A solution has a higher boiling point than its associated pure solvent does.

How many of the following are oxidation-reduction reactions? I. reaction of a metal with a nonmetal II. synthesis III. combustio

nadezda [96]

Answer : The oxidation-reduction reactions are:

I. reaction of a metal with a nonmetal

II. synthesis

III. combustion

Explanation :

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.

I. Reaction of a metal with a nonmetal :

When sodium react with chlorine gas then it react to give sodium chloride.

$2Na+Cl_2\rightarrow 2NaCl$

In this reaction, the oxidation state of sodium changes from (0) to (+1) and shows oxidation and the oxidation state of chlorine changes from (0) to (-1) and shows reduction. So, It is an oxidation-reduction reaction.

II. Synthesis reaction :

A chemical reaction where multiple substances or reactants combine to form a single product.

When hydrogen react with oxygen then it react to give water.

$2H_2+O_2\rightarrow 2H_2O$

In this reaction, the oxidation state of hydrogen changes from (0) to (+1) and shows oxidation and the oxidation state of oxygen changes from (0) to (-2) and shows reduction. So, It is an oxidation-reduction reaction.

III. Combustion reaction :

A chemical reaction in which a hydrocarbon reaction with the oxygen to give product as carbon dioxide and water.

When methane react with oxygen gas then it react to give carbon dioxide and water.

$CH_4+2O_2\rightarrow CO_2+2H_2O$

In this reaction, the carbon of methane gain oxygen and shows oxidation and the oxygen gas gain hydrogen and shows reduction. So, It is an oxidation-reduction reaction.

IV. Precipitation reaction :

It is defined as the reaction in which an insoluble salt formed when two aqueous solutions are combined.

The insoluble salt that settle down in the solution is known an precipitate.

It is a double displacement reaction. So, it is not an oxidation-reduction reaction.

V. Decomposition reaction :

A chemical reaction in which the the larger molecule decomposes to give two or more smaller molecules.

The oxidation state remains same on reactant and product side. So, it is not an oxidation-reduction reaction.

3 0

2 years ago

ch question carries 2 mark. Time Remaining : 00 : 46 : 33 Some oxides are given below. (i)Na2O (ii)NO2 (iii) CO2 (iv) MgO a) Whi

cluponka [151]

<h3>Further explanation</h3>

The basic oxide is an oxide-forming a base solution.

These oxides are mainly from group 1 alkaline and group 2-alkaline earth

If this oxide is dissolved in water it will form an alkaline solution

LO + H₂O --> L(OH)₂ ---> alkaline earth

L₂O + H₂O --> LOH --> alkaline

So the basic oxides : Na₂O and MgO

Na₂O + H₂O --> NaOH (sodium hydroxide, strong base)

MgO + H₂O --> Mg(OH)₂ (magnesium hydroxide, strong base)

The aqueous solution of CO₂ , obtained by dissolving CO₂ in water

CO₂ + H₂O --> H₂CO₃ (carbonic acid)

In general, basic oxide is obtained from metal oxide, while acid oxide is obtained from non-metal oxide

6 0

3 years ago

Which of the following describes metallic character on the periodic table? Question 1 options: A) It increases as you move right

Ilia_Sergeevich [38]

Answer:

D

Explanation:

Metallic character decreases as you move across a period in the periodic table from left to right. This occurs as atoms more readily accept electrons to fill a valence shell than lose them to remove the unfilled shell. Metallic character increases as you move down an element group in the periodic table. This is because electrons become easier to lose as the atomic radius increases, where there is less attraction between the nucleus and the valence electrons because of the increased distance between them.

3 0

2 years ago

For a process Arightwards harpoon over leftwards harpoonB, at 25 °C there is 10% of A at equilibrium while at 75 °C, there is 80

Lostsunrise [7]

This question is describing the following chemical reaction at equilibrium:

$A\rightleftharpoons B$

And provides the relative amounts of both A and B at 25 °C and 75 °C, this means the equilibrium expressions and equilibrium constants can be written as:

$K_1=\frac{90\%}{10\%}=9\\\\K_2=\frac{20\%}{80\%} =0.25$

Thus, by recalling the Van't Hoff's equation, we can write:

$ln(K_2/K_1)=-\frac{\Delta H}{R}(\frac{1}{T_2} -\frac{1}{T_1} )$

Hence, we solve for the enthalpy change as follows:

$\Delta H=\frac{-R*ln(K_2/K_1)}{(\frac{1}{T_2} -\frac{1}{T_1} ) }$

Finally, we plug in the numbers to obtain:

$\Delta H=\frac{-8.314\frac{J}{mol*K} *ln(0.25/9)}{[\frac{1}{(75+273.15)K} -\frac{1}{(25+273.15)K} ] } \\\\\\\Delta H=4,785.1\frac{J}{mol}$

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5 0

2 years ago

A 215-g sample of copper metal at some temperature is added to 26.6 g of water. The initial water temperature is 22.22 oC, and t

andrezito [222]

The initial temperature of the copper metal was 27.38 degrees.

Explanation:

Data given:

mass of the copper metal sample = 215 gram

mass of water = 26.6 grams

Initial temperature of water = 22.22 Degrees

Final temperature of water = 24.44 degrees

Specific heat capacity of water = 0.385 J/g°C

initial temperature of copper material , Ti=?

specific heat capacity of water = 4.186 joule/gram °C

from the principle of:

heat lost = heat gained

heat gained by water is given by:

q water = mcΔT

Putting the values in the equation:

qwater = 26.6 x 4.186 x (2.22)

qwater = 247.19 J

qcopper = 215 x 0.385 x (Ti-24.4)

= 82.77Ti - 2019.71

Now heat lost by metal = heat gained by water

82.77Ti - 2019.71 = 247.19

Ti = 27.38 degrees

8 0

3 years ago