Be sure to answer all parts. Nitric oxide (NO) reacts with molecular oxygen as follows: 2NO(g) + O2(g) → 2NO2(g) Initially NO and O2 are separated in two different chambers connected by a valve. When the valve is opened, the reaction quickly goes to completion. Determine what gases remain at the end and calculate their partial pressures. Assume that the temperature remains constant at 25°C. Initial conditions are as follows: NO: 3.90 L, 0.500 atm O2: 2.09 L, 1.00 atm

Answers

Answer 1

Answer:

The remain gases are [tex]o_{2}_{(g)}[/tex] and [tex]NO_{2}_{(g)}[/tex]

Pressure of [tex]O_{2}_{(g)}[/tex] [tex]1.09 atm O_{2}_{(g)}[/tex]

Pressure of [tex]NO_{2}_{(g)}[/tex] [tex]1.09 atm NO_{2}_{(g)}[/tex]

Explanation:

We have the following reaction

[tex]2NO_{(g)} +O_{2}_{(g)}\longrightarrow 2NO_{2}_{(g)}[/tex]

Now we calculate the limit reagents, to know which of the two gases is completely depleted and which one is in excess.

Excess gas will remain in the tank when the reagent limits have run out and the reaction ends.

To calculate the limit reagent, we must calculate the mols of each substance. We use the ideal gas equation

[tex]PV= nRT[/tex]

We cleared the mols

[tex]n=\frac{PV}{RT}[/tex]

PV=nrT

replace the data for each gas

Constant of ideal gases

[tex]R= 0.082\frac{atm.l}{mol.K}[/tex]

Transform degrees celsius to kelvin

[tex]25+273=298K[/tex]

[tex]NO_{g}[/tex]

[tex]n=\frac{0.500atm.3.90l}{298k.0.082\frac{atm.l}{k.mol} } \\ \\ n=0.080mol NO_{(g)} \\[/tex]

[tex]O_{2}_{(g)[/tex]

[tex]n=\frac{1atm.2.09l}{298k.0.082\frac{atm.l}{k.mol} } \\ \\ n=0.086mol O_{2}_{(g)} \\[/tex]

Find the limit reagent by stoichiometry

[tex]2NO_{(g)} +O_{2}_{(g)}\longrightarrow 2NO_{2}_{(g)}[/tex]

[tex]0.086mol O_{2}_(g).\frac{2mol NO_{(g)} }{1mol O_{2}_{(g)} } =0.17mol NO_{(g)}[/tex]

Using [tex]O_{2}_{(g)}[/tex]as the limit reagent produces more [tex]NO_{(g)}[/tex] than I have, so oxygen is my excess reagent and will remain when the reaction is over.

[tex]NO_{(g)}[/tex]

[tex]2NO_{(g)} +O_{2}_{(g)}\longrightarrow 2NO_{2}_{(g)}\\ \\ 0.080mol NO_{(g)}.\frac{1mol O_{2}_ {(g)} }{2mol NO_{(g)} } =0.04mol O_{2}_{(g)}[/tex]

Using [tex]NO_{(g)}[/tex] as the limit reagent produces less [tex]O_{2}_{(g)}[/tex] than I have, so [tex]NO_{(g)}[/tex]  is my excess reagent and will remain when the reaction is over.

Calculate the moles that are formed of [tex]NO_{2}_{(g)}[/tex]  

[tex]2NO_{(g)} +O_{2}_{(g)}\longrightarrow 2NO_{2}_{(g)}\\ \\ 0.080mol NO_{(g)}.\frac{2mol NO_{2}_ {(g)} }{2mol NO_{(g)} } =0.080mol NO_{2}_{(g)}[/tex]

We know that for all [tex]NO_{(g)}[/tex] to react, 0.04 mol [tex]O_{2}_{(g)}[/tex] is consumed.

we subtract the initial amount of [tex]O_{2}_{(g)}[/tex] less than necessary to complete the reaction. And that gives us the amount of mols that do not react.

[tex]0.086-0.04= 0.046[/tex]

The remain gases are[tex]O_{2}_{(g)}[/tex] and [tex]NO_{2}_{(g)}[/tex]

calculate the volume that gases occupy  

[tex]0.080 mol NO_{2}_{(g)} .\frac{22.4l NO_{2}_{(g)} }{1molNO_{2}_{(g)} }}  =1.79 lNO_{2}_{(g)}[/tex]

[tex]0.046 mol O_{2}_{(g)} .\frac{22.4 l O_{2}_{(g)} }{1molO_{2}_{(g)} }}  =1.03 l O_{2}_{(g)}[/tex]

Calculate partial pressures with the ideal gas equation

[tex]PV= nRT[/tex]

[tex]P=\frac{nRT}{V}[/tex]

Pressure of [tex]O_{2}_{(g)}[/tex]

[tex]P=\frac{0.046mol.0.082\frac{atm.l}{K.mol} 298K}{1.03l}= 1.09 atmO_{2}_{(g)}[/tex]

Pressure of [tex]NO_{2}_{(g)}[/tex]

[tex]P=\frac{0.080mol.0.082\frac{atm.l}{K.mol} 298K}{1.79l}= 1.09 atmNO_{2}_{(g)}[/tex]

Answer 2

Answer:

It will remain O₂ and NO₂, with partial pressures: pO₂ = 0.186 atm, and pNO₂ = 0.325 atm.

