For the decomposition of calcium carbonate, consider the following thermodynamic data (Due to variations in thermodynamic values for different sources, be sure to use the given values in calculating your answer.): ΔH∘rxn 178.5kJ/mol ΔS∘rxn 161.0J/(mol⋅K) Calculate the temperature in kelvins above which this reaction is spontaneous. Express your answer to four significant figures and include the appropriate units.

Answer:

The van't Hoff factor for sodium chloride in X is 1.9 .

Explanation:

Mass of liquid x = 950 g = 0.950 kg

When 282 g of glycine are dissolved in 950 g of a certain mystery liquid X.

Depression in freezing point of the solution [tex]\Delta T_f= 8.2^oC[/tex]

Molality of the solution = m

The van't Hoff factor for glycine (non ionic) in liquid X= i =1

[tex]m=\frac{282 g}{75.07 g/mol\times 0.950 kg}=3.9542 mol/kg[/tex]

[tex]\Delta T_f=i\times K_f\times m[/tex]

[tex]8.2^oC=1\times K_f\times 3.9542 mol/kg[/tex]

The value of molal depression constant for liquid X:

[tex]K_f=2.0737 ^oC kg/mol[/tex]..(1)

Now 282. g of sodium chloride are dissolved in the same mass of X.

Depression in freezing point of the NaCl solution [tex]\Delta T_f'= 20.0^oC[/tex]

Molality of the NaCl solution = m'

The van't Hoff factor for NaCl(ionic) in liquid X= i'

[tex]m'=\frac{282 g}{58.5 g/mol\times 0.950 kg}=5.0742 mol/kg[/tex]

[tex]\Delta T_f'=i'\times K_f\times m'[/tex]

[tex]20.0^oC=i'\times 2.0737 ^oC kg/mol\times 5.0742 mol/kg[/tex]

i = 1.9007 ≈ 1.9

The van't Hoff factor for sodium chloride in X is 1.9 .

Answer : The mass of copper deposit is, 1.98 grams

Explanation :

First we have to calculate the charge.

Formula used : [tex]Q=I\times t[/tex]

where,

Q = charge = ?

I = current = 10 A

t = time = 10 min = 600 sec      (1 min = 60 sec)

Now put all the given values in this formula, we get

[tex]Q=10A\times 600s=6000C[/tex]

Now we have to calculate the number of atoms deposited.

As, 1 atom require charge to deposited = [tex]2\times (1.6\times 10^{-19})[/tex]  

Number of atoms deposited = [tex]\frac{(6000)}{2\times(1.6\times 10^{-19})}=1.875\times 10^{22}[/tex] atoms

Now we have to calculate the number of moles deposited.

Number of moles deposited = [tex]\frac{(1.875\times 10^{22})}{(6.022\times 10^{23})}=0.03113[/tex] moles

Now we have to calculate the mass of copper deposited.

1 mole of Copper has mass = 63.5 g  

Mass of Copper Deposited = [tex]63.5\times 0.03113 =1.98g[/tex]

Therefore, the mass of copper deposit is, 1.98 grams

Answer:

For A: The molarity of solution is 0.218 M.

For B: The molarity of solution is 0.532 M.

For C: The molarity of solution is [tex]8.552\times 10^{-4}M[/tex]

Explanation:

Molarity is defined as the number of moles present in one liter of solution.

Mathematically,

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

Or,

[tex]\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}[/tex]

We are given:

Moles of [tex]LiNO_3[/tex] = 0.12 moles

Volume of the solution = 5.5 L

Putting values in above equation, we get:

[tex]\text{Molarity of }LiNO_3=\frac{0.12}{5.5L}\\\\\text{Molarity of }LiNO_3}=0.0218M[/tex]

Hence, the molarity of solution is 0.0218 M.

We are given:

Given mass of [tex]C_2H_6O[/tex] = 60.7 g

Molar mass of [tex]C_2H_6O[/tex] = 46 g/mol

Volume of the solution = 2.48 L

Putting values in above equation, we get:

[tex]\text{Molarity of }C_2H_6O=\frac{60.7g}{46g/mol\times 5.5L}\\\\\text{Molarity of }C_2H_6O}=0.532M[/tex]

Hence, the molarity of solution is 0.532 M.

We are given:

Given mass of KI = 14.2 mg = [tex]14.2\times 10^{-3}g[/tex]     (Conversion factor: [tex]1mg=10^{-3}g[/tex]

Molar mass of KI = 166 g/mol

Volume of the solution = 100 L

Putting values in above equation, we get:

[tex]\text{Molarity of KI}=\frac{14.2\times 10^{-3}g\times 1000}{166g/mol\times 100mL}\\\\\text{Molarity of KI}=8.552\times 10^{-4}M[/tex]

Hence, the molarity of solution is [tex]8.552\times 10^{-4}M[/tex]

Answer:

[tex]\boxed{\text{139 $\, ^{\circ}$C}}[/tex]

Explanation:

The question is asking, "At what temperature does the vapour pressure of water equal 3.4 atm?"

To answer this question, we can use the Clausius-Clapeyron equation:

[tex]\ln \left (\dfrac{p_{2}}{p_{1}} \right) = \dfrac{\Delta_{\text{vap}}H}{R} \left(\dfrac{1 }{ T_{1} } - \dfrac{1}{T_{2}} \right)[/tex]

Data:

p₁ = 1 atm;    T₁ = 373.15C

p₂ = 3.4atm; T₂ = ?

R  = 8.314 J·K⁻¹mol⁻¹

[tex]\Delta_{\text{vap}}H = \text{39.67 kJ//mol}[/tex]

(The enthalpy of vaporization changes with temperature. Your value may differ from the one I chose.)

Calculation:

[tex]\begin{array}{rcl}\ln \left (\dfrac{p_{2}}{p_{1}} \right)& = & \dfrac{\Delta_{\text{vap}}H}{R} \left( \dfrac{1}{T_{1}} - \dfrac{1}{T_{2}} \right)\\\\\ln \left (\dfrac{3.4}{1} \right)& = & \dfrac{39670}{8.314} \left(\dfrac{1 }{ 373.15 } - \dfrac{1}{T_{2}} \right)\\\\\ln3.4 & = & 4771 \left(\dfrac{1 }{ 373.15 } - \dfrac{1}{T_{2}} \right)\\\\1.224 & = & 12.78 - \dfrac{4771}{T_{2}}\\\\\dfrac{4771}{T_{2}} & = & 11.56\\\\\end{array}[/tex]

[tex]T_{2} & = & \dfrac{4771}{11.56} = \text{412.6 K} = \textbf{139 $\, ^{\circ}$C}\\\\\text{The maximum temperature is } \boxed{\textbf{139 $\, ^{\circ}$C}}[/tex]

The approximate maximum temperature inside the pressure cooker with the safety valve set at 3.4 atm would be 138°C.

