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Ierofanga [76]
3 years ago
12

A 0.73 m solution of chloroacetic acid, , has a ph of 1.50. calculate for the acid.

Chemistry
1 answer:
erik [133]3 years ago
6 0
PH = 0.1289<span> for </span>1.50<span> M solution of weak acid with Ka value of </span><span>.73</span>
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If the pH of a 1.00-in. rainfall over 1800 miles2 is 3.70, how many kilograms of sulfuric acid, H2SO4, are present, assuming tha
NARA [144]
There are 2.32 x 10^6 kg sulfuric acid in the rainfall. 

Solution: 
We can find the volume of the solution by the product of 1.00 in and 1800 miles2: 
     1800 miles2 * 2.59e+6 sq m / 1 sq mi = 4.662 x 10^9 sq m 
     1.00 in * 1 m / 39.3701 in = 0.0254 m  
     Volume = 4.662 x 10^9 m^2 * 0.0254 m
                  = 1.184 x 10^8 m^3 * 1000 L / 1 m3
                  = 1.184 x 10^11 Liters 

We get the molarity of H2SO4 from the concentration of [H+] given by pH = 3.70: 
     [H+] = 10^-pH = 10^-3.7 = 0.000200 M 
     [H2SO4] = 0.000100 M  
 
By multiplying the molarity of sulfuric acid by the volume of the solution, we can get the number of moles of sulfuric acid: 
     1.184 x 10^11 L * 0.000100 mol/L H2SO4 = 2.36 x 10^7 moles H2SO4 

We can now calculate for the mass of sulfuric acid in the rainfall: 
     mass of H2SO4 = 2.36 x 10^7 moles * 98.079 g/mol
                               = 2.32 x 10^9 g * 1 kg / 1000 g
                               = 2.32 x 10^6 kg H2SO4
3 0
3 years ago
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The lower your salary, the less efficient you are. This represents a ______
creativ13 [48]

Answer:

I believe that it is a negative correlation

7 0
2 years ago
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Is the bond C=O polar or non polar
rewona [7]

Answer:

Non Polar

Explanation:

Non Polar molecule because of its linear symmetric shape

6 0
3 years ago
Consider the following equilibrium: 2SO^2(g) + O2(9) = 2 SO3^(g)
saul85 [17]

Answer:

At equilibrium, the forward and backward reaction rates are equal.

The forward reaction rate would decrease if \rm O_2 is removed from the mixture. The reason is that collisions between \rm SO_2 molecules and \rm O_2\! molecules would become less frequent.

The reaction would not be at equilibrium for a while after \rm O_2 was taken out of the mixture.

Explanation:

<h3>Equilibrium</h3>

Neither the forward reaction nor the backward reaction would stop when this reversible reaction is at an equilibrium. Rather, the rate of these two reactions would become equal.

Whenever the forward reaction adds one mole of \rm SO_3\, (g) to the system, the backward reaction would have broken down the same amount of \rm SO_3\, (g)\!. So is the case for \rm SO_2\, (g) and \rm O_2\, (g).

Therefore, the concentration of each species would stay the same. There would be no macroscopic change to the mixture when it is at an an equilibrium.

<h3>Collision Theory</h3>

In the collision theory, an elementary reaction between two reactants particles takes place whenever two reactant particles collide with the correct orientation and a sufficient amount of energy.

Assume that \rm SO_2\, (g) and \rm O_2\, (g) molecules are the two particles that collide in the forward reaction. Because the collision has to be sufficiently energetic to yield \rm SO_3\, (g), only a fraction of the reactions will be fruitful.

Assume that \rm O_2\, (g) molecules were taken out while keeping the temperature of the mixture stays unchanged. The likelihood that a collision would be fruitful should stay mostly the same.

Because fewer \!\rm O_2\, (g) molecules would be present in the mixture, there would be fewer collisions (fruitful or not) between \rm SO_2\, (g) and \rm O_2\, (g)\! molecules in unit time. Even if the percentage of fruitful collisions stays the same, there would fewer fruitful collisions in unit time. It would thus appear that the forward reaction has become slower.

<h3>Equilibrium after Change</h3>

The backward reaction rate is likely going to stay the same right after \rm O_2\, (g) was taken out of the mixture without changing the temperature or pressure.

The forward and backward reaction rates used to be the same. However, right after the change, the forward reaction would become slower while the backward reaction would proceed at the same rate. Thus, the forward reaction would become slower than the backward reaction in response to the change.

Therefore, this reaction would not be at equilibrium immediately after the change.

As more and more \rm SO_3\, (g) gets converted to \rm SO_2\, (g) and \rm O_2\, (g), the backward reaction would slow down while the forward reaction would pick up speed. The mixture would once again achieve equilibrium when the two reaction rates become equal again.

5 0
2 years ago
How many Zn atoms are there in 4.65 moles of Zn?
Nesterboy [21]
So your answer would pretty much be 2.80 x 10^24. The picture is just the explanation and how you would get that answer.

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