Explanation:

First, let's identify the initial amount of each reactant using the ideal gas law:

PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas equation (0.082 atm.L/mol.K), and T is the temperature (25°C + 273 = 298 K).

n = PV/RT

NO:

n = (0.500*3.90)/(0.082*298)

n = 0.0798 mol

O₂:

n = (1.00*2.09)/(0.082*298)

n = 0.0855 mol

By the stoichiometry of the reaction, we must found which reactant is limiting and which is in excess. The limiting reactant will be totally consumed. Thus, let's suppose that NO is the limiting reactant:

2 moles of NO ------------------ 1 mol of O₂

0.0798 mol ------------------- x

By a simple direct three rule:

2x = 0.0798

x = 0.0399 mol of O₂

The number of moles of oxygen needed is lower than the number of moles in the reaction, so O₂ is the limiting reactant, and NO will be totally consumed. The number of moles of NO₂ formed will be:

2 moles of NO --------------- 2 moles of NO₂

0.0798 mol ---------------- x

By a simple direct three rule:

x = 0.0798 mol of NO₂

And the number of moles of O₂ that remains is the initial less the total that reacts:

n = 0.0855 - 0.0399

n = 0.0456 mol of O₂

The final volume will be the total volume of the containers, V = 3.90 + 2.09 = 5.99 L, so by the ideal gas law:

PV = nRT

P = nRT/V

O₂:

P = (0.0456*0.082*298)/5.99

P = 0.186 atm

NO₂:

P = (0.0798*0.082*298)/5.99

P = 0.325 atm


Related Questions

If you start with 512 grams of aluminum and 1147 grams of copper chloride to make aluminum chloride and copper, what is the limiting reagent? 2Al + 3CuCl -> 2AlCl3 + 3Cu

Answers

First you need to calculate the number of moles of aluminium and copper chloride.

number of moles = mass / molecular weight

moles of Al = 512 / 27 = 19 moles

moles of CuCl = 1147 / 99 = 11.6 moles

From the reaction you see that:

if        2 moles of Al will react with 3 moles of CuCl

then  19 moles of Al will react with X moles of CuCl

X = (19 × 3) / 2 = 28.5 moles of CuCl, way more that 11.6 moles of CuCl wich is the quantity you have. So the copper chloride is the limiting reagent.

For which H-atom wavefunction are you mostlikely to find the electron farthest from thenucleus?1.2p2.3p3.4pcorrect4.2s5.1s

Answers

Answer:

Option 3. 4p

Explanation:

The greater the principal quantum number, i.e. the main energy level, means that the electron is occupying a larger orbital (bigger radius) thus you must check which wavefunction has the bigger principal quantum number. When two electrons have the same principal quantum number, then you must check the orbital shape because the rank of the radii of the orbitals in increasing order is: s < p < d < f.

The list of choices is:

1. 2p2. 3p3. 4p4. 2s5. 1s

Hence, the wavefunction 4p (third option) is the largest one and, so, it is for it that you mostlikely would find the electron the farthest from its nucleus.

In the 4p wavefunction is mostlikely to find the electron farthest from thenucleus.

Electronic configuration

The electronic configuration of an atom determines the orbital location of electrons, based on energy level and orbital type.

The higher the energy level, the farther that electron is from the nucleus.

Thus, the wavefunction that demonstrates the electron furthest from the nucleus is 4p, which has a higher energy level.

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Calculate Δ Hrxn for the following reaction: CH4(g)+4Cl2(g)→CCl4(g)+4HCl(g) given these reactions and their ΔH values: C(s)C(s)H2(g)+++2H2(g)2Cl2(g)Cl2(g)→→→CH4(g),CCl4(g),2HCl(g),ΔH=−74.6 kJΔH=−95.7 kJΔH=−184.6 kJ Express the enthalpy in kilojoules to one decimal place.

Answers

Answer : The enthalpy of the following reaction is, -390.3 KJ

Explanation :

The given balanced chemical reactions are,

(1) [tex]C(s)+2H_2(g)\rightarrow CH_4(g)[/tex]     [tex]\Delta H_1=-74.6KJ/mole[/tex]  

(2) [tex]C(s)+2Cl_2(g)\rightarrow CCl_4(g)[/tex]     [tex]\Delta H_2=-95.7KJ/mole[/tex]

(3) [tex]H_2(g)+Cl_2(g)\rightarrow 2HCl(g)[/tex]     [tex]\Delta H_3=-184.6KJ/mole[/tex]

The final reaction of is,

[tex]CH_4(g)+4Cl_2(g)\rightarrow CCl_4(g)+4HCl(g)[/tex]       [tex]\Delta H_{rxn}=?[/tex]

Now adding reaction 2 and twice of reaction 3 and reverse of reaction 1, we get the enthalpy of of the reaction.

The expression for enthalpy for the following reaction will be,

[tex]\Delta H_{rxn}=[2\times \Delta H_3]+[-1\times \Delta H_1]+[1\times \Delta H_2][/tex]

where,

n = number of moles

Now put all the given values in the above expression, we get:

[tex]\Delta H_{rxn}=[2mole\times (-184.6KJ/mole)]+[-1mole\times (-74.6KJ/mole)]+[1\times (-95.7KJ/mole)]=-390.3KJ[/tex]

Therefore, the enthalpy of the following reaction is, -390.3 KJ

Answer:

-390.3 KJ

Explanation:

For Hess's Law, we need to get the corresponding equation below using the sequence of reactions given

By manipulating the reaction, either reversing them or multiplying/dividing them to a certain factor, we can get to the target equation as well as the total enthalpy

CH4(g) + 4Cl2(g) → CCl4(g) + 4HCl(g)

C(s) + 2H2(g) → CH4(g)     ΔH = −74.6kJ (needs to reverse)

C(s) + 2Cl2(g) → CCl4(g)    ΔH = −95.7kJ (retain)

H2(g) + Cl2(g) → 2HCl(g)   ΔH = −184.6kJ (multiply by 2 to get 4Cl2 and cancel out 4 HCl and 4 H2)

Therefore, it is -390.3 KJ

What is the maximum number of electrons that can be contained in the first level

Answers

Your answer is:

The first level (or shell) can hold up to 2 electrons.

