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3 years ago

Classify the processes as endothermic or exothermic.

Chemistry

1 answer:

joja [24]3 years ago

7 0

endothermic absorbs heat

exothermic gives heat

a. endothermic

b. exothermic

c. endothermic

d. exothermic

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What is the major defining property of gasses?.

jonny [76]

Answer:

Gases have no definite shape or volume. They are fluid, allowing particles/molecules to move freely.

The behavior of a gas is that the volume changes directly with temperature. With a constant volume, the pressure will be directly proportional to the amount of gas.

Explanation:

These are some of the properties I can think of

5 0

2 years ago

The reaction of NO2 with ozone produces NO3 in a second-order reaction overall.

Brilliant_brown [7]

Answer : The rate of reaction is,

$Rate=4.77\times 10^{-19}M/s$

The appearance of $NO_3$ is, $4.77\times 10^{-19}M/s$

Explanation :

The general rate of reaction is,

$aA+bB\rightarrow cC+dD$

Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.

The expression for rate of reaction will be :

$\text{Rate of disappearance of A}=-\frac{1}{a}\frac{d[A]}{dt}$

$\text{Rate of disappearance of B}=-\frac{1}{b}\frac{d[B]}{dt}$

$\text{Rate of formation of C}=+\frac{1}{c}\frac{d[C]}{dt}$

$\text{Rate of formation of D}=+\frac{1}{d}\frac{d[D]}{dt}$

$Rate=-\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=+\frac{1}{c}\frac{d[C]}{dt}=+\frac{1}{d}\frac{d[D]}{dt}$

From this we conclude that,

In the rate of reaction, A and B are the reactants and C and D are the products.

a, b, c and d are the stoichiometric coefficient of A, B, C and D respectively.

The negative sign along with the reactant terms is used simply to show that the concentration of the reactant is decreasing and positive sign along with the product terms is used simply to show that the concentration of the product is increasing.

The given rate of reaction is,

$NO_2(g)+O_3(g)\rightarrow NO_3(g)+O_2(g)$

The rate law expression will be:

$Rate=k[NO_2][O_3]$

Given:

Rate constant = $k=1.69\times 10^{-4}M^{-1}s^{-1}$

$[NO_2]$ = $1.77\times 10^{-8}M$

$[O_3]$ = $1.59\times 10^{-7}M$

$Rate=k[NO_2][O_3]$

$Rate=(1.69\times 10^{-4})\times (1.77\times 10^{-8})\times (1.59\times 10^{-7})$

$Rate=4.77\times 10^{-19}M/s$

The expression for rate of appearance of $NO_3$ :

$\text{Rate of reaction}=\text{Rate of appearance of }NO_3=+\frac{d[NO_3]}{dt}$

As, $\text{Rate of reaction}=4.77\times 10^{-19}M/s$

So, $\text{Rate of appearance of }NO_3=+\frac{d[NO_3]}{dt}=4.77\times 10^{-19}M/s$

Thus, the appearance of $NO_3$ is, $4.77\times 10^{-19}M/s$

7 0

3 years ago

Consider the reaction. 2 Pb ( s ) + O 2 ( g ) ⟶ 2 PbO ( s ) An excess of oxygen reacts with 451.4 g of lead, forming 367.5 g of

Art [367]

Answer : The percent yield of the reaction is, 75.6 %

Solution : Given,

Mass of Pb = 451.4 g

Molar mass of Pb = 207 g/mole

Molar mass of PbO = 223 g/mole

First we have to calculate the moles of Pb.

$\text{ Moles of }Pb=\frac{\text{ Mass of }Pb}{\text{ Molar mass of }Pb}=\frac{451.4g}{207g/mole}=2.18moles$

Now we have to calculate the moles of $PbO$

The balanced chemical reaction is,

$2Pb(s)+O_2(g)\rightarrow 2PbO(s)$

From the reaction, we conclude that

As, 2 mole of $Pb$ react to give 2 mole of $PbO$

So, 2.18 mole of $Pb$ react to give 2.18 mole of $PbO$

Now we have to calculate the mass of $PbO$

$\text{ Mass of }PbO=\text{ Moles of }PbO\times \text{ Molar mass of }PbO$

$\text{ Mass of }PbO=(2.18moles)\times (223g/mole)=486.1g$

Theoretical yield of $PbO$ = 486.1 g

Experimental yield of $PbO$ = 367.5 g

Now we have to calculate the percent yield of the reaction.

$\% \text{ yield of the reaction}=\frac{\text{ Experimental yield of }PbO}{\text{ Theoretical yield of }PbO}\times 100$

$\% \text{ yield of the reaction}=\frac{367.5g}{486.1g}\times 100=75.6\%$

Therefore, the percent yield of the reaction is, 75.6 %

3 0

4 years ago

3. The formula for table salt is NaCl. Is table salt ionic or covalent? Explain

d1i1m1o1n [39]

Answer:

it's an ionic compound

7 0

3 years ago

Part b an "empty" container is not really empty if it contains air. how may moles of nitrogen are in an "empty" two-liter cola b

Sedbober [7]

<span>6.38x10^-2 moles
First, let's determine how many moles of gas particles are in the two-liter container. The molar volume for 1 mole at 25C and 1 atmosphere is 24.465 liters/mole. So
2 L / 24.465 L/mol = 0.081749438 mol
Now air doesn't just consist of nitrogen. It also has oxygen, carbon dioxide, argon, water vapor, etc. and the total number of moles includes all of those other gasses. So let's multiply by the percentage of nitrogen in the atmosphere which is 78%
0.081749438 mol * 0.78 = 0.063764562 mol.
Rounding to 3 significant figures gives 6.38x10^-2 moles</span>

4 0

3 years ago

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