LAST ATTEMPT! Covalent compound naming !

Answer:

25.5mL of 0.100M NaOH are needed to reach buffer capacity.

Explanation:

The buffer capacity is reached when the ratio between moles of conjugate base (Sodium acetate) and moles of weak acid (Acetic acid) is 10:

Moles sodium acetate / Moles Acetic acid = 10

The reaction of acetic acid, HA, with NaOH, to produce sodium acetate, NaA is:

HA + NaOH → H2O + NaA

That means the moles of NaOH added = Moles of HA that are being subtracted and moles of NaA that are been produced.

The initial moles of each species is:

Acetic acid:

23.34mL = 0.02334L * (0.147mol / L) = 0.00343 moles Acetic Acid

Sodium Acetate:

33.66mL = 0.03366L * (0.185mol / L) = 0.00623 moles Sodium Acetate

We can write the moles of each species when NaOH is added as:

Moles sodium acetate / Moles Acetic acid = 10

0.00623 moles + X / 0.00343 moles - X = 10

Where X are moles of NaOH added

Solving for X:

0.00623 moles + X = 0.0343 moles - 10X

11X = 0.0281

X = 0.00255 moles of NaOH are needed

In Liters:

0.0255mol NaOH * (1L / 0.100mol) = 0.0255L of 0.100M NaOH are needed =

Answer:

0.3323

Explanation:

GIven that:

Density of the metal = 10.6 g/cm^3

atomic weight = 176.8 g/mol

atmic radius = 0.130 nm

values of a and c = 0.570 nm and 0.341 nm respectively

For us to determine the atomic packing factor, we need to first determine the volume of all spheres (Vs) and the volume of unit cell (Vc).

However, the number of atoms  in the unit cell (n) can be computed as:

[tex]n = \dfrac{\rho * V_c *N_A}{A} \\ \\ n = \dfrac{(10.6) * (5.7)^2 (3.41)*(10^{-24}) *(6.022*10^{23})}{176.8}[/tex]

n = 4.0

Thus, the number of atoms in the unit cell is 4

∴

The atomic paking factor (APF) is calculated by using the formula:

[tex]\dfrac{Vs}{Vc} = \dfrac{4 * \dfrac{4}{3}\pi *R^3 }{a^2 *c} \\ \\ \\ \dfrac{Vs}{Vc} = \dfrac{4 * \dfrac{4}{3}\pi *(1.30*10^{-8})^3 }{(5.70*10^{-8})^2 *(3.41*10^{-8})}[/tex]

= 0.3323

Answer:

In Haloarenes the C atom to which the X group is attached is SP2 hybridized thus it is become difficult to replace it by the Nucleophile. Since arenes and Vinyl halides are electron rich molecules due to presenceof n bonds, they repel Nucleophile attacking them.

Question:-

Why it is difficult for haloarenes to undergo nucleoplhilic subsituⁿ reaction?

Answer:-

Haloarenes are less reactive towards the nucleoplhilic substitution rxⁿ . This is due to following reasons :-

[tex]\red{\bigstar}\underline{\textsf{ Reason 1 :- Partial double bond character .}}[/tex]

Halogen atom has one lone pair, and due to presence of π - σ - lp , resonance is established in the compound ( see attachment) . Due to resonance there is a partial double bond character in the carbon halogen bond , so it is difficult to break a double bond than a single bond.

[tex]\rule{200}2[/tex]

[tex]\red{\bigstar}\underline{\textsf{ Reason 2 :- $\pi$ cloud .}}[/tex]

When a nucleoplhile comes to attack , it is repelled by the π-cloud of the benzene ring.

[tex]\rule{200}2[/tex]

[tex]\red{\bigstar}\underline{\textsf{ Reason 3 :- Phenyl cation .}}[/tex]

If somehow the halogen atoms leaves the benzene ring ,being more electronegative than carbon , it takes away the electron , thus a positive charge is left on benzene ring and the phenyl cation so formed is very unstable .

[tex]\rule{200}2[/tex]

Answer:

requiring immediate action or attention.

word similar to urgent: Acute

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:

o

Explanation:

it is not a gas because the particles do not move freely it may be a liquid or a solid partly and mostly liquidized.

Answer:

17

Explanation:

Step 1: Calculate the needed concentrations

[A]i = 1.00 mol/5.00 L = 0.200 M

[B]i = 1.80 mol/5.00 L = 0.360 M

[B]e = 1.00 mol/5.00 L = 0.200 M

Step 2: Make an ICE chart

        A(aq) + 2 B(aq) ⇄ C(aq)

I       0.200    0.360        0

C        -x           -2x         +x

E     0.200-x  0.360-2x   x

Then,

[B]e = 0.360-2x = 0.200

x = 0.0800

The concentrations at equilibrium are:

[A]e = 0.200-0.0800 = 0.120 M

[B]e = 0.200 M

[C]e = 0.0800 M

Step 3: Calculate the concentration equilibrium constant (K)

K = [C] / [A] × [B]²

K = 0.0800 / 0.120 × 0.200² = 16.6 ≈ 17

Answer:

electric charge

Explanation:

Charged particles are deflected in an electric or a magnetic field. The particles discovered by J.J. Thomson were charged particles.