Hope this helps! C:

How many grams of barium sulfate can be produced from the reaction of 2.54 grams sodium sulfate and 2.54 g barium chloride? Na2SO4(aq) + BaCl2(aq) --> BaSO4(s) + 2NaCl(aq) Report your answer to 3 decimal places.

Answers

Answer: 2.796 grams

Explanation:

[tex]Na_2SO_4+BaCl_2\rightarrow 2NaCl+BaSO_4[/tex]

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]

[tex]\text{Number of moles of sodium sulphate}=\frac{2.54g}{142g/mol}=0.018moles[/tex]

[tex]\text{Number of moles of barium chloride}=\frac{2.54g}{208g/mol}=0.012moles[/tex]

According to stoichiometry:

1 mole of [tex]BaCl_2[/tex] reacts with 1 mole of [tex]Na_2SO_4[/tex]

0.012 moles of [tex]BaCl_2[/tex] will react with=[tex]\frac{1}{1}\times 0.012=0.012moles[/tex] of [tex]Na_2SO_4[/tex]

Thus [tex]BaCl_2[/tex] is the limiting reagent as it limits the formation of product. [tex]Na_2SO_4[/tex]  is the excess reagent as (0.018-0.012)=0.006 moles are left unused.

1 mole of [tex]BaCl_2[/tex] produces 1 mole of [tex]BaSO_4[/tex]

0.012 moles of [tex]BaCl_2[/tex] will produce=[tex]\frac{1}{1}\times 0.012=0.012moles[/tex] of [tex]BaSO_4[/tex]

Mass of [tex]BaSO_4=moles\times {\text {Molar mass}}=0.012\times 233=2.796g[/tex]

Thus 2.796 grams of [tex]BaSO_4[/tex]  are produced.

The scientific principle which is the basis for balancing chemical equations is

Answers

Answer:

Law of Conservation of Mass

Explanation:

Final answer:

The Law of Conservation of Matter forms the basis for balancing chemical equations, ensuring that the number of each element is equal on both sides of the equations. This principle also serves to describe a reaction's stoichiometry, where amounts of reactants and products in a reaction are considered.

Explanation:

The scientific principle which forms the basis for balancing chemical equations is the Law of Conservation of Matter. According to this principle, the same number of each element must be represented on the reactant (input) and product (output) sides of an equation. This ensures that equations accurately reflect the reality that matter is not created or destroyed in a chemical reaction.

For instance, consider the following chemical equation: PC15 (s) + H₂O(1) →→→ POC13 (1) + 2HCl(aq). In this equation, we balance the equation by ensuring that for each of the elements involved (P, C, and H), the number of atoms of that element is equal on both sides of the equation.

Beyond simply balancing the equation, this principle also helps describe a reaction's stoichiometry, which involve the relationships between amounts of reactants and products. Coefficients from the balanced equation can be used in computations relating to reactant and product masses, molar amounts, and other quantitative properties.

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At 900.0 K, the equilibrium constant (Kp) for the following reaction is 0.345. 2SO2 + O2 (g) →2 SO3 (g) At equilibrium, the partial pressure of SO2 is 36.9 atm and that of O2 is 16.8 atm. The partial pressure of SO3 is ________ atm.

Answers

Final answer:

To calculate the partial pressure of SO3 at equilibrium for the given reaction, we can use the equilibrium constant (Kp) and the known partial pressures of SO2 and O2. By substituting values into the equation Kp = (P(SO3))^2 / (P(SO2))^2 * P(O2), we can find the partial pressure of SO3.

Explanation:

Given the equilibrium constant (Kp) of 0.345 and the partial pressures of SO2 and O2 at equilibrium as 36.9 atm and 16.8 atm respectively, we can use the equation Kp = (P(SO3))^2 / (P(SO2))^2 * P(O2). Let's substitute the known values into the equation:

Kp = (P(SO3))^2 / (36.9 atm)^2 * (16.8 atm)

Now we can solve for the partial pressure of SO3:

(P(SO3))^2 = Kp * (36.9 atm)^2 * (16.8 atm)

P(SO3) = sqrt(Kp * (36.9 atm)^2 * (16.8 atm))

Calculating this value will give us the partial pressure of SO3 at equilibrium.

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Place the following in order of increasing dipole moment.

I. BCl3 II. BIF2 III. BClF2

A) I < II < III
B) II < I < III
C) II < III < I
D) I < II = III
E) I < III < II

Answers

Final answer:

The order of increasing dipole moment is II < III < I. BCl3 has no dipole moment, BIF2 has a dipole moment, and BClF2 has a greater dipole moment compared to BIF2.

Explanation:

The order of increasing dipole moment is II < III < I.

The dipole moment depends on the electronegativity difference of the bonded atoms and the bond length. In this case, BCl3 has no dipole moment since the molecule is symmetrical and the individual dipole moments cancel out. BIF2 has a dipole moment due to the electronegativity difference between B and I atoms, and BClF2 has a greater dipole moment compared to BIF2 due to the additional electronegativity difference between B and Cl atoms.

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The correct answer is Option A. The order of increasing dipole moment is I < II < III.

BCl₃ is a symmetrical molecule with no net dipole moment because its individual B-Cl bonds cancel out. Hence, BCl₃ has the smallest dipole moment.

For BIF₂, the asymmetry is introduced as iodine and fluorine have different electronegativities, creating a net dipole moment. However, with only one I atom, the effect is moderate.

BClF₂ is more polar than BIF₂ because the different electronegativities of Cl and F combined in an asymmetrical structure increase the net dipole moment.

The volume of 7.91 M HCl needed to make 196.1 mL of 2.13 M HCl is ____.

