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 ; and O2, 163 mmHg . The mixture also contains helium gas. Part A What is the partial pressure of the helium gas? PHe P H e = nothing mmHg Request Answer Part B What mass of helium gas is present in a 10.2-L sample of this mixture at 283 K ? m m = nothing g Request Answer

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:

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:

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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Answer:

Concentration of sulfuric acid in the acid rain sample is 0.0034467 mol/L.

Explanation:

Volume of NaOH = 1.7 ml = 0.0017 L

Molarity of NaOH = 0.0811 M

Moles of NaOH = n

[tex]0.0811 M=\frac{n}{0.0017 L}[/tex]

n = 0.0001378 mol

[tex]H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O[/tex]

According to reaction, 2 mol of NaOH neutralize 1 mol of sulfuric acid.

Then 0.0001378 mol of NaOH will neutralize:

[tex]\frac{1}{2}\times 0.0001378 mol=6.8935\times 10^{-5} mol[/tex] of sulfuric acid.

Concentration of sulfuric acid in the acid rain sample: x

[tex]x=\frac{6.8935\times 10^{-5}}{0.02 L}=0.0034467 mol/L[/tex]

Concentration of sulfuric acid in the acid rain sample is 0.0034467 mol/L.

To find the concentration of sulfuric acid in the given sample of rain, we can use the concept of titration. By calculating the moles of NaOH used in the titration and using the stoichiometry of the reaction, we can determine the concentration of sulfuric acid. The concentration of sulfuric acid in this sample of rain is 0.00685 M.

To calculate the concentration of sulfuric acid in the given sample of rain, we can use the concept of titration. From the given information, we know that 1.7 mL of 0.0811 M NaOH was required to reach the end point. Since the titration reaction is between NaOH and HCl, we can use the stoichiometry of the reaction to calculate the concentration of sulfuric acid.

First, we need to find the moles of NaOH used in the titration:

Moles of NaOH = volume (L) × Molarity

Moles of NaOH = 0.0017 L × 0.0811 M

Moles of NaOH = 0.000137 mol

Since the ratio of HCl to NaOH in the titration reaction is 1:1, the moles of HCl (sulfuric acid) in the rain is also 0.000137 mol.

Finally, we can calculate the concentration of sulfuric acid in the sample:

Concentration of sulfuric acid = moles/volume

Concentration of sulfuric acid = 0.000137 mol/0.020 L

Concentration of sulfuric acid = 0.00685 M

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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:

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 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:

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 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:

[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 : 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]

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.

Hello There!

A "THEORY" summarizes a hypothesis or in some cases summarizes a group of hypotheses that have been supported with repeated testing.

A "HYPOTHESIS" is a proposed explanation that usually happens before testing when you have limited evidence.

____________________________________________________________

It is the act of conducting a test or investigation. "EXPERIMENT"

It is the act of recording an object in action. "OBSERVATION"

Answer: The expression for equilibrium constant is [tex]K_{eq}=\frac{[HOCl]^2}{[H_2O][Cl_2]^2}[/tex]

Explanation:

Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as [tex]K_{eq}[/tex]

For the general chemical equation:

[tex]aA+bB\rightleftharpoons cC+dD[/tex]

The expression for [tex]K_c[/tex] is given as:

[tex]K_c=\frac{[C]^c[D]^d}{[A]^a[B]^b}[/tex]

For the given chemical reaction:

[tex]2HgO(s)+H_2O(l)+2Cl_2(g)\rightleftharpoons 2HOCl(aq.)+HgO.HgCl_2(s)[/tex]

The expression for [tex]K_{eq}[/tex] is given as:

[tex]K_{eq}=\frac{[HOCl]^2[HgO.HgCl_2]}{[HgO]^2[H_2O][Cl_2]^2}[/tex]

The concentration of solid is taken to be 0.