Pressure cookers increase the boiling temperature of water by creating higher pressure inside the cooker. This allows food to cook faster. In the case of the particular pressure cooker mentioned with the safety valve set to vent steam if the pressure exceeds 3.4 atm the approximate maximum temperature would be the boiling temperature of water at that pressure.

According to the phase diagram for water the boiling temperature of water at 3.4 atm is approximately 138°C. Therefore the approximate maximum temperature inside the pressure cooker with the safety valve set at 3.4 atm would be 138°C.

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Answer: c. The rate of the forward reaction will increase.

Explanation:

Any change in the equilibrium is studied on the basis of Le-Chatelier's principle.

This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.

For the given equation:

[tex]2NH_3(g)\rightleftharpoons N_2(g)+3H_2(g)[/tex]  [tex]\Delta H=+ve[/tex]

This is a type of Endothermic reaction because heat is absorbed in the reaction.

When the temperature  is increased, according to the Le-Chatlier's principle, the equilibrium will shift in the direction where decrease in temperature occurs. As, this is an endothermic reaction, forward reaction will decrease the temperature. Hence, the equilibrium will shift in the right or forward direction.

In the given endothermic reaction, increasing the temperature will cause the system to adjust according to Le Châtelier's Principle. This means the system will favor the direction of the reaction that absorbs heat, causing the rate of the forward reaction to increase.

In the given reaction where ΔH is positive, the reaction is endothermic meaning it absorbs heat. When the temperature of the system is raised, the system will try to counteract that change in order to reach a new equilibrium. According to Le Châtelier's Principle, the system will favor the direction of reaction that absorbs heat. In this case, the forward reaction of 2 NH3(g) becoming N2(g) + 3 H2(g) is endothermic, thus the rate of the forward reaction will increase to absorb the added heat and reach a new equilibrium.

So, the correct answer is: c. The rate of the forward reaction will increase.

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To produce 1.00 L of 0.100 M Ba(OH)2, 401 mL of the original Ba(OH)2 solution should be diluted with water. This was calculated using the dilution equation M1V1 = M2V2.

First, let's calculate the molarity of the Ba(OH)2 solution. The formula weight of Ba(OH)2 is 171.34 g/mol, so the given 51.24 g is 0.299 moles.

The solution volume is 1.20 L, so the solution's original molarity (M1) is 0.299 mol/1.20 L = 0.249 M. Using the dilution equation M1V1 = M2V2, where M2 (final molarity) is 0.100 M and V2 (final volume) is 1.00 L, the volume of the original solution (V1) needed is V1 = (M2V2)/M1 = (0.100 M * 1.00 L) / 0.249 M = 0.401 L or 401 mL.

Therefore, 401 mL of the original Ba(OH)2 solution must be diluted with water to produce 1.00 L of 0.100 M Ba(OH)2.

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To calculate the volume of solution needed to make 1.00 L of 0.100 M Ba(OH)2, we can use the equation (initial concentration) x (initial volume) = (final concentration) x (final volume). First, calculate the moles of Ba(OH)2 from the given mass. Next, find the initial volume of the Ba(OH)2 solution using its concentration. Finally, use the equation to find the final volume of the diluted solution.

To calculate how many milliliters of the Ba(OH)2 solution must be diluted with water to make 1.00 L of 0.100 M Ba(OH)2, we can use the equation: (initial concentration) x (initial volume) = (final concentration) x (final volume). First, we convert the given mass of Ba(OH)2 into moles using the molar mass. Then, we calculate the initial volume of the Ba(OH)2 solution using its concentration. Finally, we use the equation to find the final volume of the diluted solution.

Given:

First, calculate the moles of Ba(OH)2:

Moles of Ba(OH)2 = (mass of Ba(OH)2) / (molar mass of Ba(OH)2)

Next, calculate the initial volume of the Ba(OH)2 solution:

Initial volume of Ba(OH)2 solution = (moles of Ba(OH)2) / (concentration of Ba(OH)2)

Finally, use the equation (initial concentration) x (initial volume) = (final concentration) x (final volume) to find the final volume of the diluted solution:

Final volume of diluted solution = (final concentration of Ba(OH)2 x final volume of diluted solution) / (initial concentration of Ba(OH)2)

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Answer:

78.87 liters of bromine gas at 300 °C and 735 Torr are formed.

Explanation:

[tex]5NaBr+NaBrO_3 +3H_2SO_4\rightarrow

3Br_2+3Na_2SO_4+3H_2O

[/tex]

Moles of sodium bromide = [tex]\frac{275 g}{103 g/mol}=2.6699 mol[/tex]

Moles of sodium bromate =[tex]\frac{176 g}{151 g/mol}=1.1655 mol[/tex]

According to reaction , 1 mol of sodium bromate  reacts with 5 moles of sodium bromide.  Then 1.1655 mol of sodium bromate will react with:

[tex]\frac{5}{1}\times 1.1655 mol=5.8278 mol[/tex] of sodium bromide.

This means that sodium bromide is in limiting amount the amount of  bromine gas depends upon sodium bromide.

According to reaction 5 moles of sodium bromide gives 3 moles of bromine gas.

Then 2.6699 moles of sodium bromide will give:

[tex]\frac{3}{5}\times 2.6699 mol=1.60194 mol[/tex] of bromine gas

Volume occupied by bromine gas at 300 °C and 735 Torr.

Pressure of the gas = P =735 Torr = 0.9555 atm

Temperature of the gas = T = 300°C = 573 K

n = 1.60194 mol

[tex]PV=nRT[/tex]

[tex]V=\frac{1.60194 mol\times 0.0821 atm L/ mol K\times 573 K}{0.9555atm}[/tex]

V = 78.87 L

78.87 liters of bromine gas at 300 °C and 735 Torr are formed.

Answer:

36.4 moles

Explanation:

This is a problem where a solute is added to water and this then decreases the vapor pressure of the water from what it was when it was pure.  The amount it will decrease the vapor pressure is directly related to the mole fraction of the solute (or put another way, the mole fraction of the water).

mole fraction of water x vapor pressure of water = vapor pressure of solution.  In equation form it is ...

