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3 years ago

This is chemistry: 2 questions having to do with particles

Chemistry

1 answer:

PilotLPTM [1.2K]3 years ago

6 0

Answer 1 :

Image 1 shows : Liquid phase

Image 2 shows : Gas phase

Image 3 shows : Solid phase

Image 4 shows : Solid phase

Solid phase : In this state, the molecules are closely and tightly packed. There is no vacant space present in it.

Liquid phase : In this state, the molecules are loosely packed and there is an intermediate vacant space present in it.

Gas phase : In this state, the molecules are very loosely packed and there is a lot of vacant space present in it.

Answer 2 :

Liquid to gas : Evaporation

Solid to gas : Sublimation

Gas to solid : Deposition

Solid to liquid : Melting

Liquid to solid : Freezing

Gas to liquid : Condensation

Explanation :

Evaporation : It is a type of vaporization process in which a liquid changes into gas phase by providing heat.

Sublimation : It is a process in which a solid changes directly into gas phase without passing through a liquid phase.

Deposition : It is a process in which a gas transforms directly into a solid phase without passing through a liquid phase.

Melting : It is a process in which a solid changes into liquid phase by providing heat.

Freezing : It is a process in which a liquid transform into a solid phase.

Condensation : It is a process in which a water vapor(gas) changes into liquid state.

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At 25∘C, the decomposition of dinitrogen pentoxide, N2O5(g), into NO2(g) and O2(g) follows first-order kinetics with k=3.4×10−5

Annette [7]

Answer:

4600s

Explanation:

$2N_{2}O_{5}(g) - - -> 4NO_{2}(g) +O_{2}$

For a first order reaction the rate of reaction just depends on the concentration of one specie [B] and it’s expressed as:

$-\frac{d[B]}{dt}=k[B] - - - -\frac{d[B]}{[B]}=k*dt$

If we have an ideal gas in an isothermal (T=constant) and isocoric (v=constant) process.

PV=nRT we can say that P = n so we can express the reaction order as a function of the Partial pressure of one component.

$-\frac{d[P(B)]}{P(B)}=k*dt$

$-\frac{d[P(N_{2}O_{5})]}{P(N_{2}O_{5})}=k*dt$

Integrating we get:

$\int\limits^p \,-\frac{d[P(N_{2}O_{5})]}{P(N_{2}O_{5})}=\int\limits^ t k*dt$

$-(ln[P(N_{2}O_{5})]-ln[P(N_{2}O_{5})_{o})])=k(t_{2}-t_{1})$

Clearing for t2:

$\frac{-(ln[P(N_{2}O_{5})]-ln[P(N_{2}O_{5})_{o})])}{k}+t_{1}=t_{2}$

$ln[P(N_{2}O_{5})]=ln(650)=6.4769$

$ln[P(N_{2}O_{5})_{o}]=ln(760)=6.6333$

$t_{2}=\frac{-(6.4769-6.6333)}{3.4*10^{-5}}+0= 4598.414s$

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3 years ago

Which nonmetal has the greatest potential to be oxidized

Tasya [4]

Answer:

Hydrogen

Explanation:

A reducing agent is a substance which gives up its electrons to become oxidized. Generally, metals are oxidized (reducing agents) while non-metals are reduced (oxidizing agents).

However, hydrogen which is a non-metal is usually oxidized in the presence of stronger oxidizing non-metals such as fluorine and oxygen.

Hydrogen thus, acts as a reducing agent by giving up its electrons to become oxidized. Even though among all non-metals, Hydrogen has the greatest potential to be oxidized, it is a poor reducing agent compared to reactive metals.

7 0

3 years ago

Predict which member of each pair will be more acidic. Explain your answers. (a) methanol or tert-butyl alcohol (b) 2-chloroprop

Art [367]

Answer:

A. Methanol

B. 2-chloropropan-1-ol

C. 2,2-dichloroethanol

D. 2,2-difluoropropan-1-ol

Explanation:

Primary alcohols are stronger acids than secondary alcohols which are stronger than tertiary alcohols.

This trend is so because of the stability of the alkoxide ion formed(stabilising the base, increases the acidity). A more stabilised alkoxide ion is a weaker conjugate base (dissociation of an acid in water).

By electronic factors, When there are alkyl groups donating electrons, the density of electrons on th O- will increase a d thereby make it less stable.

By stearic factors, More alkyl group bonded to the -OH would mean the bulkier the alkoxide ion which would be harder to stabilise.

Down the group of the periodic table, basicity (metallic character) decreases as we go from F– to Cl– to Br– to I– because that negative charge is being spread out over a larger volume that is electronegativity decreases down the group.

Electronegative atoms give rise to inductive effect and a decrease in indutive effects leads to a decrease in acidity. Therefore an Increasing distance from the -OH group lsads to a decrease in acidity.

From above,

A. Methanol

B. 2-chloropropan-1-ol

C. 2,2-dichloroethanol

D. 2,2-difluoropropan-1-ol

6 0

3 years ago

CH4 with pressure 1 atm and volume 10 liter at 27°C is passed into a reactor with 20% excess oxygen, how many moles of oxygen is

BaLLatris [955]

Answer : The moles of $O_2$ left in the products are 0.16 moles.

Explanation :

First we have to calculate the moles of $CH_4$ .

Using ideal gas equation:

$PV=nRT$

where,

P = pressure of gas = 1 atm

V = volume of gas = 10 L

T = temperature of gas = $27^oC=273+27=300K$

n = number of moles of gas = ?

R = gas constant = 0.0821 L.atm/mol.K

Now put all the given values in the ideal gas equation, we get:

$(1atm)\times (10L)=n\times (0.0821L.atm/mol.K)\times (300K)$

$n=0.406mole$

Now we have to calculate the moles of $O_2$ .

The balanced chemical reaction will be:

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

From the balanced reaction we conclude that,

As, 1 mole of $CH_4$ react with 2 moles of $O_2$

So, 0.406 mole of $CH_4$ react with $2\times 0.406=0.812$ moles of $O_2$

Now we have to calculate the excess moles of $O_2$ .

$O_2$ is 20 % excess. That means,

Excess moles of $O_2$ = $\frac{(100 + 20)}{100}$ × Required moles of $O_2$

Excess moles of $O_2$ = 1.2 × Required moles of $O_2$

Excess moles of $O_2$ = 1.2 × 0.812 = 0.97 mole

Now we have to calculate the moles of $O_2$ left in the products.

Moles of $O_2$ left in the products = Excess moles of $O_2$ - Required moles of $O_2$

Moles of $O_2$ left in the products = 0.97 - 0.812 = 0.16 mole

Therefore, the moles of $O_2$ left in the products are 0.16 moles.

7 0

3 years ago

How can you tell whether a fat contains primarily saturated or unsaturated fatty acids?

Romashka-Z-Leto [24]

In terms of physically observing it, if the substance is a lions based substance, like oil or margarine, depending on the state, this will inform how well the hydrocarbons can pack together well or not, or will there be bends in the ring preventing the molecules from packing well.

Liquids are unsaturated
Solids are saturated.

3 0

3 years ago