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a)
b)
c)
d)

Answer:

There are fifteen molecular orbitals in benzene filled with electrons.

Explanation:

Benzene is an aromatic compound. Let us consider the number of bonding molecular orbitals that should be present in the molecule;

There are 6 C-C σ bonds, these will occupy six bonding molecular orbitals filled with electrons.

There are 6 C-H σ bonds, these will occupy another six molecular orbitals filled with electrons

The are 3 C=C π bonds., these will occupy three bonding molecular pi orbitals.

All these bring the total number of bonding molecular orbitals filled with electrons to fifteen bonding molecular orbitals.

Answer:

ag and au are sure not to react. but hg and sn might or might not

Answer:

b. 0°C and 1 bar

Explanation:

Hello,

In this case, the STP conditions are standard temperature and pressure sets of conditions for experimental measurements to be established to allow comparisons to be made between different sets of data, it means that a specific pressure and temperature is assigned to analyze the properties of a substance. Such conditions are strictly 0°C and 1 bar because a large number of physical, chemical and thermodynamic properties are measured at them, therefore the standard entropy of a substance refers to its entropy at: b. 0°C and 1 bar.

Best regards.

Answer:

a.

[tex]m_K=8.056gK\\ \\m_{Cl_2}=4.028gCl_2[/tex]

b.

[tex]m_{Cr}=10.51gCr\\ \\m_{O_2}=4.851gO_2[/tex]

c.

[tex]m_{Sr}=13.88gSr\\\\m_{N_2}=1.479gN_2[/tex]

Explanation:

Hello,

In this case, we proceed via stoichiometry in order to compute the masses of all the reactants as shown below:

a. [tex]2K+Cl_2\rightarrow 2KCl[/tex]

[tex]m_K=15.36gKCl*\frac{1molKCl}{74.55gKCl}*\frac{2molK}{2molKCl}* \frac{39.1gK}{1molK}=8.056gK\\ \\m_{Cl_2}=15.36gKCl*\frac{1molKCl}{74.55gKCl}*\frac{1molCl_2}{2molKCl}* \frac{70.9gCl_2}{1molCl_2}=4.028gCl_2[/tex]

b. [tex]4Cr+ 3O_2\rightarrow 2Cr_2O_3[/tex]

[tex]m_{Cr}=15.36gCr_2O_3*\frac{1molCr_2O_3}{152gCr_2O_3l}*\frac{4molCr}{2molCr_2O_3}* \frac{52gCr}{1molCr_2O_3}=10.51gCr\\ \\m_{O_2}=15.36gCr_2O_3*\frac{1molCr_2O_3}{152gCr_2O_3l}*\frac{3molO_2}{2molCr_2O_3}* \frac{32gO_2}{1molCr_2O_3}=4.851gO_2[/tex]

c. [tex]3Sr(s) + N_2(g) \rightarrow Sr_3N_2[/tex]

[tex]m_{Sr}=15.36gSr_3N_2*\frac{1molSr_3N_2}{290.86gSr_3N_2}*\frac{3molSr}{1molSr_3N_2}* \frac{87.62gSr}{1molSr}=13.88gSr\\\\m_{N_2}=15.36gSr_3N_2*\frac{1molSr_3N_2}{290.86gSr_3N_2}*\frac{1molN_2}{1molSr_3N_2}* \frac{28gN_2}{1molN_2}=1.479gN_2[/tex]

Regards.

Explanation:

The equation is given as;

2 A + B + 3 C → Products

The order of the reaction refers to the extent at which the rate depends n the concentration of the reactant.

The order of reaction is experimentally obtained. It can also be obtained from the rate law of the reaction.

If the rate law is given as;

rate law = k [A]²[B][C]³

Then the order is second order with respect to A.

The order is second order with respect to A.

Given that;

2A + B + 3C → Products at 470 K

Find:

Order of reaction with respect to A

Computation:

The reaction that takes place refers to how much the rate is influenced by the reactant concentration.

The reaction order is determined empirically. This can also be derived from the reaction's rate law.

Rate law = k[A]²[B][C]³

So, The order is second order with respect to A.

