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3 years ago

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Separation of mixtures and purification of chemical substances

1 answer:

Mnenie [13.5K]3 years ago

4 0

Answer:

I don't understand what to do with this

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Three substances A, B, and C melt at 00C,500C and -1500C

Drupady [299]

Answer:

That information is better presented and analyzed in a table.

This table shows you all the information and the answers:

Substance         melting point   boiling point    room temperature    conclusion
                                    °C                  °C                      °C                (state)

A                                  0                    100                    25                    liquid

B                                  50                  200                    25                    solid
C                               -150                   10                 25                    gas

Explanation:

1) Substance A at 25° is above the melting point and below the boiling point, then it is liquid (just like water)

2) Substance B at 25°C is below the melting point, so it is solid.

3) Substance C at 25°C is above the boiling point, so it is gas.

7 0

3 years ago

E is solution of HCI of unk concentration f contain 4.8g of NaOH in 25cm point of salution

Readme [11.4K]

Answer:

18

Explanation:

no explanation sorry

5 0

3 years ago

Thx please help thx thx

Alisiya [41]

This force is opposing in nature .it resists object velocity .

5 0

3 years ago

An unknown gaseous substance has a density of 1.06 g/L at 31 °C and 371 torr. If the substance has the following percent composi

Anarel [89]

Answer:

C) C4H6 - Right answer

Explanation:

Let's combine the Ideal Gases Law with density to get the molecular formula for the unknown gas.

Density = mass / volume

1.06 g /L means that 1.06 grams of compound occupy 1 liter of volume.

P . V = n . R . T

Pressure in Torr must be converted to atm

760 Torr are 1 atm

371 Torr are __ (371 .1)/760 = 0.488 atm

0.488 atm . 1L = 1.06g/MM . 0.082 . 304K

(0.488 atm . 1L) / 0.082 . 304K = 1.06g/ MM

Mass / Molar mass = Moles → That's why the 1.06 g / MM

0.0195 mol = 1.06g / MM

1.06g/0.0195 mol = MM → 54.3 g/m

Now, let's use the composition

100 g of compound have 88.8 g of C

54.3 g of compound have ___ (54.3 . 88.8) /100 = 48 g of C

100 g of compound have 11.2 g of H

54.3 g of compound have __ (54.3 . 11.2)/100 = 6 g of H

48 g of C are included un 4 atoms

6 g of H are included in 6 atoms

4 0

3 years ago

What is the concentration (in M) of a 225ml potassium sulfate solution that contains 4.15g of potassium?

mars1129 [50]

The concentration of solution in M or mol/L can be calculated using the following formula:

$C=\frac{n}{V}$ .... (1)

Here, n is number of moles and V is volume of solution in L.

The molecular formula of potassium sulfate is $K_{2}SO_{4}$ thus, there are 2 moles of potassium in 1 mol of potassium sulfate.

1 mol of potassium will be there in 0.5 mol of potassium sulfate.

Mass of potassium is 4.15 g, molar mass is 39.1 g/mol.

Number of moles can be calculated as follows:

$n=\frac{m}{M}$

Here, m is mass and M is molar mass

Putting the values,

$n=\frac{(4.15 g}{(39.1 g/mol}=0.1061 mol$

Thus, number of moles of $K_{2}SO_{4}$ will be $0.1061\times 0.5=0.053 mol$ .

The volume of solution is 225 mL, converting this into L,

$1 mL=10^{-3}L$

Thus,

$225 mL=0.225 L$

Putting the values in equation (1),

$C=\frac{(0.053 mol}{0.225 L}=0.236 M$

Therefore, concentration of potassium sulfate solution is 0.236 M.

4 0

3 years ago