Sodium sulfate dissolves as follows: Na2SO4(s) → 2Na (aq) SO42- (aq). How many moles of Na2SO4 are required to make 1.0 L of solution in which the Na concentration is 0.10 M

Answer : The theoretical yield of [tex]CO_2[/tex] is, 112.64 grams.

Explanation : Given,

Given moles of [tex]C_8H_{18}[/tex] = 4 moles

Given moles of [tex]O_2[/tex] = 4 moles

Molar mass of [tex]CO_2[/tex] = 44 g/mole

First we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

[tex]2C_8H_{18}+25O_2\rightarrow 16CO_2+18H_2O[/tex]

From the given balanced reaction, we conclude that

As, 25 moles of [tex]O_2[/tex] react with 2 moles of [tex]C_8H_{18}[/tex]

So, 4 moles of [tex]O_2[/tex] react with [tex]\frac{2}{25}\times 4=0.32[/tex] moles of [tex]C_8H_{18}[/tex]

From this we conclude that, [tex]C_8H_{18}[/tex] is an excess reagent because the given moles are greater than the required moles and [tex]O_2[/tex] is a limiting reagent because it limits the formation of product.

Now we have to calculate the moles of [tex]CO_2[/tex].

As, 25 moles of [tex]O_2[/tex] react to give 16 moles of [tex]CO_2[/tex]

So, 4 moles of [tex]O_2[/tex] react to give [tex]\frac{16}{25}\times 4=2.56[/tex] moles of [tex]CO_2[/tex]

Now we have to calculate the mass of [tex]CO_2[/tex].

[tex]\text{Mass of }CO_2=\text{Moles of }CO_2\times \text{Molar mass of }CO_2[/tex]

[tex]\text{Mass of }CO_2=(2.56mole)\times (44g/mole)=112.64g[/tex]

Therefore, the theoretical yield of [tex]CO_2[/tex] is, 112.64 grams.

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:

P₂O₅ + 3H₂O => 2H₃PO₄

Explanation:

Nothing to explain.

The Lewis structure of PH₃ is attached to the image below. Phosphine (PH3) consists of a phosphorus atom bonded to three hydrogen atoms.

In the Lewis structure of PH₃, the central phosphorus atom (P) is surrounded by three hydrogen atoms (H). The phosphorus atom has five valence electrons, and each hydrogen atom contributes one valence electron, resulting in a total of eight valence electrons in the structure.

In the lewis structure, each line represents a single bond, and the two dots around the phosphorus atom represent the lone pair of electrons.

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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:

0.6425 moles of [tex]NH_3[/tex] is required to react with 25.7 grams of [tex]O_2[/tex].

Explanation:

mas of oxygen gas = 25.7 g

moles of oxygengas = [tex]\frac{25.7 g}{32 g/mol}=0.8031 mol[/tex]

[tex]4NH_3 (g)+5O_2 (g) \rightarrow 4 NO(g) + 6H_20 (I)[/tex]

According to reaction given above,  5 moles of oxygen gas reacts with 4 moles of ammonia gas.

Then 0.8031 moles of oxygen gas will react with :

[tex]\frac{4}{5}\times 0.8031 mol=0.6425 mol[/tex] of ammonia gas

0.6425 moles of [tex]NH_3[/tex] is required to react with 25.7 grams of [tex]O_2[/tex].

Answer:

Explanation:

1) Data:

a) Constant volume

b) Initial pressure: P₀

c) Intitial temperature: T₀ = 0.987°C = 0.987 + 273.15 K = 274.137 K

d) Final pressure: P₁ = 2 P₀

e) Final temperature: unknown, T₁

2) Applied principles:

        ⇒ P / T = constante ⇒ P₁ / T₁ = P₀ / T₀

3) Solution:

a) Solve for T₁

b) Substitute the data:

c) Converto to °C:

To keep a balloon inflated at the same volume when the pressure exerted on it is doubled, the temperature must be increased according to Gay-Lussac's Law. After performing the necessary calculations, we found that the new required temperature should be approximately 275.124°C.

This question relates to Gas Laws, specifically Gay-Lussac's Law, which states that the pressure of a gas is directly proportional to its temperature, assuming the volume stays constant. In this scenario, we need to work out the new temperature when the initial pressure is doubled.

First, let's convert the initial temperature from Celsius to Kelvin. Kelvin = Celsius + 273.15, so, 0.987°C = 274.137K.

According to Gay-Lussac's Law (P₁/T₁ = P₂/T₂ where P is pressure and T is temperature), if P2 = 2*P1 then the new temperature T₂ = 2× T₁, as the volume remains constant.