Select one:
a. 52.8
b. 728
c. 0.198
d. 3490

Answers

Answer:

a) 52.8

Explanation:

M1V1 = M2V2

(7.91 M)(x ml) = (2.13 M) (196.1 ml)

(7.91M) (xml) = 417.693 M.ml

x ml = 417.693/ 7.91

x   =  52.8

Answer:

A. 52.8

Explanation:

We make use of the dilution formula to solve this

[tex]M_1\times V_1 =M_2\times V_2[/tex]

Where ([tex]M_1[/tex] and [tex]M_2[/tex] are the initial molarity and final molarity and [tex]V_1[/tex] and [tex]V_2[/tex] are  the initial volume and final volume)

Plugging into the values in the formula  

[tex]M_1\times V_1 =M_2\times V_2[/tex]

[tex]7.91M \times V_1=2.13M \times 196.1mL[/tex]

[tex]V_1=\frac {(2.13M \times 196.1mL)}{7.91M}[/tex]

= 52.8 mL is the Answer

Of the following elements, which could be classified as a metalloid (semimetal)? Select all that apply.
(A) Mg
(B) Ti
(C) Ge
(D) Si
(E) Rn
(F) Au
(G) Bi
(H) At
(I) Br
(J) B
(K) Eu

Answers

Answer:

(C) Ge

(D) Si

(H) At

(J) B

Hope this helps!

Which atom is most likely to attract electrons in a bond?

A) K
B) Na
C) S
D) Se

Answers

C) S

Because of the trends of electronegativity. Increases across a period and decreases down a group.
Final answer:

In a bond, Sulfur (S) is most likely to attract electrons due to its higher electronegativity. This is based on the trend in the periodic table where electronegativity increases across a group from left to right and decreases down a group.

Explanation:

In a bond, elements that have high electronegativity are more likely to attract electrons. Electronegativity is a measure of the ability of an atom to attract the electrons in a bond. It increases across a period from left to right, and decreases down a group in the periodic table. Based on this premise, among the atoms listed, Sulfur (S) and Selenium (Se) have higher electronegativities than Potassium (K) and Sodium (Na).

However, between Sulfur and Selenium, Sulfur (S) is more likely to attract electrons in a bond because it is higher up in the table, given that electronegativity decreases down a group. Therefore, the answer to your question is C) Sulfur (S).

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Which of the following would increase the rate of a chemical reaction between hydrochloric acid (HCl) and solid zinc metal (Zn)?

A. Decreasing the concentration of HCl
B. Pulverizing the zinc metal into a fine powder
C. Performing the reaction at a lower temperature
D. Decreasing the amount of Zn

Answers

Answer:

B. Pulverizing the zinc metal into a fine powder

Answer: B. Pulverizing the zinc metal into a fine powder

Explanation:

Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

[tex]Zn+2HCl\rightarrow ZnCl_2+H_2[/tex]

A. Decreasing the concentration of [tex]HCl[/tex] and [tex]Zn[/tex]: Thus rate of the reaction would decrease on decreasing the concentration of hydrochloric acid and zinc.

B.  Pulverizing the zinc metal into a fine powder : If the solute particles is present in smaller size, more it can take part in the chemical reaction due to large surface area, hence increasing the rate of reaction.

C. Performing the reaction at a lower temperature. Decreasing the temperature means the energy of the particles is less and thus lesser reactants would cross the energy barrier and thus lesser will be the rate.

What temperature must a gas initially at 10 °c be brought to for the pressure to triple?

Answers

Answer:

I think that depends on the type of gas and the volume of the container.

What is the percent-by-mass concentration of acetic acid (CH3COOH) in a vinegar solution that contains 51.80 g acetic acid in a 1.000−L solution? The density of this solution is 1.005 g/mL.

Answers

Final answer:

The percent-by-mass concentration of acetic acid in the vinegar solution is approximately 5.15%, calculated by dividing the acetic acid mass by the total solution mass and then multiplying by 100.

Explanation:

To find the percent-by-mass concentration of acetic acid in vinegar, the mass of acetic acid is divided by the total mass of the solution and then multiplied by 100. First, convert the solution volume to mass using the given density. The density of the solution is 1.005 g/mL, which means 1.000 L (or 1000 mL) of solution has a mass of 1005 g (1000 mL × 1.005 g/mL). The mass of acetic acid is given as 51.80 g. Thus, the percent-by-mass concentration is calculated as (51.80 g / 1005 g) × 100%.

The calculation gives a percent-by-mass concentration of approximately 5.15%. This value represents the mass of acetic acid expressed as a percentage of the total mass of the solution.

What condition must be met for a battery to be rechargeable? Either its anode or its cathode must generate a gas as a result of the electrochemical reaction. It must generate electricity via an acid-base reaction rather than via an oxidation-reduction reaction. The battery must be open to the outside so that it can vent any internal pressure that builds up from gases within it. The electrochemical reaction of the battery must be reversible.

Answers

Answer:

The electrochemical reaction of the battery must be reversible.

Explanation:

An electrochemical cell is device in which chemical reactions produce electricity. There are two main types of electrochemical cells:  

Primary cells: In primary cells, the chemical reaction through which electric current is generated is irreversible and hence cannot be re-charged.  

Secondary cells: are those in which chemical reactions through which electricity is generated is reversible. Hence they can be re-charged.

For the chemical equation SO2(g)+NO2(g)↽−−⇀SO3(g)+NO(g) the equilibrium constant at a certain temperature is 3.80. At this temperature, calculate the number of moles of NO2(g) that must be added to 2.50 mol SO2(g) in order to form 1.00 mol SO3(g) at equilibrium.

Answers

The number of moles of [tex]NO_2[/tex] that should be added in order to form 1 mol  [tex]SO_3\\[/tex] at equilibrium is 0.263 mol.

Given,

Initial Conditions:

Moles of [tex]SO_\\2[/tex] initially = 2.50 mol

Moles of [tex]SO_3[/tex] at equilibrium = 1.00 mol

Moles of [tex]NO_2[/tex] initially = 0 mol

Moles of NO initially = 0 mol

Equilibrium constant = 3.80

Changes in Moles:

Change in moles of [tex]SO_2[/tex] = 2.50 mol (initial moles of [tex]SO_2[/tex])

Change in moles of [tex]SO_3[/tex]= 1.00 mol (moles of [tex]SO_3[/tex]at equilibrium)

The equilibrium constant expression for this reaction is:

[tex]K_c = \frac{[SO_3][NO]}{[SO_2][NO_2]}[/tex]

Where the square brackets denote the concentration of each species.

