CHEMISTRY COLLEGE

2Identify at least two factors which influence
periodic poperties of elements in groups
and periods​

Answers

Answer 1
Answer:

atomic radius

ionization potential

Answer 2
Answer: Atomic radius and Ionization Energy

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Help me question 5 ASAP

An aqueous solution is 4.44 M nitric acid and the density of the solution is 1.42 g/mL. Calculate the mole fraction of this solution.

Answers

The mole fraction of HNO3 is  0.225

Explanation:

1.Given data

Density = 1.429 /ml

Mass% = 63.01 g HNO3 / 100g of solution

The mass of 63.01 g is in 100 / 1.142 /ml of solution

Or 63.01 g in 55.7 mL

Molarity = 15.39 moles / L

Mass of water in 100g = 100 - 63.01=36.99 g

So 63.01 grams in 36.99 grams of water

So mass of HNO3 in 1000grams of water = 63.01* x 1000 / 36.99 = 1703

Moles of HNO3 in 1000g = 1703 / 63.01 = 27.03 moles

Molality = 27.03 molal (mole / Kg)

Mole fraction = Mole of HN03 / Moles of water + mole of HNO3

Mole of water = 62/ 18 = 3.44

Moles of HNO3 = 63.01 / 63.01 = 1.000

Mole fraction = 1.000 / 3.44 + 1.000 = 0.225

The mole fraction of HNO3 is  0.225

The two reactions above, show routes for conversion of an alkene into an oxirane. If the starting alkene is cis-3-hexene the configurations of the oxirane products, A and B are Product A: _______ Product B: _______ Will either of these two oxirane products rotate the plane of polarization of plane polarized light

Answers

Answer:

Product A and B : (2R,3S)-2,3-diethyloxirane and (2S,3R)-2,3-diethyloxirane.

Explanation:

A double bond is converted to an oxirane through oxidation by peracids e.g. mCPBA (meta-chloroperoxybenzoic acid).

Epoxidation can occur at both face of double bond result in formation of two stereoisomers.

Product A and B : (2R,3S)-2,3-diethyloxirane and (2S,3R)-2,3-diethyloxirane

Both A and B contain plane of symmetry. Hence, both the products are achiral. So, they do not rotate the plane of polarization of plane polarized light.

Consider two aqueous solutions of NaCl. Solution 1 is 4.00 M and solution 2 is 0.10 M. In what ratio (solution 1 to solution 2) must these solutions be mixed in order to produce a 0.86 M solution of NaCl

Answers

Answer:

The ratio of solution 1 to solution 2 is 24.20 to 100.00.

Explanation:

We will mix V₁ (L) of solution 1 with V₂ (L) of solution 2 to get the final solution.

So the mole concentration in the final solution is calculated as below, note that C₁ is the concentration of solution 1, and C₂ is the concentration of solution 2

[M] = (V_(1) C_(1) +V_(2)C_(2))/(V_(1)+V_(2)) = \frac{4 V_(1) + 0.1 V_(2)}_{V_(1)+V_(2)}}=0.86

Then we can calculate for the ratio

(V_(1))/(V_(2))=(0.86-0.10)/(4.00-0.86) =(0.76)/(3.14) or (24.20)/(100.00)

Conduct metric Titration of H_2(SO_4) and Ba(OH)_2 Write an equation (including states of matter) for the reaction between H_2(SO_4) and Ba(OH)_2 At the very start of the titration, before any titrant has been added to the beaker, what is present in the solution? What is the conducting species in this initial solution? Describe what happens as titrant is added to the beaker. Why does the conductivity of the solution decrease? What is the identity of the solid formed? What is the conducting species present in the beaker? What happens when the conductivity value reaches its minimum value (which is designated as the equivalence point for this type of titration)? What is the conducting species in the beaker? Describe what happens at additional titrant is added past the equivalence point. Why does the conductivity of the solution increase? What is the conducting species present in the beaker?

Answers

Answer:

a) H₂SO₄ + Ba(OH)₂ ⇄ BaSO₄(s) + 2 H₂O(l)

b) H₂SO₄, H⁺, HSO₄⁻, SO₄²⁻. H₂O, H⁺, OH⁻.

c) H⁺, HSO₄⁻, SO₄²⁻

d) As the titration takes place, reaction [1] proceeds to the right. The conductivity of the solution decreases because the amount of H⁺, HSO₄⁻, SO₄²⁻ decreases. The formed solid is barium sulfate BaSO₄. Since BaSO₄ is very insoluble, the main responsible for conductivity are still H⁺, HSO₄⁻ and SO₄²⁻,

e) At the equivalence point equivalent amounts of H₂SO₄ and Ba(OH)₂ react. The conducting species are Ba²⁺, SO₄²⁻, H⁺ and OH⁻.

f) After the equivalence point there is an excess of Ba(OH)₂. The ions Ba²⁺ and OH⁻ are responsible for the increase in the conductivity, being the major conducting species.

Explanation:

a) Write an equation (including states of matter) for the reaction between H₂SO₄ and Ba(OH)₂.

The balanced equation is:

H₂SO₄ + Ba(OH)₂ ⇄ BaSO₄(s) + 2 H₂O(l)   [1]

b) At the very start of the titration, before any titrant has been added to the beaker, what is present in the solution?

In the beginning there is H₂SO₄ and the ions that come from its dissociation reactions: H⁺, HSO₄⁻, SO₄²⁻. There is also H₂O and a very small amount of H⁺ and OH⁻ coming from its ionization.

