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3 years ago

How would you prepare a saturated solution of potassium nitrate?

1 answer:

5 0

Using titration method,

Pipette 25 cm3 of KOH into a conical flask. Add a few drops of methyl orange(indicator).
Then pour a beaker of Dilute nitric acid (HNO3) into a burette. Record the readings of the burette before starting.

1)While releasing some nitric acid into the conical flask, swirl the conical flask.
2)Stop swirlimg or adding nitric acid when color of the solution in the conical flask turns light pink. That will be end point. Record the reading of the burette.
3)Repeat the experiment without adding methyl orange(indicator)
4)Add the same vol of nitric acid used previously.
5)Solution in conical flask transfer over to a dish.
6)Heat the solution till it becomes saturated.

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A 2.500g sample of compound containing only Carbon and Hydrogen is found containing 2.002g of Carbon. Determine the empirical fo

Dima020 [189]

The empirical formula : CH₃

<h3>Further explanation</h3>

Given

2.5 g sample

2.002 g Carbon

Required

The empirical formula

Solution

Mass of Hydrogen :

= 2.5 - 2.002

= 0.498

Mol ratio C : H :

C : 2.002/12 = 0.167

H : 0.498/1 = 0.498

Divide by 0.167 :

C : H = 1 : 3

7 0

2 years ago

How many moles of carbon, hydrogen, and oxygen are present in a 100-g sample of ascorbic acid?

Y_Kistochka [10]

There are:

3.41 moles of C

4.54 moles of H

3.40 moles of O.

Why?

To solve the problem, the first thing that we need to do is to write the chemical formula of the ascorbic acid.

$C_{6}H_{8}O_{6}$

Now, we know that there are 100 grams of the compound, so, the masses of each element will represent the percent in the compound.

We have that:

$C_{6}=12.0107g*6=72.08g\\\\H_{8}=1.008g*8=8.064g\\\\O_{6}=15.999g*6=95.994g\\\\C_{6}H_{8}O_{6}=72.08g+8.064g+95.994g=176.138g$

To know the percent of each element, we need to to the following:

$C=\frac{72.08g}{176.138g}*100=0.409*100=40.92(percent)\\\\H=\frac{8.064g}{176.138g}*100=4.58(percent)\\\\O=\frac{95.994}{176.138g}*100=54.49(percent)$

So, we know that for the 100 grams of the compound, there are:

40.92 grams of C

4.58 grams of H

54.49 grams of O

We know the molecular masses of each element:

$C=12.0107\frac{g}{mol}\\\\H=1.008\frac{g}{mol}\\\\O=15.999\frac{g}{mol}{mol}$

Now, to calculate the number of moles of each element, we need to divide the mass of each element by the molecular mass of each element:

$C=\frac{40.92g}{12.010\frac{g}{mol}}=3.41mol\\\\H=\frac{4.58g}{1.008\frac{g}{mol}}=4.54mol\\\\O=\frac{54.49g}{15.999\frac{g}{mol}}=3.40mol$

Hence, we have that there are 3.41 moles of C, 4.54 moles of H, and 3.40 moles of O.

Have a nice day!

5 0

3 years ago

A sample of hydrated copper (II) sulfate (CuSO4•nH2O) is heated to 150°C and produces 103.74 g anhydrous copper (II) sulfate and

wariber [46]

Answer:

Explanation:

Firstly, we convert what we have to percentage compositions.

There are two parts in the molecule, the sulphate part and the water part.

The percentage compositions is as follows:

Sulphate- (103.74)/(103.74 + 58.55) × 100% = apprx 64%

The water part = 100 - 64 = 36%

Now, we divide the percentages by the molar masses.

For the CuSO4 molar mass is 64 + 32 + 4(16) = 160g/mol

For the H2O = 2(1) + 16 = 18g/mol

Now we divide the percentages by these masses

Sulphate = 64/160 = 0.4

Water = 36/18 = 2

The ratio is thus 0.4:2 = 1:5

Hence, there are 5 water molecules.

3 0

3 years ago

How many significant figures are in 1.00 L

likoan [24]

Answer:

3 significant numbers

6 0

3 years ago

Calculate the number of moles of aluminum, sulfur, and oxygen atoms in 4.00 moles of aluminum sulfate, al2(so4)3. express the nu

bagirrra123 [75]

<span>To calculate the number of moles of aluminum, sulfur, and oxygen atoms in 4.00 moles of aluminum sulfate, al2(so4)3. We will simply inspect the "number" of aluminum, sulfur, and oxygen atoms available per one mole of the compound. Here we have Al2(SO4)3, which means that for every mole of aluminum sulfate, there are 2 moles of aluminum, 3 (1 times 3) moles of sulfur, and 12 (4x3) moles of oxygen. Since we have four moles of Al2(SO4)3 given, we simply multiply 4 times the moles present per 1 mole of the compound. So we have 4x2 = 8 moles of Al, 4x3 = 12 moles of sulfur, and 4x12 = 48 moles of oxygen.

So the answer is:
8,12,48

</span>

3 0

3 years ago

Read 2 more answers