¿que usos se le da a la calorimetría en la industria o cotidianidad?

Honors Stoichiometry Activity Worksheet

vaieri [72.5K]

Explanation: Here, we will be considering 1 cup is equal to 1 mol.

2 water + sugar + lemon juice → 4 lemonade

Moles of water present in 946.36 g of water $=\frac{946.36 g}{236.59 g/mol}=4 mol$

Moles of sugar present in 196.86 g of water $=\frac{196.86 g}{225 g/mol}=0.8749 mol$

Moles of lemon juice present in 193.37 g of water $=\frac{193.37 g}{257.83 g/mol}=0.7499 mol$

Moles of lemonade in 2050.25 g of water $=\frac{2050.25 g}{719.42 g/mol}=2.8498 mol$

As we can see that number of moles of lemon juice are limited.

So, we will consider the reaction will complete in accordance with moles of lemon juice.

1 mole lemon juice reacts with 2 mol of water,then 0.7499 mol of lemon juice will react with:

$\frac{2}{1}\times 0.7499 mol = 1.4998 mol$ of water

Mass of water used = 1.4998 mol × 236.59 g/mol=354.8376 g

Water remained unused = 946.36 g - 354.8376 g =591.5223 g

1 mole lemon juice reacts with mol of sugar,then 0.7499 mol of lemon juice will react with:

$\frac{1}{1}\times 0.7499 mol = 0.7499 mol$ of water

Mass of sugar used = 0.7499 mol × 225 g/mol = 168.7275 g

Sugar remained unused = 196.86 g - 28.1325 g

1 mole of lemon juice gives 4 moles of lemonade.

Then 0.7499 mol of lemon juice will give:

$\frac{4}{1}\times 0.7499 mol=2.996 mol$ of lemonade

Mass of lemonade obtained = 2.996 mol × 719.42 g/mol = 2157.9722 g

Theoretical yield of lemonade = 2157.9722 g

Experimental yield of lemonade = 2050.25 g

Percentage yield of lemonade:

$\frac{\text{Experimental yield}}{\text{theoretical yield}}\times 100$

$\frac{2050.25 g}{2157.9722 g}\times 100=95.00\%$

The percentage yield of lemonade is 95% and ingredient which remained unused were water and sugar.

6 0

4 years ago

If Iron (Fe) gained a proton what would it be?

Mkey [24]

If iron (Fe) gained a proton it would become cobalt (CO).

8 0

3 years ago

What is added to white sugar to create brown sugar?

lesya692 [45]

Molasses is the answer

4 0

3 years ago

Assume that a milliliter of water contains 20 drops. How long, in hours, will it take you to count the number of drops

garri49 [273]

Answer:

126.18 hr

Explanation:

Data given:

1 mL of water = 20 drops

count rate = 10 drops/s

time in hours for one gallon = ?

Solution:

First we calculate number of mL (milliliter) of water in gallon

As we know

1 gallon = 3785.4 mL

As,

1 galon consist of 3785.4 mL of water, so now we count number of drops that contain 3785.4 mL of water

As we Know 1 mL water contain 20 drops then 3785.4 mL of water contain how many drops:

Apply unity formula

1 mL water ≅ 20 drops

3785.4 mL water ≅ X drops

Do cross multiplication

X drops of water = 20 drops x 3785.4 mL / 1 mL

X drops of water = 75708 drops

So, we come to know that one gallon contain 75708 drops of water and we have to calculate the time in hour to count these drops

First we calculate time in seconds

As we Know 10 drops water count in one second then how many seconds it will take to count 75708 drops

Apply unity formula

1 second ≅ 10 drops

X second ≅ 75708 drops

Do cross multiplication

X second = 1 second x 75708 drops / 10 drops

X second = 7570.8 second

So it take 7570.8 second to count 1 gallon water drops

Now convert seconds to hours

As,

60 seconds = 1 hr

7570.8 second = 7570.8 / 60 = 126.18 hr

So it take 126.18 hr to count 1 gallon water drops.

6 0

4 years ago

How many grams are there in 4 moles of cu(cn)2?

Nina [5.8K]

To answer this question, we need to convert the given number of moles of the compound copper (II) cyanide to grams by multiplying the compound's molar mass. The molar mass of copper (II) cyanide is 115.55 g/mol. Multiply 4 by this number, the answer is 462.2 grams.

8 0

3 years ago

Read 2 more answers