What does lowercase v represent in this formula?

Answer:

See explanation

Explanation:

The work function of a metal is defined as that minimum energy which is required to remove one electron from the surface of a metal when it is irradiated with a photon of light. The work function is different for different metals.

The ionization energy of a metal is the energy required to remove an electron from an atom. It depends on the position of the electron within the atom.

The work function specifically refers to the energy required to remove an electron from the conduction band of a metal. Hence, the work function is always lower than the ionization energy.

a) HNO₂ + C₂H₇NO → N₂ + C₂H₆O + H₂O

b) H₃O⁺ + F⁻ → HF + H₂O

[tex]\large\color{lime}\boxed{\colorbox{black}{Answer : - }}[/tex]

a) HNO₂ + C₂H₇NO → N₂ + C₂H₆O + H₂O

b) H₃O⁺ + F⁻ → HF + H₂O

I think the answer most be d

In chemistry like dissolves like hence hexane will dissolve in gasoline.

Dissolution stems from intermolecular interaction between solute and solvent molecules.

If this interaction can not occur, dissolution of one substance in another is impossible.

Hexane dissolves in gasoline because the both substances are non-polar and can interact with each other effectively.

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Answer:

B. 0.0000005461m

I used the method of moving the decimal.

Complete Question  

Complete Question is attached below

Answer:

Option A

[tex]D=A[/tex] And [tex]C>A[/tex]

Explanation:

From the question, we are told that:

The Chemical Reaction

 [tex]2A_{aq}+B_{aq} \leftrightarrow 3C_{aq}+2D_{aq}[/tex]

Generally, the equation for Equilibrium constant is mathematically given by

 [tex]K=\frac{C^c*D^d}{A^a*B^b}[/tex]

Therefore

 [tex]K=\frac{C^3*D^d}{A^2*B^b}[/tex]

Hence we conculde

 [tex]D=A[/tex] And [tex]C>A[/tex]

Therefore Option A

Answer:

A. filtration

Hope it helps

Answer:

[SO4²⁻] = 0.015M

Explanation:

When H2SO4 is dissolved in water, HSO4- is produced in a direct reaction as follows:

H2SO4 → HSO4- + H+

As 1 mole of H2SO4 produce 1 mole of HSO4-, the molarity of HSO4- in this first reaction is 0.034M

Now, the HSO4- is in equilibrium with SO42- and H+ as follows:

HSO4⁻ ⇄ SO4²⁻ + H⁺

Where the equilibrium constant, K, is defined as:

K = 1.2x10⁻² = [SO4²⁻] [H⁺] / [HSO4⁻]

Where [] are the equilibrium concentrations of each species in the reaction.

The equilibrium concentrations are:

[SO4²⁻] = X

[H⁺] = X

[HSO4⁻] = 0.034M - X

Where X is reaction coordinate

Replacing:

1.2x10⁻² = [X] [X] / [0.034-X]

4.08x10⁻⁴ - 1.2x10⁻²X = X²

4.08x10⁻⁴ - 1.2x10⁻²X - X² = 0

Solving for X:

X = -0.027M. False solution, there are no negative concentrations.

X = 0.015M. Right solution.

That means the equilibrium molarity of SO4²⁻,

[SO4²⁻] = X

From each of the protein structures listed, the option with the highest level of protein structure as regards with order of amino acids and overall macromolecule structure is quaternary structure. That is option D.

The protein one of the essential nutrients that is found in and consumed by mammals.

There are different types of proteins and their functions depends on their shape, structure or conformation.

The structure of proteins include:

Therefore, the option with the highest level of protein structure as regards with order of amino acids and overall macromolecule structure is quaternary structure.

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Answer:

IUPAC Name:

N-ethyl-N-methylaniline

Common Name:

Benzenamine

Answer:

pH = 3.02

Explanation:

Acetic Acid is a weak acid (HOAc) that ionizes only ~1.5% as follows:

HOAc ⇄ H⁺ + OAc⁻.

In pure water the hydronium ion concentration [H⁺] equals the acetate ion concentration [OAc⁻] and can be determined* using the formula [H⁺] = [OAc⁻] = SqrRt(Ka·[acid]) = SqrRt(1.8x10⁻⁵ x 0.0500)M = 9.5x10⁻⁴M.

By definition, pH = -log[H⁺] = -log(9.5x10⁻⁴) = 3.02

______________________________________________________

*This formula can be used to determine the [H⁺] & [Anion⁻] concentrations for any weak acid in pure water given its Ka-value and the molar concentration of acid in solution.

Answer:

H-C = C-H NaNH2 [tex]\ \to \0}[/tex] H-C = CNa

H-C = C - CH2 CH2 CH2 CH3

Explanation:

NaNH2 acts as base in this reaction. The organic products released after the reaction of carbon hydrogen atom with sodium amide. These products released after the chemical reaction when carbon and hydrogen atom reacts and NaNH2 acts as base then substitution nucleophilic reaction takes place.

