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

Answers

Answer 1

Answer:

ANSWER

The volume of the oxygen gas is 17.5 L

EXPLANATION

Given that;

The mass of oxygen gas is 12 grams

The temperature of the gas is 25 degrees Celcius

The pressure of the gas is 53 kPa

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

Step 1; Assume the gas behaves like an ideal gas

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

$\text{ PV }=\text{ nRT}$

Where

P is the pressure of the gas

V is the volume of the gas

n is number of moles of the gas

R is the universal gas constant

T is the temperature of the gas

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

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

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

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

Step 3; Convert the temperature to degree Kelvin

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

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

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

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

Hence, the volume of the oxygen gas is 17.5 L

Calculating the pH

a) 0 mL

Initially, the pH of the solution is given by the dissociation of NH₃ in water.

NH₃ + H₂O ⇄ NH₄⁺ + OH⁻ (1)

The constant of the above reaction is:

$Kb = ([NH_(4)^(+)][OH^(-)])/([NH_(3)]) = 1.76\cdot 10^(-5)$ (2)

At the equilibrium, we have:

NH₃ + H₂O ⇄ NH₄⁺ + OH⁻ (3)

0.030 M - x x x

$1.76\cdot 10^(-5)*(0.030 - x) - x^(2) = 0$

After solving for x and taking the positive value:

x = 7.18x10⁻⁴ = [OH⁻]

Now, we can calculate the pH of the solution as follows:

$pH = 14 - pOH = 14 + log(7.18\cdot 10^(-4)) = 10.86$

Hence, the initial pH is 10.86.

b) 10 mL

After the addition of HCl, the following reaction takes place:

NH₃ + HCl ⇄ NH₄⁺ + Cl⁻ (4)

We can calculate the pH of the solution from the equilibrium reaction (3).

$1.76\cdot 10^(-5)(Cb - x) - (Ca + x)*x = 0$ (5)

Finding the number of moles of NH₃ and NH₄⁺

The number of moles of NH₃ (nb) and NH₄⁺ (na) are given by:

$n_(b) = n_(i) - n_(HCl)$ (6)

$n_(b) = 0.030 mol/L*0.030 L - 0.025 mol/L*0.010 L = 6.5\cdot 10^(-4) moles$

$n_(a) = n_(HCl)$ (7)

$n_(a) = 0.025 mol/L*0.010 L = 2.5 \cdot 10^(-4) moles$

Calculating the concentrations of NH₃ and NH₄⁺

The concentrations are given by:

$Cb = (6.5\cdot 10^(-4) moles)/((0.030 L + 0.010 L)) = 0.0163 M$ (8)

$Ca = (2.5 \cdot 10^(-4) mole)/((0.030 L + 0.010 L)) = 6.25 \cdot 10^(-3) M$ (9)

Calculating the pH

After entering the values of Ca and Cb into equation (5) and solving for x, we have:

$1.76\cdot 10^(-5)(0.0163 - x) - (6.25 \cdot 10^(-3) + x)*x = 0$

x = 4.54x10⁻⁵ = [OH⁻]

Then, the pH is:

$pH = 14 + log(4.54\cdot 10^(-5)) = 9.66$

Hence, the pH is 9.66.

c) 20 mL

We can find the pH of the solution from the reaction of equilibrium (3).

Calculating the concentrations of NH₃ and NH₄⁺

The concentrations are (eq 8 and 9):

$Cb = (0.030 mol/L*0.030 L - 0.025 mol/L*0.020 L)/((0.030 L + 0.020 L)) = 8.0\cdot 10^(-3) M$

$Ca = (0.025 mol/L*0.020 L)/((0.030 L + 0.020 L)) = 0.01 M$

Calculating the pH

After solving the equation (5) for x, we have:

$1.76\cdot 10^(-5)(8.0\cdot 10^(-3) - x) - (0.01 + x)*x = 0$

x = 1.40x10⁻⁵ = [OH⁻]

Then, the pH is:

$pH = 14 + log(1.40\cdot 10^(-5)) = 9.15$

So, the pH is 9.15.

d) 35 mL

We can find the pH of the solution from reaction (3).

