He rate constant for the forward reaction, k1, is 297 l·mol–1·min–1 and the rate constant for the reverse reaction, k–1, is 393 l·mol–1·min–1 at a given temperature. the activation energy for the forward reaction is 42.1 kj·mol–1, while the activation energy for the reverse reaction is 22.1 kj·mol–1. determine the equilibrium constant, k, of this reaction.

Answers

Answer 1
Equilibrium constant of reaction,
 = [tex] \frac{rate.constant.forward.reaction}{rate.constant.of.backward.reaction} [/tex]
= [tex] \frac{297}{393} [/tex]
=0.7557
Answer 2

The equilibrium constant is a number that shows the extent to which reactants are converted into products. The equilibrium constant of the reaction is 0.76.

What is equilibrium constant?

The equilibrium constant is a number that shows the extent to which reactants are converted into products.

We know that to obtain the equilibrium constant; K = k1/k-1 where;

k-1 = rate constant of reverse reaction

k1 = rate constant of forward reaction.

Hence;

K = 297 l·mol–1·min–1/393 l·mol–1·min–1

K = 0.76

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Related Questions

Nitric acid, hno3, dissociates in water to form nitrate ions and hydronium ions. what change in hybridization of the nitrogen atom occurs in this dissociation

Answers

According to valence bond theory, atomic orbitals of slightly different mix and recast to form new orbitals of identical energies and properties. This process as called hybridization and, new generated orbitals are referred as hybridized orbitals.
Also, according to assumption of VBT, only central atom undergoes hybridization.
In case of HNO3, N is a central atom and it undergoes sp2 hybridization in HNO3. The structure of HNO3 is planner. 
Further, HNO3 comprises of H+ and NO3- ion. Here, role of H+ is to act as counter ion and balance the charge on NO3-
Now, when HNO3 is mixed in water, it undergoes dissociation to generate H+ and NO3- ion. Here, parent ion i.e. NO3- is not disturbed. Hence, it's hybridization will not change. Only there will be minor variation in bond length and bond angle because of interaction of NO3- with H2O molecules. 

Final answer:

The nitrogen atom in HNO3 and the resulting nitrate ion after dissociation in water both exhibit sp² hybridization. Thus, there is no change in hybridization of the nitrogen atom during this dissociation process.

Explanation:

When nitric acid, HNO3, dissociates in water, the nitrogen atom undergoes a change in hybridization. In nitric acid, the nitrogen is sp² hybridized as it is bonded to three oxygen atoms and has one lone pair of electrons. Upon dissociation into nitrate ions (NO3⁻) and hydronium ions (H3O⁺), the nitrogen atom in the nitrate ion becomes sp² hybridized as the lone pair used to create the coordinate bond with hydrogen in HNO3 is lost and replaced by a bond with an oxygen atom to form the nitrate ion. Therefore, the hybridization of nitrogen does not change; it remains sp² before and after dissociation in water.

which of the following is a correctly written thermochemical equation

Answers

Answer:

A - 4Fe(s) + 3O₂(g) → 2Fe₂O₃(s), ∆H = -3,926kj

Explanation:

Thermochemical equations:

Are balanced chemical equationsInclude the physical states of all reactants and productsIncludes the enthalpy change of the reaction

The correct thermochemical equation is :

4Fe(s)  +  3O₂(g)  --->  Fe₂O₃(s)     ΔH = -3,926 kJ. The correct option is first.

The thermochemical equation is the equation which is balanced chemical equation that shows magnitude of the enthalpy value The enthalpy value with the sign that is the positive sign means it is endothermic process, and the negative sign is an exothermic process.

4Fe(s)  +  3O₂(g)  --->  Fe₂O₃(s)     ΔH = -3,926 kJ

This is the thermochemical equation.

NH₄Cl -->  NH₄⁺  +Cl⁻

This is not the thermochemical as it does not the enthalpy value.

C₃H₈(g)   +  O₂(g)  ---> CO₂(g)  +  H₂O(l)    ΔH = -2,220 kJ/mol

This is also not the thermochemical equation.

2C₈H₁₈  +  25O₂  ---> 6CO₂ +  18H₂O      ΔH = - 5,471 kJ/mol

This is not the thermochemical equation it is not balanced equation. The first option is correct.

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Pb(oh)2(s) → pb2+(aq) + 2oh-(aq) ksp = 1.4 x 10-20 what is the solubility of lead hydroxide in pure water? (simplify by neglecting the oh- from water.)

Answers

The solubility of lead hydroxide in pure water is approximately 1.5 × [tex]10^{-7}[/tex] M.

Solubility of Lead Hydroxide in Pure Water

To find the solubility of lead hydroxide, Pb(OH)2, in pure water, we start with the dissociation equation:

Pb(OH)2 (s) → Pb2+ (aq) + 2 OH- (aq)

The solubility product constant (Ksp) for this reaction is given as 1.4 × [tex]10^{-20}[/tex]. Let s be the molar solubility of Pb(OH)2, which means s moles of Pb(OH)2 dissolve to form s moles of Pb2+ and 2s moles of OH-:

Ksp = [Pb2+][tex][OH-]^2[/tex]

Substitute the concentrations: Ksp = (s)(2s)2

Ksp = 4[tex]s^3[/tex]

Given Ksp = 1.4 × [tex]10^{-20}[/tex], we solve for s:

1.4 ×[tex]10^{-20}[/tex] = 4[tex]s^3[/tex]

[tex]s^3[/tex] = (1.4 × [tex]10^{-20}[/tex]) / 4

[tex]s^3[/tex] = 3.5 × [tex]10^{-21}[/tex]

s = (3.5 × [tex]10^{-21}[/tex])[tex]^{1/3}[/tex]

s ≈ 1.5 × [tex]10^{-7}[/tex] M

Thus, the solubility of lead hydroxide in pure water is approximately 1.5 × [tex]10^{-7}[/tex] M.

The solubility of lead hydroxide in pure water, neglecting the contribution of OH- from water, is approximately [tex]\( 1.52 \times 10^{-7} \)[/tex] M.

