Mixture problem 2 - linear equation

In summary, a mixture problem 2 - linear equation is a word problem that involves finding the ratio or proportion of different elements in a mixture and can be solved by creating and solving a linear equation. To approach solving this type of problem, one should identify known and unknown quantities and create a system of equations using algebraic methods. An example of a mixture problem 2 - linear equation is finding the amount of 40% and 60% acid solutions needed to make 10 liters of a 50% acid solution. Real-life applications of this type of problem can be found in fields such as chemistry and economics. While there are no specific shortcuts or tricks for solving these types of problems, careful reading and understanding of the problem, as well
  • #1
paulmdrdo1
385
0
1. A clay contains 45% silica and 10% water. Determine the percentage of silica in the clay on a dry (water-free) basis. All percentages are by weight.

my answer is 100% silica. because if the clay is water free the clay is 100% silica. is this correct?

2. A coal contains 2.4% water. After drying, the moisture-free residue contains 71.0% carbon.
Determine the percentage of carbon on the “wet basis.” All percentages are by weight.

for this I have no idea. please give me some assistance here. thanks!
 
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  • #2
paulmdrdo said:
1. A clay contains 45% silica and 10% water. Determine the percentage of silica in the clay on a dry (water-free) basis. All percentages are by weight.

my answer is 100% silica. because if the clay is water free the clay is 100% silica. is this correct?
But before drying silica and water constituted only 45% + 10% = 55% of clay. The rest 45% were other substances. Do you think they evaporated during drying?

Suppose wet clay weighs 100 kg. Then silica weighs 45 kg, water weighs 10 kg and other substances also weigh 45 kg. After drying, water evaporates, but silica and other substances remain. What share of the new total volume does silica make up?

paulmdrdo said:
2. A coal contains 2.4% water. After drying, the moisture-free residue contains 71.0% carbon.
Determine the percentage of carbon on the “wet basis.” All percentages are by weight.

for this I have no idea. please give me some assistance here. thanks!
Suppose the orginal coal weighs 100 kg. Then water weighs 2.4 kg, so dry matter, including carbon, weighs 97.6 kg. Of that dry matter carbon makes up 71%. Find the mass of carbon and divide by the original mass, i.e., 100 kg.
 
  • #3
the silica would still be 45%. Am I correct?

please I need a quick response for this. thanks!
 
  • #4
Evgeny.Makarov said:
Suppose wet clay weighs 100 kg. Then silica weighs 45 kg, water weighs 10 kg and other substances also weigh 45 kg. After drying, water evaporates, but silica and other substances remain. What share of the new total volume does silica make up?

paulmdrdo said:
the silica would still be 45%. Am I correct?
No. The total mass (sorry, I mistakenly said volume before) after evaporation is: 45 kg silica + 45 kg other substances = 90 kg. Silica makes up 45 / 90 = 0.5 of that new total mass.
 
  • #5


I would like to provide a clear and thorough explanation for both problems.

For the first problem, we need to determine the percentage of silica in the clay on a dry (water-free) basis. This means we need to calculate the percentage of silica in the clay without taking into account the water content. To do this, we can use the following equation:

% silica on dry basis = (% silica in clay - % water in clay) / (100% - % water in clay)

Substituting the values given in the problem, we get:

% silica on dry basis = (45% - 10%) / (100% - 10%) = 35% / 90% = 0.3888 = 38.88%

This means that on a dry basis, the clay contains 38.88% silica.

Now, for the second problem, we need to determine the percentage of carbon on the "wet basis." This means we need to calculate the percentage of carbon in the coal while taking into account the water content. To do this, we can use the following equation:

% carbon on wet basis = (% carbon in residue / % residue in coal) * 100%

Substituting the values given in the problem, we get:

% carbon on wet basis = (71.0% / 97.6%) * 100% = 72.95%

Therefore, on a wet basis, the coal contains 72.95% carbon. This is because the water content is included in the calculation, so the percentage of carbon is lower compared to the moisture-free residue.

I hope this explanation helps you understand the problems better. Always make sure to carefully read and understand the problem before attempting to solve it, and use the appropriate equations and units to get the correct answer.
 

FAQ: Mixture problem 2 - linear equation

What is a mixture problem 2 - linear equation?

A mixture problem 2 - linear equation is a type of word problem that involves finding the ratio or proportion of different elements in a mixture. It can be solved by setting up and solving a linear equation.

How do I approach solving a mixture problem 2 - linear equation?

To solve a mixture problem 2 - linear equation, you should first identify the known and unknown quantities in the problem. Then, create a system of equations based on the given information and use algebraic methods to solve for the unknown variables.

Can you provide an example of a mixture problem 2 - linear equation?

Sure, here's an example: A chemist has a 40% acid solution and a 60% acid solution. How many liters of each solution should be mixed together to create 10 liters of a 50% acid solution?

Let x be the amount of 40% solution and y be the amount of 60% solution. We can set up the following system of equations:

x + y = 10 (total volume of solution)

0.4x + 0.6y = 0.5(10) (ratio of acid in the final solution)

Solving this system gives x = 6 liters and y = 4 liters.

What are some real-life applications of mixture problem 2 - linear equations?

Mixture problems 2 - linear equations are commonly used in fields such as chemistry and economics. For example, a chemist may need to mix different concentrations of a solution to create a desired product, or an economist may need to find the optimal ratio of ingredients for a product to maximize profit.

Are there any shortcuts or tricks for solving mixture problem 2 - linear equations?

While there are no specific shortcuts or tricks for solving mixture problem 2 - linear equations, it is important to carefully read and understand the problem, identify the relevant information, and set up the equations correctly. Practice and familiarity with algebraic methods can also make solving these types of problems more efficient.

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