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Answer:

A. power the Calvin cycle.

Explanation:

because it helps to run theblife of plants with easily

Answer:

[tex][N_2]_{eq}=[H_2]_{eq}=0.09899M[/tex]

[tex][NO]_{eq}=0.00202M[/tex]

Explanation:

Hello,

In this case, for the given chemical reaction:

[tex]2NO\rightleftharpoons N_2+O_2[/tex]

We know the equilibrium constant and equilibrium expression:

[tex]Kc=2.4x10^3=\frac{[N_2][O_2]}{[NO]^2}[/tex]

That in terms of the reaction extent [tex]x[/tex] (ICE procedure) we can write:

[tex]2.4x10^3=\frac{x*x}{(0.2M-2*x)^2}[/tex]

In such a way, solving for [tex]x[/tex] by using a quadratic equation or solver, we obtain:

[tex]x_1=0.09899M\\x_2=0.1010M[/tex]

Clearly the solution is 0.09899M since the other value will result in a negative equilibrium concentration of NO. In such a way, the equilibrium concentrations of all the species are:

[tex][N_2]_{eq}=[H_2]_{eq}=x=0.09899M[/tex]

[tex][NO]_{eq}=0.2M-2*0.09899M=0.00202M[/tex]

Regards.

Answer:

a. Rate = k×[A]

b. k = 0.213s⁻¹

Explanation:

a. When you are studying the kinetics of a reaction such as:

A + B → Products.

General rate law must be like:

Rate = k×[A]ᵃ[B]ᵇ

You must make experiments change initial concentrations of A and B trying to find k, a and b parameters.

If you see experiments 1 and 3, concentration of A is doubled and the Rate of the reaction is doubled to. That means a = 1

Rate = k×[A]¹[B]ᵇ

In experiment 1 and to the concentration of B change from 1.50M to 2.50M but rate maintains the same. That is only possible if b = 0. (The kinetics of the reaction is indepent to [B]

Rate = k×[A][B]⁰

b. Replacing with values of experiment 1 (You can do the same with experiment 3 obtaining the same) k is:

Rate = k×[A]

0.320M/s = k×[1.50M]

Answer: The number of moles of a solute per kilogram of solvent

Explanation:

Answer:

a. Methanol remains the same

b. Methanol decreases

c. Methanol increases

d. Methanol remains the same

e. Methanol increases

Explanation:

Methanol is produced by the reaction of carbon monoxide and hydrogen in the presence of a catalyst as follows; 2H2+CO→CH3OH.

a) The presence or absence of a catalyst makes no difference on the equilibrium position of the system hence the methanol remains constant.

b) The amount of methanol decreases because the equilibrium position shifts towards the left and more reactants are formed since the reaction is exothermic.

c) If the volume is decreased, there will be more methanol in the system because the equilibrium position will shift towards the right hand side.

d) Addition of helium gas has no effect on the equilibrium position since it does not participate in the reaction system.

e) if more CO is added the amount of methanol increases since the equilibrium position will shift towards the right hand side.

Answer:

A precipitate will form, BaCO₃

Explanation:

When Ba²⁺ and CO₃²⁻ ions are in an aqueous media, BaCO₃(s), a precipitate, is produced following its Ksp expression:

Ksp = 5.1x10⁻⁹ = [Ba²⁺] [CO₃²⁻]

Where the concentrations of the ions are the concentrations in equilibrium

For actual concentrations of a solution, you can define Q, reaction quotient, as:

Q = [Ba²⁺] [CO₃²⁻]

If Q > Ksp, the ions will react producing BaCO₃, if not, no precipitate will form.

Actual concentrations of Ba²⁺ and CO₃²⁻ are:

[Ba²⁺] = [Ba(NO₃)₂] = 1.1x10⁻³ × (20.0mL / 100.0mL) = 2.2x10⁻⁴M

[CO₃²⁻] = [Na₂CO₃] = 8.4x10⁻⁴ × (80.0mL / 100.0mL) = 6.72x10⁻⁴M

100.0mL is the volume of the mixture of the solutions

Replacing in Q expression:

Q = [Ba²⁺] [CO₃²⁻]

Q = [2.2x10⁻⁴M] [6.72x10⁻⁴M]

Q = 1.5x10⁻⁷

As Q > Ksp

Answer: The volume of liquid used will be 18.055 mL.

Answer:

NO3 has 3 atoms of oxygen in it per molecule (indicated by the subscript of 3) and (NO3)2 means you have 2 NO3 molecules, meaning you double the subscript to get 6 oxygen atoms.

Explanation:

Answer:

[tex]\% m/m= 14.3\%[/tex]

Explanation:

Hello,

In this case, the by mass percent is computed as shown below:

[tex]\% m/m=\frac{m_{solute}}{m_{solute}+m_{solvent}} *100\%[/tex]

Whereas the solute is the sugar and the solvent the water, therefore, the concentration results:

[tex]\% m/m=\frac{63g}{63g+378g} *100\%\\\\\% m/m= 14.3\%[/tex]

Best regards.

Answer

Iridium(IV)Nitrite

The correct IUPAC name of the Ir(NO₂)₄ compound is Iridium(IV)Nitrite.

Whether it's in a continuous chain or just a ring, the largest chain of carbons joined by a single bond serves as the basis for IUPAC nomenclature.

A chemical compound would seem to be a substance that contains numerous similar molecules made of atoms from different elements joined by chemical bonds.

The given compound is Ir(NO₂)₄. It can be seen that 4 nitro group is attached with Ir and its coordination number is 4. Hence, the IUPAC name will be Iridium(IV)Nitrite.

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Answer:

W = -120 KJ

Explanation:

Since the piston–cylinder assembly undergoes an isothermal process, then the temperature is constant.

Thus; T1 = T2 = 400K

change in entropy; ΔS = −0.3 kJ/K

Formula for change in entropy is written as;

ΔS = Q/T

Where Q is amount of heat transferred.

Thus;

Q = ΔS × T

Q = -0.3 × 400

Q = -120 KJ

From the first law of thermodynamics, we can find the workdone from;

Q = ΔU + W

Where;

ΔU is Change in the internal energy

W = Work done

Now, since it's an ideal gas model, the change in internal energy is expressed as;

ΔU = m•C_v•ΔT

Where;

m is mass

C_v is heat capacity at constant volume

ΔT is change in temperature

Now, since it's an isothermal process where temperature is constant, then;

ΔT = T2 - T1 = 0

Thus;

ΔU = m•C_v•ΔT = 0

ΔU = 0

From earlier;

Q = ΔU + W

Thus;

-120 = 0+ W

W = -120 KJ

Answer: D - 126.4g

Explanation:  

% Yield = Actual Yield/Theoretical Yield

38% = Actual Yield/332.5

38/100 = Actual Yield/332.5

(.38)(332.5) = 126.35 g = 126.4 g Actual Yield

Answer:

is D. the correct answer

Explanation:

I'm not sure if it is. Please let me know if I'm mistaking.

Answer:

I'm not 100% sure but I think A. "None of these" is correct

Explanation:

Option B. "aq" denotes that the substance is an aqueous solution

Option C. "l" means the element is in a liquid state

Option D. "g" denotes that the substance is in a gaseous state

Hope this helped :))

The phase or state of matter is the form that is observable and is in the form of solid, liquid, or gas. The phases of the matter can be expressed as aq, l, g, s. Thus, option A is correct.

Phases of matter are distinctive and observable matters that have varying particle composition, volume, shape, symbols, etc. There are four known states of matter namely gas, liquid, plasma, and solid.

The different phases of matter have different properties like the arrangement of the particles in solid are very closed, liquids are loosely bonded, and gases have free-flowing. This makes solid rigid, liquids and gases free flow.

The phases of matter can be expressed in the equations in the brackets along the reactants and products. The solids are represented as (s), liquids (l), gases (g), and aqueous (aq).

Therefore, option A. none of these is the correct option.

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Answer:

To calculate the [OH-] in the solution we must first find the pOH

That's

pH + pOH = 14

pOH = 14 - pH

First to find the pH we use the formula

From the question

[H3O+]= 2.6 × 10^-8 M

pH = - log 2.6 × 10^-8

pH = 7.6

pH = 8

So we pOH is

pOH = 14 - 8 = 6

To find the [OH-] we use the formula

6 = - log [OH-]

Find antilog of both sides

The solution is slightly basic since it's pH is in the basic region and slightly above the neutral point 7

Hope this helps you

Answer:

Explanation:

Number of moles of propane:

=Mass in grams ÷ Relative molecular Mass

= 100/((12*3) + (1*8))

= 100 ÷ 44

= 2.2727

Mole ratio propane:carbon (IV) oxide = 1:3(from the equation)

Number of moles of CO2 = 3*2.2727 = 6.8181

Mass in grams = Relative molecular Mass * Number of moles

= 44 * 6.8181

= 299.9964 grams

A number of moles of propane:

Mass in grams ÷ Relative molecular Mass

= 100/((12*3) + (1*8))

= 100 ÷ 44

= 2.2727

Mole ratio propane:carbon (IV) oxide = 1:3(from the equation)

Number of moles of CO2 = 3*2.2727 = 6.8181

Mass in grams = Relative molecular Mass * Number of moles

=44 * 6.8181

= 299.9964 grams

Carbon dioxide is used as a refrigerant, in fireplace extinguishers, for inflating lifestyles rafts and life jackets, blasting coal, foaming rubber and plastics, selling the increased vegetation in greenhouses, and immobilizing animals earlier than slaughter, and in carbonated liquids.

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Answer:

I think long roots

Explanation:

Answer:

1. pH = 2.98

2. pH = 4.02

3. pH = 8.12

Explanation:

1. Initial molarity of benzoic acid (Molar mass: 122.12g/mol; Ka = 6.14x10⁻⁵) is:

0.235 ₓ (1mol / 122.12g) = 1.92x10⁻³ moles / 0.100L = 0.01924M

The equilibrium of benzoic acid with water is:

C6H5CO2H(aq) + H2O(l) → C6H5O-(aq) + H3O+(aq)

And Ka is defined as the ratio between equilibrium concentrations of products over reactants, thus:

Ka = 6.14x10⁻⁵ = [C6H5O⁻] [H3O⁺] / [C6H5CO2H]

The benzoic acid will react with water until reach equilibrium. And equilibrium concentrations will be:

[C6H5CO2H] = 0.01924 - X

[C6H5O⁻] = X

[H3O⁺] = X

Replacing in Ka:

6.14x10⁻⁵ = [X] [X] / [0.01924 - X]

1.1815x10⁻⁶ - 6.14x10⁻⁵X = X²

1.1815x10⁻⁶ - 6.14x10⁻⁵X - X² = 0

Solving for X:

X = -0.0010→ False solution. There is no negative concentrations

X = 0.0010567M → Right solution.

pH = - log [H3O⁺] and as [H3O⁺] = X:

pH = - log [0.0010567M]

2.

pH of a buffer is determined using H-H equation (For benzoic acid:

pH = pka + log [C6H5O⁻] / [C6H5OH]

pKa = -log Ka = 4.21 and [] could be understood as moles of each chemical

The benzoic acid reacts with NaOH as follows:

C6H5OH + NaOH → C6H5O⁻ + Na⁺ + H₂O

That means NaOH added = Moles C6H5O⁻ And C6H5OH = Initial moles (1.92x10⁻³ moles - Moles NaOH added)

7.00mL of NaOH 0.108M are:

7x10⁻³L ₓ (0.108 mol / L) = 7.56x10⁻⁴ moles NaOH = Moles C₆H₅O⁻

And moles C6H5OH = 1.92x10⁻³ moles - 7.56x10⁻⁴ moles = 1.164x10⁻³ moles C₆H₅OH

Replacing in H-H equation:

pH = 4.21 + log [7.56x10⁻⁴ moles] / [ 1.164x10⁻³ moles]

3. At equivalence point, all C6H5OH reacts producing C6H5O⁻. The moles are 1.164x10⁻³ moles

Volume of NaOH to reach equivalence point:

1.164x10⁻³ moles ₓ (1L / 0.108mol) = 0.011L. As initial volume was 0.100L, In equivalence point volume is 0.111L and concentration of C₆H₅O⁻ is:

1.164x10⁻³ moles / 0.111L = 0.01049M

Equilibrium of  C₆H₅O⁻ with water is:

C₆H₅O⁻(aq) + H₂O(l) ⇄  C₆H₅OH(aq) + OH⁻(aq)

Kb = [C₆H₅OH] [OH⁻]/ [C₆H₅O⁻]

Kb = kw / Ka = 1x10⁻¹⁴ / 6.14x10⁻⁵ = 1.63x10⁻¹⁰

Equilibrium concentrations of the species are:

C₆H₅O⁻ = 0.01049M - X

C₆H₅OH = X

OH⁻ = X

Replacing in Kb expression:

1.63x10⁻¹⁰ = X² / 0.01049- X

1.71x10⁻¹² - 1.63x10⁻¹⁰X - X² = 0

Solving for X:

X = -1.3x10⁻⁶ → False solution

X = 1.3076x10⁻⁶ → Right solution

[OH⁻] =  1.3076x10⁻⁶

as pOH = -log [OH⁻]

pOH = 5.88

And pH = 14 - pOH

Answer: d=2000 g/L

Explanation:

Density is mass/volume. The units are g/L. Since we are given mass and volume, we can divide them to find density. First, we need to convert kg to g.

[tex]10kg*\frac{1000g}{1kg} =10000 g[/tex]

Now that we have grams, we can divide to get density.

[tex]d=\frac{10000g}{5 L}[/tex]

d=2000g/L

Option A is correct. Oxygen-16 occurs most naturally among these isotopes.

The three stable isotopes that make up natural oxygen are O-16, O-17, and 0-18, with O-16 having the highest natural abundance (99.762%). The standard atomic weight ranges between [15.99903, 15.99977], depending on the terrestrial source .

Oxygen-16 (16O) is a stable oxygen isotope with 8 neutrons and 8 protons in its nucleus. Its mass is 15.99491461956 u. Oxygen-16 is the most abundant isotope of oxygen, accounting for 99.762% of natural oxygen abundance. 16O has a high relative and absolute abundance because it is a primary product of stellar evolution and a primordial isotope, which means it can be produced by stars that were originally made entirely of hydrogen. The majority of 16O is produced in stars at the end of the helium fusion process; the triple-alpha process produces 12C, which catches an additional 4He to produce 16O. The process of burning neon generates an additional 16O. Oxygen-16 is twice as powerful.

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Answer:

Propilen là một chất khí không màu với mùi giống như dầu mỏ. Propilen nhẹ hơn nước và tan rất ít trong nước 0.61 g/ . Không hòa tan trong các dung môi phân cực như nước, chỉ tan trong dung môi không phân cực hay ít phân cực.. Propilen không có tính dẫn điện.

Answer:

the mas is .291 g

Explanation:

the mass of a object does not change. so when added the substance the beaker. you had the mass of both objects together. you know the mass of the beaker and you know the mass of both. since mass does not change. the beakers mass is still 74.605g. the mass of both objects is 74.896. all you have to do is subtract the mass of the beaker from the total mass. 74.896 - 74.605 equals .291g. so the mass of the unknown substance Is .291g

Answer:

0 g.

Explanation:

Hello,

In this case, since the reaction between methane and oxygen is:

[tex]CH_4+2O_2\rightarrow CO_2+2H_2O[/tex]

If 0.963 g of methane react with 7.5 g of oxygen the first step is to identify the limiting reactant for which we compute the available moles of methane and the moles of methane consumed by the 7.5 g of oxygen:

[tex]n_{CH_4}=0.963gCH_4*\frac{1molCH_4}{16gCH_4}=0.0602molCH_4\\ \\n_{CH_4}^{consumed}=7.5gO_2*\frac{1molO_2}{32gO_2}*\frac{1molCH_4}{2molO_2} =0.117molCH_4[/tex]

Thus, since oxygen theoretically consumes more methane than the available, we conclude the methane is the limiting reactant, for which it will be completely consumed, therefore, no remaining methane will be left over.

[tex]left\ over=0g[/tex]

Regards.