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3 years ago

If a solid piece of naphthalene is heated and remains at 80°C until it is completely melted, you know that 80ºC is the

Chemistry

2 answers:

White raven [17]3 years ago

8 0

<h2>Answer : Freezing and Melting point </h2><h3>Explanation : </h3>

The solid piece of naphthalene when heated is observed to melt after it attains the temperature of 80° C and then this 80° C is called as it melting point. After 80° C naphthalene will start to melt and get transformed into liquid state. Also if this naphthalene is allowed to cool down after 80° C it will be the freezing point of naphthalene.

Rom4ik [11]3 years ago

6 0

Answer: Melting point of the naphthalene is 80 $^oC$ .

Explanation:

Melting point of a substance is temperature at which given solid changes its state from solid to liquid. It is also referred as freezing point of the substance.

Freezing point of substance is a temperature at which given substance changes its state from liquid to solid.

Here, the solid piece of naphthalene is heated and remains at 80°C until it is completely melted.

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Directions: Each set of lettered choices below refers to the numbered statements immediately following it. Select the one letter

maksim [4K]

Answer : The only reaction (C) that shows that the same reactant undergoes both oxidation and reduction.

Explanation :

Disproportionation reaction : It is defined as the reaction in which the same reactant undergoes both oxidation and reduction reaction. It is a redox reaction.

Redox reaction or Oxidation-reduction reaction : It is defined as the reaction in which the oxidation and reduction reaction takes place simultaneously.

Oxidation reaction : It is defined as the reaction in which a substance looses its electrons. In this, oxidation state of an element increases. Or we can say that in oxidation, the loss of electrons takes place.

Reduction reaction : It is defined as the reaction in which a substance gains electrons. In this, oxidation state of an element decreases. Or we can say that in reduction, the gain of electrons takes place.

(A) The given balanced reaction is,

$H_2SeO_4(aq)+2Cl^-(aq)+2H^+(aq)\rightarrow H_2SeO_3(aq)+Cl_2(g)+H_2O(l)$

This reaction is a redox reaction but not disproportionation reaction because in this reaction there are two reactants in which chlorine shows oxidation and selenium shows reduction.

(B) The given balanced reaction is,

$S_8(s)+8O_2(g)\rightarrow 8SO_2(g)$

This reaction is a redox reaction but not disproportionation reaction because in this reaction there are two reactants in which sulfur shows oxidation and oxygen shows reduction.

(C) The given balanced reaction is,

$3Br_2(aq)+6OH^-(aq)\rightarrow 5Br^-(aq)+BrO_3^-(aq)+3H_2O(l)$

This reaction is a disproportionation reaction because in this reaction only one reactant bromine that shows both oxidation and reduction reaction.

(D) The given balanced reaction is,

$Ca^{2+}(aq)+SO_4^{2-}(aq)\rightarrow CaSO_4(s)$

This reaction is a combination reaction in which the two reactant react to give a single product. There is no changes in the oxidation state of calcium and sulfate.

(E) The given balanced reaction is,

$PtCl_4(s)+2Cl^-(aq)\rightarrow PtCl_6^{2-}(aq)$

This reaction is a combination reaction in which the two reactant react to give a single product. There is no changes in the oxidation state of platinum and chlorine.

Hence, the only reaction (C) that shows that the same reactant undergoes both oxidation and reduction.

6 0

3 years ago

How many milliliters of a 17% benzalkonium chloride stock solution would be needed to prepare a liter of a 1:200 solution of ben

Nuetrik [128]

Here is the complete question.

Benzalkonium Chloride Solution ------------> 250ml

Make solution such that when 10ml is diluted to a total volume of 1 liter a 1:200 is produced.

Sig: Dilute 10ml to a liter and apply to affected area twice daily

How many milliliters of a 17% benzalkonium chloride stock solution would be needed to prepare a liter of a 1:200 solution of benzalkonium chloride?

(A) 1700 mL

(B) 29.4 mL

(D) 294 mL

Answer:

(B) 29.4 mL

Explanation:

1 L = 1000 mL

1:200 solution implies the $\frac{weight}{volume}$ in 200 mL solution.

200 mL of solution = 1g of Benzalkonium chloride

1000 mL will be $\frac{1000mL}{200mL}=\frac{1g}{xg}$

200mL × 1g = 1000 mL × x(g)

x(g) = $\frac{200mL*1g}{1000mL}$

x(g) = 0.2 g

That is to say, 0.2 g of benzalkonium chloride in 1000mL of diluted solution of 1;200 is also the amount in 10mL of the stock solution to be prepared.

∴ $\frac{10mL}{250mL}=\frac{0.2g}{y(g)}$

y(g) = $\frac{250mL*0.2g}{10mL}$

y(g) = 5g of benzalkonium chloride.

Now, at 17% $\frac{weight}{volume}$ concentrate contains 17g/100ml:

∴ the number of milliliters of a 17% benzalkonium chloride stock solution that is needed to prepare a liter of a 1:200 solution of benzalkonium chloride will be;

= $\frac{17g}{5g} = \frac{100mL}{z(mL)}$

z(mL) = $\frac{100mL*5g}{17g}$

z(mL) = 29.41176 mL

≅ 29.4 mL

Therefore, there are 29.4 mL of a 17% benzalkonium chloride stock solution that is required to prepare a liter of a 1:200 solution of benzalkonium chloride

4 0

4 years ago

Who is generally credited with developing the process of using heat?

galina1969 [7]

Answer:

Robert Boyle

Explanation:

I think it's that

6 0

3 years ago

A student attempts to separate 4.656 g of a sand/salt mixture just like you did in this lab. After carrying out the experiment,

nikitadnepr [17]

Answer:

Explanation:

a ) Total mixture = 4.656 g

Sand recovered = 2.775 g

percent composition of sand in the mixture

= (2.775 g / 4.656 g ) x 100

= 59.6 % .

b )

Total of sand and salt recovered = 2.775 g + .852 g = 3.627 g .

Total mixture = 4.656 g

percent recovery = (3.627 / 4.656 ) x 100

= 77.9 % .

4 0

3 years ago

Delocalization which occurs through overlap between stigma and p orbitals

elixir [45]

Answer:

Single covalent bonds that form between nuclei are created from the "head-to-head" overlap of orbitals and are called stigma (s) bonds. ... Another type of bond, a pi (p) bond is formed when two p orbitals overlap. Pi bonds are found in double and triple bond structures

Explanation:

Delocalization happens when electric charge is spread over more than one atom. For example, bonding electrons may be distributed among several atoms that are bonded together.

7 0

3 years ago