Psolution = Xsolvent x P0solvent

We can solve for Xsolvent which is the mole fraction of the water.

Xsolvent = Psolution/Posolvent

Xsolvent = 19.8 torr/23.7 torr = 0.835

Since there are 6.00 moles of solute, we can find total moles present in solution.

x moles water/6.00 moles solute + x moles water = 0.835

x = 30.4 moles of water

Total moles present in solution = 6.00 moles + 30.4 moles = 36.4 moles

Final answer:

The total moles in the solution can be determined using Raoult's Law. After calculating the mole fraction of water based on the vapor pressures, this fraction is related to the moles of solute to find the total moles in solution, which is approximately 36.45 moles.

Explanation:

To find the total moles in the aquatic solution, we use Raoult's Law, which states that the vapor pressure of a solution is directly proportional to the mole fraction of solvent present. We start with the provided vapor pressures: 19.8 torr for the solution and 23.7 torr for pure water. Using the formula P₂ = X₂ * P°₂, where P₂ is the vapor pressure of the solvent in the solution, P°₂ is the vapor pressure of the pure solvent, and X₂ is the mole fraction of the solvent, we can first solve for X₂.

X₂ = P₂ / P°₂ = 19.8 torr / 23.7 torr = 0.8354 (mole fraction of water). Since there are 6.00 moles of solute, the mole fraction of solute X₁ is 1 - X₂ = 0.1646. The mole fraction is a ratio of moles of one component over total moles, so if we let x be the total moles, X₁ = 6.00 moles / x moles. Setting these equal and solving for x gives us x = 6.00 moles / 0.1646 = 36.45 moles as the total amount present in solution.

Thus, there are approximately 36.45 moles in the solution comprising of water and the nonvolatile solute.

Answer : The balanced chemical equation is,

[tex]2NH_3(g)+3N_2O(g)\rightarrow 4N_2(g)+3H_2O(g)+210kcal[/tex]

Explanation :

Balanced chemical equation : It is defined as the number of atoms of individual elements present on the reactant side must be equal to the number of atoms of individual elements present on product side.

The given unbalanced chemical reaction is,

[tex]NH_3(g)+N_2O(g)\rightarrow N_2(g)+H_2O(g)+105kcal[/tex]

This chemical reaction is an unbalanced reaction because in this reaction, the number of atoms of individual elements are not balanced.

In order to balanced the chemical reaction, the coefficient 2 is put before the [tex]NH_3[/tex], the coefficient 3 is put before the [tex]N_2O\text{ and }H_2O[/tex] and the coefficient 4 is put before the [tex]N_2[/tex].

The energy evolved in this reaction = [tex]105Kcal\times 2=210Kcal[/tex]

Thus, the balanced chemical reaction will be,

[tex]2NH_3(g)+3N_2O(g)\rightarrow 4N_2(g)+3H_2O(g)+210kcal[/tex]

The balanced equation for the reaction between NH₃(g) and N₂O(g) to form N₂(g) and H₂O(g) with an energy term is 4 NH₃(g) + 6 N₂O(g) rightarrow 5 N₂(g) + 6 H₂O(g) + 420 kcal, which shows that 4 moles of NH₃ react to release 420 kcal of energy.

To write a balanced equation for the reaction between NH₃(g) and N₂O(g) that results in the formation of N₂(g) and H₂O(g) with an energy term in kcal, we first need to establish the correct stoichiometry of the reactants and products. Since the question states that 105 kcal of energy are evolved per mole of NH₃(g) that reacts, the energy term must be included as a product (since the reaction is exothermic).

Given the reaction:
NH₃(g) + N₂O(g) rightarrow N₂(g) + H₂O(g)
we can balance it as follows:
4 NH₃(g) + 6 N₂O(g) rightarrow 5 N₂(g) + 6 H₂O(g) + 420 kcal
This indicates that 4 moles of NH₃ react with 6 moles of N2O to produce 5 moles of N₂, 6 moles of H₂O, and release 420 kcal of energy (since 4 moles of NH₃ react, x 105 kcal/mol = 420 kcal).

Answer : The theoretical yield of HCl is, 56.1735 grams

Explanation : Given,

Mass of [tex]BCl_3[/tex] = 60 g

Mass of [tex]H_2O[/tex] = 37.5 g

Molar mass of [tex]BCl_3[/tex] = 117 g/mole

Molar mass of [tex]H_2O[/tex] = 18 g/mole

Molar mass of [tex]HCl[/tex] = 36.5 g/mole

First we have to calculate the moles of [tex]BCl_3[/tex] and [tex]H_2O[/tex].

[tex]\text{Moles of }BCl_3=\frac{\text{Mass of }BCl_3}{\text{Molar mass of }BCl_3}=\frac{60g}{117g/mole}=0.513moles[/tex]

[tex]\text{Moles of }H_2O=\frac{\text{Mass of }H_2O}{\text{Molar mass of }H_2O}=\frac{37.5g}{18g/mole}=2.083moles[/tex]

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

[tex]BCl_3(g)+3H_2O(l)\rightarrow H_3BO_3(s)+3HCl(g)[/tex]

From the balanced reaction we conclude that

As, 1 mole of [tex]BCl_3[/tex] react with 3 mole of [tex]H_2O[/tex]

So, 0.513 moles of [tex]BCl_3[/tex] react with [tex]3\times 0.513=1.539[/tex] moles of [tex]H_2O[/tex]

From this we conclude that, [tex]H_2O[/tex] is an excess reagent because the given moles are greater than the required moles and [tex]BCl_3[/tex] is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of [tex]HCl[/tex].

As, 1 mole of [tex]BCl_3[/tex] react to give 3 moles of [tex]HCl[/tex]

So, 0.513 moles of [tex]BCl_3[/tex] react to give [tex]3\times 0.513=1.539[/tex] moles of [tex]HCl[/tex]

Now we have to calculate the mass of [tex]HCl[/tex].

[tex]\text{Mass of }HCl=\text{Moles of }HCl\times \text{Molar mass of }HCl[/tex]

[tex]\text{Mass of }HCl=(1.539mole)\times (36.5g/mole)=56.1735g[/tex]

Therefore, the theoretical yield of HCl is, 56.1735 grams

Answer:

Explanation:

To find the maximum amount of copper (in grams) allowed in 100 g of water use the maximum amount ratio (1.3 mg / kg)  and set a proportion with the unknown amount of copper (x) and the amount of water (100 g):

First, convert 100 g of water to kg: 100 g × 1 kg / 1000 g = 0.1 kg.