So, T₂ (in Kelvin) = 2 × T₁ = 2 × 274.137K = 548.274K

To convert T2 back to Celsius, we subtract 273.15 from the Kelvin temperature: T₂ (in Celsius) = 548.274K - 273.15 = 275.124°C.

So, if the pressure exerted on the balloon is doubled, the temperature needs to be approximately 275.124°C to keep the balloon inflated at the same volume.

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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:

What are the solutes

Explanation:

Answer: the correct answer is option D.<3

Explanation:

e2020

Answer: Option (d) is the correct answer.

Explanation:

According to Le Chaltelier's principle, when there occurs any change in an equilibrium reaction then the equilibrium will shift in a direction that will oppose the change.

This means that when pressure is applied on reactant side with more number of moles then the equilibrium will shift on product side that has less number of moles.

For example, [tex]NO_{2}(g) + CO(g) \rightleftharpoons NO(g) + CO_{2}(g)[/tex]

Since here, there are same number of moles on both reactant and product side. So, when volume is decreased at a constant temperature in this system then there will occur no change in the equilibrium state.

Thus, we can conclude that in the given when volume of the system is decreased at constant temperature, then no shift will occur.

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:

t(1/2) = 3.5 min

Explanation:

From A = A₀e⁻ᵏᵗ => k = ln(A/A₀)/-t = [ln(1/16)/-14]min⁻¹ = 0.198 min ⁻¹

=> t(1/2) = 0.693/k = (0.693/0.198)min = 3.5min

The half-life of the polonium is  3.5 minutes. This can be determined by radioactive decay.

Given information,

Final amount = 1/16

Initial amount = 1

Time elapsed = 14 minutes

Radioactive decay: N = N₀ × (1/2)^(t / T)

Where:

N = Final amount of the radioactive substance

N₀ = Initial amount of the radioactive substance

t = Time elapsed

T = Half-life of the radioactive substance

1/16 = 1 × (1/2)^(14.0 / T)

(1/2)⁴ = (1/2)^(14.0 / T)

Since the base (1/2) is the same on both sides, exponents can be equated:

4 = 14.0 / T

T = 14.0 / 4

T = 3.5 minutes

Therefore, the corresponding half-life of polonium is 3.5 minutes.

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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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1. Answer:  For sulfur to attain a noble gas configuration, it must gain TWO ELECTRONS.

EXPLANATION:

The noble gases are the only group of elements that have complete number of electrons in their outermost shells, this make them to be stable hence they don't normally participate in chemical reactions. Other elements participate in chemical reactions in order to form bonds with one another with the aim of attaining noble gas octet structure. The octet structure states that a stable element must have eight (8) electrons in its outermost shell.

The atomic number of sulfur is 16 and its electronic configuration is 2,8,6. The electronic configuration shows that sulfur has 6 electrons in its outermost shell, thus, it needs two more electron in order to have 8 electrons in its outermost shell.  

2. Answer: When sulfur gain two electrons it will attain the electronic configuration of ARGON.

EXPLANATION

Sulfur needs two electrons to attain the octet structure. When sulfur gain two electrons, the total number of electrons in its atoms will become 16 + 2, which is equal to 18 and its electronic configuration will be 2, 8, 8. The atomic number of the noble gas argon is also 18, thus, sulfur will attain the electronic structure of argon if it gains two electrons.

Answer:

Explanation:

Sulfur is a non-metal that belongs the group VI  on the periodic table. Their general valence shell configuration is ns²np⁴. This group has six electrons in their valence or outermost shell. To attain nobility, the would require eight electrons to complete their octet.

The valence shell of sulfur belongs to the L-orbital and it has a maximum capacity of eight electrons.

To complete the octet of the outermost shell, S would require two electrons from a donating or sharing atom.

If sulfur gains two electrons, it would perfectly resembles Argon(18)

The electronic configuration would be: 1S²2S²2P⁶3S²3P⁶

Using the Ideal Gas Law, one can find out the volume of a gas under given conditions by substituting the values of pressure, number of moles, and temperature into the equation and performing the relevant calculation.

Given the parameters of the problem, we can use the Ideal Gas Law to calculate the volume of gas within the balloon.

The Ideal Gas Law is represented by the equation PV = nRT, where:

To convert the given temperature from Celsius to Kelvin, we add 273 to the Celsius temperature (-5 °C + 273 = 268K).

We can now substitute the given and derived values into the Ideal Gas Law: PV = nRT into 0.700V = 4.80*0.081*268, then calculate V = (4.80*0.081*268) / 0.700 to find the volume of the gas.

