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By pipet, 11.00 mL of a 0.823 MM stock solution of potassium permanganate (KMnO4) was transferred to a 50.00-mL volumetric flask and diluted to the calibration mark. Determine the molarity of the resulting solution. A stock solution of potassium permanganate (KMnO4) was prepared by dissolving 13.0g KMnO4 with DI H2O in a 100.00-mL volumetric flask and diluting to the calibration mark. Determine the molarity of the solution Molarity= O.822 M

Answers

Answer 1

Answer:

Answer:

1) 0.18106 M is the molarity of the resulting solution.

2) 0.823 Molar is the molarity of the solution.

Explanation:

1) Volume of stock solution = $V_1=11.00 mL$

Concentration of stock solution = $M_1=0.823 M$

Volume of stock solution after dilution = $V_2=50.00 mL$

Concentration of stock solution after dilution = $M_2=?$

$M_1V_1=M_2V_2$ ( dilution )

$M_2=(0.823 M* 11.00 mL)/(50 ,00 mL)=0.18106 M$

0.18106 M is the molarity of the resulting solution.

Molarity of the solution is the moles of compound in 1 Liter solutions.

$Molarity=\frac{\text{Mass of compound}}{\text{Molar mas of compound}* Volume (L)}$

Mass of potassium permanganate = 13.0 g

Molar mass of potassium permangante = 158 g/mol

Volume of the solution = 100.00 mL = 0.100 L ( 1 mL=0.001 L)

$Molarity=(13.0 g)/(158 g/mol* 0.100 L)=0.823 mol/L$

0.823 Molar is the molarity of the solution.

Answer 2

Answer:

Final answer:

To determine the molarity of the resulting solution, we can use the formula M1V1 = M2V2. Plugging in the given values, we find that the molarity of the resulting solution is 0.180 MM.

Explanation:

To determine the molarity of the resulting solution, we need to use the formula:

M1V1 = M2V2

Where M1 is the molarity of the stock solution, V1 is the volume of the stock solution used, M2 is the molarity of the resulting solution, and V2 is the final volume of the resulting solution.

Using the given values, we have:

M1 = 0.823 MM

V1 = 11.00 mL

V2 = 50.00 mL

Substituting these values into the formula, we can find the molarity of the resulting solution.

M2 = (M1 * V1) / V2

Plugging in the values:

M2 = (0.823 MM * 11.00 mL) / 50.00 mL = 0.180 MM

The molarity of the resulting solution is 0.180 MM.

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A 2.00 L container of gas has a pressure of 1.00 atm at 300 K. The temperature of the gas is halved to 150K, and the measured pressure of the same 2.00 Liter sample is 0.420 atm. Which of the following is the best explanation for these observations? A. Pressure is proportional to temperature for a fixed volume of gas. B. The molecules of the gas occupy a significant portion of the volume. C. The molecules of the gas have negligible volume of their own. D. The molecules have significant attractive forces at 150 K. E. The gas is closer to an ideal gas at 150 K.

Answers

The best explanation for the observation is that, the Pressure is proportional to temperature for a fixed volume of gas. (Option A)

From the question given above, the following data were obtained:

Initial volume (V₁) = 2 L

Initial pressure (P₁) = 1 atm

Initial temperature (T₁) = 300 K

Final temperature (T₂) = 150 K

Final volume (V₂) = 2 L = constant

Final pressure (P₂) = 0.420 atm

From the above, we can see that the volume is constant.

Applying the combine gas equation, we can conclude as follow:

P₁V₁ / T₁ = P₂V₂ / T₂

V₁ = V₂

P₁ / T₁ = P₂ / T₂

P/T = constant

P = constant × T

Thus, we can conclude that the pressure is proportional to the temperature at constant volume. This simply implies that the pressure will increase if the temperature increase and it will also decrease if the temperature decreases.

The correct answer to the question is Option A.

Learn more about gas laws: brainly.com/question/9631148

Answer:

Explanation:

PV=nRT

PV/nT

V/T -> (1)/(300)=(x)/(150)

x=.420

Which of the following acids (listed with pKa values) and their conjugate base would form a buffer with a pH of 8.10?(A) HC7H5O2, pKa = 4.19
(B) HF, pKa = 3.46
(C) HClO, pKa = 7.54
(D) HCN, pKa = 9.31
(E) HClO2, pKa = 1.96

Answers

Answer:

The buffer of pH 8.10 will be formed by the HClO having pKa value of 7.54.

Explanation:

Buffer is defined as the substance that can withstand the changes in the solution due to addition of acid or base. Buffer acts to neutralize the small amounts of acids or base when added.

Buffer is composed of two parts:

A weak acid and its conjugate base and a weak base with its conjugate acid.

According to the Bronsted-Lowry theory,acids are the substances that release $\text {H^(2) }$ $\text H^(+)$ , whereas the substances that accept bases.

Similarly, when an acid loses a proton ( $\text H^(+)$ ) $\text {H^(2) }$ $\text {H^(+) }$ , it is converted into the conjugate base, such that the conjugate acid-base pair of HClO is $\text {H}^(+)/\text{ClO}^(-)$ .

The buffer range for a substance is either one unit more or one unit less than the pKa value of the given substance. Thus, to obtain the pH value of 8.10, the acid with pKa value of 7.54 will be selected. The pH range of $\text {H}^(+)/\text{ClO}^(-)$ will have the buffer range between 6.54 and 8.54. Thus, pH of 8.10 will be formed by

Hence, rest of the options are incorrect because the buffer range for given acids will not be close the given value of pH as 8.10.