Calculating the concentrations of NH₃ and NH₄⁺

$Cb = (0.030 mol/L*0.030 L - 0.025 mol/L*0.035 L)/((0.030 L + 0.035 L)) = 3.85\cdot 10^(-4) M$

$Ca = (0.025 mol/L*0.035 L)/((0.030 L + 0.035 L)) = 0.0135 M$

Calculating the pH

After solving the equation (5) for x, we have:

$1.76\cdot 10^(-5)(3.85\cdot 10^(-4) - x) - (0.0135 + x)*x = 0$

x = 5.013x10⁻⁷ = [OH⁻]

Then, the pH is:

$pH = 14 + log(5.013\cdot 10^(-7)) = 7.70$

So, the pH is 7.70.

e) 36 mL

Finding the number of moles of NH₃ and NH₄⁺

$n_(b) = 0.030 mol/L*0.030 L - 0.025 mol/L*0.036 L = 0$

$n_(a) = 0.025 mol/L*0.036 L = 9.0 \cdot 10^(-4) moles$

Since all the NH₃ reacts with the HCl added, the pH of the solution is given by the dissociation reaction of the NH₄⁺ produced in water.

At the equilibrium, we have:

NH₄⁺ + H₂O ⇄ NH₃ + H₃O⁺

Ca - x x x

$Ka = (x^(2))/(Ca - x)$

$Ka(Ca - x) - x^(2) = 0$ (10)

Calculating the acid constant of NH₄⁺

We can find the acid constant as follows:

$Kw = Ka*Kb$

Where Kw is the constant of water = 10⁻¹⁴

$Ka = (1\cdot 10^(-14))/(1.76 \cdot 10^(-5)) = 5.68 \cdot 10^(-10)$

Calculating the pH

The concentration of NH₄⁺ is:

$Ca = (9.0 \cdot 10^(-4) moles)/((0.030 L + 0.036 L)) = 0.0136 M$

After solving the equation (10) for x, we have:

x = 2.78x10⁻⁶ = [H₃O⁺]

Then, the pH is:

$pH = -log(H_(3)O^(+)) = -log(2.78\cdot 10^(-6)) = 5.56$

Hence, the pH is 5.56.

f) 37 mL

Now, the pH is given by the concentration of HCl that remain in solution after reacting with NH₃ (HCl is in excess).

Calculating the concentration of HCl

$C_(HCl) = (0.025 mol/L*0.037 L - 0.030 mol/L*0.030 L)/((0.030 L + 0.037 L)) = 3.73 \cdot 10^(-4) M = [H_(3)O^(+)]$

Calculating the pH

$pH = -log(H_(3)O^(+)) = -log(3.73 \cdot 10^(-4)) = 3.43$

Therefore, the pH is 3.43.

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Answer:

a)10.87

b)9.66

c)9.15

d)7.71

e) 5.56

f) 3.43

Explanation:

tep 1: Data given

Volume of 0.030 M NH3 solution = 30 mL = 0.030 L

Molarity of the HCl solution = 0.025 M

Step 2: Adding 0 mL of HCl

The reaction: NH3 + H2O ⇔ NH4+ + OH-

The initial concentration:

[NH3] = 0.030M [NH4+] = 0M [OH-] = OM

The concentration at the equilibrium:

[NH3] = 0.030 - XM

[NH4+] = [OH-] = XM

Kb = ([NH4+][OH-])/[NH3]

1.8*10^-5 = x² / 0.030-x

1.8*10^-5 = x² / 0.030

x = 7.35 * 10^-4 = [OH-]

pOH = -log [7.35 * 10^-4]

pOH = 3.13

pH = 14-3.13 = 10.87

Step 3: After adding 10 mL of HCl

The reaction:

NH3 + HCl ⇔ NH4+ + Cl-

NH3 + H3O+ ⇔ NH4+ + H2O

Calculate numbers of moles:

Moles of NH3 = 0.030 M * 0.030 L = 0.0009 moles

Moles HCl = 0.025 M * 0.010 L = 0.00025 moles

Moles NH4+ = 0 moles

Number of moles at the equilibrium:

Moles NH3 = 0.0009 -0.00025 =0.00065 moles

Moles HCl = 0

Moles NH4+ = 0.00025 moles

Concentration at the equilibrium:

[NH3]= 0.00065 moles / 0.040 L = 0.01625M

[NH4+] = 0.00625 M

pOH = pKb + log [NH4+]/[NH3]

pOH = 4.75 + log (0.00625/0.01625)

pOH = 4.34

pH = 9.66

Step 3: Adding 20 mL of HCl

Calculate numbers of moles:

Moles of NH3 = 0.030 M * 0.030 L = 0.0009 moles

Moles HCl = 0.025 M * 0.020 L = 0.00050 moles

Moles NH4+ = 0 moles

Number of moles at the equilibrium:

Moles NH3 = 0.0009 -0.00050 =0.00040 moles

Moles HCl = 0

Moles NH4+ = 0.00050 moles

Concentration at the equilibrium:

[NH3]= 0.00040 moles / 0.050 L = 0.008M

[NH4+] = 0.01 M

pOH = pKb + log [NH4+]/[NH3]

pOH = 4.75 + log (0.01/0.008)

pOH = 4.85

pH = 14 - 4.85 = 9.15

Step 4: Adding 35 mL of HCl

Calculate numbers of moles:

Moles of NH3 = 0.030 M * 0.030 L = 0.0009 moles

Moles HCl = 0.025 M * 0.035 L = 0.000875 moles

Moles NH4+ = 0 moles

Number of moles at the equilibrium:

Moles NH3 = 0.0009 -0.000875 =0.000025 moles

Moles HCl = 0

Moles NH4+ = 0.000875 moles

Concentration at the equilibrium:

[NH3]= 0.000025 moles / 0.065 L = 3.85*10^-4M

[NH4+] = 0.000875 M / 0.065 L = 0.0135 M

pOH = pKb + log [NH4+]/[NH3]

pOH = 4.75 + log (0.0135/3.85*10^-4)

pOH = 6.29

pH = 14 - 6.29 = 7.71

Step 5: adding 36 mL HCl

Calculate numbers of moles:

Moles of NH3 = 0.030 M * 0.030 L = 0.0009 moles

Moles HCl = 0.025 M * 0.036 L = 0.0009 moles

Moles NH4+ = 0 moles

Number of moles at the equilibrium:

Moles NH3 = 0.0009 -0.0009 =0 moles

Moles HCl = 0

Moles NH4+ = 0.0009 moles

[NH4+] = 0.0009 moles / 0.066 L = 0.0136 M

Kw = Ka * Kb

Ka = 10^-14 / 1.8*10^-5

Ka = 5.6 * 10^-10

Ka = [NH3][H3O+] / [NH4+]

Ka =5.6 * 10^-10 = x² / 0.0136

x = 2.76 * 10^-6 = [H3O+]

pH = -log(2.76 * 10^-6)

pH = 5.56

Step 6: Adding 37 mL of HCl

Calculate numbers of moles:

Moles of NH3 = 0.030 M * 0.030 L = 0.0009 moles

Moles HCl = 0.025 M * 0.037 L = 0.000925 moles

Moles NH4+ = 0 moles

Number of moles at the equilibrium:

Moles NH3 = 0.0009 -0.000925 =0 moles

Moles HCl = 0.000025 moles

Concentration of HCl = 0.000025 moles / 0.067 L = 3.73 * 10^-4 M

pH = -log 3.73*10^-4= 3.43

A battery can provide a current of 4.60 A at 3.40 V for 2.50 hr. How much energy (in kJ) is produced? 1st attempt kJ Energy

Answers

Answer:

The energy produced equals 140.760 kJ

Explanation:

The relation between power, current and voltage is

$Power=Current* Voltage$

Applying the given values in the relation above we get

$Power=4.60* 3.40=15.64W$

Now Since $Power=(Energy)/(Time)\n\nEnergy=Power* Time$

Again applying the calculated values we get

$Energy=15.64* 2.50* 3600=140760Joules=140.76kJ$

What is the molarity of the following solutions?a. 19.5 g NaHCO3 in 460.0 ml solution
b. 26.0 g H2SO4 in 200.0 mL solution
c. 15.0 g NaCl dissolved to make 420.0 mL solution

Answers

Answer:

a) NaHCO3 = 0.504 M

b) H2SO4 = 1.325 M

c) NaCl = 0.610 M

Explanation:

Step 1: Data given

Moles = mass / molar mass

Molarity = moles / volume

a. 19.5 g NaHCO3 in 460.0 ml solution

Step 1: Data given

Mass NaHCO3 = 19.5 grams

Volume = 460.0 mL = 0.460 L

Molar mass NaHCO3 = 84.0 g/mol

Step 2: Calculate moles NaHCO3

Moles NaHCO3 = 19.5 grams / 84.0 g/mol

Moles NaHCO3 = 0.232 moles

Step 3: Calculate molarity

Molarity = 0.232 moles / 0.460 L

Molarity = 0.504 M

b. 26.0 g H2SO4 in 200.0 mL solution

Step 1: Data given

Mass H2SO4 = 26.0 grams

Volume = 200.0 mL = 0.200 L

Molar mass H2SO4 = 98.08 g/mol

Step 2: Calculate moles H2SO4

Moles H2SO4 = 26.0 grams / 98.08 g/mol

Moles H2SO4 = 0.265 moles

Step 3: Calculate molarity

Molarity = 0.265 moles / 0.200 L

Molarity =1.325 M

c. 15.0 g NaCl dissolved to make 420.0 mL solution

Step 1: Data given

Mass NaCl = 15.0 grams

Volume = 420.0 mL = 0.420 L

Molar mass NaCl = 58.44 g/mol

Step 2: Calculate moles NaCl

Moles NaCl = 15.0 grams / 58.44 g/mol

Moles NaCl = 0.256 moles

Step 3: Calculate molarity

Molarity = 0.256 moles / 0.420 L

Molarity =0.610 M

A system absorbs 12 J of heat from the surroundings; meanwhile, 28 J of work is done on the system. What is the change of the internal energy ΔEth of the system?

Answers

Answer: The value of change in internal energy of the system is, 40 J.

Explanation : Given,

Heat absorb from the surroundings = 12 J

Work done on the system = 28 J

First law of thermodynamic : It is a law of conservation of energy in which the total mass and the energy of an isolated system remains constant.

As per first law of thermodynamic,

$\Delta U=q+w$

where,

$\Delta U$ = internal energy = ?

q = heat absorb from the surroundings

w = work done on the system

Now put all the given values in this formula, we get the change in internal energy of the system.

$\Delta U=12J+28J$

$\Delta U=40J$

Therefore, the value of change in internal energy of the system is, 40J.

Mg(OH)2 is a sparingly soluble salt with a solubility product, Ksp, of 5.61×10−11. It is used to control the pH and provide nutrients in the biological (microbial) treatment of municipal wastewater streams. What is the ratio of solubility of Mg(OH)2 dissolved in pure H2O to Mg(OH)2 dissolved in a 0.180 M NaOH solution? Express your answer numerically as the ratio of molar solubility in H2O to the molar solubility in NaOH.