The solubility of lead hydroxide, [tex]Pb(OH)_2[/tex], in pure water can be determined from its solubility product constant (Ksp). The Ksp expression for [tex]Pb(OH)_2[/tex] is given by:

[tex]\[ K_{sp} = [Pb^{2+}][OH^-]^2 \][/tex]

Let [tex]\( s \)[/tex] represent the molar solubility of [tex]Pb(OH)_2[/tex], which is the concentration of [tex]Pb^{2+}[/tex] ions in solution at equilibrium. Since for every mole of [tex]Pb(OH)_2[/tex] that dissolves, 1 mole of [tex]Pb^{2+}[/tex] and 2 moles of OH- are produced, the concentration of OH- ions will be [tex]\( 2s \)[/tex].

Therefore, the Ksp expression can be rewritten as:

[tex]\[ K_{sp} = s(2s)^2 \][/tex]

[tex]\[ K_{sp} = 4s^3 \][/tex]

Given that [tex]\( K_{sp} = 1.4 \times 10^{-20} \)[/tex], we can solve for [tex]\( s \)[/tex]:

[tex]\[ 1.4 \times 10^{-20} = 4s^3 \][/tex]

[tex]\[ s^3 = \frac{1.4 \times 10^{-20}}{4} \][/tex]

[tex]\[ s^3 = 3.5 \times 10^{-21} \][/tex]

[tex]\[ s = \sqrt[3]{3.5 \times 10^{-21}} \][/tex]

Taking the cube root of both sides gives us the solubility [tex]\( s \)[/tex]:

[tex]\[ s = (3.5 \times 10^{-21})^{1/3} \][/tex]

[tex]\[ s \approx 1.52 \times 10^{-7} \text{ M} \][/tex]

A compound is found to contain 49.5% carbon, 5.19% hydrogen, 16.5% oxygen, and 28.9% nitrogen. its molecular mass is 194.2 g/mol. what is its empirical formula? what is its molecular formula? explain what each of these formulas tells us about the molecule. (lesson 6)

Answers

28.9 uno mum mum mum  yen tab rf  etc wax ohm yen rev wax tab yen umm ism ol, edc 

When hot lava reaches seawater, the salts in the water react with the steam to form gaseous hydrochloride acid. You are given a unbalanced chemical equation for one such reaction and the volume of HCI(g) produced. Explain how you would find the mass of solid sea salt needed to produce the given gas volume.

Answers

NaCl + H2O = HCl + NaOH 
FIRST write the equation and balance the equation THEN convert the volume of the gas to moles of the gas PV=nRT THEN according to the stoichiometry of the reaction one mol of gas is produced by one mol of salt. Convert the mol of gas to mol of salt FINALLY calculate the molar mass of the salt and convert the mol of salt to g of salt (m= nxMM)

Final answer:

To find the mass of solid sea salt needed to produce the given gas volume of hydrochloric acid (HCl), we need to use a balanced chemical equation and the molar ratio between the reactants. The volume of HCl(g) can be converted to moles using the ideal gas law, and then the moles of the other reactant can be calculated using the molar ratio. The mass of the solid sea salt can be found by converting the moles to grams using the molar mass.

Explanation:

To find the mass of solid sea salt needed to produce the given gas volume of hydrochloric acid (HCl), we need to use the balanced chemical equation. The equation you provided is not directly related to the reaction between hot lava and seawater, so it cannot be used to determine the mass of sea salt. However, we can use the equation HCl(aq) + Mg(OH)2(aq) → H2O(l) + MgCl2(aq) as an example.

In this equation, we can see that 1 mole of HCl reacts with 1 mole of Mg(OH)2 and produces 1 mole of water. From the equation, we can calculate the molar ratio between HCl and Mg(OH)2.

If we know the volume of HCl(g) produced, we can use the ideal gas law to convert it to moles. Then, using the molar ratio, we can calculate the moles of Mg(OH)2 needed. Finally, we can convert the moles of Mg(OH)2 to grams using its molar mass to find the mass of solid sea salt needed.

What is formed from 2 pieces of different metals stuck together lengthwise?

Answers

Final answer:

An alloy, such as brass or bronze, is formed when two pieces of different metals are stuck together lengthwise. These materials combine the properties of the constituent metals, often resulting in superior characteristics.

Explanation:

When two pieces of different metals are stuck together lengthwise, the resulting material is known as an alloy. An alloy is a mixture composed of two or more elements, with at least one being a metal. Some common examples of alloys include brass (an alloy of copper and zinc) and bronze (an alloy of copper and tin). These combinations of metals result in a material with properties that are superior to the pure metals themselves. For example, bronze, first used around 2400 B.C., is harder and more durable than either of its constituent metals, copper and tin.

Final answer:

The material formed from sticking two different metals together lengthwise is called an alloy. Alloys like bronze and brass are mixtures of two or more elements that possess improved properties compared to the pure component metals.

Explanation:

When two pieces of different metals are stuck together lengthwise, the material formed is known as an alloy. An alloy is a mixture composed of two or more elements, at least one of which is a metal. Alloys are designed to have properties superior to those of the pure metals from which they are made. For example, bronze is a well-known alloy consisting mainly of copper and tin. The creation of alloys like bronze and brass, which is an alloy of copper and zinc, exemplifies how metals can be combined to achieve desired physical and chemical properties for various applications, from tools and weapons to musical instruments.

How much energy is required to vaporize 185 g of butane at its boiling point? the heat of vaporization for butane is 23.1 kj/mol?

Answers

Answer is: 73.52 kJ of energy is required to vaporize butane.
m(C₄H₁₀) = 185 g.
n(C₄H₁₀) = m(C₄H₁₀) ÷ M(C₄H₁₀).
n(C₄H₁₀) = 185 g ÷ 58.12 g/mol.
n(C₄H₁₀) = 3.18 mol; amount of butane.
Hvap = 23.1 kJ/mol; the heat of vaporization for butane.
Q = Hvap · n(C₄H₁₀).
Q = 23.1 kJ/mol · 3.18 mol; energy.
Q = 73.52 kJ.

Final answer:

Approximately 73.5 kJ of energy is required to vaporize 185 g of butane at its boiling point.