When these charged particles are passed through electric and magnetic fields, deflection occurs depending on the nature of the charge.

A positive charge is deflected towards the negative part of an electric field or the south pole of a magnetic field.

A negative charge is selected towards the positive end of an electric field or the north pole of a magnetic field.

Answer:

When atoms approach one another closely, the electron clouds interact with each other and with the nuclei. If this interaction is such that the total energy of the system is lowered, then the atoms bond together to form a molecule.

Explanation:

Answer:

The lewis structure for NO₃⁻ is shown in the attachment below

For the central nitrogen atom:

The number of lone pairs = 0

The number of single bonds = 2  

The number of double bonds= 1

Explanation:

The lewis structure for NO₃⁻ is shown in the attachment below.

From the Lewis structure

For the central nitrogen atom:

The number of lone pairs = 0

The number of single bonds = 2  

The number of double bonds= 1

Explanation:

The standard entropy change of a reaction has a positive value. This reaction results in an increase in entropy.

Positive entropy means the system has increased its degree of disorderness.

Answer:

a) N2

b) They all have the same average kinetic energy.

Explanation:

At a given temperature, the speed of a gas molecule depends on its relative molecular mass. The heavier the gas, the lesser its average velocity at a given temperature. On that basis, N2 molecules are the fastest moving gas molecules.

At a particular temperature, all gases have the same average kinetic energy.

Answer:

21.3 g. Option B

Explanation:

The reaction is:

4HF(g) + SiO₂(s) → SiF₄(g) + 2H₂O(g)

We analysed it and it is correctly balanced.

4 moles of hydrogen fluoride react to 1 mol of silicon dioxide in order to produce 1 mol of silicon fluoride and 2 moles of water vapor.

We determine molar mass of each reactant:

HF → 1.01 g/mol + 19 g/mol = 20.01 g/mol

SiO₂ → 16 g/mol . 2 + 28.09 g/mol = 60.09 g/mol

We convert mass to moles: 16 g . 1 mol /60.09g = 0.266 moles of glass

Ratio is 1:4. 1 mol of glass react to 4 moles of HF

Our 0.266 moles may react to (0.266 . 4) / 1 = 1.07 moles of gas

We convert moles to mass: 1.07 mol . 20.01 g/mol = 21.3 g

Answer:

by the own's formula energy sublevels are 2 the power of n or principal quantum number this means 2 the power of 4 equal to 32

Explanation:

The given chemical reaction is:

[tex]2 N_2 (g) + O_2(g) -> 2 N_20 (g) delta Hrxn= +163.2 kJ[/tex]

When two moles of nitrogen reacts with oxygen, it requires 163.2kJ of energy.

When 5.00 mol of nitrogen requires how much energy?

[tex]5.00 mol x \frac{163.2 kJ }{2 mol} \\=408 kJ[/tex]

Hence, the answer is 408 kJ of heat energy is required.

Answer:

it is propane

C3H8 it is propane

The Full name of the given compound is Propane.

Propane is a three-carbon alkane with the molecular formula C₃H₈.

It is a gas at standard temperature and pressure, but compressible to a transportable liquid.

In the figure given ;

In the given figure, there is single bond present between Carbon and Hydrogen. Hence, The Full name of the given compound is Propane.

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Answer:

2KHCO

3

+H

2

SO

4

→K

2

SO

4

+2CO

2

+2H

2

O

Answer:

The two carbon atoms share a total of six electrons with each other.

Explanation:

Looking at the structure of the molecule H-C≡C-H as shown in the question, we will notice that there exists a triple bond between the two carbon atoms.

Each bond between the two carbon atoms represents two electrons shared. Since there are three bonds between the two carbon atoms, then a total of six electrons were shared between the two carbon atoms hence the answer chosen above.

Answer:

Systems move from ordered behavior to more random behavior.

Explanation:

Entropy refers to the degree of disorderliness in a system. The second law of thermodynamics can be restated in terms of entropy as follows; “any spontaneous process in any isolated system always results in an increase in the entropy of that system.''(science direct)

According to this law, systems tend towards a more disorderly behaviour (increase in entropy) hence the answer given above.