Now, set the proportion:

Solve for x:

Convert mg to grams:

Answer: 0.00013 g of copper.

The maximum amount of copper allowed in 100g of drinking water is 0.13mg or 0.00013g.

The maximum allowable level of copper in drinking water, as stated, is 1.3 mg/kg. To convert this to grams per 100 g (or equivalently, mg per 100 kg), we use the same value as copper is allowed in the ratio of 1.3 mg for every kg of water. As 1 kg is the same as 1000 g, if we have 100 g of water, we simply divide by 10 to find the allowable quantity of copper.

Therefore, in 100 g of water, the maximum amount of copper allowed will be 0.13 mg or 0.00013 g.

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Answer : The correct net ionic equation will be,

[tex]Zn(s)+2H^+(aq)\rightarrow Zn^{2+}(aq)+H_2(g)[/tex]

Explanation :

Spectator ions : It is defined as the same number of ions present on reactant and product side that do not participate in the reactions.

In the net ionic equations, we are not include the spectator ions in the equations.

The given balanced ionic equation will be,

[tex]Zn(s)+2HBr(aq)\rightarrow ZnBr_2(aq)+H_2(g)[/tex]

The ionic equation in separated aqueous solution will be,

[tex]Zn(s)+2H^+(aq)+2Br^-(aq)\rightarrow Zn^{2+}(aq)+2Br^-(aq)+H_2(g)[/tex]

In this equation, [tex]Br^-[/tex] ion is the spectator ions.

By removing the [tex]Br^-[/tex], spectator ions from the balanced ionic equation, we get the net ionic equation.

Hence, the net ionic equation will be,

[tex]Zn(s)+2H^+(aq)\rightarrow Zn^{2+}(aq)+H_2(g)[/tex]

Answer:

For Part A: The partial pressure of Helium is 218 mmHg.

For Part B: The mass of helium gas is 0.504 g.

Explanation:

We are given:

[tex]p_{CO_2}=245mmHg\\p_Ar}=119mmHg\\p_{O_2}=163mmHg\\P=745mmHg[/tex]

To calculate the partial pressure of helium, we use the formula:

[tex]P=p_{CO_2}+p_{Ar}+p_{O_2}+p_{He}[/tex]

Putting values in above equation, we get:

[tex]745=245+119+163+p_{He}\\p_{He}=218mmHg[/tex]

Hence, the partial pressure of Helium is 218 mmHg.

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

PV = nRT

or,

[tex]PV=\frac{m}{M}RT[/tex]

where,

P = Pressure of helium gas = 218 mmHg

V = Volume of the helium gas = 10.2 L

m = Mass of helium gas = ? g

M = Molar mass of helium gas = 4 g/mol

R = Gas constant = [tex]62.3637\text{ L.mmHg }mol^{-1}K^{-1}[/tex]

T = Temperature of helium gas = 283 K

Putting values in above equation, we get:

[tex]218mmHg\times 10.2L=\frac{m}{4g/mol}\times 62.3637\text{ L.mmHg }mol^{-1}K^{-1}\times 283K\\\\m=0.504g[/tex]

Hence, the mass of helium gas is 0.504 g.

A gas mixture with a total pressure of 745 mmHg contains CO₂ (245 mmHg), Ar (119 mmHg), O₂ (163 mmHg) and He (218 mmHg). 0.504 g of helium occupy 10.2 L at 283 K and 218 mmHg.

A gas mixture with a total pressure of 745 mmHg contains each of the following gases at the indicated partial pressures: CO₂, 245 mmHg; Ar, 119 mmHg; O₂, 163 mmHg; He, unknown partial pressure.

The total pressure is equal to the sum of the partial pressures.

[tex]P = pCO_2 + pAr + pO_2 + pHe\\\\pHe = P - pCO_2 - pAr - pO_2 = 745 mmHg - 245 mmHg - 119 mmHg - 163 mmHg = 218 mmHg[/tex]

Helium occupies 10.2 L at 218 mmHg and 283 K. We can calculate the moles of helium using the ideal gas equation.

[tex]P \times V = n \times R \times T\\\\n = \frac{P \times V}{R \times T} = \frac{218mmHg \times 10.2 L}{(62.4mmHg/mol.K) \times 283K} = 0.126 mol[/tex]

Finally, we will convert 0.126 moles of helium to grams using its molar mass (4.00 g/mol).

[tex]0.126 mol \times \frac{4.00g}{mol} = 0.504 g[/tex]

A gas mixture with a total pressure of 745 mmHg contains CO₂ (245 mmHg), Ar (119 mmHg), O₂ (163 mmHg) and He (218 mmHg). 0.504 g of helium occupy 10.2 L at 283 K and 218 mmHg.

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Answer:

CO(g) + H₂O(g) <=> CO₂(g) + H₂(g), (volume is decreased) .. No effect.

PCl₃(g) + Cl₂(g) <=> PCl₅(g) , (volume is increased)               .. Shift left.

CaCO₃(s) <=> CaO(s) + CO₂(g) , (volume is increased)         .. Shift right.

Explanation:

Le Châtelier's principle states that when there is an dynamic equilibrium, and this equilibrium is disturbed by an external factor, the equilibrium will be shifted in the direction that can cancel the effect of the external factor to reattain the equilibrium.

CO(g) + H₂O(g) <=> CO₂(g) + H₂(g)            (volume is decreased)

So, decreasing the volume will have no effect on the reaction.

PCl₃(g) + Cl₂(g) <=> PCl₅(g) , (volume is increased)

so, increasing the volume will shift the reaction left.

CaCO₃(s) <=> CaO(s) + CO₂(g) , (volume is increased)

so, increasing the volume will shift the reaction right.

Answer: The percent by mass of perchloric acid in the mixture is 22.92 %.

Explanation:

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

We are given:

Volume of potassium hydroxide = 20.4mL = 0.0204 L   (Conversion factor: 1 L = 1000 mL)

Molarity of the solution = 0.922 moles/ L

Putting values in above equation, we get:

[tex]0.922mol/L=\frac{\text{Moles of potassium hydroxide}}{0.0204L}\\\\\text{Moles of potassium hydroxide}=0.0188mol[/tex]

[tex]HClO_4+KOH\rightarrow KClO_4+H_2O[/tex]

By Stoichiometry of the reaction:

1 mole of potassium hydroxide reacts with 1 mole of perchloric acid.

