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Leni [432]
2 years ago
14

Which among the materials have a varying amount of components are they compounds or mixtures​

Chemistry
2 answers:
ruslelena [56]2 years ago
8 0

Answer:

Mixture Have variable composition and can be separated into component parts by physical methods. Mixtures contain more than one kind of molecule, and their properties depend on the relative amount of each component present in the mixture.

Elodia [21]2 years ago
7 0

Answer:

Matter can be broken down into two Zalo categories: pure substances and mixtures. Pure substances are further broken down into elements and compounds. Mixtures are physically combined structures that can be separated into their original components.

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The molal boiling-point and freezing-point constants are different for different solvents.
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Answer:

I think the answer is true, sorry if I am wrong

Explanation:

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3 years ago
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Cci(4) will be soluble in
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LIKE DISSOLVES LIKE. Since Ccl4 is non-polar, it'll be soluble in any non-polar solvent. Hope this helps you!
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3 years ago
graduated cylinder is filled with water to a volume of 6.2 ML. an irregular shaped plastic object weighing 1.2 g is placed in th
Aleks04 [339]

Answer:

A. Density of object = 0.86 g/mL

B. The object will float in the water.

Explanation:

The following data were obtained from the question:

Volume of water = 6.2 mL

Mass (m) of object = 1.2 g

Volume of water + Object = 7.59 mL

Density of object =?

Density of water = 1 g/mL

Next, we shall determine the volume of the object. This can be obtained as follow:

Volume of water = 6.2 mL

Volume of water + Object = 7.59 mL

Volume of object =?

Volume of object = (Volume of water + Object) – (Volume of water)

Volume of object = 7.59 – 6.2

Volume of object = 1.39 mL

Therefore, the volume of the object is 1.39 mL

A. Determination of the density of the object.

Mass (m) of object = 1.2 g

Volume (V) of object = 1.39 mL

Density (D) of object =?

Density = mass /volume

Density = 1.2/1.39

Density of object = 0.86 g/mL

B. Determination of whether the object will float or sink.

Density of object = 0.86 g/mL

Density of water = 1 g/mL

From the above, we can see that the density of water is greater than that of the object. This implies that the object is lighter than water. Therefore, the object will float in the water.

8 0
3 years ago
The heats of combustion of ethane (C2H6) and butane (C4H10) are 52 kJ/g and 49 kJ/g, respectively. We need to produce 1.000 x 10
LekaFEV [45]

Answer :

(1) The number of grams needed of each fuel (C_2H_6)\text{ and }(C_4H_{10}) are 19.23 g and 20.41 g respectively.

(2) The number of moles of each fuel (C_2H_6)\text{ and }(C_4H_{10}) are 0.641 moles and 0.352 moles respectively.

(3) The balanced chemical equation for the combustion of the fuels.

C_2H_6+\frac{7}{2}O_2\rightarrow 2CO_2+3H_2O

C_4H_{10}+\frac{13}{2}O_2\rightarrow 4CO_2+5H_2O

(4) The number of moles of CO_2 produced by burning each fuel is 1.28 mole and 1.41 mole respectively.

The fuel that emitting least amount of CO_2 is C_2H_6

Explanation :

<u>Part 1 :</u>

First we have to calculate the number of grams needed of each fuel (C_2H_6)\text{ and }(C_4H_{10}).

As, 52 kJ energy required amount of C_2H_6 = 1 g

So, 1000 kJ energy required amount of C_2H_6 = \frac{1000}{52}=19.23g

and,

As, 49 kJ energy required amount of C_4H_{10} = 1 g

So, 1000 kJ energy required amount of C_4H_{10} = \frac{1000}{49}=20.41g

<u>Part 2 :</u>

Now we have to calculate the number of moles of each fuel (C_2H_6)\text{ and }(C_4H_{10}).

Molar mass of C_2H_6 = 30 g/mole

Molar mass of C_4H_{10} = 58 g/mole

\text{ Moles of }C_2H_6=\frac{\text{ Mass of }C_2H_6}{\text{ Molar mass of }C_2H_6}=\frac{19.23g}{30g/mole}=0.641moles

and,

\text{ Moles of }C_4H_{10}=\frac{\text{ Mass of }C_4H_{10}}{\text{ Molar mass of }C_4H_{10}}=\frac{20.41g}{58g/mole}=0.352moles

<u>Part 3 :</u>

Now we have to write down the balanced chemical equation for the combustion of the fuels.

The balanced chemical reaction for combustion of C_2H_6 is:

C_2H_6+\frac{7}{2}O_2\rightarrow 2CO_2+3H_2O

and,

The balanced chemical reaction for combustion of C_4H_{10} is:

C_4H_{10}+\frac{13}{2}O_2\rightarrow 4CO_2+5H_2O

<u>Part 4 :</u>

Now we have to calculate the number of moles of CO_2 produced by burning each fuel to produce 1000 kJ.

C_2H_6+\frac{7}{2}O_2\rightarrow 2CO_2+3H_2O

From this we conclude that,

As, 1 mole of C_2H_6 react to produce 2 moles of CO_2

As, 0.641 mole of C_2H_6 react to produce 0.641\times 2=1.28 moles of CO_2

and,

C_4H_{10}+\frac{13}{2}O_2\rightarrow 4CO_2+5H_2O

From this we conclude that,

As, 1 mole of C_4H_{10} react to produce 4 moles of CO_2

As, 0.352 mole of C_4H_{10} react to produce 0.352\times 4=1.41 moles of CO_2

So, the fuel that emitting least amount of CO_2 is C_2H_6

5 0
3 years ago
What are the products produced when an acid and base react together
SashulF [63]
Both products will start to cancel the acidity and how strong the base is if they are mixed. If the acid is stronger than the base then it will be an acidic product and visa versa if the base is stronger than the acid.
4 0
3 years ago
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