On using the equilibrium constant expression to solve for the change in moles of [tex]NO_2[/tex]:

[tex]\Delta n_{NO_2} = \frac{[SO_2][NO_2]}{[SO_3]}\times \frac{[NO]}{K_c}[/tex]

Substituting the values:

[tex]\rm \Delta n_{NO_2} = \frac {2.50\times 0}{1} \times \frac{1}{3.80}\\\Delta n_{NO_2} = 0.263 \ mol[/tex]

The number of moles that should be added in order to form 1 mol  [tex]SO_3\\[/tex] at equilibrium is 0.263 mol.

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Final answer:

According to the given chemical equation, for every mole of SO3 formed, a mol of NO2 is used up. Therefore, to form 1.00 mol of SO3 from 2.50 mol of SO2, you need to add 1.00 mol of NO2 to reach equilibrium.

Explanation:

This calculation, in essence, requires an application of Le Chatelier's Principle. For the chemical equation SO2(g)+NO2(g)↽−−⇀SO3(g)+NO(g) the equilibrium constant, Kc, at a certain temperature is 3.80. This implies that at equilibrium, the ratio of the concentrations of products to reactants (when raised to their stoichiometric coefficients) is 3.80.

To form 1.00 mol of SO3(g) from 2.50 mol of SO2(g), it means that according to the equation, 1 mol NO2(g) will be required. Furthermore, for the NO(g) produced, we do not have to worry about it since it is not involved in the later equilibrium calculation.

With Le Chatelier's Principle, when you increase the amount of one of the reactant (in this case NO2(g)), the equilibrium will shift to the right side to try and 'use up' that extra NO2 added, this will cause more SO3(g) to be produced and reach a new equilibrium state.

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Select the phrase that best describes scientific findings communicated through popular media.
A) generally free of technical jargon
B) peer reviewed
C) commonly used by most scientists
D) extremly reliable

Answers

Answer:

The correct answer is option A) generally free of technical jargon

Explanation:

Hello!

Let's solve this!

When we read popular magazines or newspapers, we see scientific findings. They generally do not use scientific or technical vocabulary so that more people can understand it. They are also not the most reliable sources, since to look for reliable information, we have to look for scientific journals.

After the analysis we conclude that the correct answer is option A) generally free of technical jargon

Use dimensional analysis to convert 14.5mi/hr to km/s

Answers

Answer:

0.006 48 km/s

Explanation:

1. Convert miles to kilometres

14.5 mi × (1.609 km/1 mi) = 23.33 km

2. Convert hours to seconds

1 h × (60 min/1h) × (60 s/1 min) = 3600 s

3. Divide the distance by the time

14.5 mi/1 h = 23.3 km/3600 s = 0.006 48 km/s

1) How old is a bone in which the Carbon-14 in it has undergone 8 half-lives?

Select one:
a. 45600
b. 91200
c. 91200
d. 11400

2) In the process of radiocarbon dating, the fixed period of radioactive decay used to determine age is called the

Select one:
a. exponent.
b. half-life.
c. isotope.
d. nucleus.

3) A certain byproduct in nuclear reactors, 210Po, decays to become 206Pb. After a time period of about 276 days, only about 25% of an original sample of 210Po remains. The remainder has decayed to 206Pb. Determine the approximate half-life of 210Po.

Select one:
a. 138 days
b. 276 days
c. 414 days
d. 552 days


Answers

1) How old is a bone in which the Carbon-14 in it has undergone 8 half-lives?

Using the graph form the picture you count 8 times the halving of C¹⁴ and you arrive at 45600 years.

2) In the process of radiocarbon dating, the fixed period of radioactive decay used to determine age is called the half-life.

3) A certain byproduct in nuclear reactors, 210Po, decays to become 206Pb. After a time period of about 276 days, only about 25% of an original sample of 210Po remains. The remainder has decayed to 206Pb. Determine the approximate half-life of 210Po.

What the problem is telling you is that at 276 days only 25% original sample remains. If you divide the number of days by two the quantity of original sample will be multiplied by two, and you will have 138 days and 50% of original sample. This is the answer because the the half-life of a isotope is the time in which 50% of original quantity of radioactive atoms will disintegrate.

Which is the part of an experiment that serves as the point of comparison for the results?

hypothesis
independent variable
constant
control

Answers

Hypothesis because this is with the “if..then..because” something that’s being explained. Hope this helps :)

Determine the molarity of a solution formed by dissolving 97.7 g libr in enough water to yield 750.0 ml of solution.

Answers

Answer:

1.5 M.

Explanation:

Molarity (M) is defined as the no. of moles of solute dissolved in a 1.0 L of the solution.

M = (no. of moles of LiBr)/(Volume of the solution (L).

∵ no. of moles of LiBr = (mass/molar mass) of LiBr = (97.7 g)/(86.845 g/mol) = 1.125 mol.

Volume of the solution = 750.0 mL = 0.75 L.

∴ M = (no. of moles of luminol)/(Volume of the solution (L) = (1.125 mol)/(0.75 L) = 1.5 M.

Hello!

Determine the molarity of a solution formed by dissolving 97.7 g LiBr in enough water to yield 750.0 ml of solution.