So, 0.0188 moles of potassium hydroxide will react with = [tex]\frac{1}{1}\times 0.0188=0.0188mol[/tex] of perchloric acid.

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

Moles of perchloric acid = 0.0188 moles

Molar mass of perchloric acid = 100.46 g/mol

Putting values in above equation, we get:

[tex]0.0188mol=\frac{\text{Mass of perchloric acid}}{100.46g/mol}\\\\\text{Mass of perchloric acid}=1.88g[/tex]

[tex]\text{Mass percent}=\frac{\text{Mass of the solute}}{\text{Mass of solution}}\times 100[/tex]

We are given:

Mass of perchloric acid = 1.88 g

Mass of solution = 8.20 g

Putting values in above equation, we get:

[tex]\text{Mass percent of perchloric acid}=\frac{1.88g}{8.20g}\times 100\\\\\text{Mass percent of perchloric acid}=22.92\%[/tex]

Hence, the percent by mass of perchloric acid in the mixture is 22.92 %.

Answer : The mass of [tex]PbI_2[/tex] precipitate produced will be, 9.681 grams.

Explanation : Given,

Molarity of NaI = 0.210 M

Volume of solution = 0.2 L

Molar mass of [tex]PbI_2[/tex] = 461.01 g/mole

First we have to calculate the moles of [tex]NaI[/tex].

[tex]\text{Moles of }NaI=\text{Molarity of }NaI\times \text{Volume of solution}=0.210M\times 0.2L=0.042moles[/tex]

Now we have to calculate the moles of [tex]PbI_2[/tex].

The balanced chemical reaction is,

[tex]Pb(ClO_3)_2(aq)+2NaI(aq)\rightarrow PbI_2(s)+2NaClO_3(aq)[/tex]

From the balanced reaction we conclude that

As, 2 moles of [tex]NaI[/tex] react to give 1 mole of [tex]PbI_2[/tex]

So, 0.042 moles of [tex]NaI[/tex] react to give [tex]\frac{0.042}{2}=0.021[/tex] moles of [tex]PbI_2[/tex]

Now we have to calculate the mass of [tex]PbI_2[/tex].

[tex]\text{Mass of }PbI_2=\text{Moles of }PbI_2\times \text{Molar mass of }PbI_2[/tex]

[tex]\text{Mass of }PbI_2=(0.021mole)\times (461.01g/mole)=9.681g[/tex]

Therefore, the mass of [tex]PbI_2[/tex] precipitate produced will be, 9.681 grams.

The balanced equation for the reaction is Pb(ClO3)2(aq) + 2NaI(aq) -> PbI2(s) + 2NaClO3(aq). Using stoichiometry and the molar mass of PbI2, we calculate the mass of the precipitate (PbI2) when 1.50 L of concentrated Pb(ClO3)2 is mixed with 0.200 L of 0.210 M NaI.

The balanced equation for your reaction is Pb(ClO3)2(aq) + 2NaI(aq) -> PbI2(s) + 2NaClO3(aq). Lead iodide (PbI2) is the precipitate formed in the reaction. Using stoichiometry, the molar ratio between Pb(ClO3)2 and PbI2 is 1:1 which means for each mole of Pb(ClO3)2, we get one mole of PbI2. So, the first step is to determine the number of moles in 1.50 L of concentrated Pb(ClO3)2 and 0.200 L of 0.210 M NaI. The reaction is expected to go to completion which means that all the reactants would be used up and the limiting reactant will determine the amount of the precipitate formed. In this reaction, NaI is the limiting reactant. Once you know the number of moles of the limiting reactant, you can determine the mass of the precipitate using the molar mass of PbI2.

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Answer:

[tex]\boxed{\text{1.96 g}}[/tex]

Explanation:

[tex]K = \dfrac{c_{\text{org}}}{c_{\text{aq}}} = 9.73[/tex]

1. First extraction

Let x = mass of compound extracted into ethyl acetate. Then

[tex]\begin{array}{rcl}9.73 & = & \dfrac{x/10.0}{(9.65 - x)/80.0}\\\\9.73\times (9.65 - x)/80.0  & = & x/10.0\\9.73\times (9.65 - x) & = & 8.00x\\93.89 - 9.73x & = & 8.00x\\17.73x & = & 93.89\\x & = & 5.30\\\end{array}[/tex]

So, 5.30 g are extracted into the ethyl acetate.

Mass remaining in water = (9.65 – 5.30) g = 4.35 g

2. Second extraction

[tex]\begin{array}{rcl}9.73 & = & \dfrac{x/10.0}{(4.35 - x)/80.0}\\\\9.73\times (4.35 - x)/80.0  & = & x/10.0\\9.73\times (4.35 - x) & = & 8.00x\\42.33 - 9.73x & = & 8.00x\\17.73x & = & 42.33\\x & = & 2.39\\\end{array}\\\text{So, 2.39 g are extracted into the ethyl acetate.}\\\text{ Mass remaining in water = (4.35 -2.39) g } = \boxed{\textbf{1.96 g}}[/tex]

Given a partition coefficient of 9.73 between water and ethyl acetate for an organic compound, about 0.083 grams of the compound will remain in the aqueous layer after extraction with two 10 mL portions of ethyl acetate.

This question asks how much of an organic compound will remain in water after extraction with ethyl acetate, given a known partition coefficient. The partition coefficient (9.73) is the ratio of the concentrations of the solute in each of the two solvents. Initially, the entire 9.65 grams of the compound is dissolved in 80.0 mL of water.

When the first 10.0 mL of ethyl acetate is added, the amount of solute in the water layer will be 1/(1+9.73) times the initial amount. Therefore, after the first extraction, 0.093 or 9.3% of the original amount will remain in the water, which equals to 0.093 * 9.65 g = 0.898 grams.

After the second extraction, again 9.3% of what was left after the first extraction will remain in the water, which is 0.093 * 0.898 g = 0.083 grams. So, after the two extractions with ethyl acetate, there will be approximately 0.083 grams of the compound remaining in the water.

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Answer:

See explanation ...