For Further Reference:

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Answer:

Explanation:

A buffer is a solution that can resist abrupt changes in pH when acids or bases are added. It is formed by two components:

A weak acid and its conjugate base.
A weak base and its conjugateacid.

In this case, acid and base are defined according to Bronsted-Löwry theory, which states that acids are substances that release H⁺ and bases are substances that accept H⁺. Therefore, when an acid loses an H⁺ transforms into its conjugated base. For example, HF/F⁻ is a conjugate acid-base pair.

In buffers, when an acid is added, it reacts with the base to diminish its amount:

F⁻ + H⁺ ⇄ HF

Also in buffers, when a base is added, it reacts with the acid to diminish its amount:

HF + OH⁻ = F⁻ + H₂O

The optimum pH range of work of a buffer system (known as buffer range) is between 1 unit less and 1 unit more of pH than its pKa.

So, the buffer formed by HClO/ClO⁻ works optimally in the pH range 6.54-8.54. Since pH = 8.10 is in that interval, this would be the optimal choice.

g Acetic acid is diluted with water to make a solution of vinegar. You have a sample of vinegar that contains 16.7 g of acetic acid. Determine the number of moles of acetic acid in the vinegar sample.

Answers

Answer:

0.278 mol

Explanation:

Step 1: Given and required data

Mass of acetic acid (m): 16.7 g

Chemical formula of acetic acid: CH₃COOH (C₂H₄O₂)

Step 2: Calculate the molar mass (M) of acetic acid

We will use the following expression.

M(C₂H₄O₂) = 2 × M(C) + 4 × M(H) + 2 × M(O)

M(C₂H₄O₂) = 2 × 12.01 g/mol + 4 × 1.01 g/mol + 2 × 16.00 g/mol = 60.06 g/mol

Step 3: Calculate the number of moles (n) of acetic acid

We will use the following expression.

n = m/M

n = 16.7 g/(60.06 g/mol) = 0.278 mol

Which TWO properties are characteristic of iconic compounds?brittleness
ductility
high melting point
low boiling point
malleability

Answers

Answer : The correct options are, brittleness and high melting point

Explanation :

Ionic compound : Ionic compounds are the compounds which are formed when a metal cation bonded with non-metal anion. The metal cation and non-metal anion bonded with an electrostatic force of attraction.

The properties of ionic compounds are :

Ionic compounds are brittle and hard. They breaks easily into small pieces.

They have high melting point and boiling point.

They conduct electricity in liquid state not in solid state.

Hence, the brittleness and high melting point properties are the characteristic of ionic compounds.

Some characteristics of Ionic compounds by Mimiwhatsup: brittle, high melting point, conducts electricity when molten or dissolved in water.

Suppose 0.10 mol of Cu(NO_3)_2 and 1.50 mol of NH_3 are dissolved in water and diluted to a total volume of 1.00 L. Calculate the concentrations of Cu(NH_3)_4^2+ and of Cu^2+ at equilibrium.

Answers

Explanation:

It is known that the coefficients change in concentration and in the exponents. Hence, the reaction equation will be as follows.

$Cu^(2+)(aq) + 4NH_(3)(aq) \rightarrow Cu(NH_(3))^(2)_(4)(aq)$

According to the ICE table,

$Cu^(2+)(aq) + 4NH_(3)(aq) \rightarrow Cu(NH_(3))^(2)_(4)(aq)$

Initial : 0.10 1.50 0

Change : -x -4x +x

Equilibrium: 0.10 - x 1.50 - 4x x

Hence, the mass action expression is as follows.

$K_(f) = ([Cu(NH3)^(2+)_(4)])/([Cu^(2+)][NH_(3)]_(4))$

= $(x)/((0.10 - x)(1.50 - 4x)^(4))$

As, the value of is huge, it means that the reaction is very product favored. Hence, we need to find the limiting reactant first and then we get to know what x should be.

In the given reaction ammonia is the limiting reactant, because there is less than 4 times the ammonia as the copper cation. Thus, we expect it to run out first, and so, x is approximately equal to 0.25 M.

So, putting the given values into the above equation as follows.

$1.03 * 10^(13) = (0.25)/((0.10 - 0.25)(1.50 - 4x))^(4)$

From here

$[NH_(3)] = 1.50 - 4x = ((2.33)/(1.03 * 10^(13)))^{(1)/(4)$

= M

Therefore, we can "re-solve" for x to get and verify that it is still ≈0.250 M.

x = $[Cu(NH_(3))^(2+)_(4)]$

= $\frac{1.50 - 2.31284 * 10{-4}}{4}]$

= 0.37491425 M

Thus, we can conclude that concentration of ( $Cu^(2+)$ ) is 0.37491425 M.

9. The first step in industrial nitric acid production is the catalyzed oxidation of ammonia. Without a catalyst, a different reaction predominates: 4NH3(g) + 3O2(g) ⇔ 2N2(g) + 6H2O(g) When 0.0120 mol gaseous NH3 and 0.0170 mol gaseous O2 are placed in a 1.00 L container at a certain temperature, the N2 concentration at equilibrium is 2.20×10-3 M. Calculate Keq for the reaction at this temperature.