So, the expression for [tex]K_{eq}[/tex] is given as:

[tex]K_{eq}=\frac{[HOCl]^2}{[H_2O][Cl_2]^2}[/tex]

Answer:

Explanation:

Describing a solution as concentrated tells that the solution has a relative large concentration, but it is a qualitative description, not a quantitative one, so this does not tell really how concentrated the solution is. This is, the term concentrated is a kind of vague; it just lets you know that the solution is not very diluted, but, as said initially, that there is a relative large amount (concentration) of solute.

One conclusion, of course, is that the solute is soluble: else the solution were not concentrated.

On the other hand, the terms saturated and supersaturated to define a solution are specific.

A saturated solution has all the solute that certain amount of solvent can contain, at a given temperature. A supersaturated solution has more solute dissolved than the saturated solution at the same temperature; superstaturation is a very unstable condition.

From above, there is no way that you can conclude whether a solution is supersaturated or not from the statement that a solution is concentrated, so the answer is none of the above.

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: 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:

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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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.

Answer:

537.68 torr.

Explanation:

where, P is the pressure of the gas in atm.

V is the volume of the gas in L.

n is the no. of moles of the gas in mol.

R is the general gas constant,

T is the temperature of the gas in K.

(P₁T₂) = (P₂T₁).

P₁ = 485 torr, T₁ = 40°C + 273 = 313 K,

P₂ = ??? torr, ​T₂ = 74°C + 273 = 347 K.

∴ P₂ = (P₁T₂)/(P₁) = (485 torr)(347 K)/(313 K) = 537.68 torr.

By using Gay-Lussac's Law and converting Celsius to Kelvin, we find that the new pressure of the gas is approximately 537.8 torr when the temperature increases to 74°C.

To find the new pressure (P₂) of the gas when its temperature changes, we use the ideal gas law, specifically Gay-Lussac's Law, which states that for a given amount of gas at constant volume, the pressure of the gas is directly proportional to its absolute temperature.

First, we need to convert the temperatures from Celsius to Kelvin:

Gay-Lussac's Law formula is:

P₁/T₁ = P₂/T₂

We rearrange the formula to solve for P2:

P₂ = P₁ * (T₂/T₁)

Substitute the known values:

P₂ = 485 torr * (347.15 K / 313.15 K) ≈ 537.8 torr

The new pressure of the gas inside the container is approximately 537.8 torr.

The equilibrium constant (Kc) for the reaction CO(g) + Cl2(g) ⇆ COCl2(g) is calculated as 204 based on the given equilibrium concentrations.

The question is asking us to calculate the equilibrium constant, Kc, for the reaction of carbon monoxide and molecular chlorine to form carbonyl chloride (also known as phosgene). This reaction is represented by the equation CO(g) + Cl2(g) ⇆ COCl2(g). The equilibrium constant is calculated using the concentrations of the reactants and products at equilibrium.

In this case, we have [CO] = 0.0210 M, [Cl2] = 0.0450 M, and [COCl2] = 0.204 M. To find Kc, we use the formula:

Kc = [COCl2] / ([CO] * [Cl2])

Substituting the given concentrations into the formula, we get:

Kc = 0.204 / (0.0210 * 0.0450) = 204

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The equilibrium constant at the given temperature is  [tex]215.74.[/tex]

The equilibrium constant [tex](K_c)[/tex] for the reaction between carbon monoxide [tex](CO)[/tex] and molecular chlorine [tex](Cl_2)[/tex] to form carbonyl chloride [tex](COCl_2)[/tex] can be calculated using the equilibrium concentrations provided:

[tex]\[ [CO] = 0.0210 \text{ M}, \quad [Cl_2] = 0.0450 \text{ M}, \quad [COCl_2] = 0.204 \text{ M} \][/tex]

The balanced chemical equation is:

[tex]\[ CO(g) + Cl_2(g) \rightleftharpoons COCl_2(g) \][/tex]

The equilibrium constant expression (K_c) for this reaction is:

[tex]\[ K_c = \frac{[COCl_2]}{[CO][Cl_2]} \][/tex]

Substituting the given equilibrium concentrations into the expression:

[tex]\[ K_c = \frac{[0.204]}{[0.0210][0.0450]} \][/tex]

 [tex]\[ K_c = \frac{0.204}{0.0210 \times 0.0450} \][/tex]

[tex]\[ K_c = \frac{0.204}{0.000945} \][/tex]

[tex]\[ K_c \approx 215.74 \][/tex]

 Therefore, the equilibrium constant at the given temperature is approximately 215.74.