Explanation:

5.36g NH₄Cl + 25ml(1M NaOH) => (5.36g/53g) NH₄Cl + 0.025L(1M NaOH)

=> 0.101mole NH₄Cl + 0.025mole NaOH

=> (0.101mole/0.025L) NH₄Cl + (0.025mole/0.025L) NaOH

=> 4.045M NH₄Cl + 1.000M NaOH  

=> 4.045M NH₄⁺ + 4.045M Clˉ + 0.025M Na⁺ + 0.025M OHˉ

=> 0.025M NH₄OH + (4.045 – 0.025)M NH₄⁺ + 0.025M Na⁺ + 4.045M Clˉ

=> 0.025M NH₄OH + 4.02M NH₄⁺ + 0.025M Na⁺ + 4.045M Clˉ

∴0.025M NH₄OH + 4.02M NH₄⁺ is the buffer solution. 0.025M Na⁺ and 4.045M Clˉ are not reactive.  

pH of buffer solution:  

           NH₄OH    ⇄      NH₄⁺   +    OHˉ

C(i)     0.025M            4.02M         0M

ΔC          -x                      +x             +x

C(eq)  (0.025-x)M     (4.02+x)M       x

          ≅ 0.025M       ≅ 4.02M          

Kb = [NH₄⁺][OHˉ]/[NH₄OH] => [OHˉ] = Kb[NH₄OH]/[NH₄⁺]

[OHˉ] = [(1.8 x 10ˉ⁵)(0.025)/(4.02)]M = 1.12 x 10ˉ⁷

=>  pOH = -log[OHˉ] = -log(1.12 x 10ˉ⁷) = 6.95

=> pH = 14 – pOH = 14 – 6.95 =7.04 (buffer solution is neutral)

pH of buffer solution after adding 3ml of 0.034M HCl:

… moles HCl added = 0.003L(0.034M) = 1.02 x 10ˉ⁴mole

… Molarity HCl added = 1.02 x 10ˉ⁴mole/(25 + 3)ml = 1.02 x 10ˉ⁴/0.028L =3.643 x 10ˉ³M HCl

                 NH₄OH          =>              NH₄⁺    +            OHˉ

C(i)          0.025M                           4.02M          1.12 x 10ˉ⁷M ~ (0)M*

ΔC    -3.643 x 10ˉ³M             +3.643 x 10ˉ³M                  +x

C(eq)       0.0214M                       4.0236M                       x**          

*Starting concentration of OHˉ is negligible and is assumed to be zero.  

** concentration of OHˉ after adding HCl  (expect a more acidic system).

[OHˉ](new) = Kb[NH₄OH]/[NH₄⁺]

[OHˉ] = [(1.8 x 10ˉ⁵)(0.0124)/(4.0236)]M = 5.55 x 10ˉ⁸M

=>  pOH = -log[OHˉ] = -log(5.55 x 10ˉ⁸) = 7.26

=> pH = 14 – pOH = 14 – 7.26 =6.74 (solution is slightly acidic after adding acid)  => pH shifts from 7.04 to 6.74 upon addition of 3ml(0.034M HCl).

5.36g NH₄Cl + 25ml(1M NaOH) => (5.36g/53g) NH₄Cl + 0.025L(1M NaOH)

= 0.101mole NH₄Cl + 0.025mole NaOH

= (0.101mole/0.025L) NH₄Cl + (0.025mole/0.025L) NaOH

= 4.045M NH₄Cl + 1.000M NaOH  

= 4.045M NH₄⁺ + 4.045M Clˉ + 0.025M Na⁺ + 0.025M OHˉ

= 0.025M NH₄OH + (4.045 – 0.025)M NH₄⁺ + 0.025M Na⁺ + 4.045M Clˉ

= 0.025M NH₄OH + 4.02M NH₄⁺ + 0.025M Na⁺ + 4.045M Clˉ

Therefore, 0.025M NH₄OH + 4.02M NH₄⁺ is the buffer solution. 0.025M Na⁺ and 4.045M Clˉ are not reactive.  

This is an aqueous solution consisting of a mixture of a weak acid and it's conjugate base, or vice versa.

A. pH of buffer solution:  

          NH₄OH    ⇄      NH₄⁺   +    OHˉ

C(i)     0.025M            4.02M         0M

ΔC          -x                      +x             +x

C(eq)  (0.025-x)M     (4.02+x)M       x

       ≅ 0.025M       ≅ 4.02M          

Kb = [NH₄⁺][OHˉ]/[NH₄OH] => [OHˉ] = Kb[NH₄OH]/[NH₄⁺]

[OHˉ] = [(1.8 x 10ˉ⁵)(0.025)/(4.02)]M = 1.12 x 10ˉ⁷

=  pOH = -log[OHˉ] = -log(1.12 x 10ˉ⁷) = 6.95

B. pH = 14 – pOH = 14 – 6.95 =7.04 (buffer solution is neutral)

C. pH of buffer solution after adding 3ml of 0.034M HCl:

moles HCl added = 0.003L(0.034M) = 1.02 x 10ˉ⁴mole

Molarity HCl added = 1.02 x 10ˉ⁴mole/(25 + 3)ml = 1.02 x 10ˉ⁴/0.028L =3.643 x 10ˉ³M HCl

                NH₄OH          =>              NH₄⁺    +            OHˉ

C(i)          0.025M                           4.02M          1.12 x 10ˉ⁷M ~ (0)M*

ΔC    -3.643 x 10ˉ³M             +3.643 x 10ˉ³M                  +x

C(eq)       0.0214M                       4.0236M                       x**          

Starting concentration of OHˉ is negligible and is assumed to be zero.  

Concentration of OHˉ after adding HCl  (expect a more acidic system).

[OHˉ](new) = Kb[NH₄OH]/[NH₄⁺]

[OHˉ] = [(1.8 x 10ˉ⁵)(0.0124)/(4.0236)]M = 5.55 x 10ˉ⁸M

pOH = -log[OHˉ] = -log(5.55 x 10ˉ⁸) = 7.26

= pH = 14 – pOH = 14 – 7.26 =6.74 (solution is slightly acidic after adding acid)  => pH shifts from 7.04 to 6.74 upon addition of 3ml(0.034M HCl).

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I believe pH = -log[H+]

Also, 14 = pH + pOH

Therefore pH = 14 - pOH

pH = 14 - 4.01

pH = 9.99

9.99 = -log[H+]

Solve for H+

Final answer:

The H+ concentration of an aqueous solution with a pOH of 4.01 at 25 °C is approximately 1.0 × 10^-10 M.

