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Gnoma [55]
3 years ago
9

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

Chemistry
1 answer:
Brilliant_brown [7]3 years ago
7 0

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

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Old thermometers contained very small amounts of mercury. The mercury in the photo has a melting point of -38.8 degrees Celsius.
Nikolay [14]

Answer:

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Explanation:

This one makes the most sense out of the answers.

7 0
3 years ago
1) When 2.38g of magnesium is added to 25.0cm of 2.27 M hydrochloric acid, hydrogen gas is released.
Andrej [43]

Answer:

a. HCl.

b. 0.057 g.

c. 1.69 g.

d. 77 %.

Explanation:

Hello!

In this case, since the reaction between magnesium and hydrochloric acid is:

Mg+2HCl\rightarrow MgCl_2+H_2

Whereas there is 1:2 mole ratio between them.

a) Here, we can identify the limiting reactant as that yielded the fewest moles of hydrogen gas product via the 1:1 and 2:1 mole ratios:

n_{H_2}^{by\  HCl}=0.025L*2.27\frac{molHCl}{1L}*\frac{1molH_2}{2molHCl}  =0.0284molH_2\\\\n_{H_2}^{by\  Mg}=2.38gMg*\frac{1molMg}{24.3gMg}*\frac{1molH_2}{1molMg}=0.0979molH_2

Thus, since hydrochloric yields fewer moles of hydrogen than magnesium, we realize it is the limiting reactant.

b) Here, we use the molar mass of gaseous hydrogen (2.02 g/mol) to compute the mass:

m_{H_2}=0.0284molH_2*\frac{2.02gH_2}{1molH_2}=0.057gH_2

c) Here, we compute the mass of magnesium associated with the yielded 0.0248 moles of hydrogen:

m_{Mg}^{reacted}=0.0284molH_2*\frac{1molMg}{1molH_2}*\frac{24.3gMg}{1molMg}  =0.690gMg

Thus, the mass of excess magnesium turns out:

m_{Mg}^{excess}=2.38g-0.690g=1.69gMg

d) Finally, we compute the percent yield, considering 0.044 g is the actual yield and 0.057 g the theoretical yield:

Y=\frac{0.044g}{0.057g} *100\%\\\\Y=77\%

Best regards!

8 0
3 years ago
A chemist prepares a solution of potassium dichromate by measuring out of potassium dichromate into a volumetric flask and filli
riadik2000 [5.3K]

Answer:

0.297 mol/L

Explanation:

<em>A chemist prepares a solution of potassium dichromate by measuring out 13.1 g of potassium dichromate into a 150 mL volumetric flask and filling the flask to the mark with water. Calculate the concentration in mol/L of the chemist's potassium dichromate solution. Be sure your answer has the correct number of significant digits.</em>

<em />

Step 1: Calculate the moles corresponding to 13.1 g of potassium dichromate

The molar mass of potassium dichromate is 294.19 g/mol.

13.1 g × (1 mol/294.19 g) = 0.0445 mol

Step 2: Convert the volume of solution to L

We will use the relationship 1 L = 1000 mL.

150 mL × (1 L/1000 mL) = 0.150 L

Step 3: Calculate the concentration of the solution in mol/L

C = 0.0445 mol/0.150 L = 0.297 mol/L

4 0
2 years ago
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