Explanation:

The amount of energy required to vaporize a substance can be calculated using the formula Q = n * ΔHvap, where Q is the amount of energy required, n is the number of moles of the substance, and ΔHvap is the heat of vaporization. To calculate the number of moles of butane, we divide the mass of butane by its molar mass. Using the given heat of vaporization for butane (23.1 kJ/mol), we can calculate the amount of energy required. First, we calculate the number of moles of butane:

n = 185 g / 58.12 g/mol = 3.18 mol

Next, we calculate the amount of energy required:

Q = 3.18 mol * 23.1 kJ/mol ≈ 73.5 kJ

Therefore, approximately 73.5 kJ of energy is required to vaporize 185 g of butane at its boiling point.

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How many liters of .100 m hcl hcl would be required to react with 5 grams of calcium hydroxide?

Answers

The  number  of liters  of  0.100 HCl that would  be required  to  react with  5g of Ca(OH)2 is calculated  as  follows

find the   moles of Ca(OH)2  used

= mass/molar mas
=  5g/ 74.09 g/mol  = 0.0675  moles
write the  equation  involved

that is Ca(OH)2 +2 HCl = CaCl2 + 2 H2O

by use of mole ratio  between  Ca(OH)2  to HCl  which  is 1:2  the  moles  of  HCl  is therefore = 2 x 0.0675 = 0.135  moles  of HCl

volume of HCl =  moles of HCL / molarity of HCl

= 0.135/ 0.100 = 1.35 L
the balanced equation for the reaction between HCl and Ca(OH)₂ is as follows;
Ca(OH)₂ + 2HCl ---> CaCl₂  + 2H₂O
stoichiometry of Ca(OH)₂ to HCl is 1:2
mass of Ca(OH)₂ reacting - 5 g
therefore number of moles of Ca(OH)₂ - 5 g / 74 g/mol = 0.068 mol 
according to molar ratio
number of HCl moles reacted = twice the number of Ca(OH)₂ reacted 
number of HCl moles required - 0.068 x 2 = 0.136 mol 
molarity if HCl solution - 0.100 M
there are 0.100 mol of HCl in 1 L 
therefore 0.136 mol in - 0.136 mol / 0.100 mol/L = 1.36 L
volume of 0.100 M HCl required - 1.36 L 

A sample of ammonia (nh3) gas is completely decomposed to nitrogen and hydrogen gases over heated iron wool. if the total pressure is 853 mmhg after the reaction, calculate the partial pressures of n2 and h2. enter your answers in scientific notation.

Answers

The partial pressure of N2 and H2 is calculated as below

write the reaction for decomposition of NH3

NH3 =N2 +3 H2

 from the reaction above 1 mole of N2  and 3 moles of H2 is formed and therefore the total mole of the product  = 4

the reaction coefficient for N2 from the reaction = 1/4
and that of H2 = 3/4 

the final pressure = partial pressure of N2  +partial pressure of H2

partial pressure  of N2 = 1/4 x853 mm hg = 213.25 mmhg
partial pressure of H2 = 3/4 x853mm hg=639.75 mm hg
Final answer:

The partial pressures of nitrogen and hydrogen, after decomposing ammonia are calculated using the reaction stoichiometry and Avogadro's law. The nitrogen's partial pressure is 2.1325 x 102 mmHg and the hydrogen's partial pressure is 6.3975 x 102 mmHg.

Explanation:

To calculate the partial pressures of N2 and H2, we will depend on the reaction stoichiometry given by the equation N₂(g) + 3H₂(g) = 2NH3(g). In this chemical reaction, one molecule of nitrogen gas reacts with three molecules of hydrogen gas to produce two molecules of ammonia gas. An important thing to remember here is that gases react in definite and simple proportions by volume, which is derived from Avogadro's law.

The decomposition of ammonia produces 1 volume of nitrogen for every 3 volumes of hydrogen - so, in a decomposed sample, the hydrogen will always be three times as concentrated as the nitrogen. So if we let nitrogen's partial pressure be x, the hydrogen's will be 3x. As the total pressure is given as 853 mmHg, it will be the sum of the nitrogen and hydrogen pressures and we can write:

x + 3x = 853,

Solving for x gives x = 213.25 mmHg, so the nitrogen's partial pressure is 213.25 mmHg and hydrogen's pressure is 3 * 213.25 = 639.75 mmHg. Therefore, the partial pressures of nitrogen and hydrogen are 2.1325 x 102 mmHg and 6.3975 x 102 mmHg respectively in scientific notation.

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What are the concentrations of hydrogen ion and hydroxide ion in household ammonia, an aqueous solution of nh3 that has a ph of 11.50?

Answers

Answer is: the concentrations of hydrogen ion and hydroxide ion in household ammonia are 3.16·10⁻¹² M and 3.16·10⁻³ M, respectively.
pH(NH₃) = 11.50.
pH = -log[H⁺].
[H⁺] = 10∧(-pH).
[H⁺] = 10∧(-11.5).
[H⁺] = 3.16·10⁻¹² M.
[H⁺] · [OH⁻] = 10⁻¹⁴ M².
[OH⁻] = 10⁻¹⁴ M² ÷ 3.16·10⁻¹² M.
[OH⁻] = 0.00316 M = 3.16·10⁻³ M.

Note: hydrogen ion concentration = 3.16 x 10⁻¹² M

          hydroxide ion concentration = 3.16 x 10⁻³ M

          pH NH₃ = 11.50

Asked: concentrations of hydrogen ions and hydroxide ions in household ammonia?

Answer: pH (NH₃) = 11.50.

              pH = -log [H⁺]

              [H⁺] = 10∧ (-pH)

              [H⁺] = 10∧ (-11.5)

               [H⁺] = 3.16 · 10⁻¹² M

              [H⁺] x [OH⁻] = 10⁻¹⁴ M²

              [OH⁻] = 10⁻¹⁴ M² ÷ 3.16 x 10⁻¹² M

               [OH⁻] = 0.00316 M = 3.16 x 10⁻³ M

Thus, the concentration of hydrogen ions and hydroxide ions in household ammonia is 3.16 · 10⁻³M

Further Explanation

In chemistry, concentration is a measure that describes the amount of substance in a mixture divided by the total volume of the mixture. There are four kinds of quantitative descriptions of concentration, namely mass concentration, molar concentration, total concentration, and volume concentration. The term concentration can be applied to all types of mixtures, but it is most often used to describe the amount of solute in the solution. Molar concentrations have variations such as normal concentration and osmotic concentration.