Answer:

a) E2

b) SN2

c) SN2

Explanation:

A substitution reaction involves replacement of an atom or group in a molecule by another atom or group. An elimination reaction is the loss of two atoms from the same molecule leading to the formation of a multiple bond in the molecule.

We must note that primary alkyl halides never undergo SN1/E1 reactions. However, the presence of a strong bulky base such as tert BuO- , E2 reactions predominate. In the presence of strong bases such as OH^- and good nucleophiles such as I^-, SN2 mechanism predominates.

Answer:

3.310308*10^26

Explanation:

nO=9nFe2(SO3)3=9*60.1=540.9 moles

number of atoms: 540.9*6.02*10^23

Answer:

23592U+10n→14454Xe+9038Sr+210n

Explanation:

a nuclear reaction for the neutron-induced fission of U−235 to form Xe−144 and Sr−90.

Explanation:

here's the answer to your question

2.0 L

Answer:

Solution given:

no. of mole(n)=3.0mole

molarity(M)=1.5M

we have

Volume of a substance is equal to the ratio of its no of mole to its molarity.

By this we get a relation:

Volume=no.of mole/molarity

substituting value

Volume=3.0/1.5=2

The required volume is 2 litre.

If K is greater than 1, then products are favored

The mass of radon-222 that will remain after 15.2 days given that it was originally 6 g is 0.375 g (Option B)

This is the time taken for half a substance to decay.

We'll begin our calculation by calculating the number of half-lives that has elapsed after 15.2 days. This is illustrated below:

n = t / t½

n = 15.2 / 3.8

n = 4

Thus, 4 half-lives has elapsed.

The amount of radon-222 remaining can be obtained as illustrated below:

N = N₀ / 2ⁿ

N = 6 / 2⁴

N = 6 / 16

N = 0.375 g

Thus, the amount of radon-222 remaining after 15.2 days is 0.375 g

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Answer:

An atom is the smallest particle of an element that can take part in chemical reaction.

Explanation:

hope it will help u Amri

Answer:

d

Explanation:

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

We measured the Fe(II) reduction of one of the Co(III) complexes by water at a rate of about 0.545 kJ/mol (to three significant figures).

Calculating a Reaction's Activation Energy A reaction's rate is influenced by the temperature at which it is carried out. The molecules travel more quickly and clash more frequently as the temperature rises. Moreover, the molecules contain greater kinetic energy.

We can use the Arrhenius equation to calculate the reaction's activation energy:

k = A × exp(-Ea/RT)

When the activation energy Ea, the rate constant k, the gas constant R, and the temperature T in Kelvin are all present.

Finding the natural logarithm of the equation's two sides results in:

ln(k) = ln(A) - (Ea/RT)

This equation can be rearranged to take a linear form:

ln(k) = (-Ea/R) × (1/T) + ln(A)

y = mx + b, where (1/T) is x, (-Ea/R) is the slope, and ln(A) is the y-intercept, has the form of a linear equation.

We can get the slope of the line using the given data:

slope = (-Ea/R) = (ln(k2/k1)) / (1/T2 - 1/T1)

where the rate constants for temperatures T1 and T2, respectively, are k1 and k2.

substituting the specified values:

k1 = 0.054s⁻¹ at 293 K

k2 = 0.100s⁻¹ at 298 K

T1 = 293 K

T2 = 298 K

slope = (-Ea/R)

= (ln(0.100/0.054)) / (1/298 - 1/293)

= 65.5 kJ/mol

Therefore, the activation energy of the reaction is:

Ea = slope * R = 65.5 kJ/mol × 8.314 J/mol-K = 545 J/mol

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An indicator usually signals the endpoint of a neutralization reaction by undergoing a color change. They aid in discovering the point of equivalence of a titration.

The kind of indicator used depends on the nature of acid/base reacted.

In the case of  CH3NH2 with HBr which  strong acid and weak base titration, suitable indicators include; bromophenol blue, methyl  orange, methyl red, and chlorophenol blue.

In the case of HNO3 with NaOH, this is a strong acid, strong base titration hence phenolphthalein, methyl red, chlorophenol, and bromothymol blue cresol red blue are suitable indicators.

In the case of  HNO2 with KOH, this a weak acid, strong base titration and the suitable indicators are cresol red and  phenolphthalein.  

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Answer:

Fe(NO3)3, Cr(NO3)3, Co(NO3)3

Explanation:

According to the question, no precipitate is observed when HCl was added. This means that we must rule out AgNO3.

Again, the sulphides of Cu^2+, Bi^3+ are soluble in acidic medium but according to the question, the sulphides do not precipitate at low pH hence Cu(NO3)2 and Bi(NO3)3 are both ruled out.

The sulphides of Fe^3+, Cr^3+ and Co^3+ all form precipitate in basic solution hence Fe(NO3)3, Cr(NO3)3, Co(NO3)3 may be present.

The presence of Ca(NO3)2 and KNO3 may be confirmed by flame tests.