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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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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:

.186 M

Explanation:

Remember, the equivalence point is reached when the number of moles of the two reactants is the same.

1. For this problem, obtain the number of moles of NaOH by multiplying the concentration and volume given.

2. Once obtain, the number of moles must be the same for the weak acid.

3. With step two in mind, simply solve for molarity by dividing the value obtained and 25 mL.

Calculation.

1. moles of NaOH = (.175)(.0266) = 4.66e-3

2. moles of Acetic acid = 4.66e-3

3. Concentration of acetic acid: (4.66e-3)/(.025) = .186 M

The concentration of acetic acid in the sample is 0.186 M.

The concentration of acetic acid in the sample can be determined using the concept of equivalence point in a titration. From the given information, we know that 26.6 mL of a 0.175 M NaOH solution is needed to reach the equivalence point when titrating a 25.0 mL sample of acetic acid solution.

The equation for the reaction between acetic acid (CH3COOH) and sodium hydroxide (NaOH) is:

CH3COOH + NaOH → CH3COONa + H2O

Using the balanced equation, we can determine the moles of acetic acid in the sample and then calculate its concentration:

Moles of NaOH = concentration × volume (in liters)

Moles of acetic acid = moles of NaOH

Concentration of acetic acid = moles of acetic acid / volume of acetic acid (in liters)

Substituting the given values, we get:

Concentration of acetic acid = (0.175 M) × (0.0266 L) / (0.025 L) = 0.186 M

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 net ionic equation for the given reaction is [tex]2H^+(aq.)+SO_3^{2-}(aq.)\rightarrow H_2O(l)+SO_2(g)[/tex]

Explanation:

Net ionic equation of any reaction does not include any spectator ions.

Spectator ions are the ions which do not get involved in a chemical equation. It is also defined as the ions that are found on both the sides of the chemical reaction when it is present in ionic form.

The chemical equation for the reaction of hydrochloric acid and potassium sulfite is given as:

[tex]2HCl(aq.)+K_2SO_3(aq.)\rightarrow 2KCl(aq.)+SO_2(g)+H_2O(l)[/tex]

Ionic form of the above equation follows:

[tex]2H^+(aq.)+2Cl^-(aq.)+2K^+(aq.)+SO_3^{2-}(aq.)\rightarrow 2K^+(aq.)+2Cl^-(aq.)+SO_2(g)+H_2O(l)[/tex]

As, potassium and chloride ions are present on both the sides of the reaction, thus, it will not be present in the net ionic equation.

The net ionic equation for the above reaction follows:

[tex]2H^+(aq.)+SO_3^{2-}(aq.)\rightarrow SO_2(g)+H_2O(l)[/tex]

Hence, the net ionic equation for the given reaction is written above.

The net ionic equation for the reaction between hydrochloric acid and potassium sulfite is H+ (aq)+ SO3^2- (aq) → H+ (aq) + SO3^2- (aq), following the solubility trends of sulfates and sulfites under standard conditions.

The reaction between excess hydrochloric acid (HCl) and potassium sulfite (K2SO3) is a typical acid-base neutralization reaction. In the initial step, potassium sulfite dissociates into its ions in the aqueous solution:

K2SO3 (aq) → 2K+ (aq) + SO3^2- (aq)

Hydrochloric acid, being a strong acid, also dissociates completely:

HCl (aq) → H+ (aq) + Cl- (aq)

The hydrogen ion from the acid then reacts with the sulfite ion to form sulfuric acid and water, creating a net ionic equation :

2H+ (aq) + SO3^2- (aq) → H2SO3 (aq)

Because of the solubility trends of sulfates and sulfites under standard conditions, the sulfuric acid produced also dissociates into ions:

H2SO3 (aq) → 2H+ (aq) + SO3^2- (aq)

Therefore, the overall net ionic equation is:

H+ (aq)+ SO3^2- (aq) → H+ (aq) + SO3^2- (aq)

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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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The reason why the entropy of a system increases when it is compressed or when a liquid evaporates is because these processes increase the disorder of the system.

When a gas is compressed, the molecules are forced closer together, which means that there are more ways for them to be arranged. Similarly, when a liquid evaporates, the molecules are released from the liquid state and enter the gas state, which means that there are even more ways for them to be arranged.

The reason why the entropy of a system may decrease when it is cooled and expanded is because these processes can decrease the disorder of the system. When a system is cooled, the molecules move more slowly, which means that there are fewer ways for them to be arranged. Similarly, when a system is expanded, the molecules have more space to move around, which also means that there are fewer ways for them to be arranged.