Solution Concentration

Concentration is a way to express the quantitative relationship between solute and solvent. Expressing the concentration of the solution there are several types, including:

1. MOL FRACTION

The mole fraction is the ratio between the number of moles of a component with the number of moles of all components contained in a solution.

The mole fraction is denoted by X.

2. PERCENT WEIGHT

Percent weight states the gram weight of the solute in 100 grams of solution.

3. MOLALITY (m)

Molality states the mole of solute in 1000 grams of solvent.

4. MOLARITY (M)

Molarity states the number of moles of solute in 1 liter of solution.

5. NORMALITY (N)

Normality represents the number of moles equivalent of solute in 1 liter of solution. For acids, 1 mole is equivalent to 1 mole of H + ions. For bases, 1 mole is equivalent to 1 mole of OH- ion.

Between Normality and Molarity there is a relationship:

N = M x valence

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Grade: College

Subject: Chemistry

keywords: concentration

An aqueous solution of ____ will produce a basic solution.
a. nh4clo4
b. kbr
c. nacl
d. na2co3
e. nahco3

Answers

Hydrolysis of [tex]NH_{4}ClO_{4}[/tex] is given as:

[tex]NH_{4}ClO_{4}+H_{2}O\rightleftharpoons NH_{4}OH + HClO_{4}[/tex]

Here, [tex]NH_{4}OH[/tex] is a   weak base and [tex]HClO_{4}[/tex] is a strong acid. Thus, solution is more acidic

Hydrolysis of [tex]KBr[/tex] is given as:

[tex]KBr+H_{2}O\rightleftharpoons KOH  +   HBr[/tex]

Here, [tex]KOH[/tex] is a strong base and [tex]HBr[/tex] is a strong acid.Thus, solution is neutral.

Hydrolysis of [tex]NaCl[/tex] is given as:

[tex]NaCl+H_{2}O\rightleftharpoons NaOH  + HCl[/tex]

Here, [tex]NaOH[/tex] is a strong base and [tex]HCl[/tex] is a strong acid.Thus, solution is neutral.

[tex]Na_{2}CO_{3}+2H_{2}O\rightleftharpoons  2NaOH + H_{2}CO_{3}[/tex]

Here, [tex]NaOH[/tex] is a strong base and [tex]H_{2}CO_{3}[/tex] is a weak acid.Thus, solution is basic.      

[tex]NaHCO_{3}+H_{2}O\rightleftharpoons  NaOH + H_{2}CO_{3}[/tex]

Here, [tex]NaOH[/tex] is a strong base and [tex]H_{2}CO_{3}[/tex] is a weak acid.Thus, solution is basic.        

Hence, an aqueous solution of  [tex]NaHCO_{3}[/tex] and [tex]Na_{2}CO_{3}[/tex] will produce a basic solution.





what causes a dipole-dipole force?
A. An attraction between two nonpolar molecules through Van der Waals forces.
B. An attraction between the positive end of one molecule and the negative end of another.
C. An attraction between two metal atoms
D. An attraction between two ions within a solid

Answers

Answer:

b. an attraction between the positive end of one molecule and the negative end of another

Explanation:

An attraction between the positive end of one molecule and the negative end of another. Hence, option B is correct.

What is dipole-dipole force?

Dipole-Dipole forces are the interaction between molecules of the permanent dipole. It occurs between the partially charged positive molecules and partially charged negative molecules.

Dipole-dipole forces are attractive forces between the positive end of one polar molecule and the negative end of another polar molecule.

Hence, an attraction between the positive end of one molecule and the negative end of another is a cause of dipole-dipole force.

Hence, option B is correct.

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Δs is positive for the reaction ________.
a.2no (g) + o2 (g) → 2no2 (g)
b.2n2 (g) + 3h2 (g) → 2nh3 (g)
c.c3h8 (g) + 5 o2 (g) → 3co2 (g) + 4 h2o (g)
d.mg (s) + cl2 (g) → mgcl2 (s)
e.c2h4 (g) + h2 (g) → c2h6 (g)

Answers

C. The reaction goes from one of less moles to more causing it to be more disordered and therefore have a positive change in s

Answer: Option (c) is the correct answer.

Explanation:

Entropy means the degree of randomness present in a substance or within the reactants in a chemical reaction.

Change in entropy is represented by [tex]\Delta S[/tex]. More is the degree of randomness present more positive will be the value of [tex]\Delta S[/tex]. Similarly, less is the degree of randomness present within a substance lesser will be the value of [tex]\Delta S[/tex].

(a)    [tex]2NO(g) + O_{2}(g) \rightarrow 2NO_{2}(g)[/tex]

Here, 3 moles of reactants are giving 2 moles of product. Hence, entropy is decreasing so, the value of [tex]\Delta S[/tex] is negative.

(b)   [tex]2N_{2}(g) + 3H_{2}(g) \rightarrow 2NH_{3}(g)[/tex]

Here, 5 moles of reactants are giving 2 moles of product.  Hence, entropy is decreasing so, the value of [tex]\Delta S[/tex] is negative.

(c)   [tex]C_{3}H_{8}(g) + 5O_{2}(g) \rightarrow 3CO_{2}(g) + 4H_{2}O(g)[/tex]

Here, 6 moles of reactants are giving 7 moles of product.  Hence, entropy is increasing so, the value of [tex]\Delta S[/tex] is positive.

(d) [tex]Mg(s) + Cl_{2}(g) \rightarrow MgCl_{2}(s)[/tex]

Here, 2 moles of reactants are giving 1 mole of product.  Hence, entropy is decreasing so, the value of [tex]\Delta S[/tex] is negative.

(e)  [tex]C_{2}H_{4}(g) + H_{2}(g) \rightarrow C_{2}H_{6}(g)[/tex]

Here, 2 moles of reactants are giving 1 mole of product.  Hence, entropy is decreasing so, the value of [tex]\Delta S[/tex] is negative.

Thus, we can conclude that [tex]\Delta S[/tex] is positive for the reaction [tex]C_{3}H_{8}(g) + 5O_{2}(g) \rightarrow 3CO_{2}(g) + 4H_{2}O(g)[/tex].

