The shape of a molecule is determined by:
A. All of these
B. The number of electron clouds around the atom.
C. The number of bonds.
D. Mutual repulsion between electrons.

Answer:

111.95mL of HNO3 are needed to prepare the buffer

Explanation:

We can solve this equation using H-H equation for bases:

pOH = pKb + log [HA+] / [A]

Where pOH is the pOH of the solution

pOH = 14 - pH = 14 - 8.970 = 5.03

pKb is the pKb of NH3 = 4.74

[HA+] could be taken as moles of NH4+

[A] as moles of NH3

The NH3 reacts with nitric acid, HNO3, as follows:

NH3 + HNO3 → NH4+ + NO3-

That means the moles of HNO3 added = X = Moles of NH4+ produced

And moles of NH3 are initial moles NH3 - X

Initial moles of NH3 are:

0.125L * (0.374mol/L) = 0.04675 moles NH3

Replacing in H-H equation:

pOH = pKb + log [HA+] / [A]

5.03 = 4.74 + log [X] / [0.04675-X]

0.29 = log [X] / [0.04675-X]

1.95 =  [X] / [0.04675-X]

0.0912 - 1.95X = X

0.0912 = 2.95X

X = 0.0309 moles

We need to add 0.0309 moles of HNO3. From a solution that is 0.276M:

0.0309 moles of HNO3 * (1L / 0.276moles) = 0.11195L of HNO3 are needed

In mL:

111.95mL of HNO3 are needed to prepare the buffer

Answer:

5.0 × 10⁻⁶

Explanation:

Step 1: Write the balanced equation for the solution of chromium(III) iodate

Cr(IO₃)₃(s) ⇄ Cr³⁺(aq) + 3 IO₃⁻(aq)

Step 2: Calculate the solubility product constant (Ksp)

To relate Ksp and the solubility (S), we will make an ICE chart.

        Cr(IO₃)₃(s) ⇄ Cr³⁺(aq) + 3 IO₃⁻(aq)

I                               0                 0

C                             +S              +3S

E                                S               3S

The solubility product constant is:

Ksp = [Cr³⁺] × [IO₃⁻]³ = S × (3S)³ = 27 S⁴ = 27 × (2.07 × 10⁻²)⁴ = 5.0 × 10⁻⁶

Answer:

Neither Tech A nor B is correct

Explanation:

Combustion is a chemical reaction that occurs when a chemical molecule(s) interacts quickly with oxygen and produces heat.

When hydrocarbon undergoes a complete combustion reaction, they produce water and CO2.

Tech B is also incorrect because the main purpose of a catalytic converter is to accelerate and speed up the chemical reaction rates, Hence, they are not involved in chemical reaction formation. Catalytic converters are utilized as a control device in exhaust emission to lessen the effect of toxic gas fumes.

Answer:

radical-initiated

Explanation:

Radical-initiated polymerization is unpredictable and difficult to control. The reaction proceeds indiscriminately and produces shortened chains, loops, and branches that create holes in the polymer. This reduces its mass to volume ratio.

Answer:

Hence the Volume of Gas = 3.04 L.

Explanation:  

pressure of dry gas = 765.3 - 24.3 = 741 mmhg  

Temperature of gas = 24.4+273.15 = 297.55 k  

No of mol of gas = 0.498/4 = 0.1245 mol  

R = gas constant = 0.0821 l.atm.k-1.mol-1  

From ideal-gas equation

PV = nRT  

(741/760) x v = 0.1245 x 0.0821 x 297.55  

V = Volume of Gas = 3.04 L

Answer:

74%

Explanation:

Step 1: Write the balanced equation

2 C₆H₁₀(l) + 17 O₂(g) ⇒ 12 CO₂(g) + 10 H₂O(g)

Step 2: Determine the limiting reactant

The theoretical mass ratio (TMR) of C₆H₁₀ to O₂ is 164:544 = 0.301:1.

The experimental mass ratio (EMR) of C₆H₁₀ to O₂ is 115:199 = 0.578:1.

Since EMR > TMR, the limiting reactant is O₂.

Step 3: Calculate the theoretical yield of H₂O

The theoretical mass ratio of O₂ to H₂O 544:180.