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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:

C

Explanation:

Hydronium (H₃O+) is the same as H+ (aq) because of the same net charge. The acidic property of a solvent or solution is governed the amount of H+ ion in it. The higher the H+ ions the lower the pH. In pure water H₃O+ and OH- are in equal amount. More of OH- ions turn the solution basic.

Answer:

The answer is 357.85 K.

Explanation:

In order to convert temperatures from degrees Celsius to degrees Kelvin, use the following formula:

temperature in degrees Kelvin = temperature in degrees Celsius + 273.15.

(Oftentimes, 273 is used instead of 273.15; it depends on the discretion of the teacher or student.)

The temperature given in degrees Celsius = 84.7° C

Therefore, the temperature in degrees Kelvin (K) = 84.7 + 273.15 = 357.85 K.

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]

[tex]\frac{number of moles}{Volume}[/tex]Answer:

Explanation:

Given parameters:

Concentration of acid = 0.150M

Volume of base = 35.7mL = 0.0357L

Concentration of base = 0.108M

Unknown:

Volume of acid = ?

The balanced equation of the reactions is given as:

     Na₂CO₃  + 2HNO₃ → 2NaNO₃ + CO₂ +  H₂O

To find the unkown volume of the acid, we work from the known parameters of the base to the unknown volume of the acid.

Solution

Concentraction is given as:

                        Molarity = [tex]\frac{Number of moles}{Volume}[/tex]

We first find the number of moles of the base used in the reaction. From the number of moles, we can obtain the volume of the acid used.

Number of moles of base = Molarity x volume of base

                                   = 0.0357 x 0.108 = 0.00386moles

From the balanced reaction equation, we know that:

      1 mole of the base reacted with 2 moles of the acid

0.00386 moles of the base would completely react with 0.0077moles

From this, we can now obtain the volume of acid used:

Volume of acid used = [tex]\frac{number of moles of acid}{concentration of acid}[/tex]

Volume of acid = [tex]\frac{0.0077}{0.15}[/tex] = 0.0514L = 51.41mL

The volume in mL of 0.150 M HNO₃ that will completely react with the given Na₂CO₃ is 51.4 mL

To determine the volume of HNO₃ that will completely react with the given Na₂CO₃

From the balanced chemical equation,

Na₂CO₃(aq) + 2HNO₃(aq) → 2NaNO₃(aq) + CO₂(g) + H₂O(l)

This means that 1 mole of Na₂CO₃ reacts with 2 moles of HNO₃ to produce 2 moles of NaNO₃, 1 mole of CO₂ and 1 mole of H₂O

Now, we will determine the number of moles of Na₂CO₃ present in the reaction

From the question,

Volume of Na₂CO₃ = 35.7 mL = 0.0357 L

Concentration of Na₂CO₃ = 0.108 M

From the formula

Number of moles = Concentration × Volume

∴ Number of moles of Na₂CO₃ = 0.108M × 0.0357 L

Number of moles of Na₂CO₃ = 0.0038556 moles

Therefore, 0.0038556 moles of Na₂CO₃ reacted in the reaction

According to the balanced equation,

1 mole of Na₂CO₃ reacts will react completely with 2 moles of HNO₃

∴ 0.0038556 moles of Na₂CO₃ will react with 2 × 0.0038556 moles of HNO₃

Number of moles of HNO₃ = 2 × 0.0038556 moles = 0.0077112 moles

Now, we will determine the volume of 0.150 M HNO₃ that will give this number of moles

From

Number of moles = Concentration × Volume

[tex]Volume = \frac{Number of moles }{Concentration}[/tex]

[tex]Volume = \frac{0.0077112}{0.150}[/tex]

Volume = 0.051408 L = 51.408 mL ≅ 51.4 mL

Hence, the volume in mL of of HNO₃ that will completely react with the given Na₂CO₃ is 51.4 mL

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Answer:

[tex]\boxed{\text{47.8 mL of NaOH}}[/tex]

Explanation:

For simplicity, let's write the formula of citric acid as H₃A.