Answer:

since the concentration of limiting reactant are the same for both nickel and hydrochloric acid, they both will produce the same amount if Nickel Chloride

7 mols of [tex]NiCl_2[/tex] formed in a reaction of 7 mol of Ni and 14 mol of HCl.

The moles of Ni = 7 mol

The mols of HCl = 14 mol

It is required to calculate moles of [tex]NiCl_2[/tex]

A mole corresponds to the mass of a substance that contains [tex]6.023 \times 10^{23}[/tex]particles of the substance. The mole is the SI unit for the amount of a substance. Its symbol is mol.

The reaction of Ni and HCl occurs as

[tex]Ni +2 HCl \to NiCl_2 + H_2[/tex]

Since the concentration of limiting reactant are the same for both nickel and hydrochloric acid, they both will produce the same amount of Nickel Chloride.

If we have 7 mols of Ni and 14 mols of HCl, then when the 7 moles of Ni has reacted, we will still have 7 moles of HCl unreacted.

So, the moles of [tex]NiCl_2[/tex] formed equal to the moles of Ni and HCl reacted.

Therefore, 7 mols of [tex]NiCl_2[/tex] formed in a reaction of 7 mol of Ni and 14 mol of HCl.

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Water, mercury chloride and nitrogen oxide.

Water, mercury chloride and nitrogen oxide will present in the precipitate when we slowly add a solution of mercury(II) nitrate to a solution of aqueous hydrochloric acid having halide ions both in equal concentrations. The equation of this reaction is Hg2(NO3)2 + 4 HCl ----> 2 HgCl2 + 2 H2O + 2 NO so it is concluded that from this reaction we get precipitate of water, mercury chloride and nitrogen oxide.

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Answer:

c. 2,2-dichloropentane.

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to firstly draw the structure of the reactant, pent-1-yne:

[tex]CH\equiv C-CH_2-CH_2-CH_2[/tex]

Now, we infer the halogen is added to the carbon atom with the most carbon atoms next to it, in this case, carbon #2, in order to write the following product:

[tex]CH\equiv C-CH_2-CH_2-CH_2+2HCl\rightarrow CH_3- CCl_2-CH_2-CH_2-CH_2[/tex]

Whose name is 2,2-dichloropentane.

Regards!

Answer:

[tex]\boxed {\boxed {\sf 0.75 \ mol \ Li}}[/tex]

Explanation:

We are asked to convert 5.2 grams of lithium to moles of lithium.

To convert from grams to moles, we need the molar mass. This is the measurement of the mass in 1 mole of a substance. It can be found on the Periodic Table because it is the same value as the atomic mass, but the units are grams per mole instead of atomic mass units.

Look up the molar mass of lithium.

Create a ratio using the molar mass of lithium.

[tex]\frac { 6.94 \ g \ Li}{ 1 \ mol \ Li}[/tex]

Multiply by the value we are converting: 5.2 grams of lithium.

[tex]5.2 \ g \ Li *\frac { 6.94 \ g \ Li}{ 1 \ mol \ Li}[/tex]

Flip the ratio so the units of grams of lithium cancel.

[tex]5.2 \ g \ Li *\frac{ 1 \ mol \ Li} { 6.94 \ g \ Li}[/tex]

[tex]5.2 *\frac{ 1 \ mol \ Li} { 6.94 }[/tex]

[tex]\frac{5.2} { 6.94 } \ mol \ LI[/tex]

[tex]0.749279538905 \ mol \ Li[/tex]

The original measurement of grams (5.2) has 2 significant figures, so our answer must have the same. For the number we calculated, that is the hundredth place. The 9 in the thousandths place to the right tells us to round the 4 up to a 5.

[tex]0.75 \ mol \ Li[/tex]

5.2 grams of lithium is equal to 0.75 moles of lithium atoms.

Answer:

11.76 g/cm^3

Explanation:

Mass of empty flask and stopper = 64.232g

Mass of flask filled with water = 153.617 g

Mass of water = 153.617 g - 64.232g = 89.385 g

Mass of flask, stopper and metal = 143.557 g

Mass of metal = 143.557 g - 64.232g = 79.325 g

Mass of water, flask, stopper and metal = 226.196 g

Mass of water = 226.196 g - 143.557 g = 82.639 g

Since mass of water =volume of water

Volume occupied by metal = 89.385 cm^3 - 82.639 cm^3 = 6.746 cm^3

Density of metal = mass/volume = 79.325 g/6.746 cm^3

= 11.76 g/cm^3

Answer:

See explanation and image attached

Explanation:

Methyl butanoate is an ester. Esters have the general molecular formula, RCOOR where the two Rs may represent the same or different alkyl groups.

Methyl butanoate is has a fruity odor, smelling like apples or pineapples fragrance. It is also called methyl butyrate.

The structure of the compound is shown in the image attached to this answer.

Answer:

A BEC ( Bose - Einstein condensate ) is a state of matter of a dilute gas of bosons cooled to temperatures very close to absolute zero is called BEC.

Examples - Superconductors and superfluids are the two examples of BEC.

Explanation: