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SCORPION-xisa [38]
3 years ago
10

Ethanol and water appear the saame to the naked eye at 25C. Which properties can help a chemist distinguish between? denisty, ma

lting point, specific heat capacity, mass.
Chemistry
1 answer:
Oxana [17]3 years ago
7 0

Explanation:

As it is known that there are two types of properties. These are extensive and intensive.

Extensive properties : Properties that depend on the size or amount of system. For example, mass, volume etc.

Intensive properties : Properties that do not depend on the size or amount of system. For example, density, melting point, specific heat capacity etc.

On the basis of these properties water and ethanol are distinguished as follows.

  • Density of water is 997 kg/m^{3} whereas density of ethanol is 789 kg/m^{3}. Both these liquids can be separated by intensive properties.
  • Melting point of water is zero degree celsius whereas melting point of ethanol is -114.1 degree celsius.
  • Specific heat capacity of water is 4.184 J/g ^{o}C whereas specific heat capacity of ethanol is 2.46 J/g ^{o}C.
  • Mass of the given liquids cannot be differentiated because they will keep on changing depending on the quantity required. As mass is an extensive property, therefore, it is difficult to differentiate between the two liquids.

Thus, we can conclude that properties like density, melting point, specific heat capacity can help a chemist distinguish between ethanol and water.

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Use Lewis theory to determine the chemical formula for the compound formed between Mg and Br. Group of answer choices Mg3Br2 Mg2
IceJOKER [234]

Answer:

MgBr2

Explanation:

(Mg^2+) + (Br^1-)

For every Mg we need 2 Br to balance out the compound.

5 0
2 years ago
38.25 grams of silicon is combined with 14.33 grams of nitrogen gas. How many grams of silicon nitride can be formed if nitrogen
zhuklara [117]
3Si + 2N2 --> Si3N4 (as given) 

n(Si) = m/MM = 38.25/28.085 = 1.3619 mol
n(N2) = 14.33/2*14.007 = 0.5115 mol

Therefore, N2 is limiting and Si is in excess 
The molar ratio of 2N2:Si3N4 is 2:1 
So, 0.0575 mol of silicon nitride is formed (dividing 0.5115 by 2) 

m of silicon nitride= n*mm = 0.0575*140.283 = 8.06627... g 
= 8.066g (4 significant figures) 

(hopefully it is right, but double check in case i did something wrong) :) 
6 0
3 years ago
In the titration of 50. 0 mL of 0. 400 M HCOOH with 0. 150 M LiOH, how many mL of LiOH are required to reach the equivalence poi
mart [117]

The volume of the 0.15 M LiOH solution required to react with 50 mL of 0.4 M HCOOH to the equivalence point is 133.3 mL

<h3>Balanced equation </h3>

HCOOH + LiOH —> HCOOLi + H₂O

From the balanced equation above,

The mole ratio of the acid, HCOOH (nA) = 1

The mole ratio of the base, LiOH (nB) = 1

<h3>How to determine the volume of LiOH </h3>
  • Molarity of acid, HCOOH (Ma) = 0.4 M
  • Volume of acid, HCOOH (Va) = 50 mL
  • Molarity of base, LiOH (Mb) = 0.15 M
  • Volume of base, LiOH (Vb) =?

MaVa / MbVb = nA / nB

(0.4 × 50) / (0.15 × Vb) = 1

20 / (0.15 × Vb) = 1

Cross multiply

0.15 × Vb = 20

Divide both side by 0.15

Vb = 20 / 0.15

Vb = 133.3 mL

Thus, the volume of the LiOH solution needed is 133.3 mL

Learn more about titration:

brainly.com/question/14356286

8 0
2 years ago
The activation energy of a reaction is 56.9 kj/mol and the frequency factor is 1.5×1011/s. Part a calculate the rate constant of
Vanyuwa [196]

We will use Arrehenius equation

lnK = lnA  -( Ea / RT)

R = gas constant = 8.314 J / mol K

T = temperature = 25 C = 298 K

A = frequency factor

ln A = ln (1.5×10 ^11) = 25.73

Ea = activation energy = 56.9 kj/mol = 56900 J / mol

lnK = 25.73 - (56900 / 8.314 X 298) = 2.76

Taking antilog

K = 15.8



5 0
3 years ago
2 uniones iónicas y 2 covalentes, realizando las estructuras de Lewis correspondiente y averiguar la función del compuesto armad
posledela

Answer:

Cloruro de sodio y fluoruro de sodio.

Dióxido de carbono y monóxido de hidrógeno.

Explicación:

El cloruro de sodio y el fluoruro de sodio son los compuestos que tienen enlaces iónicos. Estos compuestos iónicos se utilizan para diferentes actividades de nuestra vida diaria. El cloruro de sodio se usa para cocinar y el fluoruro de sodio se usa en la pasta de dientes para limpiar nuestros dientes. El dióxido de carbono y el monóxido de hidrógeno son compuestos que tienen enlaces covalentes. El dióxido de carbono se usa en refrescos / refrescos y algunos otros líquidos que se pueden usar en la vida diaria. El monóxido de hidrógeno es el agua pura que bebemos todos los días en nuestra vida diaria y es muy importante para nuestra supervivencia.

7 0
3 years ago
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