Ask question

Ask question

All categories

2 years ago

6

When filtered through a funnel into a flask, a mixture of substances X, Y and Z gets separated as below: - X stays in the funnel

and Y and Z filter through and collect in the flask. X and Y are solids and Z is water. Which of the following can you conclude about the properties of X and Y?

1 answer:

alisha [4.7K]2 years ago

6 0

Z is the solvent, Y is soluble in water while X is insoluble in water.

<h3>Filtration</h3>

Filtration is a method of separation of substances based on particle size. Only a particular particle size can pass through the filter. The substance that remains in the filter is the residue while the substances that passes through the filter is called the filtrate.

From the observation in the question Z is the solvent, Y is soluble in water while X is insoluble in water.

Learn more about separation of mixtures: brainly.com/question/863988

You might be interested in

Which of the following oils would consume the greatest number of equivalents of hydrogen when subject to catalytic hydrogenation

Murljashka [212]

Answer: Option (B) is the correct answer.

Explanation:

When a fatty acid contains high number of double bonds then its unsaturation will also be high and hence, it will consume greater number of equivalents of hydrogen.

In corn oil, there are no unsaturated sites are present.

In olive oil, there is one unsaturated site with majority of oleic acid. In olive oil, there are more than 70% of total unsaturated oils.

In lard oil, there are around 60% of unsaturated oils.

In herring oil, there are highest number of saturated fatty acids and lowest polyunsaturated acids.

Thus, we can conclude that out of the given options, olive oils would consume the greatest number of equivalents of hydrogen when subject to catalytic hydrogenation.

5 0

3 years ago

If 75 grams of oxygen react, how many grams of aluminum are required?

german

Answer:

84.24 g

Explanation:

Given data:

Mass of oxygen = 75 g

Mass of Al required to react = ?

Solution:

Chemical equation:

4Al + 3O₂ → 2Al₂O₃

Number of moles of oxygen:

Number of moles = mass/ molar mass

Number of moles = 75 g/ 32 g/mol

Number of moles = 2.34 mol

Now we will compare the moles of oxygen with Al.

O₂ : Al

3 : 4

2.34 : 4/3×2.34 = 3.12 mol

Mass of Al required:

Mass = number of moles × molar mass

Mass = 3.12 mol × 27 g/mol

Mass = 84.24 g

5 0

3 years ago

Student follow up

sleet_krkn [62]

Answer:

Magnesium and calcium belong to the second group i. e. alkaline earth metals. They are known as earth metals because they are extracted from the earth. They are very reactive elements. Their reactivity increases when we go from top to bottom because the outermost electrons goes farther from the nucleus i. e. atomic radius increases so less energy is needed for its removal.

8 0

3 years ago

1. The solubility of lead(II) chloride at some high temperature is 3.1 x 10-2 M. Find the Ksp of PbCl2 at this temperature.

solniwko [45]

Answer:

1) The solubility product of the lead(II) chloride is $1.2\times 10^{-4}$ .

2) The solubility of the aluminium hydroxide is $1.6\times 10^{-10} M$ .

3)The given statement is false.

Explanation:

1)

Solubility of lead chloride = $S=3.1\times 10^-2M$

$PbCl_2(aq)\rightleftharpoons Pb^{2+}(aq)+2Cl^-(aq)$

S 2S

The solubility product of the lead(II) chloride = $K_{sp}$

$K_{sp}=[Pb^{2+}][Cl^-]^2$

$K_{sp}=S\times (2S)^2=4S^3=4\times (3.1\times 10^{-2})^3=1.2\times 10^{-4}$

The solubility product of the lead(II) chloride is $1.2\times 10^{-4}$ .

2)

Concentration of aluminium nitrate = 0.000010 M

Concentration of aluminum ion = $1\timed 0.000010 M=0.000010 M$

Solubility of aluminium hydroxide in aluminum nitrate solution = $S$

$Al(OH)_3(aq)\rightleftharpoons Al^{3+}(aq)+3OH^-(aq)$

S 3S

The solubility product of the aluminium nitrate = $K_{sp}=1.0\times 10^{-33}$

$K_{sp}=[Al^{3+}][OH^-]^3$

$1.0\times 10^{-33}=(0.000010+S)\times (3S)^3$

$S=1.6\times 10^{-10} M$

The solubility of the aluminium hydroxide is $1.6\times 10^{-10} M$ .

3.

$Molarity=\frac{Moles}{Volume (L)}$

Mass of NaCl= 3.5 mg = 0.0035 g

1 mg = 0.001 g

Moles of NaCl = $\frac{0.0035 g}{58.5 g/mol}=6.0\times 10^{-5} mol$

Volume of the solution = 0.250 L

$[NaCl]=\frac{6.0\times 10^{-5} mol}{0.250 L}=0.00024 M$

1 mole of NaCl gives 1 mole of sodium ion and 1 mole of chloride ions.

$[Cl^-]=[NaCl]=0.00024 M$

Moles of lead (II) nitrate = n

Volume of the solution = 0.250 L

Molarity lead(II) nitrate = 0.12 M

$n=0.12 M]\times 0.250 L=0.030 mol$

1 mole of lead nitrate gives 1 mole of lead (II) ion and 2 moles of nitrate ions.

$[Pb^{2+}]=[Pb(NO_2)_3]=0.030 M$

$PbCl_2(aq)\rightleftharpoons Pb^{2+}(aq)+2Cl^-(aq)$

Solubility of lead(II) chloride = $K_{sp}=1.2\times 10^{-4}$

Ionic product of the lead chloride in solution :

$Q_i=[Pb^{2+}][Cl^-]^2=0.030 M\times (0.00024 M)^2=1.7\times 10^{-9}$

$Q_i$ ( no precipitation)

The given statement is false.

3 0

3 years ago

Help please need to find the frequency of the wave

Lapatulllka [165]

E=hf h=6.63*10^-34 f=8.66*10^14 E=6.63*10^-34*8.66*10^14=57.4*10^-20 joules

3 0

2 years ago