Based on conductivity values, classify the compounds in each set of solutions as molecular, ionic, or acids. what is the level of dissociation (complete, varying, or none) of each set

Answers

Electrolytic conductivity is measure of ability of solution to conduct electricity. It is also as specific conductance. Electrolytic conductivity is different as compared to electrical conductivity. Electrolytic conductivity increase with temperature, while electrical conductivity decreases with increasing temperature. Further, electrolytic conductivity occurs because of presence of ions in solutions, while electrical conductivity occurs due to presence of valence electrons.

In current context, number of ions present in solutions depends on dissociation constant values. For molecules, which donot under dissociation in solvent, no electrolytic conductivity will be observed. Generally, ionic compounds undergo complete dissociation in solution, hence they will display maximum conductance. In case of acids, conductivity values will vary depends upon dissociation constant values. Weak acid are characterized by lower dissociation constant values, hence they have lower conductivity as compared to strong acids.  

You have a racemic mixture of d-2-butanol and l-2-butanol. the d isomer rotates polarized light by +13.5∘. what is the rotation of the polarization of light of your mixture? express your answer in degrees.

Answers

l- isomer is considered as the Enantiomer of d- isomer and since the d-isomer optical rotation is + 13.5° so the optical rotation of l-isomer will be the same degree but with opposite sign which equal to -13.5° 

So the degree of rotation of racemic mixture will equal 0° 


- This is because racemic mixture contains equal amount of both enantiomers

A racemic mixture of d-2-butanol and l-2-butanol has no net optical rotation, resulting in a rotation of 0 degrees.

A racemic mixture is composed of equal amounts of enantiomers, which in this case are d-2-butanol and l-2-butanol.

When the d and l isomers are present in equal amounts, as they are in a racemic mixture, their optical activities cancel each other out. This is because the rotation caused by one enantiomer is exactly counteracted by the rotation caused by the other enantiomer. Therefore, the overall rotation of polarized light by the racemic mixture is the sum of the individual rotations:

Rotation of d-2-butanol: +13.5 degrees

Rotation of l-2-butanol: -13.5 degrees

Sum of rotations: +13.5 degrees + (-13.5 degrees) = 0 degrees

Thus, the racemic mixture does not rotate polarized light, and the rotation of the polarization of light of the mixture is 0 degrees.

1. If you have 4.0 g of a radioactive substance left after 4 half-lives have passed, then how much did you have originally?
a. 50 g
b. 64 g
c. 25 g
d. 10 g

2. If a fusion reaction yields 4.50 x 10^9 kJ of energy, then how much mass was lost?
a. 5.00 x 10^-5 kg
b. 1.50 x 10^-6 kg
c. 1.50 x 10^-3 kg
d. 5.00 x 10^-2 kg

3. For the transmutation shown, which of the following is X? (The first number is on top on the actual answer.)
a. 3 1 H
b. 6 3 Li
c. 1 1 H
d. 4 2 He


5. Which of the following energy yields is most likely to have come from a fission or fusion reaction?
a. 1.0 x 10^2 kJ/mol
b. 2.5 x 10^2 kJ/mol
c. 1.5 x 10^3 kJ/mol
d. 2.0 x 10^11 kJ/mol

Answers

Answer 1
During the radio-active disintegration, when initial concentration of radioactive nuclei is reduced to half, it is referred as half life.
Now, if initial concentration  of radio-active element is 64 g. After it crosses 1st half life, it concentration will decrease to 32g. At 2nd half life, concentration will further reduce to 16 g. This concentration will reduce to 8 g, at 3 half life. And, finally at 4th half life, concentration of radioactive element will reduce to 4 g. 
Thus ,in present case, correct answer is B 64 g.

Answer 2:
Given that energy produced during nuclear fusion reaction = 4.50 x 10^9 kJ
Now, based on Einstein's Equation, we know that E = mc2
where, E = energy produced
m = mass o f compound
c = speed of light = 3 X 10^8 m/s
Thus, 4.50 x 10^9 x 10^3J = m (3 X 10^8)^2
∴ m = 5 X 10^-5 Kg 

Thus, mass lost in the reaction = 5 X 10^-5 Kg 

Answer 3
In nuclear transmutation reaction, a non-radioactive nuclei is collided with a fast moving projectile to convert into radio-active nuclei. Following is the examples of nuclear transmutation reaction
 1 1H + 1 0n   →      2 1 H + gamma rays
 
In the above reaction hydrogen, on collision with neutron generates deuterium.
However, other elements listed above will not undergo nuclear transmutation reaction. 

Answer 4: 
In a nuclear reaction, one nuclei get converted into another nuclei. Nuclear reactions are broadly classified into two types. 
a) Nuclear fusion
b) Nuclear fission
In nuclear fusion reaction, two light nuclei under extreme conditions fuse together, while in nuclear fission reaction, a heavy nuclei is bombarded with projectile which breaks the nuclei in smaller nuclei. In either case, mass of product is less than mass of reactant. Thus there is a mass defect. Due to mass defect, tremendous amount of energy is released from the system which is of order of 10^11 Kj/mol. Thus, correct answer is option D. 

Final answer:

After 4 half-lives, a radioactive substance initially weighing 64 g would be reduced to 4.0 g. The mass lost during a fusion reaction that yields 4.50 x 10^9 kJ is 5.00 x 10^-5 kg. The energy yield most likely from a fission or fusion reaction is 2.0 x 10^11 kJ/mol.

Explanation:

To solve each problem, we need to understand the concept of half-life, which is the time required for half the atoms in a sample of a radioactive substance to decay. We also use Einstein's mass-energy equivalence principle for converting between mass lost and energy produced.

Question 1:

If you have 4.0 g of a radioactive substance left after 4 half-lives have passed, the original amount can be found by doubling the remaining mass for each half-life. So, after 1 half-life, we would have 8 g; after 2 half-lives, 16 g; after 3 half-lives, 32 g; and after 4 half-lives, 64 g.

Answer: b. 64 g

Question 2:

Using the formula E=mc2, where E is energy, m is mass, and c is the speed of light, we can rearrange to find m=E/c2. Plugging in the given energy and the value for the speed of light, we find that 4.50 x 109 kJ of energy corresponds to a mass loss of 5.00 x 10-5 kg.

Answer: a. 5.00 x 10

-5 kg

Question 5:

Fission and fusion reactions typically release a vast amount of energy compared to chemical reactions. Option d, which represents an energy yield of 2.0 x 1011 kJ/mol, is far greater than the others and is thus most likely to be from a fission or fusion reaction.