Explanation:

To determine the hydronium ion concentration (H+) of an aqueous solution with a given pOH, we use the relationship that at 25 °C the sum of pH and pOH is 14. Given a pOH of 4.01, we calculate the pH as 14 - 4.01 = 9.99. We then find the H+ concentration by taking the inverse log of the pH.

H+ concentration = 10-pH = 10-9.99 ≈ 1.0 × 10-10 M.

This calculation gives us the H+ concentration in moles per liter (M), which is the standard unit for concentration in chemistry.

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Answer:

4190.22 L = 4.19 m³.

Explanation:

2P₂ + 5O₂ ⇄ 2P₂O₅.

It is clear that 2 mol of P₂ react with 5 mol of O₂ to produce 2 mol of P₂O₅.

no. of moles of P₂O₅ = mass/molar mass = (6920 g)/(283.88 g/mol) = 24.38 mol.

Using cross multiplication:

5 mol of O₂ is needed to produce → 2 mol of P₂O₅, from stichiometry.

??? mol of O₂ is needed to produce → 24.38 mol of P₂O₅.

∴ The no. of moles of O₂ needed = (5 mol)(24.38 mol)/(2 mol) = 60.95 mol.

where, P is the pressure of the gas in atm (P = 606.1 mm Hg/760 = 0.8 atm).

V is the volume of the gas in L (V = ??? L).

n is the no. of moles of the gas in mol (n = 60.95 mol).

R is the general gas constant (R = 0.0821 L.atm/mol.K),

T is the temperature of the gas in K (396.90°C + 273 = 669.9 K).

∴ V of oxygen needed = nRT/P = (60.95 mol)(0.0821 L.atm/mol.K)(669.9 K)/(0.8 atm) = 4190.22 L/1000 = 4.19 m³.

Answer:

[tex]\boxed{240 - 230e^{-\frac{t}{40}}}[/tex]

Explanation:

[tex]\text{Let A = mass of salt after t min}\\\text{and }r_{i} = \text{rate of salt coming into talk}\\\text{and }r_{o}$ =\text{rate of salt going out of tank}[/tex]

1. Set up an expression for the rate of change of salt concentration.

[tex]\dfrac{\text{d}A}{\text{d}t} = r_{i} - r_{o}\\\\r_{i} = \dfrac{\text{6 L}}{\text{1 min}} \times \dfrac{\text{1 g}}{\text{1 L}} = \text{6 g/min}\\\\r_{o} = \dfrac{\text{6 L}}{\text{1 min}} \times \dfrac {A\text{ g}}{\text{240 L}} =\dfrac{x}{40}\text{ g/min}\\\\\dfrac{\text{d}A}{\text{d}t} = 6 - \dfrac{x}{40}[/tex]

2. Integrate the expression

[tex]\dfrac{\text{d}A}{\text{d}t} = \dfrac{240 - x}{40}\\\\\dfrac{\text{d}A}{240 - A} = \dfrac{\text{d}t}{40}\\\\\int \frac{\text{d}A}{240 - A} = \int \frac{\text{d}t}{40}\\\\-\ln |240 - A| = \frac{t}{40} + C[/tex]

3. Find the constant of integration

[tex]-\ln |240 - A| = \frac{t}{40} + C\\\\\text{At $t$ = 0, $A$ = 10, so}\\\\-\ln |240 - 10| = \frac{0}{40} + C\\\\C = -\ln 230[/tex]

4. Solve for A as a function of time.

[tex]\text{The integrated rate expression is}-\ln |240 - A| = \frac{t}{40} - \ln 230\\\\\text{Solve for } A\\\\\ln|240 - A| = \ln 230 - \frac{t}{40}\\\\|240 - A| = 230e^{-\frac{t}{40}}\\\\240 - A = \pm 230e^{-\frac{t}{40}}\\\\x = 240 \pm 230e^{-\frac{t}{40}}\\\\A(0) = 10 \text{ so we choose the negative sign}\\\\x = \boxed{\mathbf{240 - 230e^{-\frac{t}{40}}}}[/tex]

The diagram shows A as a function of time. The mass of salt in the tank starts at 10 g and increases asymptotically to 240 g.

Answer:

81% Yield

Explanation:

2CO        +               O₂                                =>     2CO₂

Excess                3.70g O₂                          =>     8.25g CO₂ (actural yield)

               (3.70g O₂)/(32g O₂/mol O₂)

                   = 0.1156 mol O₂                       =>  2(0.1156) mol CO₂

                                                                          = 10.175g (Theoretical Yield)

%Yield = (Actual Yield / Theoretical Yield)100%  

= (8.25g/10.175g)100% = 81% Yield

Answer: The moles of hydrogen gas that can be formed are 0.18 moles.

Explanation:

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

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]   ....(1)

Given mass of sodium metal = 13.08 g

Molar mass of sodium metal = 23 g/mol

Putting values in above equation, we get:  

[tex]\text{Moles of sodium metal}=\frac{13.08g}{23g/mol}=0.57mol[/tex]

Given mass of hydrochloric acid = 13.08 g

Molar mass of hydrochloric acid = 36.5 g/mol

Putting values in above equation, we get:  

[tex]\text{Moles of hydrochloric acid}=\frac{13.08g}{36.5g/mol}=0.36mol[/tex]

For the given chemical equation:

[tex]2Na+2HCI\rightarrow 2NaCl+H_2[/tex]

By Stoichiometry of the reaction:

2 moles of hydrochloric acid reacts with 2 moles of sodium metal.

So, 0.36 moles of hydrochloric acid will react with = [tex]\frac{2}{2}\times 0.36=0.36moles[/tex] of sodium metal.

As, given amount of sodium metal is more than the required amount. Thus, it is considered as an excess reagent.

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

By Stoichiometry of the above reaction:

2 moles of hydrochloric acid is producing 1 moles of hydrogen gas.

So, 0.36 moles of hydrochloric acid will produce = [tex]\frac{1}{2}\times 0.36=0.18moles[/tex] of hydrogen gas.

Hence,  the moles of hydrogen gas that can be formed are 0.18 moles.

Answer: Option (C) is the correct answer.

Explanation:

According to Arrhenius, bases are the species which when dissolved in water will give hydroxide ions, that is, [tex]OH^{-}[/tex].

For example, [tex]NaOH + H_{2}O \rightarrow Na^{+} + OH^{-}[/tex]

Arrhenius acids are the species which when dissolved in water will give hydrogen ions, that is, [tex]H^{+}[/tex].