199 g O₂ × 180 g H₂O/544 g O₂ = 65.8 g H₂O

Step 4: Calculate the percent yield of H₂O

%yield = (experimental yield/theoretical yield) × 100%

%yield = (49 g/65.8 g) × 100% = 74%

Answer:

Percentage yield of H₂O = 74.24%

Explanation:

The balanced equation for the reaction is given below:

2C₆H₁₀ + 17O₂ —> 12CO₂ + 10H₂O

Next, we shall determine the masses of C₆H₁₀ and O₂ that reacted and the mass of H₂O produced from the balanced equation. This is can be obtained as follow:

Molar mass of C₆H₁₀ = 82 g/mol

Mass of C₆H₁₀ from the balanced equation = 2 × 82 = 164 g

Molar mass of O₂ = 32 g/mol

Mass of O₂ from the balanced equation = 17 × 32 = 544 g

Molar mass of H₂O = 18 g/mol

Mass of H₂O from the balanced equation = 10 × 18 = 180 g

SUMMARY:

From the balanced equation above,

164 g of C₆H₁₀ reacted with 544 g of O₂ to produce 180 g of H₂O.

Next, we shall determine the limiting reactant. This can be obtained as follow:

From the balanced equation above,

164 g of C₆H₁₀ reacted with 544 g of O₂.

Therefore, 115 g of C₆H₁₀ will react to produce = (115 × 544)/164 = 381 g of O₂.

From the calculations made above, we can see that a higher mass (i.e 381 g) of O₂ than what was given (i.e 199 g) is needed to react with 115 g of C₆H₁₀.

Therefore, O₂ is the limiting reactant and C₆H₁₀ is the excess reactant.

Next, we shall determine the theoretical yield of H₂O. This can be obtained by using the limiting reactant as shown below:

From the balanced equation above,

544 g of O₂ reacted to produce 180 g of H₂O.

Therefore, 199 g of O₂ will react to produce = (199 × 180)/544 = 66 g of H₂O.

Thus, the theoretical yield of H₂O is 66 g.

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

Actual yield of H₂O = 49 g

Theoretical yield of H₂O = 66 g

Percentage yield of H₂O =?

Percentage yield = Actual yield /Theoretical yield × 100

Percentage yield of H₂O = 49/66 × 100

Percentage yield of H₂O = 74.24%

Answer:

True

Explanation:

When Methyl Pentene is introduced in a chemical reaction with dichloromethane then the major product will be bromomethylpentene. There can be small amount of bromo methyl pentene than the amount of methyl pentene introduced for reaction.

Answer:

The products will be favored at equilibrium.

Explanation:

The balanced chemical equation for the reaction is the following:

2 COF₂ (g) + ⇌ CO₂(g) + CF₄(g)

The reactant is COF₂ (left side) and the products are CO₂ and CF₄ (right side).

The equilibrium constant is given by the ratio between the partial pressures (P) of products and reactants, because they are in the gas phase. Thus, the expression of the equilibrium constant is the following:

[tex]Kp = \frac{P(CO_{2}) P(CF_{4}) }{P(COF_{2} )^{2} } = 2.2 x 10^{6}[/tex]

Since Kp>>>>1 ⇒ (P(CO₂) x P(CF₄)) > (P(COF₂))²

So, the partial pressures of the products (CO₂ and CF₄) are higher than the partial pressure of the reactant (COF₂).

Therefore, products will be favored at equilibrium at 298 K.

Answer:

Both substances undergo substitution reactions.

Explanation:

Let us go back to the idea of aromaticity. Aromatic substances are said to possess (4n + 2) π electrons according to Huckel rule.

Aromatic substances are unusually stable and the aromatic ring can not be destroyed by addition reactions.

Since both benzene and cyclooctatetraene are both aromatic, they do not undergo addition reactions whereby the aromatic ring is destroyed. They both undergo substitution reaction in which the aromatic ring is maintained.

Answer:

C₂H₄O₂

Explanation:

Step 1: Divide each percentage by the atomic mass of the element

C: 40.00/12.01 = 3.331

H: 6.67/1.01 = 6.60

O: 53.33/16.00 = 3.333

Step 2: Divide all the numbers by the smallest one

C: 3.331/3.331 = 1

H: 6.60/3.331 ≈ 2

O: 3.333/3.331 ≈ 1

The empirical formula is CH₂O, with a molecular weight of 12 g/mol + 2 × 1 g/mol + 16 g/mol = 30 g/mol. The molecular weight of the compound must be a product of 30, such as 60 (between 55 and 62 g/mol). Since we have to multiply by 2 (30 to 60) to get to the molecular weight of the compound, we also have to multiply the empirical formula by 2 to get the chemical formula of the compound.

CH₂O × 2 = C₂H₄O₂

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:

The right answer is "105.17 ppb".

Explanation:

According to the question,

The amount of [tex]Pb^{2+}[/tex] in ppb will be:

= [tex]0.5076\times 10^{-6}\times 207.2\times 106[/tex]

= [tex]105.17 \ ppb[/tex]

Thus, the amount of [tex]Pb^{2+}[/tex] is above action limit.