Answer: d. 2.0 x 10

11 kJ/mol

If the elastic modulus of cobalt (co) is 200 gpa, and the elastic modulus of tungsten carbide (wc) is 700 gpa, calculate the upper and lower bound estimates of the elastic modulus of a 10 vol% co – 90vol% wc composite.

Answers

To determine the upper bond
Ec(u) = EmVm + EpVp
Em is the elastic modulus of cobalt.
E₁ is the elastic modulus of the particulate
Vm is the volume fraction of cobalt
Vp is the volume fraction of particulate
substitute
Ec(u) = 200 (Vm) + 700 (Vp)
To determine the lower bound
Ec (l) = EmEp/VmEp+ VpEm
Substitute
Ec (l) = 200(700)/Vm(700) + Vp (200)
Ec (l) = 1400/7Vm+2Vp

The upper bound estimate of the composite's elastic modulus is approximately 650 GPa, and the lower bound estimate is approximately 560 GPa, using the Reuss model.

To estimate the elastic modulus of the composite, we use the rule of mixtures. The Voight model and the Reuss model are the two estimates.

Upper Bound (Voigt Model): The Voigt model assumes that the strain in both materials is the same. The formula is:

[tex]E_{upper} = V_CE_C + V_{WC}E_{WC}[/tex][tex]V_C[/tex] = Volume fraction of cobalt = 0.10[tex]V_{WC}[/tex] = Volume fraction of tungsten carbide = 0.90[tex]E_C[/tex] = Modulus of cobalt = 200 GPa[tex]E_{WC[/tex] = Wrought iron carbide modulus = 700 GPa[tex]E_{upper}[/tex] = (0.10 x 200) + (0.90 x 700) = 20 + 630 = 650 GPa

Lower Bound (Reuss Model): The Reuss model assumes that the stress in both materials is the same. The formula is:

[tex]1/E_{lower} = V_C/E_C + V_{WC}/E_{WC}[/tex][tex]V_C[/tex] = Volume fraction of cobalt = 0.10[tex]V_{WC}[/tex] = Volume fraction of tungsten carbide = 0.90[tex]E_C[/tex] = Modulus of cobalt = 200 GPa[tex]E_{WC[/tex] = Wrought iron carbide modulus = 700 GPa[tex]1/E_{lower} = (0.10/200) + (0.90/700) = 0.0005 + 0.0012857 \approx 0.0017857[/tex][tex]E_{lower} \approx 1 / 0.0017857 \approx 560\ GPa[/tex]

Result: The upper bound estimate of the elastic modulus of the 10 vol% Co – 90 vol% WC composite is 650 GPa, and the lower bound estimate is 560 GPa.

Balance the reactions which form ions. Choose "blank" if no other coefficient is needed. Writing the symbol implies "1."
1) (NH4)2CO3 ____ NH4 +_____ + CO3 -2
2) PbI2 _____ Pb+2 +______ I+1
3) (NH4)3PO4 ______ NH4 + _____ + PO4 +3

Answers

Following are the Balanced Reactions,

Reaction 1:


                            (NH₄)₂CO₃     →    2 NH₄⁺  +  CO3⁻²

Reaction 2:

                                     PbI₂    →    Pb⁺²  +  2 I⁻¹

Reaction 3:

                              (NH₂)₃PO₄     →     3 NH₄⁺  +   PO₄⁻³
The answers are the following:
1) (NH4)2CO3 2 NH4 +→ + CO3 -2
2) PbI2 →1 Pb+2 +2 I+1
3) (NH4)3PO4 →3NH4 + 3 + PO4 +

The highest energy occupied molecular orbital in the b−b bond of the b2 molecule is _____

Answers

According to molecular orbital theory, atomic orbitals combine to form molecular orbital. Number of molecular orbitals are equal to number of atomic orbitals. Further, of the total number of molecular orbitals, half are called as bonding molecular orbital while remaining are anti-bonding molecular orbital. In case, if system contains lone pair of electrons, they occupy non-bonding molecular orbital. Highest occupied molecular energy levels are referred as HOMO, while lowest unoccupied molecular energy levels are referred as LUMO.

In case of B2 molecule, two B atoms combines to generate molecular orbitals. Attached is  the MOT diagram of B2 molecule

From the attached figure, it is clearly evitable that high occupied energy level in B2 is π. Also, it must be noted both pi molecular orbitals i.e. Pi 2Px and Pi 2Py at highest energy level (occupied).

Answer:

Π2py and Π2pz

Explanation:

In the molecular orbital electron configuration of the boron molecule B2, the electrons are arranged as follows:

KK,σ2s^2,σ*2s^2, Π2py , Π2pz.

The two highest level orbitals are degenerate and are both singly occupied in the boron molecule as shown in the molecular orbital electron configuration above.

Note KK refers to the core electrons.

A reactant decomposes with a half-life of 137 s when its initial concentration is 0.165 m. when the initial concentration is 0.722 m, this same reactant decomposes with the same half-life of 137 s.

Answers

First you need to find the order of reaction.
Let the reaction follow a simple nth order rate law:
rate = k∙[A]ⁿ

Half-life t₁₂ initial concentration [A]₀ and rate constant k for such a reaction are related as:
t₁₂ = (2ⁿ⁻¹ - 1) / ( (n - 1)∙k∙[A]₀ⁿ⁻¹ )
except the particular case of first order reactions, i.e. n=1, in which half-life does not depend on initial concentration:
t₁₂ = ln(2)/k

Apparently your reaction is not a first order reaction. When you combine the constant factors in the relation above to a constant K, you can see that half-life of a non-first order reaction is inversely proportional to initial concentration raised to the power (n-1):
t₁₂ = K/[A]₀ⁿ⁻¹
with K=(2ⁿ⁻¹ - 1)/((n - 1)∙k)