For example, [tex]CH_{3}COOH + H_{2}O \rightleftharpoons H_{3}O^{+} + CH_{3}COO^{-}[/tex]

In water, when an acid loses a hydrogen ion then the specie formed is known as conjugate base.

Here, [tex]CH_{3}COO^{-}[/tex] is the conjugate base of [tex]CH_{3}COOH[/tex].

Similarly, species which accept the hydrogen ion result in the formation of conjugate acid.

Hence, [tex]H_{3}O^{+}[/tex] is the conjugate acid of [tex]H_{2}O[/tex].

Thus, we can conclude that [tex]CH_{3}COO^{-}[/tex] is the conjugate base.

Taking into account the Brønsted-Lowry acid-base theory, CH₃COO₋ is a conjugate base.

The Brønsted-Lowry acid-base theory (or the Brønsted-Lowry theory) identifies acids and bases based on whether the species accepts or donates protons or H⁺.

According to this theory, acids are proton donors while bases are proton acceptors. That is, an acid is a species that donates an H⁺ proton while a base is a chemical species that accepts an H⁺ proton from the acid.

So, reactions between acids and bases are H⁺ proton transfer reactions, causing the acid to form its conjugate base and the base to form its conjugated acid by exchanging a proton.

In other words, a conjugate base is an ion or molecule resulting from the acid that loses the proton, while a conjugate acid is an ion or molecule resulting from the base that gains the proton:

acid + base ⇄ conjugate base + conjugate acid

In this case, when an acid loses a hydrogen ion then the specie formed is known as conjugate base.

Then, CH₃COO₋ is the conjugate base of CH₃COOH.

The correct statement is option C) CH₃COO₋ is a conjugate base.

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Answer:

Cu + Fe 3  Pb 2 +

Explanation:

the most reactive metal is the strongest reducing agent but weakest oxidizing agent. And therefore copper being the least  reactive turns to be the strongest oxidizing agent followed by iron then lead.

Answer:

130ml of HCl(36%) in 4.90L solution => pH = 1.50

Explanation:

Need 4.90L of HCl(aq) solution with pH = 1.5.

Given pH = 1.5 => [H⁺] = 10⁻¹·⁵M = 0.032M in H⁺

[HCl(36%)] ≅ 12M in HCl

(M·V)concentrate = (M·V)diluted

12M·V(conc) = 0.032M·4.91L

=> V(conc) needed = [(0.032)(4.91)/12]Liters = 0.0130Liters or 130 ml.

Mixing Caution => Add 131 ml of HCl(36%) into a small quantity of water (~500ml) then dilute to the mark.

Answer:

Explanation:

1) In a sublevel for which l = 0, there are _1__ orbital(s), and the maximum number of electrons that can be accommodated is _2__.

l represents the azimuthal quantum numbers and l = n-1. Where n is the principal quantum number. The principal quantum number(n) gives the main energy level in which the orbital is located.

For l = 0, n = 1

To find the maximum number of electrons, we use 2n². Where n is 1 for the given problem.

2) In a principle energy level for which n = 3, the maximum number of electrons that can be accommodated is _18__.

We simply evaluate using 2n², since n = 3, the maxium number of electrons would be 2(3)² = 18 electrons

3) Give the appropriate values of n and l for an orbital of 3p: n = _3__, and l = _1__.

3p denotes the principal quantum number(n) and the azimuthal quantum number(l)

n = 3

l = 1

An atom with principal quantum number of 3 will have azimuthal numbers of 0,1 and 2. This shows that six electrons can fill the three degenerate orbitals.

The orbital designation is given as:

                 for 0                        3s

                       1                         3p

                       2                         3d

This is why the 3p orbital will have an azimuthal number of 1.

In a sublevel with l = 0, there is one orbital and a maximum capacity of 2 electrons. In a principle energy level with n = 3, the maximum number of electrons that can be accommodated is 18. For an orbital of 3p, the values of n and l are 3 and 1, respectively.

For a sublevel with l = 0, there is only one orbital with a maximum capacity of 2 electrons. In a principle energy level with n = 3, the maximum number of electrons that can be accommodated is given by 2n², so in this case, it would be 2(3)² = 18 electrons. For an orbital of 3p, the corresponding values of n and l would be n = 3 and l = 1.

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Answer: Option (a) is the correct answer.

Explanation:

When atoms which are chemically combine to each other through sharing of electrons, that is, forming covalent bonds with each other then the compound formed is known as a molecular compound.

For example, [tex]SF_{6}[/tex] is a molecular compound.

Generally, non-metals lead to the formation of molecular compounds.

On the other hand, when an electron is transferred from one atom to another then compound formed is known as ionic compound.

For example, [tex]Al_{2}O_{3}[/tex] is an ionic compound.

Generally, metals and no-metals on chemically combining together leads to the formation of ionic compounds.

Thus, we can conclude that sulfur, fluorine is the pair of elements which is most apt to form a molecular compound with each other.

The pair of elements that is most apt to form a molecular compound with each other is sulfur and fluorine.

A molecular compound is formed between two nonmetals. Molecular compounds have a covalent bond between the atoms in the compound.

If we look at the options listed, the only group that contains two nonmetals is  (a) sulfur, fluorine.

These two elements form a molecular compound.

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Answer:

For 1: 3.3 moles of oxygen gas is required.

For 2: 14 moles of hydrogen gas is required.

For 3: 1.5 moles of oxygen gas is required.

Explanation:

The chemical reaction of oxygen and hydrogen to form water follows:

[tex]O_2+2H_2\rightarrow 2H_2O[/tex]

By Stoichiometry of the above reaction:

2 moles of hydrogen gas reacts with 1 mole of oxygen gas.

So, 6.6 moles of hydrogen gas will react with = [tex]\frac{1}{2}\times 6.6=3.3mol[/tex] of oxygen gas.

Hence, 3.3 moles of oxygen gas is required.

By Stoichiometry of the above reaction:

1 mole of oxygen gas reacts with 2 moles of hydrogen gas.

So, 7 moles of oxygen gas will react with = [tex]\frac{2}{1}\times 7=14mol[/tex] of hydrogen gas.

Hence, 14 moles of hydrogen gas is required.

By Stoichiometry of the above reaction:

2 moles of water is formed from 1 mole of oxygen gas.

So, 3.0 moles of water will be formed from = [tex]\frac{1}{2}\times 3.0=1.5mol[/tex] of oxygen gas.

Hence, 1.5 moles of oxygen gas is required.