Which of these scenarios involve a reaction that is at equilibrium? a. the reaction is producing reactants and products at an equal rate.
b. the reaction is producing more reactants than products.
c. the reaction is only producing products
d. the reaction is producing more products than reactants.

Answers

Answer 1

Answer: at equilibrium the reaction is producing reactants and product at an equal rate so option A is correct hope this helps

Answer 2

Answer:

Answer:

a. the reaction is producing reactants and products at an equal rate. is the correct answer.

Explanation:

If a reaction is at equilibrium then the reaction rates for both the reverse and forward directions are equal.

catalyst that increases the reaction rate of a reaction, but it does not change the equilibrium state for that reaction.

Thus at the equilibrium stage, the concentration of the reactants and the product remain constant.

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Which is the electronic configuration for oxygen?1s2 2s2 2p1
1s2 2s2 2p3
1s2 2s2 2p4
1s2 2s2 2p6

Answers

Answer:

1s², 2s², 2p⁴

Explanation:

Electronic configuration is the distribution of electrons of an atom in atomic orbitals. As we know Oxygen is a non metal and is present in group 6 and period 2 with atomic number 8. The atomic number in fact specifies the number of protons. Hence, for a neutral oxygen atom there must be 8 electrons to balance the charges of protons.

These 8 electrons are distributed in two main energy levels i.e. n = 1 and 2 and sub energy levels i.e. s and p. According to certain rules like Aufbau Principle, Pauli's Exclusion Principle and Hund's Rule the electronic configuration for eight electrons is as,

1s², 2s², 2px², 2p¹, 2p¹

Final answer:

The electronic configuration for oxygen is 1s2 2s2 2p4.

Explanation:

The electron configuration for oxygen is 1s2 2s2 2p4.

In the electron configuration, the numbers represent the energy levels (1s, 2s, 2p), and the superscripts represent the number of electrons in each orbital. The electron configuration follows the Aufbau principle, which states that electrons fill the lowest energy levels first.

In the case of oxygen, there are 8 electrons in total. The first two electrons fill the 1s orbital, the next two fill the 2s orbital, and the remaining four fill the 2p orbital (with two electrons each in the three 2p orbitals).

Learn more about Electronic Configuration here:

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In which sample is the average kinetic energy of the particles greatest?(1) 10. mL of HCl(aq) at 25°C
(2) 15 mL of HCl(aq) at 20.°C
(3) 10. mL of H2O(l) at 35°C
(4) 15 mL of H2O(l) at 30.°C

Answers

Answer: The correct answer is option 3.

Explanation: Average kinetic energy is defined as the energy of motion of the particles of a system. It is a measure of Kelvin Temperature.

Mathematically,

$K=(3)/(2)(R)/(N_A)T$

where,

K = kinetic energy of the molecules measured in joules.

R = Gas constant = $8.314Jmol^(-1)K^(-1)$

$N_A$ = Avogadro's number = $6.022* 10^(23)atoms/mol$

T = Temperature in kelvins.

Kinetic energy is directly proportional to the temperature of the system.

$\text{Kinetic energy}\propto \text{Temperature}$

More the temperature, more will be the kinetic energy of the molecules.

Hence, the correct answer is Option 3.

The question asks about the average kinetic energy, so it is related only to temperature. Then the answer is (3) with the highest temperature.

Which compound has the highest percent composition by mass of strontium?(1) SrC12 (2) SrI2 (3) SrO (4) SrS

Answers

The answer is (3). The number of Sr is the same so if the compound has the smallest gram formula mass, it has the highest percent composition by mass of strontium. So the answer is (3).

Answer: option (3) SrO, with 84.56%

Explanation:

You have to calculate the mass of strontium and the mass of the compound, for each of the four compounds, and compute the percent:

percent of strontium = (mass of strontium / mass of the chemical formula) × 100 %.

These are the atomic masses of the elements that you have to use:

Atomic mass of Sr: 87.62 g/mol

Atomic mass of Cl: 35.453 g/mol

Atomic mass of I: 126.904 g/mol

Atomic mass of O: 15.999 g/mol

Atomic mass of S: 32.065 g/mol

(1) SrCl₂:

i) mass of Sr: 87.62 g/mol

ii) mass of Cl₂: 2×35.453 g/mol = 70.906 g/mol

iii) mass of SrCl₂: 87.62 g/mol + 70.906 g/mol = 158.526 g/mol

iv) Sr % = [87.62 / 158.526 ] × 100 = 55.27%

(2) SrI₂

Following the same procedure:

i) Sr % = [87.62 / (87.62 + 2× 126.904) ] × 100 = 25.66%

(3) SrO

i) Sr % = [87.62 / (87.62 + 15.999) ] × 100 = 84.56%

(4) SrS

i) Sr % = [87.62 / (87.62 + 32.065) ] × 100 = 73.21%

The cell potential of the following electrochemical cell depends on the gold concentration in the cathode half-cell:________ Pt(s)
|H2(g,1atm)
|H+(aq,1.0M)
|Au3+(aq,?M)
|Au(s).
What is the concentration of Au3+ in the solution if Ecell is 1.23 V ?