We have the following data:  

M (Molarity) =? (in mol / L)

m1 (mass of the solute) = 97.7 g

V (solution volume) = 750 ml → V (solution volume) = 0.75 L

MM (molar mass of LiBr)

Li = 6.941 u

Br = 79.904 u

---------------------------

MM (molar mass of LiBr) = 6.941   + 79.904

MM (molar mass of LiBr) = 86.845 g/mol

Now, let's apply the data to the formula of Molarity, let's see:

[tex]M = \dfrac{m_1}{MM*V}[/tex]

[tex]M = \dfrac{97.7}{86.845*0.75}[/tex]

[tex]M = \dfrac{97.7}{65.13375}[/tex]

[tex]M = 1.49999... \to \boxed{\boxed{M \approx 1.5\:mol/L}}\:\:\:\:\:\:\bf\green{\checkmark}[/tex]

________________________

________________________

*** Another way to solve is to find the number of moles (n1) and soon after finding the molarity (M), let's see:

[tex]n_1 = \dfrac{m_1\:(g)}{MM\:(g/mol)}[/tex]

[tex]n_1 = \dfrac{97.7\:\diagup\!\!\!\!\!g}{86.845\:\diagup\!\!\!\!\!g/mol}[/tex]

[tex]n = 1.12499.. \to \boxed{n_1 \approx 1.125\:mol}[/tex]

[tex]M = \dfrac{n_1\:(mol)}{V\:(L)}[/tex]

[tex]M = \dfrac{1.125\:mol}{0.75\:L}[/tex]

[tex]\boxed{\boxed{M = 1.5\:mol/L}}\:\:\:\:\:\:\bf\blue{\checkmark}[/tex]

_____________________

[tex]\bf\purple{I\:Hope\:this\:helps,\:greetings ...\:Dexteright02!}\:\:\ddot{\smile}[/tex]

A 0.8870 g sample of a mixture of NaCl and KCl is dissolved in water, and the solution is then treated with an excess of AgNO3 to yield 2.142 g of AgCl. Calculate the percent by mass of each compound in the mixture

Answers

Answer : The percent by mass of NaCl and KCl are, 92.22 % and 7.78 % respectively.

Explanation :

As we know that when a mixture of NaCl and KCl react with excess [tex]AgNO_3[/tex] then the silver ion react with the chloride ion in both NaCl and KCl to form silver chloride.

Let the mass of NaCl be, 'x' grams and the mass of KCl will be, (0.8870 - x) grams.

The molar mass of NaCl and KCl are, 58.5 and 74.5 g/mole respectively.

First we have to calculate the moles of NaCl and KCl.

[tex]\text{Moles of }NaCl=\frac{\text{Mass of }NaCl}{\text{Molar mass of }NaCl}=\frac{xg}{58.5g/mole}=\frac{x}{58.5}moles[/tex]

[tex]\text{Moles of }KCl=\frac{\text{Mass of }KCl}{\text{Molar mass of }KCl}=\frac{(0.8870-x)g}{74.5g/mole}=\frac{(0.8870-x)}{74.5}moles[/tex]

As, each mole of NaCl and KCl gives one mole of chloride ions.

So, moles of chloride ions in NaCl = [tex]\frac{x}{58.5}moles[/tex]

Moles of chloride ions in KCl = [tex]\frac{(0.8870-x)}{74.5}moles[/tex]

The total moles of chloride ions = [tex]\frac{x}{58.5}moles+\frac{(0.8870-x)}{74.5}moles[/tex]

Now we have to calculate the moles of AgCl.

As we know that, this amount of chloride ion is same as the amount chloride ion present in the AgCl precipitate. That means,

Moles of AgCl = Moles of chloride ion = [tex]\frac{x}{58.5}moles+\frac{(0.8870-x)}{74.5}moles[/tex]

Now we have to calculate the moles of AgCl.

The molar mass of AgCl = 143.32 g/mole

[tex]\text{Moles of }AgCl=\frac{\text{Mass of }AgCl}{\text{Molar mass of }AgCl}=\frac{2.142g}{143.32g/mole}=0.0149moles[/tex]

Now we have to determine the value of 'x'.

Moles of AgCl = [tex]\frac{x}{58.5}moles+\frac{(0.8870-x)}{74.5}moles[/tex]

0.0149 mole = [tex]\frac{x}{58.5}moles+\frac{(0.8870-x)}{74.5}moles[/tex]

By solving the term, we get the value of 'x'.

[tex]x=0.818g[/tex]

The mass of NaCl = x = 0.818 g

The mass of KCl = (0.8870 - x) = 0.8870 - 0.818 = 0.069 g

Now we have to calculate the mass percent of NaCl and KCl.

[tex]\text{Mass percent of }NaCl=\frac{\text{Mass of }NaCl}{\text{Total mass of mixture}}\times 100=\frac{0.818g}{0.8870g}\times 100=92.22\%[/tex]

[tex]\text{Mass percent of }KCl=\frac{\text{Mass of }KCl}{\text{Total mass of mixture}}\times 100=\frac{0.069g}{0.8870g}\times 100=7.78\%[/tex]

Therefore, the percent by mass of NaCl and KCl are, 92.22 % and 7.78 % respectively.

Final answer:

The percentages of NaCl and KCl in the chemical mixture are 97.97% and 125.53% respectively, but these percentages raise issues with the experimental results as KCl exceeds 100%. Conversion from grams to moles and vice versa is essential in this calculation.

Explanation:

To answer your question, let's start by figuring out the amount of NaCl and KCl in the mixture. When mixed with AgNO3, both these compounds yield AgCl. The given mass of AgCl is 2.142 g, which we can convert into moles using its molar mass (143.32 g/mol), yielding 0.0149 mol of AgCl.

The molar amounts of NaCl and KCl in the original mixture are also equal to this value because each NaCl or KCl molecule yields one AgCl molecule. Now, let's convert these molar amounts back into grams using the molar masses of NaCl (58.44 g/mol) and KCl (74.56 g/mol), yielding 0.869 g of NaCl and 1.113 g of KCl.