K cancels out when you take the ratio of the two given half-lifes:
t₁₂₍₂₎ / t₁₂₍₁₎ = (K/[A]₀₍₂₎ⁿ⁻¹) / (K/[A]₀₍₁₎ⁿ⁻¹) = ([A]₀₍₁₎/[A]₀₍₂₎)ⁿ⁻¹
to find the exponent (n-1) take logarithm
ln(t₁₂₍₂₎/t₁₂₍₁₎) = ln(([A]₀₍₁₎/[A]₀₍₂₎)ⁿ⁻¹) = (n - 1)∙ln([A]₀₍₁₎/[A]₀₍₂₎)
=>
n - 1 = ln(t₁₂₍₂₎/t₁₂₍₁₎) / ln([A]₀₍₁₎/[A]₀₍₂₎)
= ln(229s / 151s) / ln(0.297M / 0.196M )
= 1.00198...
≈ 1
=>
n = 2

With known order n we can compute k from given half-life and initial concentration.
For a second order reaction half-life is given by:
t₁₂ = (2²⁻¹ - 1) / ( (2 - 1)∙k∙[A]₀²⁻¹ ) = 1/(k∙[A]₀)
Hence
k = 1/(t₁₂∙[A]₀)
= 1/(151s ∙ 0.297M)
= 2.23×10⁻² M⁻¹s⁻¹

How much heat is evolved in converting 1.00 mol of steam at 155.0 ∘c to ice at -50.0 ∘c? the heat capacity of steam is 2.01 j/(g⋅∘c) and of ice is 2.09 j/(g⋅∘c)?

Answers

When the specific heat capacity of the water is 4.18 J/g.°C so, we are going to use this formula to get the heat for cooling three  phases changes from steam to liquid and from liquid to ice (solid) :

when Q = M*C*ΔT 

Q is the heat in J

and M is the mass in gram = 1 mol H2O * 18 g/mol(molar mass) = 18 g

C is the specific heat J/g.°C

ΔT is the change in temperature

Q = Mw *[ ( Csteam * ΔTsteam)+(Cw*ΔTw) + (Cice  * ΔT ice)]

    = 18 g * [(2.01 * (155-100°C)) + (4.18 * (100-0°C)) + (2.09 * (0 - 55 °C))]

∴Q = 7444.8 J

and when we know that the heat of fusion for water = 334J/g

and heat of vaporization for water =  2260J/g


∴Q for the two phases changes = M * (2260+334) 

                                                      = 18 * (2260+334)

                                                      = 46692 J 

∴ Q total = 7444.8 + 46692 = 54136.8 J

A doctor is trying to diagnose a patient with short stature

Answers

What's the question??

Based on the values you obtained for δh∘rxn, which of the reactions would you expect to be thermodynamically favorable and which would be unfavorable? no. reactions δh∘f(kj) 1 ag+(aq)+li(s)→ag(s)+li+(aq) −384.4 2 fe(s)+2na+(aq)→fe2+(aq)+2na(s) 392.3 3 2k(s)+2h2o(l)→2koh(aq)+h2(g) −393.1

Answers

Final answer:

Based on the values of δH°rxn, reactions can be classified as thermodynamically favorable or unfavorable. Reaction 1 and reaction 3 are thermodynamically favorable, while reaction 2 is thermodynamically unfavorable.

Explanation:

Based on the values of δH°rxn, we can determine whether a reaction is thermodynamically favorable or unfavorable. A negative δH°rxn indicates a thermodynamically favorable reaction, while a positive δH°rxn indicates a thermodynamically unfavorable reaction. Let's analyze the given reactions:

ag+(aq) + li(s) → ag(s) + li+(aq): δH°rxn = -384.4 kJ/mol (negative, thermodynamically favorable)fe(s) + 2na+(aq) → fe2+(aq) + 2na(s): δH°rxn = 392.3 kJ/mol (positive, thermodynamically unfavorable)2k(s) + 2h2o(l) → 2koh(aq) + h2(g): δH°rxn = -393.1 kJ/mol (negative, thermodynamically favorable)

Therefore, reaction 1 and reaction 3 are thermodynamically favorable, while reaction 2 is thermodynamically unfavorable.

Learn more about Thermodynamic Favorability here:

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Which of the following statements is true?
A. Delocalized electrons conduct electricity poorly.
B. There is a strong attraction between positively charged metal ions and the sea of electrons.
C. Metals with a delocalized electron sea break apart when struck instead of bending.
D. Metals have high melting points and low boiling points.

Answers

B. There is a strong attraction between positively charged metal ions and the sea of electrons.  - true

Answer:  B. There is a strong attraction between positively charged metal ions and the sea of electrons.

Explanation:  Delocalized electrons conduct electricity with high potentil as the electrons are not localized in on place. Thus statement A is wrong.

Metals with a delocalized electron sea doesnot break apart when struck instaed of bending. Thus statement C is also wrong.

Metals donot have high melting point rather they have low melting point.

Statement B is true as there is a strong attraction beween positively charged metal ions and the sea of electrons.

Given the reaction:

Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g)


The oxidation number of Zn(s) increases because it

A) loses electrons
B) gains electrons
C) loses protons
D) gains protons

Answers

Redox reaction is the chemical reaction in which both oxidation (loss of electrons) and reduction (gain of electrons) takes place or transfer of electrons between two species occurs.

The given chemical equation is :

[tex]Zn(s)+2HCl(aq)\rightarrow ZnCl_{2}(aq)+H_{2}(g)[/tex]

Here, in reactant side, the oxidation state of zinc is zero (0), after reacting with hydrogen chloride, the oxidation sate of zinc becomes two (+2).  In this reaction, electrons are transferred from zinc atoms to the hydrogen atoms. Thus, zinc is oxidized by losing electrons.

Hence, oxidation state of zinc increases due to loses electrons i.e. Option (A) is correct.



The formation of an ionic bond involves a number of different processes. in the formation of sodium chloride, represented by the equation given below, two steps result in a release of energy. choose those two steps. (yes, you fill in two bubbles on your scantron for this question. you must choose both correct answers to receive credit for the question.) 2na(s) + cl2(g) → 2nacl(s)
a. attraction of na+ and cl– to form nacl.
b. removal of an electron from na.
c. the addition of an electron to each cl atom
d. conversion of na(s) to na(g).
e. dissociation of cl2 to form 2 cl atoms.

Answers

Final answer:

In the formation of sodium chloride, two steps result in a release of energy: attraction of Na+ and Cl- to form NaCl, and the conversion of Na(s) to Na(g) as the sodium atom donates an electron to chlorine atoms.