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:

The answer is "170.9 mm Hg".

Explanation:

[tex]\text{Mass of acetone = volume} \times density[/tex]

                          [tex]= 70.0 \times 0.791\\\\ = 55.37\ g\\[/tex]

[tex]\text{Moles of acetone} = \frac{mass}{molar\ mass}\\\\[/tex]

                            [tex]=\frac{55.37}{58.08}\\\\ = 0.9533\ mol[/tex]

[tex]\text{Mass of ethyl acetate = volume} \times density[/tex]

                                   [tex]= 73.0 \times 0.900\\\\ = 65.7\ g[/tex]  

[tex]\text{Moles of ethyl acetate = mass} \times\ molar\ mass[/tex]

                                    [tex]= \frac{65.7}{88.105} \\\\= 0.7457\ mol[/tex]

[tex]\text{Mole fraction of acetone x(acetone)} = \frac{0.9533}{(0.9533 + 0.7457)}\ = 0.5611\\\\[/tex] [tex]\text{Mole fraction of ethyl acetate x(ethyl acetate)} =\frac{0.7457}{(0.9533 + 0.7457) }= 0.4389[/tex]

Applying Raoult's law: [tex]\text{Vapor pressure = x(acetone)P(acetone) + x(ethyl acetate)P(ethyl acetate)}\\\\= 0.5611 \times 230.0 + 0.4389 \times 95.38\\\\ = 170.9\ mm \ Hg\\[/tex]

The solvent for an organic reaction is prepared by mixing 70.0 mL of acetone (C3H6O) with 75.0 mL of ethyl acetate (C4H8O2).

The vapor pressure of the stored mixture is: 170.03 mmHg

In the given information, there is some information that is still missing.

The parameters that we are being given include:

The  first step we need to take is to determine the mass and number of moles  of each compound (i.e. for acetone and ethyl acetate)

For us to do that:

We need the density of acetone and ethyl acetate, which is not given:

Assuming that at a standard condition of vapour pressure:

Then;

Using the relation:

[tex]\mathbf{Density = \dfrac{Mass}{volume}}[/tex]

Mass of acetone = Density of acetone × volume of acetone

Mass of acetone = 0.791  g/mL  × 70.0 mL

Mass of acetone = 55.37 g

Mass of ethyl acetate = Density of ethyl acetate  × volume of ethyl acetate

Mass of ethyl acetate = 0.900 g/mL  ×  75.0 mL

Mass of ethyl acetate = 67.5 g

At standard conditions;

Now, using the formula for calculating the numbers of moles which can be expressed as:

[tex]\mathbf{Number \ of \ moles = \dfrac{mass}{molar \ mass}}[/tex]

For acetone:

[tex]\mathbf{Number \ of \ moles = \dfrac{55.37 \ g}{58.08 \ g/mol}}[/tex]

[tex]\mathbf{Number \ of \ moles =0.95334 \ mol}[/tex]

For ethyl acetate:

[tex]\mathbf{Number \ of \ moles = \dfrac{67.5 \ g}{88.11 \ g/mol}}[/tex]

[tex]\mathbf{Number \ of \ moles =0.76609 \ mol}[/tex]

Now, we will determine the mole fraction of each compound.

The mole fraction describes the ratio a certain constituent of a mixture to the total amount of all the constitutent in the mixture.

Using the formula:

[tex]\mathbf{mole \ fraction = \dfrac{n_A}{n_A+n_B+...n_N}}[/tex]

For Acetone:

[tex]\mathbf{mole \ fraction = \dfrac{0.95334}{0.95334+0.76609}}[/tex]

[tex]\mathbf{mole \ fraction =0.5545 }[/tex]

For ethyl acetate:

[tex]\mathbf{mole \ fraction = \dfrac{0.76609}{0.76609+0.95334}}[/tex]

[tex]\mathbf{mole \ fraction =0.4455}[/tex]

Finally, we can compute determine the vapour pressure of the stored mixture using Raoult's Law.

Raoult's Law posits that the constituent of a partial pressure in a mixture of a liquid is proportional to the mole fraction of that constituent in the mixture provided the temperature is constant.

