Arithmetic Word Problems, 8

Category: Arithmetic

"Published in Newark, California, USA"

You collect baseball cards. Suppose you start out with 15. Since her father makes baseball cards, Cindy decides to quadruple your baseball cards. Since you're nice, you give Mike 10 baseball cards. Tiffany gives you another 12 baseball cards. Pearl gives you another 23 baseball cards. How many baseball cards do you have at the end?  

Solution:

The given word problem is about the application of four basic operations. At the start, you have 15 baseball cards.  

If Cindy quadruples your baseball cards, then you have 15 x 4 = 60 baseball cards.

If you give 10 baseball cards to Mike, then you have 60 - 10 = 50 baseball cards.

If Tiffany gives you 12 baseball cards, then you have 50 + 12 = 62 baseball cards.

Finally, if Pearl gives you 23 baseball cards, then you have 62 + 23 = 85 baseball cards at the end. 

Arithmetic Word Problems, 7

Category: Arithmetic

"Published in Newark, California, USA"

You collect baseball cards. Suppose you start out with 15. Maria takes half of one more than the number of baseball cards you have. Since you're nice, you give Peter 1 baseball card. Since his father makes baseball cards, Paul decides to triple your baseball cards. How many baseball cards do you have at the end?

Solution:

The given word problem is about the application of four basic operations. At the start, you have 15 baseball cards. 

One more than the number of baseball cards you have is 15 + 1 = 16 baseball cards. Half of 16 is 8. 

If Maria takes half of one more than the number of baseball cards you have, then you have 15 - 8 = 7 balls.

If you give 1 baseball card to Peter, then you have = 7 - 1 = 6 baseball cards.

Finally, if Paul tripled your baseball cards, then you have 6 x 3 = 18 baseball cards at the end.   

Arithmetic Word Problems, 6

Category: Arithmetic

"Published in Newark, California, USA"

You collect balls. Suppose you start out with 7. John takes half of one more than the number of balls you have. Since his mother makes balls, Peter decides to triple your balls. How many balls do you have at the end?

Solution:

The given word problem is about the application of four basic operations. At the start, you have 7 balls. 

One more than the number of balls you have is 7 + 1 = 8 balls. Half of 8 is 4.

If John takes half of one more than the number of balls you have, then you have 7 - 4 = 3 balls. 

Finally, if Peter tripled your balls, then you have 3 x 3 = 9 balls at the end. 

Arithmetic Word Problems, 5

Category: Arithmetic

"Published in Newark, California, USA"

You collect pens. Suppose you start out with 7. Mike gives you another 22 pens. Since her father makes pens, Cindy decides to double your pens. Since you're nice, you give Sharon 19 pens. How many pens do you have at the end? 

Solution:

The given word problem is about the application of four basic operations. At the start, you have 7 pens. 

If Mike gives you 22 pens, then you have 7 + 22 = 29 pens.

If Cindy doubles your pens, then you have 29 x 2 = 58 pens.

Since you're nice, you give Sharon 19 pens. Therefore, you have 58 - 19 = 39 pens left at the end.

Arithmetic Word Problems, 4

Category: Arithmetic

"Published in Vacaville, California, USA"

You collect coins. Suppose you start out with 11. Since you're nice, you give Sharon 3 coins and you give Steven 4 coins. How many coins do you have at the end?

Solution:

The given word problem is about the application of four basic operations. At the start, you have 11 coins. 

If you give Sharon 3 coins since you're nice, then you have 11 - 3 = 8 coins left. 

Finally, if you give Steven 4 coins since you're nice, then you have 8 - 4 = 4 coins left at the end.
 

Arithmetic Word Problems, 3

Category: Arithmetic

"Published in Vacaville, California, USA"

You collect coins. Suppose you start out with 11. Since his mother makes coins, Miguel decides to double your coins. Since you're nice, you give Mary 12 coins. How many coins do you have at the end?  Solution: The given word problem is about the application of four basic operations. At the start, you have 11 coins.  If Miguel doubles the number of coins you have, then you have 11 x 2 = 22 coins.  Since you're nice, then you give 12 coins to Mary. Therefore, you have 22 - 12 = 10 coins left at the end.

Arithmetic Word Problems, 2

Category: Arithmetic

"Published in Vacaville, California, USA"

You collect baseball cards. Suppose you start out with 16. Since you're nice, you give John 3 baseball cards. How many baseball cards do you have at the end?

Solution:

The given word problem is about the application of four basic operations. At the start, you have 16 balls. 

If you give 3 baseball cards to John, then you will have 16 - 3 = 13 baseball cards left at the end. 