H₂SO₄(aq) ⇄ H⁺(aq) + HSO₄⁻(aq)

HSO₄⁻(aq) ⇄ H⁺(aq) + SO₄²⁻(aq)

H₂O(l)  ⇄ H⁺(aq) + OH⁻(aq)

c) What is the conducting species in this initial solution?

The main responsible for conductivity are the ions coming from H₂SO₄: H⁺, HSO₄⁻, SO₄²⁻.

d) Describe what happens as titrant is added to the beaker. Why does the conductivity of the solution decrease? What is the identity of the solid formed? What is the conducting species present in the beaker?

As the titration takes place, reaction [1] proceeds to the right. The conductivity of the solution decreases because the amount of H⁺, HSO₄⁻, SO₄²⁻ decreases. The formed solid is barium sulfate BaSO₄. Since BaSO₄ is very insoluble, the main responsible for conductivity are still H⁺, HSO₄⁻ and SO₄²⁻,

e) What happens when the conductivity value reaches its minimum value (which is designated as the equivalence point for this type of titration)? What is the conducting species in the beaker?

At the equivalence point equivalent amounts of H₂SO₄ and Ba(OH)₂ react. Only BaSO₄ and H₂O are present, and since they are weak electrolytes, there is a small amount of ions to conduct electricity. The conducting species are Ba²⁺ and SO₄²⁻ coming from BaSO₄ and H⁺ and OH⁻ coming from H₂O.

f) Describe what happens at additional titrant is added past the equivalence point. Why does the conductivity of the solution increase? What is the conducting species present in the beaker?

After the equivalence point there is an excess of Ba(OH)₂. The ions Ba²⁺ and OH⁻ are responsible for the increase in the conductivity, being the major conducting species.

Final answer:

The chemical reaction between H2SO4 and Ba(OH)2 forms BaSO4 and water, reducing conductivity by reducing the number of free ions. Beyond the equivalence point, the conductivity increases due to the dissociated ions from the excess Ba(OH)2 in the solution.

Explanation:

Chemical Reaction and Metric Titration

Firstly, the equation representing the reaction between sulfuric acid (H2SO4) and barium hydroxide (Ba(OH)2) is:


Ba(OH)2 (aq) + H2SO4 (aq) → BaSO4 (s) + 2H2O (l)

In the beginning, the solution only contains H2SO4 with its dissociated ions serving as the conducting species. As titrant (Ba(OH)2) is added, they react to form BaSO4, a solid precipitate reducing the number of free ions in the solution, thus decreasing conductivity. At the equivalence point, all H2SO4 has reacted, and conductivity reaches its minimum as there are lesser free ions for conduction. If additional titrant is added past the equivalence point, conductivity increases due to excess Ba(OH)2's dissociated ions that increase ion concentration in solution.

Learn more about Metric Titration here:

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What volume will 12 g of oxygen gas (O2) occupy at 25 °C and a pressure of 53 kPa?

Answers

ANSWER

The volume of the oxygen gas is 17.5 L

EXPLANATION

Given that;

The mass of oxygen gas is 12 grams

The temperature of the gas is 25 degrees Celcius

The pressure of the gas is 53 kPa

To find the volume of the oxygen gas, follow the steps below

Step 1; Assume the gas behaves like an ideal gas

Therefore, apply the ideal gas equation to find the volume of the gas

\text{ PV }=\text{ nRT}

Where

P is the pressure of the gas

V is the volume of the gas

n is number of moles of the gas

R is the universal gas constant

T is the temperature of the gas

Step 2: Find the number of moles of the oxygen gas using the below formula

\text{ mole }=\text{ }\frac{\text{ mass}}{\text{ molar mass}}

Recall, that the molar mass of the oxygen gas is 32 g/mol

\begin{gathered} \text{ mole }=\text{ }\frac{12}{\text{ 32}} \n \text{ mole }=\text{ 0.375 mol} \end{gathered}

Step 3; Convert the temperature to degree Kelvin

\begin{gathered} \text{ T }=\text{ t }+\text{ 273.15} \n \text{ t }=\text{ 25}\degree C \n \text{ T }=25\text{ }+\text{ 273.15} \n \text{ T }=\text{ 298.15K} \end{gathered}

Step 4; Substitute the given data into the formula in step 1

Recall, that R is 8.314 L kPa K^-1 mol^-1

\begin{gathered} \text{ 53 }*\text{ V }=\text{ 0.375}*\text{ 8.314}*\text{ 298.15} \n \text{ 53V }=\text{ 929.557} \n \text{ Divide both sides by 53} \n \text{ }\frac{\cancel{53}V}{\cancel{53}}\text{ }=\text{ }(929.557)/(53) \n \text{ V }=\text{ }(929.557)/(93) \n \text{ V }=\text{ 17.5 L} \end{gathered}

Hence, the volume of the oxygen gas is 17.5 L

Which of the following describes the correct order for relative solubility of minerals in sedimentary rocks? a. Evaporate minerals, quartz, and calcite all have the same relative solubility.
b. Evaporate minerals are more soluble than calcite and quartz.
c. Evaporate minerals are more soluble than quartz and less soluble than calcite.
d. Evaporate minerals are less soluble than quartz and calcite.

Answers

Answer:

Evaporate minerals are more soluble than calcite and quartz.

Explanation:

Evaporate minerals are the water soluble minerals which at higher concentration precipitate out and crystallized forming rocks.

example of chemicals present are:

chlorides and sulphates.

Quartz is silica (very less soluble, or insoluble)

Calcite is calcium carbonate, again an insoluble salt.

Thus

Evaporate minerals are more soluble than calcite and quartz.
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