Answers

Answer:

The ration of the molar solubility is 165068.49.

Explanation:

The solubility reaction of the magnesium hydroxide in the pure water is as follows.

$Mg(OH)_(2)\Leftrightarrow Mg^(2+)(aq)+2(OH)^(-)(aq)$

$[Mg^(2+)][OH^(-)]$

Initial 0 0

Equili +S +2S

Final S 2S

$K_(sp)=[Mg^(2+)][OH^(-)]$

$5.61* 10^(-11)=(S)(2S)^(2)$

$S=((5.61* 10^(-11))/(4))^(1/3)=2.41* 10^(-4)M$

Solubility of $Mg(OH)_(2)$ in 0.180 M NaOH is a follows.

$Mg(OH)_(2)\Leftrightarrow Mg^(2+)(aq)+2(OH)^(-)(aq)$

$[Mg^(2+)][OH^(-)]$

Initial 0 0

Equili +S +2S

Final S 2S+0.180M

$K_(sp)=[Mg^(2+)][OH^(-)]$

$5.61* 10^(-11)=(S)(2S+0.180)^(2)$

$S=1.46* 10^(-9)M$

$Ratio\,of\,solubility=(2.41* 10^(-4))/(1.46* 10^(-9))=165068.49$

Therefore, The ration of the molar solubility is 165068.49.

5. As a sample of matter is heated, its particles *O A are unaffected
O B. move more quickly
O C. move more slowly
D. stop moving

Answers

Answer:

B. move more quickly

Explanation:

As a sample of matter is heated, we know that the particles begins to move more quickly.

This is because the temperature of a substance is directly proportional to the average kinetic energy of a system.

As the temperature increases, the body gains more kinetic energy
This is translated to the particles of the medium.
Then they begin to move quickly and very fast enough.

Final answer:

Upon heating a sample of matter, the particles tend to move more quickly. This is because the increase in temperature boosts the kinetic energy of the particles, thereby making them move faster.

Explanation:

When a sample of matter is heated, it causes the particles of the matter to move more quickly. This phenomenon is due to the increase in temperature, which leads to an increase in the kinetic energy of the particles. Kinetic energy refers to the energy that an object possesses due to its motion. Hence, as heat supplies energy, it causes atoms and molecules in a matter to move faster. As a result, solids expand and liquids and gases become less dense as their particles spread out to absorb the heat.

Learn more about Kinetic Energy here:

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A weather system moving through the American Midwest produced rain with an average pH of 5.02. By the time the system reached New England, the rain it produced had an average pH of 4.66. How much more acidic was the rain falling in New England?

Many drugs are sold as their hydrochloric salts (r2nh2+cl−), formed by reaction of an amine (r2nh) with hcl. part 1 out of 4 draw the major organic product formed from the formation of acebutolol with hcl. acebutolol is a β blocker used to treat high blood pressure. omit any inorganic counterions.

. Which law of motion relates the action of a stretched rubber band

Pharmacists sometimes measure medicines in the unit of "grains," where 1 gr = 65 mg. The label on a bottle of aspirin reads "Aspirin, 5 gr." How many mg of aspirin are in a tablet containing 5 gr?

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

Answers

Related Questions

Answers

Calculating the pH

a) 0 mL

b) 10 mL

Finding the number of moles of NH₃ and NH₄⁺

Calculating the concentrations of NH₃ and NH₄⁺

Calculating the pH

c) 20 mL

Calculating the concentrations of NH₃ and NH₄⁺

Calculating the pH

d) 35 mL

Calculating the concentrations of NH₃ and NH₄⁺

Calculating the pH

e) 36 mL

Finding the number of moles of NH₃ and NH₄⁺

Calculating the acid constant of NH₄⁺

Calculating the pH

f) 37 mL

Calculating the concentration of HCl

Calculating the pH

Answers

Answers

Answers

Answers

Answers

Final answer:

Explanation:

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