Arithmetic Word Problems

Category: Arithmetic

"Published in Vacaville, California, USA"

You collect balls. Suppose you start out with 16. Mike gives you another 17 balls. How many balls do you have at the end?

Solution:

The given word problem is about the application of four basic operations. At the start, you have 16 balls.

If Mike gives you another ball, you will have 16 + 17 = 33 balls at the end. 

Chemical Equilibrium of Gases, 2

Category: Chemical Engineering Math

"Published in Newark, California, USA"

At 3000K and 1 atm, CO₂ is 40% dissociated to CO and O₂. Calculate its percentage dissociation when the pressure is increased to 2 atm.

Solution:

Consider the chemical reaction above as follows

For gases, the equilibrium constant of the above reaction is

At the start of the reaction, the total pressure of CO₂ which is the total pressure of gas mixture is equal to1 atm. At equilibrium, the partial pressure of gases are as follows:
 

 

Hence, the equilibrium constant for the reaction above is

 

For 1 atm, if x is the partial pressure of CO₂ dissociated, then the partial pressure of gases are as follows:

 

If the pressure is increased to 2 atm, then the partial pressure of gases will be multiplied by 2. Substitute the value of partial pressure of gases at the chemical equilibrium equation above, we have

 

Since the above equation cannot be factored by factoring and even synthetic division, then we have to do the trial and error method until we get zero at the right side of the equation. The best way is to use Excel program in solving for the value of x because we can input the formula and we can assign any values of x in the spreadsheet as follows

Photo by Math Principles in Everyday Life

From the Excel spreadsheet, the value of x which is the partial pressure of CO₂ dissociated is 0.33648572 atm since the right side of the equation is very close to zero.

Therefore, the percent of dissociation of CO₂ is  

Ideal Gas Law Problems, 6

Category: Chemical Engineering Math

"Published in Vacaville, California, USA"

PCl_5(g) dissociates into PCl_3(g) and Cl_2(g) when heated at 250°C and 1 atm. If the density of the gas mixture at equilibrium is 4.4 g/L, what is the fraction of PCl_5(g) dissociated?

Solution:

Consider the chemical reactions above as follows

At equilibrium, there are PCl_5(g), PCl_3(g) and Cl_2(g) gas mixture in 1 liter. If the volume of the reaction at equilibrium is 1 liter, then the number of moles of gas mixture at equilibrium is

At the start of the reaction, the volume is also 1 liter. The number of moles of PCl_5(g) which is also the number of moles of gas mixture at the start of the reaction is

Hence, the amount of PCl_5(g) dissociated or converted into PCl_3(g) and Cl_2(g) is 

Therefore, the fraction of PCl_5(g) dissociated is 
 

 

Dalton's Law of Partial Pressure, 4

Category: Chemical Engineering Math

"Published in Newark, California, USA"

A solvent-water mixture is to be distilled at 95°C. The vapor pressure of the solvent at this temperature is 130 mm Hg and that of water is 640 mm Hg. The solvent is immiscible in water and has a molecular weight of 150. What is the weight of the solvent in kilograms that will be carried over in the distillate with 200 kg of water?

Solution:

From the given word problem, it is about Dalton's Law of Partial Pressure because it involves the mixture of gas vapors which are solvent and water. In this problem, the molecular weight of a solvent is unknown but the partial pressure of solvent and water vapor are given. From the given partial pressures, we can solve for the mole fraction of a solvent vapor as follows

If W is the weight of a solvent and its molecular weight is 150 kg/kgmole, therefore the weight of a solvent that will be carried over in the distillate with 200 kg water is

Dalton's Law of Partial Pressure, 3

Category: Chemical Engineering Math

"Published in Newark, California, USA"

If ethanol and methanol are mixed 50/50 by weight at 60°C, and the solution assumed ideal, what is the composition of the vapor above the solution?

Solution:

From the given word problem, it is about Dalton's Law of Partial Pressure because it involves the mixture of gas vapors. Ethanol and methanol are mixed in equal proportion or amount. 

Basis: 100 grams of the mixture

Let's consider first the liquid phase. The number of moles of ethanol and methanol in the solution are
 

 

Next, consider the gas or vapor phase. From Saturated Vapor Pressure Data, the vapor pressure of pure ethanol at 60°C is 0.461395 atm and for the pure methanol is 0.832997 atm. Hence, the partial pressure of ethanol is

and the partial pressure of methanol is

 

 

 

Therefore, the % ethanol in the vapor is
 

 

 

 

 

and the % methanol in the vapor is