The correct format for the answer is:

[tex]\[ \boxed{K_c \approx 215.74} \][/tex]

This value indicates that at equilibrium, the concentration of carbonyl chloride is much higher than the concentrations of carbon monoxide and chlorine gas, suggesting that the reaction strongly favors the formation of [tex]COCl_2[/tex] under these conditions.

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: 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:

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:

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.

Explanation:

[tex]2 N_2O_5(g)\rightarrow 4 NO_2(g)+O_2(g)[/tex]

Rate of the reaction ,k= [tex]3.38\times 10^{-5} s^{-1}[/tex]

Half life of the [tex]N_2O_5=t_{\frac{1}{2}}[/tex]

[tex]t_{\frac{1}{2}}=\frac{0.693}{k}=\frac{0.693}{3.38\times 10^{-5} s^{-1}}[/tex](first order kinetics)

[tex]t_{\frac{1}{2}}=20,502.958 seconds[/tex]

Half life of the [tex]N_2O_5[/tex] is 20,502.958 seconds.

Integrated rate equation for first order kinetics in gas phase is given as:

[tex]k=\frac{2.303}{t}\log\frac{p_o}{2p_o-p}[/tex]

p= pressure of the gas at given time t.

[tex]p_o[/tex] = Initial pressure of the gas

(i) When, t = 50 sec

[tex]p_o=500 torr[/tex]

[tex]3.38\times 10^{-5} s^{-1}=\frac{2.303}{50 s}\log\frac{500 Torr}{2(500 Torr)-p}[/tex]

p = 500.49 Torr

(ii)When, t = 20 min = 1200 sec

[tex]p_o=500 torr[/tex]

[tex]3.38\times 10^{-5} s^{-1}=\frac{2.303}{1200 s}\log\frac{500 Torr}{2(500 Torr)-p}[/tex]

p = 519.83 Torr

Final answer:

The half-life of N2O5 is 20443 seconds. The pressure after 50 seconds is 476.83 Torr and the pressure after 20 minutes is also 476.83 Torr.

Explanation:

The rate constant for the first-order decomposition of N2O5 in the reaction 2 N2O5(g) → 4 NO2(g) + O2(g) is kr=3.38×10−5 s−1 at 25 °C. To find the half-life of N2O5, we can use the formula for first-order reactions:

t1/2 = ln(2) / k,

where t1/2 is the half-life and k is the rate constant. Plugging in the values, we get:

t1/2 = ln(2) / (3.38×10−5 s−1) = 20443 seconds.

To calculate the pressure after a certain time, we can use the ideal gas law:

PV = nRT,

where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature. Since the volume and number of moles are constant, we can rearrange the equation to:

P = (nRT) / V.

To find the pressure after a certain time, we need to know the number of moles at that time. Using the initial pressure, we can calculate the number of moles using the ideal gas law:

n = PV / RT.

Let's calculate the pressure at different times:

(i) After 50 seconds:

n = (500 Torr) / (0.0821 L·atm/(mol·K) × 298 K) = 19.39 moles

P = (19.39 moles × 0.0821 L·atm/(mol·K) × 298 K) / 1 L = 476.83 Torr

(ii) After 20 minutes (1200 seconds):

n = (500 Torr) / (0.0821 L·atm/(mol·K) × 298 K) = 19.39 moles

P = (19.39 moles × 0.0821 L·atm/(mol·K) × 298 K) / 1 L = 476.83 Torr

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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