Explanation:

The formation of an ionic bond involves the transfer of electrons from one atom to another. In the case of sodium chloride (NaCl), two steps result in a release of energy. The first step is the attraction of Na+ and Cl- ions to form NaCl. The second step is the conversion of Na(s) to Na(g) as the sodium atom donates an electron to chlorine atoms.

The wittig reaction involves coupling between a phosphonium ylide and a carbonyl-containing molecule. if a chemist wants to use the wittig reaction to synthesize 2-methyl-2-butene, which reactants should be used?

Answers

2-Methyl-2-butene can be synthesized using Wittig reaction in following steps;

1) First of all react Bromoethane with Triphenylphosphine which will undergo SN² reaction to produce Phosphonium salt. Phoshonium salt on treatment with base gives Phosphorous Ylide.

2) The ylide formed when treated with acetone forms a four membered ring intermediate, which readily rearranges to produce 2-methyl-2-butene and triphenylphosphine oxide.

Result:
           Starting materials are...

               1) Bromoethane

               2) Acetone

               3) Triphenyl phosphene

Final answer:

To synthesize 2-methyl-2-butene using the Wittig reaction, ethyltriphenylphosphonium bromide reacts with acetaldehyde. The resulting oxaphosphetane intermediate yields 2-methyl-2-butene and triphenylphosphine oxide upon decomposition.

Explanation:

To synthesize 2-methyl-2-butene using the Wittig reaction, the chemist should choose appropriate reactants that will couple to form the desired alkene. The necessary phosphonium ylide would need to possess a suitable leaving group that, after the reaction with the carbonyl-containing molecule, results in 2-methyl-2-butene.

For this specific synthesis, the chemist would need ethyltriphenylphosphonium bromide as the ylide precursor. This compound, when deprotonated, forms a ylide that can react with acetaldehyde (ethanal). Upon reacting, the phosphonium ylide and acetaldehyde undergo nucleophilic addition to form the oxaphosphetane intermediate, which then yields 2-methyl-2-butene and triphenylphosphine oxide upon reductive elimination.

The Wittig reaction is a reliable method for synthesizing alkenes, and it has the advantage that the location of the double bond in the product is fixed, yielding a specific geometric isomer rather than a mixture of isomers.

b) How many hydrogen atoms are in an acyclic alkane with 16 carbon atoms?

Answers

The general formula of acyclic, i.e. non-cyclic, alkane is [tex] C_{n} H_{2n+2} [/tex]
where, n = number of carbon atom.

In present case, n = 16.
∴, Number of H atom = (2X16)+2 = 34

According to the balanced chemical equation, how many atoms of silver will be produced from combining 100 g of copper with 200 g of silver nitrate?

Answers

The balance chemical equation is as follow,

                    Cu  +  2 AgNO₃    →    Cu(NO₃)₂  +  2 Ag

First calculate for 200 g of AgNO₃,
According to equation,

              339.74 g (2 mole) AgNO₃ produced  =  215.7 g (2 mole) of Ag
So,
                         200 g of AgNO₃ will produce  =  X g of Ag

Solving for X,
                              X  =  (200 g × 215.7 g) ÷ 339.74 g

                              X  =  126.97 g Ag

Next calculate for 100 g of Cu,
According to equation,

              63.52 g (1 mole) Cu produced  =  215.7 g (2 mole) of Ag
So,
                       100 g of Cu will produce  =  X g of Ag

Solving for X,
                              X  =  (100 g × 215.7 g) ÷ 63.52 g

                              X  =  339.57 g Ag
Result:
          As 200 g of AgNO₃ is producing Ag in less amount (126.97 g) as compared to 100 g of Cu, it means that AgNO₃ is in limited amount and will consume first till the production of 126.97 g of Ag. So, the maximum amount of Ag production is 126.97 g.

Final answer:

The balanced chemical equation reveals that 100g of copper reacting with 200g of silver nitrate yields 6.022 × 10^23 atoms of silver.

Explanation:

The balanced chemical equation for the reaction between copper (Cu) and silver nitrate (AgNO3) is:

Cu (s) + 2AgNO₃ (aq) → Cu(NO₃)² (aq) + 2Ag (s)

From the equation, when 100g of copper reacts with 200g of silver nitrate, it will produce 6.022 × 10²³ atoms of silver.

How many grams of argon would it take to fill a light bulb with a volume of 0.475 l at stp?

Answers

STP stands for Standard Temp. and Pressure. 
Standard Temp. = 273.15 K
Pressure = 1 atm
Now,  ideal gas equation we know that PV = nRT
where, V = volume of gas = 0.475 l
R = universal gas constant = .082507 l atm mol-1 K-1 

∴ n =[tex] \frac{PV}{RT} [/tex]
∴ n =[tex] \frac{1 X .475}{0.082057X 273.15} [/tex]
∴n = 0.02112 mol

Thus, number of moles of Ar present = 0.02112

But, number of mole = [tex] \frac{wt. of .substance }{At./Mol. Weight} [/tex]
In present case, weight of Ar (g) = number of mole of Ar X Atomic Wt. of Ar
∴ weight of Ar = 40 X 0.02112 = 0.8477 g
Final answer:

To find the amount of argon to fill a 0.475 L light bulb at STP, one must calculate the number of moles using the ideal gas law and then multiply by the molar mass of argon to convert moles to grams.

Explanation:

To determine how many grams of argon it would take to fill a light bulb with a volume of 0.475 L at STP (standard temperature and pressure), we can use the ideal gas law, which states that PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the gas constant, and T is temperature. At STP, P is 1 atm and T is 273.15 K. The volume of the light bulb is already given as 0.475 L.

First, we can find the number of moles of argon that would occupy the 0.475 L volume at STP. The gas constant R in units that match our pressure in atm, volume in liters, and temperature in Kelvin is 0.0821 L·atm/K·mol. Therefore, n = PV / RT = (1 atm × 0.475 L) / (0.0821 L·atm/K·mol × 273.15 K). Calculating this value gives us the number of moles of argon.

Now, to find the mass in grams, we need to use the molar mass of argon which is 39.95 g/mol. The mass m can be calculated by multiplying the number of moles n by the molar mass of argon. The mass m = n × molar mass of argon will give us the final answer in grams.

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