However, these do not add up to the original sample weight (0.887g). This can be due to experimental errors or purity issues. Nevertheless, to calculate the percentage mass of each compound in the mixture, you can use the formula: (observed mass / total mass) * 100%, hence, the percentages are: (0.869 g / 0.8870 g) * 100% = 97.97% NaCl and (1.113 g / 0.8870 g) * 100% = 125.53% KCl.

Yet, these results seem incorrect due to the KCl percentage being higher than 100%, suggesting a need for a re-check of experimental results or re-calculation.

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Select all that apply: The melting points of organic compounds are usually lower than those of inorganic compounds. The boiling points of organic compounds are usually lower than those of inorganic compounds. The flammability of organic compounds is usually lower than that of inorganic compounds.

Answers

Answer:

The melting points of organic compounds are usually lower than those of inorganic compounds.

The boiling points of organic compounds are usually lower than those of inorganic compounds.

Explanation:

Organic compounds are compounds of carbon with exception of simple ones like oxides, carbides and carbonates. Inorganic compounds are other compounds excluding organic compounds.

Inorganic compounds due to the nature of their bonds typically have higher melting and boiling points compared to organic compounds.

Organic compounds are also highly flammable and would easily burn compared to inorganic ones.

How can the value of Ksp be related to the molar solubility of a compound? View Available Hint(s) How can the value of Ksp be related to the molar solubility of a compound? When a common ion is present, the solubility of the compound decreases, and this is reflected in a lower value of Ksp. The square of a compound's molar solubility equals the value of Ksp for the compound. The value of Ksp equals the concentration of the compound in a saturated solution, which can be converted to the molar solubility using the molar mass. The molar solubility can be used to calculate the concentrations of ions in solution, which in turn are used to calculate Ksp.

Answers

Answer:

The molar solubility can be used to calculate the concentrations of ions in solution, which in turn are used to calculate Ksp.

Explanation:

Consider a slightly soluble solid with formula M₃X₂. Its solubility product expression is

[tex]\begin{array}{rcccc}M_{3}X_{2}(s) & \rightleftharpoons&3M^{2+}(aq) & + & 2X^{3-}(aq)\\& & 3s & &2s\\\\K_\text{sp}& = & [3s]^{3}[2s]^{2}&= & 108s^{5}\\\\\end{array}[/tex]

Thus, the molar solubility can be used to calculate the concentrations of ions in solution, which in turn are used to calculate Ksp.

A is wrong. The solubility product constant is a constant. It does not change in the presence of a common ion.

B is wrong. It is correct only for compounds with formula MX.

C is wrong. Ksp does not equal the concentration of the compound in solution.

The value of Ksp is related to the molar solubility in that molar solubility helps calculate the ion concentrations in a solution, which are used to calculate Ksp. The presence of a common ion affects both solubility and Ksp through the common ion effect.

The value of Ksp, or the solubility product constant, is closely related to the molar solubility of a compound, which is a measure of how much of the compound can dissolve in a given amount of solvent to form a saturated solution. The connection between Ksp and molar solubility can be understood through a series of steps:

First, molar solubility is calculated by converting the solubility of a compound (often given in g/L) to moles per liter using the molar mass of the compound.Next, the dissociation equation of the compound is used to determine the concentration of each ion in the solution.Finally, these ion concentrations are used in the Ksp expression to calculate the solubility product constant.

Furthermore, if a common ion is present in the solution, it affects the solubility of the compound and thus the Ksp value. This is a demonstration of the common ion effect, where the presence of a common ion decreases the solubility and Ksp of the compound according to Le Chatelier's principle.

Therefore, understanding the relationship between Ksp and molar solubility is essential for predicting the solubility of compounds and determining the possible concentrations of ions in solution.

one method for generating chlorine gas is by reacting potassium permanganate and hydrochloric acid. how many liters of Cl2 at 40 C and a pressure of 1.05 atm can be produced by the reaction of 6.23 g KMnO4 with 45.0 ml of 6.00 m HCl?

Answers

Answer: The volume of chlorine gas produced in the reaction is 2.06 L.

Explanation:

For potassium permanganate:

To calculate the number of moles, we use the equation:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]

Given mass of potassium permanganate = 6.23 g

Molar mass of potassium permanganate = 158.034 g/mol

Putting values in above equation, we get:

[tex]\text{Moles of potassium permanganate}=\frac{6.23g}{158.034g/mol}=0.039mol[/tex]

For hydrochloric acid:

To calculate the moles of hydrochloric acid, we use the equation:

[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}[/tex]

Molarity of HCl = 6.00 M

Volume of HCl = 45.0 mL = 0.045 L   (Conversion factor: 1 L = 1000 mL)

Putting values in above equation, we get:

[tex]6.00mol/L=\frac{\text{Moles of HCl}}{0.045L}\\\\\text{Moles of HCl}=0.27mol[/tex]

For the reaction of potassium permanganate and hydrochloric acid, the equation follows:

[tex]2KMnO_4+16HCl\rightarrow 2MnCl_2+5Cl_2+2KCl+8H_2O[/tex]

By Stoichiometry of the reaction:

16 moles of hydrochloric acid reacts with 2 moles of potassium permanganate.

So, 0.27 moles of hydrochloric acid will react with = [tex]\frac{2}{16}\times 0.27=0.033moles[/tex] of potassium permanganate.

As, given amount of potassium permanganate is more than the required amount. So, it is considered as an excess reagent.

Thus, hydrochloric acid is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

16 moles of hydrochloric acid reacts with 5 moles of chlorine gas.

So, 0.27 moles of hydrochloric acid will react with = [tex]\frac{5}{16}\times 0.27=0.0843moles[/tex] of chlorine gas.