1. Balanced chemical equation.

[tex]\rm H$_{3}$A + 3NaOH $\longrightarrow$ Na$_{3}$A + 3H$_{2}$O[/tex]

2. Moles of H₃A

[tex]\text{Moles of H$_{3}$A} =\text{ 0.306 g H$_{3}$A} \times \dfrac{\text{1 mol H$_{3}$A}}{\text{192.12 g H$_{3}$A }} = 1.593 \times 10^{-3} \text{ mol H$_{3}$A}[/tex]

3. Moles of NaOH.

[tex]\text{Moles of NaOH} = 1.593 \times 10^{-3} \text{ mol H$_{3}$A} \times \dfrac{\text{3 mol NaOH} }{\text{1 mol H$_{3}$A}}\\= 4.778 \times 10^{-3}\text{ mol NaOH}[/tex]

4. Volume of NaOH

[tex]V = 4.778 \times 10^{-3}\text{ mol NaOH}\times \dfrac{\text{1 L NaOH}}{\text{0.1000 mol NaOH}} = \text{0.047 78 L NaOH} =\textbf{47.8 mL NaOH}\\\\\text{The student will have to use }\boxed{\textbf{47.8 mL of NaOH}}[/tex]

The volume of NaOH solution is "47.8 mL".

Given values are:

Concentration of NaOH ,

Mass,

The equation,

→ [tex]H_3C_6 H_5 O_7 + 3 NaOH \rightarrow Na_3C_6H_5O_7+ 3 H_2O[/tex]

Now,

→ The moles of [tex]H_3C_6 H_5 O_7[/tex] will be:

= [tex]\frac{mass}{molar \ mass \ of \ H_3C_6 H_5 O_7}[/tex]

= [tex]\frac{0.306}{192.124}[/tex]

= [tex]0.001593 \ mol[/tex]

→ Moles of NaOH will be:

= [tex]3\times moles \ of \ H_3C_6 H_5 O_7[/tex]

= [tex]3\times 0.001593[/tex]

= [tex]0.004778 \ mol[/tex]

hence,

→ The volume of NaOH will be:

= [tex]\frac{moles}{concentration \ of \ NaOH}[/tex]

= [tex]\frac{0.004778}{0.1000}[/tex]

= [tex]0.0478 \ L[/tex]

or,

= [tex]47.8 \ mL[/tex]

Thus the above response is right.

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Answer:

Explanation:

1) Calculate the mass of carbon (C)

2) Calculate the number of moles of C

3) Calculate the number of moles of hydrogen atoms, H:

4) Calculate the number of moles of oxygen atoms, O:

5) Find the mole ratios:

Summary of moles:

Divide every amount by the smallest number, which is 0.184:

Multiply by 3 to round to integer numbers:

Multiply by 2 to round to integer numbers:

Use the mole ratios as superscripts to write the empirical formula

Just as a reference, you can search in internet and find that one compound with that empirical formula is citric acid.

Final answer:

To find the empirical formula, convert the mass and number of atoms to moles, find the mole ratio among the elements by dividing by the smallest number of moles, and then use multipliers to reach whole numbers. The empirical formula for the compound in the question is approximately C3H4O3.

Explanation:

To determine the empirical formula of the unknown sample, we must first convert the given percentages and number of atoms to moles. The sample contains 37.51% carbon, which means there are 2.22 g of carbon in the 5.91 g sample. Using the molar mass of carbon (12.01 g/mol), we calculate the moles of carbon:

2.22g C × (1 mol C / 12.01 g C) = 0.185 moles C

Next, we find the moles of hydrogen, given that there are 1.4830 × 1023 hydrogen atoms. Since 1 mole of any substance contains Avogadro's number of atoms (6.022 × 1023), we have:

1.4830 × 1023 atoms H × (1 mol H / 6.022 × 1023 atoms H) = 0.246 moles H

Similarly, for oxygen atoms, there are 1.2966 × 1023 oxygen atoms:

1.2966 × 1023 atoms O × (1 mol O / 6.022 × 1023 atoms O) = 0.215 moles O

With the moles of each element known, we can now find the mole ratio to get the empirical formula by dividing each element's moles by the smallest amount of moles present among the elements.

Smallest moles = 0.185 moles C

Mole ratio of carbon: 0.185 moles C / 0.185 = 1Mole ratio of hydrogen: 0.246 moles H / 0.185 = 1.33Mole ratio of oxygen: 0.215 moles O / 0.185 = 1.16

To get whole number ratios, we multiply each mole ratio by a common factor to reach the smallest whole numbers which, in this case, is approximately 3 (since 1.33 is close to 4/3 and 1.16 is close to 1, we select 3 as a multiplier to approximate hydrogen as 4 and oxygen as 1).

Mole ratio of carbon: 1 × 3 = 3Mole ratio of hydrogen: 1.33 × 3 ≈ 4Mole ratio of oxygen: 1.16 × 3 ≈ 3

The empirical formula for this compound is approximately C3H4O3.