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2 years ago

Question 14 of 32

Chemistry

1 answer:

ser-zykov [4K]2 years ago

6 0

The equilibrium expression for the reaction; C(s) + O₂(g) ------->CO₂(g) is [CO2]/[C] [O2] option C.

<h3>What is the equilibrium expression?</h3>

The equilibrium expression shows how much of the reactants are converted into products. If the equilibrium constant is large and positive, most of the reactants have been converted into products.

Thus, the equilibrium expression for the reaction; C(s) + O₂(g) ------->CO₂(g) is [CO2]/[C] [O2] option C.

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Which acid/base pair will give an equivalence point that cannot be predicted solely by a general knowledge of acid and base stre

sleet_krkn [62]

The acid/base pair that give equivalence point that Cannot be predicted by general knowledge is NaOH and HCI ONH.

An Acid is a substances that is corrosive in nature and turn blue lithmus paper to red which it react with base to produce salt and water.

Acid dissolve metals.

Base is a substance that turn red lihthmus paper to blue and react with acid to produce salt and water.

Therefore, The acid/base pair that give equivalence point that Cannot be predicted by general knowledge is NaOH and HCI ONH.

The question is incomplete as the options were not given. The options were gotten from another website.

Select the correct answer below:

ONaOH and HCI ONH,

HC ONH, and CH, COOH

NaOH and Christmas, COOH

Learn more about acid and base below.

brainly.com/question/2506771

8 0

2 years ago

For the following reactions, write a balanced equation using half-reactions and calculate the voltage to be expected.

Iteru [2.4K]

a) $2Na(s) +2H_2O(I)$ → $2Na^+(aq) + OH^-(aq)+ H_2(g)$

b) $2Ag (s) +2H^(aq)$ → $2 Ag^+ (aq) +H_2(g)$

<h3>What are half-reactions?</h3>

The half-reaction method is a way to balance redox reactions. It involves breaking the overall equation down into an oxidation part and a reduction part.

$2Na(s) +2H_2O(I)$ → $2Na^+(aq) + OH^-(aq)+ H_2(g)$

$E^0 cell = E^0 (reduction) - E^0 (oxidation)$

= $E^0(\frac{H_2O}{H_2}, OH^-) -E^0(Na^+/Na)$

= -0.83 - (-2.71) =1.88V

$2Ag (s) +2H^(aq)$ → $2 Ag^+ (aq) +H_2(g)$

$E^0cell$ = $E^0(H^+/H_2) -E^0(Ag^+/Ag)$

$E^0cell$ =-0. - (0.8) =-0.8V

Learn more about the half-reactions here:

https://brainly.in/question/18053421

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8 0

2 years ago

In general, what is the effect of increased temperature on the solubility of a gas?

Alecsey [184]

The physical explanation is that increasing temperature increases the kinetic energy of the gas molecules. Hence, their random motion breaks more intermolecular bonds and the gas is less dissolved in the solvent. In contrast, solid solutes in water have increased solubility with increased temperatures.

5 0

2 years ago

Read 2 more answers

Calculate the initial rate for the formation of C at 25 ∘C, if [A]=0.50M and [B]=0.075M.Express your answer to two significant f

N76 [4]

The question is incomplete, here is the complete question:

Calculate the initial rate for the formation of C at 25°C, if [A]=0.50 M and [B]=0.075 M. Express your answer to two significant figures and include the appropriate units.Consider the reaction

A + 2B ⇔ C

whose rate at 25°C was measured using three different sets of initial concentrations as listed in the following table:

The table is attached below as an image.

<u>Answer:</u> The initial rate for the formation of C at 25°C is $2.25\times 10^{-2}Ms^{-1}$

<u>Explanation:</u>

Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

$A+2B\rightleftharpoons C$

Rate law expression for the reaction:

$\text{Rate}=k[A]^a[B]^b$

where,

a = order with respect to A

b = order with respect to B

Expression for rate law for first trial:

$5.4\times 10^{-3}=k(0.30)^a(0.050)^b$ ....(1)

Expression for rate law for second trial:

$1.1\times 10^{-2}=k(0.30)^a(0.100)^b$ ....(2)

Expression for rate law for third trial:

$2.2\times 10^{-2}=k(0.50)^a(0.050)^b$ ....(3)

Dividing 2 by 1, we get:

$\frac{1.1\times 10^{-2}}{5.4\times 10^{-3}}=\frac{(0.30)^a(1.00)^b}{(0.30)^a(0.050)^b}\\\\2=2^b\\b=1$

Dividing 3 by 1, we get:

$\frac{2.2\times 10^{-2}}{5.4\times 10^{-3}}=\frac{(0.50)^a(0.050)^b}{(0.30)^a(0.050)^b}\\\\4.07=2^a\\a=2$

Thus, the rate law becomes:

$\text{Rate}=k[A]^2[B]^1$ ......(4)

Now, calculating the value of 'k' by using any expression.

Putting values in equation 1, we get:

$5.4\times 10^{-3}=k[0.30]^2[0.050]^1\\\\k=1.2M^{-2}s^{-1}$

Calculating the initial rate of formation of C by using equation 4, we get:

$k=1.2M^{-2}s^{-1}$

[A] = 0.50 M

[B] = 0.075 M

Putting values in equation 4, we get:

$\text{Rate}=1.2\times (0.50)^2\times (0.075)^1\\\\\text{Rate}=2.25\times 10^{-2}Ms^{-1}$

Hence, the initial rate for the formation of C at 25°C is $2.25\times 10^{-2}Ms^{-1}$

8 0

3 years ago

A 50 L cylinder is filled with argon gas to a pressure of 10130.0 kPa at 300°C. How many moles of argon gas does the cylinder co

KengaRu [80]

In this problem, we need to use the ideal gas law. The following is the formula used in ideal gas law: PV = nRT, where n refers to the moles and R is the gas constant.

Given
P = 10130.0 kPa
V = 50 L
T = 300 degree celcius + 273.15 = 573.15 K
R = 8.314 L. kPa/K.mol

Solution
To get the moles which represent the "n" in the formula, we need to rearrange the equation.

PV = nRT PV
---- ------ ---> n = --------
RT RT RT

10130.0 kPa x 50 L
n= ---------------------------------------------
8.314 L. kPa/K.mol x 573.15 K
506,500
= ----------------------------
4,765.17 mol K

=106.29 mol Ar

So the moles of argon gas is 106.29 moles

8 0

2 years ago

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