To calculate the volume of gas, we use the equation given by ideal gas equation:

[tex]PV=nRT[/tex]

where,

P = pressure of the gas = 1.05 atm

V = Volume of gas = ? L

n = Number of moles = 0.0843 mol

R = Gas constant = [tex]0.0820\text{ L atm }mol^{-1}K^{-1}[/tex]

T = temperature of the gas = [tex]40^oC=[40+273]K=313K[/tex]

Putting values in above equation, we get:

[tex]1.05atm\times V=0.0843\times 0.0820\text{ L atm }mol^{-1}K^{-1}\times 313K\\\\V=2.06L[/tex]

Hence, the volume of chlorine gas produced in the reaction is 2.06 L.

Final answer:

The question involves a stoichiometry problem in chemistry, where we calculate the volume of chlorine gas from the reaction of potassium permanganate with hydrochloric acid. We identify the limiting reactant and then use the ideal gas law to find the volume under the given conditions of temperature and pressure.

Explanation:

The student's question concerns a chemical reaction between potassium permanganate (KMnO4) and hydrochloric acid (HCl) to produce chlorine gas (Cl2). This is a stoichiometry problem where we will calculate the volume of chlorine gas generated at specific conditions using the ideal gas law. To find the number of moles of chlorine gas that can be produced, we first determine the limiting reactant. Then, using the ideal gas law (PV=nRT), we can calculate the volume at the given temperature and pressure.

First, we must find the reaction equation:
KMnO4 + 16HCl → 2KCl + 5Cl2 + 2MnCl2 + 8H2O

Next, we calculate the moles of KMnO4 and HCl. Following this, we determine the limiting reactant and use it to calculate moles of Cl2 produced. Lastly, we use the ideal gas law (PV=nRT, with R=0.0821 L·atm/(mol·K)) to find the volume of Cl2 at 40°C and 1.05 atm.

The volume of 7.91 M HCl needed to make 196.1 mL of 2.13 M HCl is ____.

Select one:
a. 52.8
b. 728
c. 0.198
d. 3490

Answers

Answer:

a. 52.8

Explanation:

To find the number of moles of HCl we use the relation M₁V₁=M₂V₂

where M₁ is the initial molarity, M₂ the new molarity, V₁ the initial volume used, and V₂ the final volume obtained.

M₁=7.91 M

M₂=2.13 M

V₁=?

V₂=196.1 mL

Replacing these values in the relationship.

M₁V₁=M₂V₂

7.91 M× V₁=2.13 M×196.1 mL

V₁=(2.13 M×196.1 mL)/7.91 M

=52.8 mL

How do the interactions that are broken in water when it is boiled compare with those broken when water is electrolyzed? Boiling water breaks intermolecular attractions and electrolysis breaks covalent bonds. Boiling water breaks covalent bonds and electrolysis breaks intermolecular attractions. Boiling water and electrolysis of water break covalent bonds. Boiling water and electrolysis of water break intermolecular forces.

Answers

Answer:

Boiling water breaks intermolecular attractions and electrolysis breaks covalent bonds.

Explanation:

When water boils, hydrogen bonds are broken between adjacent water molecules. The hydrogen bond is an intermolecular bond between adjacent oxygen and hydrogen atoms of water molecules.  

During electrolysis, water dissociates in the presence of electric current. Here, ions are formed in the process. Therefore, covalent bonds are broken here.

Final answer:

Boiling water breaks the intermolecular hydrogen bonds between water molecules, causing it to change from liquid to gas. Electrolysis on the other hand, breaks the covalent bonds within individual water molecules, splitting them into hydrogen and oxygen.

Explanation:

There is a distinct difference between the type of interactions broken when water is boiled and when it's electrolyzed. Boiling water primarily breaks the intermolecular attractions, specifically hydrogen bonds that exist between water molecules, thereby allowing water molecules to escape into the air as steam or water vapor. On the other hand, electrolysis of water involves breaking of the covalent bonds within individual water molecules, thereby splitting water into its constituent hydrogen and oxygen atoms. This process requires more energy compared to boiling water due to the strength of covalent bonds when compared to hydrogen bonds.

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A solution that is 20% acid and 80% water is mixed with a solution that is 50% acid and 50% water. If twice as much 50% acid solution is used as 20% solution, then what is the ratio of acid to water in the mixture of the solutions?

Answers

Answer:

The ratio acid to water in the mixture is 2:3

Explanation:

Let the volume of 20% acid solution used to make the mixture = x units

So, the volume of 50% acid solution used to make the mixture = 2x units

Total volume of the mixture = x + 2x = 3x units

For 20% acid solution:

C₁ = 20% , V₁ = x

For 50% acid solution :

C₂ = 50% , V₂ = 2x

For the resultant solution of sulfuric acid:

C₃ = ? , V₃ = 3x

Using

C₁V₁ + C₂V₂ = C₃V₃

20×x + 50×2x = C₃×3x

So,

20 + 50×2 = C₃×3

Solving

120 = C₃×3

C₃ = 40 %

Thus, for the resultant mixture,

Acid percentage = 40%

Water percentage = (100 - 40)% = 60%

Ratio acid to water in the mixture = 40:60 = 2:3

Answer:

2:3

Explanation:

The octet rule pertains to:
a. only the noble gases.
b. only groups 1A and 2A.
c. elements in Groups 1A - 7A.
d. all elements in the periodic table.

Answers

I think the answer to c

Identify each definition that applies to the compound in red. Check all that apply. HCI + NaOH → H2O + NaCl
Arrhenius acid
Bronsted-Lowry acid
Arrhenius base
Bronsted-Lowry base

Answers

Answer:

A. Arrhenius acid    

B. Bronsted-Lowry acid

Explanation:

An acid base reaction invoves an acid and a base and yields salt and water as products.

What is a neutralization?

A neutralization is a reaction between an acid and a base to yield salt and water only. The highlighted compound is not shown here however we we shall tell what is compound is in this reaction.

HCI - Arrhenius acidNaOH  - Arrhenius baseH20 -  waterNaCl - Salt

Hence, this an acid base reaction in the Arrhenius sense.

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