes them to stop brea

Answers

Acetaminophen is used to treat patients with aches and fever. It is safe to give a dose of 4 grams in 24 hours period.

It is an effective medicine which does not require much time to provide relief to the patient.

Usually tylenol is the best medicine which consists of Acetaminophen in appropriate quantity to provide relief from pain and fever.

The reporting for the doses introduced to the patient must be routine since it is not an emergency case.

The correct answer is a. Routine

Learn more at https://brainly.com/question/24335381

If you had a cup full of methanol and a pool full of methanol, would the mass change?

Answers

Answer:

the mass does not change

Explanation:

Explain the working principle of a clinical thermometer ​

Answers

Answer:

A clinical thermometer is a thermometer used to measure human body temperature. Most made in the 20th century are mercury-in-glass thermometers. They are accurate and sensitive, having a narrow place where the mercury level rises very fast. A kink in the tube stops the mercury level from falling on its own.

An equilibrium mixture of N2, H2, and NH3 at 700 K contains 0.036 M N2 and 0.15 M H2. At this temperature, Kc for the reaction N2(g) + 3 H2(g)<=> 2NH3(g) is 0.29.
What is the concentration of NH3?

Answers

Answer:

5.94×10¯³

Explanation:

The following data were obtained from the question:

Concentration of N2, [N2] = 0.036 M

Concentration of H2, [H2] = 0.15 M

Equilibrium constant (Kc) = 0.29 M

Concentration of NH3, [NH3] =.…?

The equation for the reaction is given below:

N2(g) + 3H2(g) <=> 2NH3(g)

Thus, we can determine the concentration of NH3 by using the equilibrium expression for the reaction.

This is illustrated below:

The equilibrium constant for a reaction is simply defined as the ratio of the concentration of the product raised to their coefficient to the concentration of the reactant raised to their coefficient.

The equilibrium constant for the above equation is given below:

Kc = [NH3]² / [N2] [H2]³

Inputting the value of Kc, [N2], and [H2] the value of [NH3]can be obtained as follow:

Kc = [NH3]² / [N2] [H2]³

0.29 = [NH3]²/ 0.036 × 0.15³

Cross multiply

[NH3]² = 0.29 × 0.036 × 0.15³

[NH3]² = 3.5235×10¯⁵

Take the square root of both side

[NH3] = √(3.5235×10¯⁵)

[NH3] = 5.94×10¯³

Therefore, the concentration of NH3, [NH3] is 5.94×10¯³ M.

Please tell the answer​

Answers

Answer:

see the photo

Explanation:

it was the answer

What is the pOH of a solution at 25.0∘C with [H3O+]=4.8×10−6 M?

Answers

Answer:

8.68

Explanation:

pOH = 8.68

all you need is contained in the sheet

Answer:

Approximately [tex]8.68[/tex].

Explanation:

The [tex]\rm pOH[/tex] of a solution can be found from the hydroxide ion concentration [tex]\rm \left[OH^{-}\right][/tex] with the following equation:

[tex]\displaystyle \rm pOH = -\log_{10} \rm \left[OH^{-}\right][/tex].

On the other hand, the ion-product constant of water, [tex]K_{\text{w}}[/tex], relates the hydroxide ion concentration [tex]\rm \left[OH^{-}\right][/tex] of a solution to its hydronium ion concentration [tex]\rm \left[{H_3O}^{+}\right][/tex]:

[tex]K_\text{w} = \rm \left[{H_3O}^{+}\right] \cdot \rm \left[OH^{-}\right][/tex].

At [tex]25 \; ^\circ \rm C[/tex], [tex]K_{\text{w}} \approx 1.0 \times 10^{-14}[/tex]. For this particular [tex]25 \; ^\circ \rm C[/tex] solution, [tex]\rm \left[{H_3O}^{+}\right] = 4.8 \times 10^{-6}\; \rm mol \cdot L^{-1}[/tex].

Hence the [tex]\rm \left[OH^{-}\right][/tex] of this solution:

[tex]\begin{aligned}\left[\mathrm{OH}^{-}\right] &= \frac{K_\text{w}}{\rm \left[{H_3O}^{+}\right]} \\ &= \frac{1.0 \times 10^{-14}}{4.8 \times 10^{-6}}\; \rm mol\cdot L^{-1} \approx 2.08333 \times 10^{-9}\; \rm mol\cdot L^{-1}\end{aligned}[/tex].

Therefore, the [tex]\rm pOH[/tex] of this solution would be:

[tex]\begin{aligned}\rm pOH &= -\log_{10} \rm \left[OH^{-}\right] \\ &\approx -\log_{10} \left(4.8 \times 10^{-6}\right) \approx 8.68\end{aligned}[/tex].

Note that by convention, the number of decimal places in [tex]\rm pOH[/tex] should be the same as the number of significant figures in [tex]\rm \left[OH^{-}\right][/tex].

For example, because the [tex]\rm \left[{H_3O}^{+}\right][/tex] from the question has two significant figures, the [tex]\rm \left[OH^{-}\right][/tex] here also has two significant figures. As a result, the [tex]\rm pOH[/tex] in the result should have two decimal places.

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