Answer:

[tex]x=54g[/tex]

Explanation:

From the question we are told that:

Content of Sterling silver:

Let x be sterling silver

Silver [tex]S=0.9x[/tex]

Copper [tex]C=0.1x[/tex]

Total  silver available [tex]T=48.3[/tex]

Generally the equation for Total amount to be made is mathematically given by

[tex]T=\frac{x*90}{100}[/tex]

[tex]x=\frac{48.3*100}{90}[/tex]

[tex]x=54g[/tex]

Answer:

See explanation and image attached

Explanation:

Aromatic compounds undergo electrophilic aromatic substitution reactions in which the aromatic ring is maintained.

Substituted benzenes may be more or less reactive towards electrophilic aromatic substitution than benzene depending on the nature of the substituent present in the ring.

Substituents that activate the ring towards electrophilic substitution such as -OCH3 are ortho-para directing.

The major products of the bromination of anisole are p-bromoanisole and o-bromoanisole. The resonance structures leading to these products are shown in the image attached.

Answer:

XSO₄

Explanation:

XCl₂ is the chloride of metal X. The sum of the charges of the cation and the anion must be zero because the salt is electrically neutral. The charge of  the cation of X is:

1 × X + 2 × Cl = 0

1 × X + 2 × (-1) = 0

X = +2

X has a charge +2 and sulphate (SO₄²⁻) a charge -2. The neutral salt they form is XSO₄.

Answer:

The amount of CH3OH present in the mixture would decrease

Explanation:

According to Le Cha-telier's principle, when a reaction is in equilibrium and one of the constraints that influence the rate of reactions is applied, the equilibrium would shift so as to neutralize the effects of the constraint.

In this case, looking at the equation of the reaction:

2H2 + CO <------> CH3OH + heat

the total number of moles on the reactant's (left hand) side is 3 (2+1) while on the product's (right hand) side, it is 1. If the pressure on the system is increased, more CH3OH (and less of H2 and CO) will be produced because its side has the lower number of moles out of the two sides.

If the pressure on the system is otherwise lowered, more of H2 and CO would be produced while the amount of CH3OH present would gradually decrease.

Answer:

495 J

Explanation:

When the hot pot was set on the counter to cool, heat energy was lost from the pot. Note that according to the first law of thermodynamics, heat is neither created nor destroyed.

This implies that, the heat energy lost from the pot must be gained by the surrounding air. Therefore, if 495 J of energy is lost from the pot, then 495 J of energy is gained by the surrounding air.

Answer:

Top layer is Organic (CH2Cl2 and product)

Explanation:

In a solvent mixture, there are usually two phases, the organic phase and the aqueous phase.

It is usual that the organic phase is almost always less dense than the aqueous phase hence the organic phase tend to remain on top of the aqueous phase.

Hence, the top layer is expected to be the organic CH2Cl2 and product.

Answer:

Hence, 15.99 g of solid Aluminum Sulfate should be added in 250 mL of Volumetric flask.

Explanation:

To make 0.187 M of Aluminum Sulfate solution in a 250 mL (0.250 L) Volumetric flask  

The molar mass of Aluminum Sulfate = 342.15 g/mol  

Using the molarity formula:-  

Molarity = Number of moles/Volume of solution in a liter  

Number of moles = Given weight/ molar mass  

Molarity = (Given weight/ molar mass)/Volume of solution in liter  

0.187 M = (Given weight/342.15 g/mol)/0.250 L  

Given weight = 15.99 g  

Answer:

21.4g of HBr is the minimum mass that could be left over.

Explanation:

Based on the reaction:

HBr + NaOH → NaBr + H2O

1 mole of HBr reacts per mole of NaOH

To solve this question we need to find the moles of both reactants. If moles NaOH > moles HBr, the difference in moles represents the minimum moles of HBr that could be left over because this reaction is 1:1. Using the molar mass we can find the minimum mass of HBr that could be left over, as follows:

Moles NaOH -40.0g/mol-

17g * (1mol/40.0g) = 0.425 moles NaOH

Moles HBr -Molar mass: 80.91g/mol-

55.8g * (1mol/80.91g) = 0.690 moles HBr

The difference in moles is:

0.690 moles - 0.425 moles =

0.265 moles of HBr could be left over

The mass is:

0.265 moles * (80.91g/mol) =

Answer:

1:2:1 is the correct ratio of carbon hydrogen to oxygen in glucose.

Answer:

160 g/mol

Explanation:

Step 1: Calculate the molarity of the solution

We will use the following expression.

π = M × R × T

where,

M = π / R × T

M = 1.84 atm / (0.0821 atm.L/mol.K) × 298 K = 0.0752 mol/L

Step 2: Calculate the moles of solute in 550 mL (0.550 L)

0.550 L × 0.0752 mol/L = 0.0413 mol

Step 3: Calculate the molecular weight of the nonelectrolyte

0.0413 moles weigh 6.60 g.

6.60 g/0.0413 mol = 160 g/mol