When Hg2+ concentration is 6.35x10^-4 M, the observed cell potential at 298K for an electrochemical cell with the following reaction is 1.499V. What is the Cr3+ concentration?

3H^2+ (aq) + 2Cr(s)= 3Hg(l) + 2Cr^3+(aq)

Answer:  Groundwater and surface water are connected. When groundwater is overused, the lakes, streams, and rivers connected to groundwater can also have their supply diminished. Land subsidence occurs when there is a loss of support below ground. This is most often caused by human activities, mainly from the overuse of groundwater, when the soil collapses, compacts, and drops.

Hope it Helps. :)

Answer:

3

Explanation:

The electron shells are labelled as K,L,M,N,O,P, and Q or 1,2,3,4,5,6, and 7.

As we go from innermost shell outwards, this number denotes the number of subshell in the shell. Electrons in outer shells have higher average energy and travel farther from the nucleus than those in inner shells.

Hence, M shell contains s,p and d subshells.

Answer:

rip there isnt a photo

Explanation:

i do know a lot about cells tho lol

Van der Waals forces hold molecules together by inducing temporary dipoles that attract each other. That is option C

Van Der Waals forces are example of those intermolecular forces which are weaker than ionic and covalent bonds that exists between molecules.

Van Der Waals forces was postulated by a Dutch physicist known as Van Der Waals. He postulated the existence of weak, short-range forces of attraction, which are independent of normal bonding forces, between non-polar molecules. He came to this conclusion after studying the  of real gases at low temperatures and high pressures that:

This is how weak Van Der Waals forces are set up. Therefore, option C is CORRECT

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Explanation:

Add water, Na2CO3 dissolves, filter, PbCO3 stays in the paper and dissolved Na2CO3 goes through as the solution. Dry the PbCO3 and you have the dry solid.

OR

Add water to dissolve then filter to obtain PbCo3 as you're residue and Na2Co3 as the filtrate. Dry the insoluble PbCo3 between filter papers and you obtain solid PbCo3

Answer:

Carbohydrates, proteins, and fats are biological macromolecules that are made up of chemical elements which are inherent to chemistry.

Chemistry explain how these macromolecules are bonded together at the molecular level and give an explanation for their behavior.

Explanation:

Answer :

volume of a gas = weight * 22.4 l / gram molecular weight

volume of o2 = ?

weight given = 20.5 g

gram molecular weight of oxygen = 32 (because of 2 oxygen atoms )

volume of oxygen = 20.5 * 22.4 / 32

volume of oxygen = 14.35 liters  

Explanation:

hope this helps you

if wrong just correct me

Answer:

0.0095 moles of acid were neutralized by the antiacid

Explanation:

The antiacid is a base that neutralize the acid in stomach. To find the moles of acid neutralized we need to find the moles of acid added initially. This acid is added in excess, then, the moles of NaOH added reacts to neutralize the moles of acid in excess. The difference between initial moles of HCl and moles of NaOH needed to titrate the excess = Moles of HCl that were neturalized by the antiacid as follows:

Moles HCl added:

42.00mL = 0.04200L * (0.250mol/L) = 0.0105 moles HCl

Moles NaOH to titrate the excess:

10.00mL = 0.01000L * (0.10mol/L) = 0.0010 moles NaOH = Moles HCl in excess.

Moles of acid that were neutralized:

0.0105 moles - 0.0010 moles =

Answer:

B. Bohr’s model electrons cannot exist between orbits, but in the electron cloud model, the location of the electrons cannot be predicted.

AND

C. The modern model explains all available data about atoms; Bohr’s model does not.

Explanation:

The answers are right on Edge. :)

Answer:

b and c

Explanation:

my assignment was 100% 2022

The number of replaceable electrons in an atom is called its valency.

Examples

Thanks !

☺️☺️☺️☺️☺️☺️☺️

Answer:

the combining capacity if an atom is know as valency.

the property of an element that determines the number of other atimd with an aton if the element can combine.

Answer:

A, B, C and D

Explanation:

It can result for all of the choices mentioned.

Not following safety and procedure for handling hazardous material results in illness, death, injury, and HIPAA violation. Thus, all options are correct.

The hazardous material safety policy and measures are the standards set by HIPAA for the safety and precautionary measures that have been followed for reducing personal risk.

The hazardous materials have been chemicals, gases, flammables, and explosives. The inappropriate handling and not following the standard procedure results in illness, injury, death, and HIPAA violation. Thus, all the options are correct.

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Answer:

The bond between sodium sulfate is an ionic bond since it's a bond between a metal and non metals however the bond between sulfur and oxygen is a covalent bond since the two are non metals and the other reason that makes this an ionic bond is that there is both losing and gaining of electrons..

I hope this helps

Answer:

a. Decrease

b. Increase

c. Increase

d. No effect

Explanation:

Glycolysis is present in muscle cells which converts glucose to pyruvate, water and NADH. It produces two molecules of ATP. Cellular respiration produces more molecules of ATP from pyruvate in mitochondria. Glycolysis increases in pyruvate kinase.

a. Loss of binding site for fructose 1,6-bisphosphate in pyruvate kinase: Decrease

b. Loss of allosteric binding site for ATP in pyruvate kinase: No effect

c. Loss of allosteric binding site for AMP in phosphofructokinase: Increase

d. Loss of regulatory binding site for ATP in phosphofructokinase: Increase

A. An important substrate in the glycolysis pathway is fructose 1,6-bisphosphate. It stimulates pyruvate kinase, an essential enzyme in glycolysis. The amount of pyruvate kinase that is activated will decrease if the fructose 1,6-bisphosphate binding site in pyruvate kinase is eliminated. As a result the rate of glycolysis in the muscle cells will probably decrease.

B. The allosteric ATP binding site of pyruvate kinase controls how active the enzyme is. However, pyruvate kinase is not significantly regulated by ATP in muscle cells. Therefore, it is unlikely that deletion of the ATP-binding allosteric site in pyruvate kinase would have no effect on the rate of glycolysis in muscle cells.

C. The rate-limiting enzyme in glycolysis, phosphofructokinase, is activated from all forms by AMP. It increases the rate of glycolysis by stimulating the activity of phosphofructokinase. If the allosteric binding site for AMP is eliminated, phosphofructokinase activation will be reduced. As a result, the rate of glycolysis in muscle cells will decrease.

D. Phosphofructokinase is inhibited allosterically by ATP. It regulates the rate of glycolysis by a feedback mechanism. High ATP concentrations cause phosphofructokinase to bind to its regulatory site, limiting its activity and delaying glycolysis. If the regulatory binding site for ATP is eliminated, the inhibitory action of ATP on phosphofructokinase would be lost. As a result, muscle cells will glycolysis at a faster rate.

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Answer:

b

Explanation:

deposition is when water turns from gas to solid. b is the only one that fits

Deposition is frost forms when cold evening temperatures convert the humidity in the air to thin layers of ice on the ground.

Deposition is a process that involves collection of large mass or when mean distance between molecules are reduced. It can also be explained as gathering of substances together to form a larger mass.

Therefore, the phenomenon explained in the given example about deposition is frost forms when cold evening temperatures convert the humidity in the air to thin layers of ice on the ground.

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Answer:

1 cell wall

2 yes

3 chloroplast

4 photosynthesis

5 Deoxyribonucleic acid (I believe)

hope this helped a little and pls mark brainiest if it did :)

Explanation:

The cell wall is a rigid layer that is found outside the cell membrane and surrounds the cell, providing structural support and protection.

Solution :

A substrate is defined as the chemical species that are being observed in the chemical reaction where the substrate reacts with a reagent and forms a product. It can also be referred to the surface where some other chemical reactions are performed.

Stereochemistry is defined as the study of relative spatial arrangement of the atoms which forms the structure of the molecules and their respective manipulations.

In the context, the products including the stereochemical designations for any chirality centers starting from the  (R)-3-methylhex-1-yne as the substrate are attached below.  

Explanation:

here's the answer to your question about

Answer:

how many mols of nh4cl are present in 279.0ml of a 0.975 M nh4cl soution?

Answer:

0.2M H2C6H5O7 < 0.2M H2C2O4

Explanation:

A weak acid/base ionizes to a very small extent in water. Hence, if we say that a substance is a weak acid/base, its percentage of ionization in solution is very little.

More volume of a very weak acid is required to neutralize a strong base. Since NaOH is a strong base, the weaker acid among the duo will require more volume for neutralization.

Since H2C6H5O7 is a weaker acid than H2C2O4, equal concentration of the both acids will require less volume of H2C2O4 than H2C6H5O7 to neutralize 0.50 M NaOH.

H₂C₆H₅O₇ is a weaker acid than H₂C₂O₄, and will require the least volume of 0.50 M NaOH to be neutralized.

H₂C₆H₅O₇ < H₂C₂O₄

The strength of an acid is related to the value of its dissociation constant, Ka or its pKa (negative logarithm of Ka)

Strong acids have high Ka values or low pKa value, whereas weak acids have low Ka values and high pKa values.

Between two acids, the acid with a higher Ka or lower pKa values is the stronger acid.

Acids are classified as either strong or weak depending on how well it ionizes in solution to produce hydrogen ions.

Strong acids ionizes completely to produce hydrogen ions.

Weak acid ionizes partially to a varying degrees in water to produce hydrogen ions.

In neutralization reactions between acids and bases, stronger acids will require the most volume of base or alkali in order to be neutralized.

H₂C₂O₄ has a Ka value of 5.9 x 10⁻² and a pKa value of 1.23

H₂C₆H₅O₇ has a Ka value of 8.4 x 10⁻⁴ and a pKa value of 3.08

Hence H₂C₂O₄ is a stronger acid than H₂C₆H₅O₇

For equal molar concentrations of the two acids, H₂C₂O₄ will produce more hydrogen ions than H₂C₆H₅O₇, and thus, will require more volume base (0.50 M NaOH) to be neutralized.

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Answer:

See Explanations

Explanation:

pH =-log[H₃O⁺] = -log[H⁺]

pOH = -log[OH⁻]

For weak acids [H⁺] = SqrRt(Ka·[Acid])

For weak bases [OH⁻] = SqrRt(Kb·[Base])

pH + pOH = 14

__________________________________________

A. Given 0.18M CH₃NH₂; Kb = (4.4 x 10⁻⁴)* => pH = 11.95

CH₃NH₂ + H₂O => CH₃NH₃OH ⇄ CH₃NH₃⁺ + OH⁻;

[OH⁻]  = SqrRt(Kb·[weak base]) = SqrRt(4.4 x 10⁻⁴ x 0.18)M = 8.97 x 10⁻³M

=> pOH = -log[OH⁻] = -log(8.93x10⁻³) = -(-2.05) = 2.05

=> pH = 14 - pOH = 14 - 2.05 = 11.95.

*Kb values for most ammonia derivatives in water can be found online by searching 'Kb-values for weak bases'. Kb-values for methyl amine and methylammonium chloride are both 4.4x10⁻⁴.

___________________________________________________

B. Given 0.18M CH₃NH₃Cl

In water ... CH₃NH₃Cl => CH₃NH₃⁺ + Cl⁻; Kb(CH₃NH₃Cl) = 4.4 x 10⁻⁴

Cl⁻ + H₂O => No Rxn (i.e.; no hydrolysis occurs) ... Cl⁻ does not react with H₂O.

Hydrolysis Reaction of Methylammonium Ion:

CH₃NH₃⁺ + H₂O => CH₃NH₄OH ⇄ CH₃NH₄⁺ + OH⁻

Ka' x Kb = Kw => Ka' = Kw/Kb = 10⁻¹⁴/4.4 x 10⁻⁴ = 2.27 x 10⁻¹¹                                   Ka' = [CH₃NH₄⁺][OH⁻]/[CH₃NH₄OH] = (x)(x)/(0.18M) = (x²/0.18M) = 2.27 x 10⁻¹¹ => x = [OH⁻] = SqrRt(2.27x10⁻¹¹ x 0.18)M = 2.02 x 10⁻⁶M => pOH = -log(2.02 x 10⁻⁶) = -(-5.69) = 5.69 => pH = 14 - pOH = 14 - 5.69 = 8.31.

*note => the general nature of halide interactions would increase acidity (lower pH) of the halogenated compound.

C. A mixture of 0.18M CH₃NH₂ and 0.18M CH₃NH₃Cl          

Mixture of 0.18M CH₃NH₂ + 0.18M CH₃NH₃Cl

In Water ...

=> 0.18M CH₃NH₃OH + 0.18M CH₃NH₃Cl

=> 0.18M CH₃NH₃⁺ + 0.1M OH⁻ + 0.18M CH₃NH₃⁺ + 0.18M Cl⁻

=> 0.36M CH₃NH₃⁺ + 0.18M OH⁻ + 0.18M Cl⁻

-----------------------------------------------------------

Ka'(CH₃NH₃⁺) x Kb(CH₃NH₂) = Kw => Ka'(CH₃NH₃⁺) = Kw/Kb(CH₃NH₂)

=> Ka'(CH₃NH₃⁺) = (10⁻¹⁴/4.4x10⁻⁴) = 2.27x10⁻¹¹

----------------------------------------------------------

From the 0.36M CH₃NH₃⁺

=>       CH₃NH₃⁺ + H₂O  ⇄ CH₃NH₄⁺ + OH⁻

C(eq)   0.36M        ----              x             x     (<= at equilibrium after mixing)

Ka'(CH₃NH₃⁺) = [CH₃NH₄⁺][OH⁻]/[CH₃NH₃⁺] = x²/(0.36M)

=> x = [OH⁻] = SqrRt(Ka'(CH₃NH₃⁺)·0.36M) = SqrRt(2.27x10⁻¹¹/0.36) = 0.0126M

=> Total [OH⁻] = 0.0126M + 0.18M = 0.1926M from hydrolysis process

=> final solution mix is therefore, 0.1926M in OH⁻ + 0.18M in Cl⁻

--------------------------------------------------------

∴pH = -log[H⁺] = -log(5.192x10⁻¹⁴) = -(-13.29) = 13.29 for solution mix

The acid and base dissociation constant and the 0.18 M of CH₃NH₂ and

CH₃NH₃Cl and the mixture give the following approximate values;

A. The pH value of the 0.18 M CH₃NH₂ is 11.93

B. The pH value of the 0.18 M CH₃NH₃Cl is 5.69

C. The pH value of the mixture is 10.644

A. 0.18 M CH₃NH₂

The solution is presented as follows;

CH₃NH₂ + H₂O → CH₃NH₃⁺ + OH⁻

Let x represent the number of moles of CH₃NH₃⁺ and OH⁻ produced, we

have;

The number of moles of CH₃NH₂ remaining = 0.18 - x

Which gives;

[tex]K_b = \mathbf{\dfrac{[CH_3NH_3^+][OH^-]}{[CH_3NH_2]}}[/tex]

[tex]K_b[/tex] for CH₃NH₂ = 4.167 × 10⁻⁴

Therefore;

[tex]4.167 \times 10^{-4} = \mathbf{\dfrac{x \times x}{0.18 - x}}[/tex]

4.167 × 10⁻⁴ × (0.18 - x) = x²

4.167 × 10⁻⁴ × (0.18 - x) - x² = 0

Which gives;

x = [OH⁻] = 8.455 × 10⁻³

pH = 14 + log[OH⁻]

Which gives;

pH = 14 + log(8.455 × 10⁻³) ≈ 11.93

B.  0.18 M CH₃NH₃Cl

The solution is presented as follows;

CH₃NH₃⁺ → CH₃NH₂ + H⁺

Let x represent the number of moles of CH₃NH₂ and H⁺ produced,

respectively, we have;

The number of moles of CH₃NH₃⁺ remaining = 0.18 - x

Which gives;

[tex]K_a = \mathbf{\dfrac{[CH_3NH_2][H^+]}{[CH_3NH_3^+]}}[/tex]

Kₐ for CH₃NH₃Cl = 2.27 × 10⁻¹¹

Therefore;

[tex]2.27\times 10^{-11} = \dfrac{x \times x}{0.18 - x}[/tex]

2.27 × 10⁻¹¹ × (0.18 - x) = x²

2.27 × 10⁻¹¹ × (0.18 - x) - x² = 0

Which gives;

x = [H⁺] ≈ 2.02 × 10⁻⁶

pH = -log[H⁺]

Which gives;

pH = -log(2.02 × 10⁻⁶) ≈ 5.69

C. For the mixture of 0.18 M CH₃NH₂ and 0.18 M of CH₃NH₃Cl, we have;

Based on the Henderson-Hasselbalch equation, we have;

[tex]pH = \mathbf{ pKa + log\dfrac{[Conjugate \ base]}{[acid ]}}[/tex]

Which gives;

[tex]pH = -log\left(2.27 \times 10^{-11} \right)+ log\dfrac{0.18}{0.18} \approx \underline{10.644}[/tex]

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Answer:

The 3d and 4s orbitals are completely filled, and the 4p orbital is partially filled.

Explanation:

The correct answer is: The 3d and 4s orbitals are completely filled, and the 4p orbital is partially filled.

The d orbital contains 10 electrons, the s orbital takes 2 electrons and the p orbital takes six electrons.

The orbital in chemistry is defined as a region in space where there is a high probability of finding an electron. There are s, p, d, f orbitals in chemistry which correspond to sharp, principal, diffuse and fundamental.

The electronic configuration of arsenic is 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p3.

From this electronic configuration, we can see that the 4s and 3d orbitals are half filled while the 4p orbital is half filled.

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Answer:

boiling point elevation - colligative property

color - non-colligative property

freezing point depression - colligative property

vapor pressure lowering - colligative property

density - non-colligative property

Explanation:

A colligative property is a property that depends on the number of particles present in the system.

Freezing point depression, boiling point elevation and vapour pressure lowering are all colligative properties of solutions.

Colour and density do not depend on the number of particles present hence they are not colligative properties.

The boiling point elevation, freezing point depression, and vapor pressure lowering are colligative properties. And color and density are non-colligative properties.

Explanation:

So, from this, we can conclude that boiling point elevation, freezing point depression, and vapor pressure lowering are colligative properties. And color and density are non-colligative properties.

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Answer:

State functions represent quantities or properties of a thermodynamic system, while non-state functions represent a process during which the state functions change. For example, the state function PV is proportional to the internal energy of an ideal gas, but the work W is the amount of energy transferred as the system performs work.

Explanation:

Answer:

0.1309 mol

Explanation:

From the given information:

The metal ion, two ions of [tex]M^{+}[/tex] reacted with Cl⁻ to form [tex]MCl_n[/tex] i.e. the compound formed is [tex]MCl_2[/tex].

The concentration of the metal ion formed [tex][M^+][/tex] = 8.279 M

The concentration of the chlorine ion formed [tex][Cl^-][/tex] = 2 × 8.279 M

= 16.558 M

∴

We know that:

[tex]\mathsf{Molarity = \dfrac{no \ of \ moles }{volume (mL)}}[/tex]

The number of moles of [tex][Cl^-][/tex] = [tex]16.558 \ mol.L^{-1} \times 7.903 \ mL \times \dfrac{1 \ L}{1000 \ mL}[/tex]

= 0.1309 mol

Answer:

B

Explanation:

b/c copper is readuction agent

The most likely redox reaction that would occur if silver and copper metal were added to a solution that contained silver and copper ions is   [tex]\rm Cu^{2+} + 2Ag \rightarrow Cu + 2Ag^+[/tex]. The correct answer is option C.

Redox reaction is a reaction in which reduction and oxidation takes place simultaneously.

In this reaction:

[tex]\rm Cu^{2+} + 2Ag \rightarrow Cu + 2Ag^+[/tex]

Copper metal has a higher reduction potential than silver metal, which means that it will be oxidized to [tex]\rm Cu^{2+}[/tex] ions before silver metal is oxidized to  [tex]\rm Ag^+[/tex] ions.

The [tex]\rm Cu^{2+}[/tex] ions in the solution will then react with the silver metal to form [tex]\rm Ag^+[/tex] ions and Copper metal. This reaction is an example of a displacement reaction, where a more reactive metal removes a less reactive metal from its compound.

Therefore, option C. [tex]\rm Cu^{2+} + 2Ag \rightarrow Cu + 2Ag^+[/tex] is the correct answer.

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Explanation:

From the question given above, the following data were obtained:

Half-life (t½) = 9.7 days

Time (t) = 60 days

Next, we shall determine the number of half-lives that has elapsed. This can be obtained as follow:

Half-life (t½) = 9.7 days

Time (t) = 60 days

Number of half-lives (n) =?

n = t / t½

Finally, we shall determine the percentage remaining. This can be obtained as follow:

Let the original amount be N₀

Let the amount remaining be N

Number of half-lives (n) = 60 / 9.7

N = N₀ / 2ⁿ

Divide both side by N₀

N/N₀ = 1/2ⁿ

N/N₀ = 1 / 2⁽⁶⁰÷⁹•⁷⁾

N/N₀ = 0.0137

Multiply by 100 to express in percentage

N/N₀ = 0.0137 × 100

Therefore, the percentage remaining after 60 days is 1.37%

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Answer:

Molarity is 0.075 M.

Explanation:

Moles:

[tex]{ \tt{ = \frac{65.5}{RFM} }}[/tex]

RFM of potassium sulphate :

[tex]{ \tt{ = (39 \times 2) + 32 + (16 \times 4)}} \\ = 174 \: g[/tex]

substitute:

[tex]{ \tt{moles = \frac{65.5}{174} = 0.376 \: moles}}[/tex]

In volume of 5.00 l:

[tex]{ \tt{5.00 \: l = 0.376 \: moles}} \\ { \tt{1 \: l = ( \frac{0.376}{5.00} ) \: moles}} \\ { \tt{molarity = 0.075 \: mol \: l {}^{ - 1} }}[/tex]

The structure of amylase deteriorates due to high temperature of the solution.

This experiment shows that the structure of amylase deteriorates due to high temperature which prevents this amylase from performing its function properly.

At high temperatures the amylase will break starch down very slowly or not at all due to denaturation of the enzyme's active site due to which it can't perform its function properly so we can conclude that high temperature denatures amylase enzyme.

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Nakula's conclusion was- "My results show that boiling destroys amylase"

Amylase is an enzyme that breaks down starch molecules into sugar molecules. He boiled the solution in tube P, and when he checked tube P for sugar, there wasn't any. He didn't boil the solution in tube Q and he found sugar in it.

Amylase had broken down starch molecules to sugar molecules in tube Q. Tube P's solution had been boiled, and this showed that when he boiled it, it destroyed the amylase, that is why the starch molecules hadn't been broken down into sugar molecules.

Answer:

d

Explanation:

since it is much convenient since the email will not get lost and it's contents will not be forgotten