Answers

Answer: The concentration of $Au^(3+)$ is $1.096* 10^(-6)$

Explanation:

The given chemical cell follows:

$Pt(s)|H_2(g,1atm)|H^+(aq,1.0M)||Au^(3+)(aq,?M)|Au(s)$

Oxidation half reaction: $H_2(g,1atm)\rightarrow 2H^(+)(aq,1.0M)+2e^-;E^o_(2H^(+)/H_2)=0.0V$ ( × 3)

Reduction half reaction: $Au^(3+)(aq,?M)+3e^-\rightarrow Au(s);E^o_(Au^(3+)/Au)=1.50V$ ( × 2)

Net cell reaction: $3H_2(g,1atm)+2Au^(3+)(aq,?M)\rightarrow 6H^(+)(aq,1.0M)+2Au(s)$

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the $E^o_(cell)$ of the reaction, we use the equation:

$E^o_(cell)=E^o_(cathode)-E^o_(anode)$

Putting values in above equation, we get:

$E^o_(cell)=1.50-(0.0)=1.50V$

To calculate the EMF of the cell, we use the Nernst equation, which is:

$E_(cell)=E^o_(cell)-(0.059)/(n)\log ([H^(+)]^6)/([Au^(3+)]^2)$

where,

$E_(cell)$ = electrode potential of the cell = 1.23 V

$E^o_(cell)$ = standard electrode potential of the cell = +1.50 V

n = number of electrons exchanged = 6

$[H^(+)]=1.0M$

$[Au^(3+)]=?M$

Putting values in above equation, we get:

$1.23=1.50-(0.059)/(6)* \log(((1.0)^6)/([Au^(3+)]^2))$

$[Au^(3+)]=-1.0906* 10^(-6),1.096* 10^(-6)$

Neglecting the negative value because concentration cannot be negative.

Hence, the concentration of $Au^(3+)$ is $1.096* 10^(-6)$

Predict the sign of the entropy change, ΔS∘, for each of the reaction displayed.Ba2+ (aq) + SO42- (aq) --> BaSO4 (s)

2NO2 (g) --> N2 (g) + 2O2 (g)

C5H12 (g) + 8O2 (g) --> 5CO2 (g) + 6H2O (g)

2NaClO3 (s) --> 2NaCl (s) + 3O2 (g)

2Na (s) + Cl2 (g) --> 2NaCl (s)

CH3OH (l) --> CH3OH (g)

Answers

Answer:

(i) ΔS° < 0 or negative

(ii) ΔS° > 0 or positive

(iii) ΔS° > 0 or positive

(iv) ΔS° > 0 or positive

(v) ΔS° < 0 or negative

(vi) ΔS° > 0 or positive

Explanation:

Entropy is a state function and extensive property of the system.

It's the measurement of degree of randomness..

Entropy of the system decreases as follows

Gas > Liquid > Amorphous solid > Crystalline solid.

In case for chemical reaction

(i) Ba²⁺ (aq) + SO₄²⁻ (aq) --> BaSO₄ (s)

Total entropy change ΔS° < 0 or negative

During this reaction aqueous i.e liquid phase converted into solid phase.

So randomness decreases and hence entropy also decreases.

(ii) 2 NO₂ (g) --> N₂ (g) + 2 O₂ (g) ΔS° > 0 or positive

Since no. of gaseous moles increases from reactants to products.

(iii) C₅H₁₂ (g) + 8 O₂ (g) --> 5 CO₂ (g) + 6 H₂O (g) ΔS° > 0 or positive

Since no. of gaseous moles increases from reactants to products.

(iv) 2 NaClO₃ (s) --> 2 NaCl (s) + 3 O₂ (g) ΔS° > 0 or positive

Since no. of gaseous moles increases from reactants to products.

(v) 2 Na (s) + Cl₂ (g) --> 2 NaCl (s) ΔS° < 0 or negative

Since no. of gaseous moles decreases from left to right so entropy also decreases.

(vi) CH₃OH (l) --> CH₃OH (g) ΔS° > 0 or positive

Because during this phase transition Liquid to gaseous randomness increases and hence entropy also increases.

HELP ME with THIS I am time i have a little bit of time and still have more questions !!! CHEMISTRY!!

Answers

Answer:

1. is B 2. is C or B

Explanation:

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Which of these scenarios involve a reaction that is at equilibrium? a. the reaction is producing reactants and products at an equal rate. b. the reaction is producing more reactants than products. c. the reaction is only producing products d. the reaction is producing more products than reactants.

Answers

Related Questions

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Answer:

Explanation:

Final answer:

Explanation:

Learn more about Electronic Configuration here:

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Which of these scenarios involve a reaction that is at equilibrium? a. the reaction is producing reactants and products at an equal rate.
b. the reaction is producing more reactants than products.
c. the reaction is only producing products
d. the reaction is producing more products than reactants.