∴ For the stored mixture = Vapor pressure of acetone + vapour pressure of ethyl acetate.

where:

For acetone;

Vapor pressure = 0.5545 × 230 mmHg

Vapour pressure = 127.54 mmHg

For ethyl acetate:

Vapour pressure = 0.4455 × 95.38 mmHg

Vapour pressure ==42.49 mmHg

Thus, the vapor pressure of the stored mixture is

= (127.54 + 42.49 ) mmHg

= 170.03 mmHg

Therefore, we can conclude that the vapour pressure of the stored mixture is 170.03 mmHg

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Answer:

Spontaneous processes are ones that occur quickly and have a low activation energy. - False -

Spontaneous processes always require an input of energy to overcome the activation energy, but always react quickly - False

Spontaneous processes can occur slowly, but always have a low activation- false

Spontaneous reactions can react slowly and can have a high activation energy - True

Spontaneous processes always react slowly and always have a high activation energy- False

Explanation:

A spontaneous reaction is reaction that proceeds on its own without us having to do a thing at all!

A spontaneous reaction may be fast or slow depending on the activation energy of the reaction. A spontaneous reaction having a high activation energy will be slow. However, if the spontaneous reaction has a low activation energy then it will be fast.

We have to note here that a spontaneous reaction proceeds without a prolonged input energy. Sometimes energy may be supplied to the reaction at the beginning for instance in the case of the combustion of hydrocarbons.

So, spontaneous processes are not necessarily fast. Some of them may have a very high activation energy such as in the rusting of iron hence they are slow.

The question is incomplete, the complete question is shown in the image attached to this answer

Answer:

2,3,3-trimethylhexane

Explanation:

IUPAC nomenclature provides a universally acceptable method of naming organic compounds from its structure.

According to this system of nomenclature;

A comma is used to separate two numbers.

A hyphen is used to separate a number from a letter.

Applying these rules, the name of the compound shown in the question should be written as 2,3,3-trimethylhexane.

Answer:

we don't understand why humans have only 46 chromosomes

Answer:

46 chromosomes is what we don't understand

Answer: Volume occupied by given neon sample at standard condition is 123.84 L.

Explanation:

Given: [tex]V_{1}[/tex] = 105 L,    [tex]T_{1} = 27^{o}C = (27 + 273) K = 300 K[/tex],     [tex]P_{1}[/tex] = 985 torr

At standard conditions,

[tex]T_{2}[/tex] = 273 K,     [tex]P_{2}[/tex] = 760 K,        [tex]V_{2}[/tex] = ?

Formula used to calculate the volume is as follows.

[tex]\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}[/tex]

Substitute the values into above formula as follows.

[tex]\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}\\\frac{985 torr \times 105 L}{300 K} = \frac{760 torr \times V_{2}}{273 K}\\V_{2} = \frac{94116.75}{760} L\\= 123.84 L[/tex]

Thus, we can conclude that volume occupied by given neon sample at standard condition is 123.84 L.

Answer:

[tex]C_t=0.165M[/tex]

Explanation:

From the question we are told that:

Slope [tex]K=0.056 M-1 s -1[/tex]

initial Concentration [tex]C_1=2.2M[/tex]

Time [tex]t=100[/tex]

Generally the equation for Raw law is mathematically given by

[tex]\frac{1}{C}_t=kt+\frac{1}{C}_0[/tex]

[tex]\frac{1}{C}_t=0.056*100+\frac{1}{2.2}_0[/tex]

[tex]C_t=0.165M[/tex]

The second-order reaction is the reaction that depends on the reactants of the first or the second-order reaction. The concentration after 100 seconds will be 0.165 M.

The specific rate constant (k) of the second-order reaction is given in L/mol/s or per M per s. It is the proportionality constant that gives the relation between the concentration and the rate of the reaction.

Given,

Slope (k)= 0.056 per M per s

Initial concentration of the reactant [tex](\rm C_{1})[/tex] = 2.2 M

Time (t) = 100 seconds

The concentration of the reaction after 100 seconds can be given by,

[tex]\rm \dfrac{1}{C_{t}} = kt + \dfrac{1}{C_{1}}[/tex]

Substitute values in the above equation:

[tex]\begin{aligned} \rm \dfrac{1}{C_{t}} &= 0.056 \times 100 + \dfrac{1}{2.02}\\\\&= 0.165 \;\rm M\end{aligned}[/tex]

Therefore, after 100 seconds the concentration is 0.165 M.

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Answer:

This question is asking to find the new temperature

The answer for the final temperature is 429.73K

Explanation:

Using Charles law equation as follows:

V1/T1 = V2/T2

Where;

V1 = initial volume (L)

V2 = final volume (L)

T1 = initial temperature (K)

T2 = final temperature (K)

According to this question;

V1 = 3.0L

V2 = 4.4L

T1 = 20°C = 20 +273 = 293K

T2 = ?

