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2 years ago

Based on a specific example of titration method. How do titration methods play a role in agriculture industry? Discuss

Chemistry

1 answer:

jolli1 [7]2 years ago

8 0

Answer:

Titration is an analytical technique that is widely used in the food industry. It allows food manufacturers to determine the quantity of a reactant in a sample. For example, it can be used to discover the amount of salt or sugar in a product or the concentration of vitamin C or E, which has an effect on product colour.

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A low-pressure weather system comes into the city of Denver. The atmospheric pressure is 693 mmHg.693 mmHg. If 78.0%78.0% of dry

scoundrel [369]

Answer:

540.54 mmHg

Explanation:

We know that the partial pressure of a substance is defined as; Mole fraction * total pressure.

If the total amount of gases in the atmosphere is 100%, the mole fraction of nitrogen gas is now

78/100 = 0.78

Thus, partial pressure of nitrogen gas = 0.78 * 693 = 540.54 mmHg

6 0

2 years ago

Which begins as chemical energy and can be transformed into electricity?

Umnica [9.8K]

Sunlight i think thats the answer

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3 years ago

What does the amount of kinetic energy an object has depend on?

Nastasia [14]

The answer is Velocity and potential energy. Kinetic energy is the total energy of a system or an object in motion and requires movement.

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3 years ago

Which of these types of changes is a physical change?

Cloud [144]

Answer:

answer is c

Explanation: cause there breaking it

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2 years ago

A student placed 10.5 g of glucose (C6H12O6) in a volumetric fla. heggsk, added enough water to dissolve the glucose by swirling

aniked [119]

<u>Answer:</u> The mass of glucose in final solution is 0.420 grams

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$ .........(1)

Initial mass of glucose = 10.5 g

Molar mass of glucose = 180.16 g/mol

Volume of solution = 100 mL

Putting values in equation 1, we get:

$\text{Initial molarity of glucose}=\frac{10.5\times 1000}{180.16\times 100}\\\\\text{Initial molarity of glucose}=0.583M$

To calculate the molarity of the diluted solution, we use the equation:

$M_1V_1=M_2V_2$

where,

$M_1\text{ and }V_1$ are the molarity and volume of the concentrated glucose solution

$M_2\text{ and }V_2$ are the molarity and volume of diluted glucose solution

We are given:

$M_1=0.583M\\V_1=20.0mL\\M_2=?M\\V_2=0.5L=500mL$

Putting values in above equation, we get:

$0.583\times 20=M_2\times 500\\\\M_2=\frac{0.583\times 20}{500}=0.0233M$

Now, calculating the mass of final glucose solution by using equation 1:

Final molarity of glucose solution = 0.0233 M

Molar mass of glucose = 180.16 g/mol

Volume of solution = 100 mL

Putting values in equation 1, we get:

$0.0233=\frac{\text{Mass of glucose in final solution}\times 1000}{180.16\times 100}\\\\\text{Mass of glucose in final solution}=\frac{0.0233\times 180.16\times 100}{1000}=0.420g$

Hence, the mass of glucose in final solution is 0.420 grams

3 0

3 years ago