Using V1/T1 = V2/T2

3/293 = 4.4/T2

Cross multiply

293 × 4.4 = 3 × T2

1289.2 = 3T2

T2 = 1289.2 ÷ 3

T2 = 429.73K

Answer:

The concentration of KOH is  0.06137 M

Explanation:

Step 1: Data given

Molar mass of KHP = 204.22 g/mol

Mass of KHP = 0.4877 grams

Volume of water = 50 mL

Volume of KOH solution = 38.91 mL

Step 2: The balanced equation

C8H5KO4 + KOH ⇒ C8H4K2O4+ H2O

Step 3: Calculate number of moles of KHP

Moles = Mass / molar mass

Moles KHP = 0.4877 grams / 204.22 g/mol

Moles KHP = 0.002388 moles

Step 4: Calculate moles of KOH

For 1 mol KHP we need 1 mol KOH to produce 1 mol C8H4K2O4 and 1 mol H2O

For 0.002388 moles KHP we need 0.002388 moles KOH

Step 5: Calculate the concentration of KOH

Concentration = moles / volume

Concentration of KOH = 0.002388 moles / 0.03891 L

Concentration of KOH = 0.06137 M

The concentration of KOH is  0.06137 M

Answer:

362g

Explanation:

heat lost by metal= heat gained by acetic acid

tfs are the same so you cando delta T

convert Cal/gc to J/gc

thectgod ig follow

Answer:

an if/then statement

Explanation:

Answer:

66.67%

Explanation:

From the given information:

mass of cyclohexane = 2.9949 grams

density of cyclohexane = 0.779 g/mL

Recall that:

Density = mass/volume

∴

Volume = mass/density

So, the volume of cyclohexane = 2.9949 g/ 0.779 g/mL

= 3.8445 mL

Also,

mass of propylbenzene = 1.6575 grams

density of propylbenzene = 0.862 g/mL

Volume of propylbenzene =  1.6575 g/ 0.862 g/mL

= 1.9229 mL

The volume % composition of cyclohexane from the mixture is:

[tex]= (\dfrac{v_{cyclohexane}}{v_{cyclohexane}+v_{propylbenzene}})\times 100[/tex]

[tex]= (\dfrac{3.8445}{3.8445+1.9229})\times 100[/tex]

[tex]= (\dfrac{3.8445}{5.7674})\times 100[/tex]

= 66.67%

Answer:

C-O: polar covalent

Mg-F: ionic

Cl-Cl: nonpolar covalent

Explanation:

Ionic bonds are formed between an atom of a metallic element and another atom of a non-metallic element. Thus, Mg-F is an ionic bond, in which Mg is the metal and F is the nonmetal.

Covalent bonds are formed between two non-metallic elements. So, C-O and Cl-Cl are covalent bonds, because C, O, and Cl are nonmetals.

In C-O, the atom of oxygen (O) has more electronegativity than the atom of carbon (C). Thus, O will attract the electrons with more strength and a difference in charge will be established between the two bonded atoms. So, this covalent bond is polar.

In Cl-Cl, both atoms have the same electronegativity because they are from the same chemical element (Cl). Thus, this bond is nonpolar.

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:

AgNO3 + NaOH = AgOH + NaNO3.

Explanation:

Balancing Strategies: In this reaction, the products are initially NaNO3 + AgOH. However the AgOH would break down into Ag2O and H2O. This would give us NaNO3 + Ag2O + H2O as our products for the overall reaction.

Balancing Strategies: In this reaction, the products are initially NaNO3 + AgOH. However the AgOH would break down into Ag2O and H2O. This would give us NaNO3 + Ag2O + H2O as our products for the overall reaction.However, the equation balanced here is the initial reaction which produces AgOH and NaNO3.

Answer: Hello your question is incomplete attached below is the missing image

answer:

Reduction half-reaction: Zn(s) ⇒ Zn⁺² (aq)  + 2e⁻ ( occurs at anode )

Oxidation half-reaction; CO²⁺(aq) + 2e⁻ ⇒ CO (s)  ( occurs at the cathode )

Overall cell reaction ; Zn(s)  + CO²⁺(aq)  ⇒ Zn⁺² (aq) +  CO (s)

Explanation:

stating the standard reduction potentials

E° zn²⁺/zn = -0.76 v

E°Co²⁺ / Co = -0.28 v

since ; -0.76 v  < -0.28 v.    Zn will be oxidized while Co²⁺ will be reduced .

Reduction half-reaction: Zn(s) ⇒ Zn⁺² (aq)  + 2e⁻ ( occurs at anode )

Oxidation half-reaction;  CO²⁺(aq) + 2e⁻ ⇒ CO (s)  ( occurs at the cathode )

hence the

Overall cell reaction ;

Zn(s)  + CO²⁺(aq)  ⇒ Zn⁺² (aq) +  CO (s)

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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