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Ainat [17]
3 years ago
6

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

t it is the only acid contributing to the pH? For sulfuric acid, Ka1 is very large and Ka2 is 0.012.
Chemistry
2 answers:
ArbitrLikvidat [17]3 years ago
7 0

<u>Ans: Mass of H2SO4 present = 231.5*10⁴ kg</u>

Height of rainfall = 1.00 inch = 0.0254 m

Area of rainfall = 1800 miles²

1 mile² = 2.59*10⁶ m²

Therefore area = 1800*2.59*10⁶ = 4.662*10⁹ m²

Volume (V) of rainfall = area (A) * height(h) = 0.0254 * 4.662*10⁹ = 1.184*10⁸ m³

1 m³ = 1000 L

V = 1.184*10¹¹ L

It is given that the pH = 3.70

pH = -log[H+]

[H+] = 10^-pH = 10⁻³.⁷⁰ = 1.995*10⁻⁴M

Now:

Molarity of H2SO4 = moles of H2SO4/volume

moles = 1.995*10⁻⁴moles.L-1 * 1.184*10¹¹ L = 2.362*10⁷ moles

Molar mass of H2SO4 = 98 g/mol

Mass of H2SO4 = 2.362*10⁷ moles * 98 g/mol = 231.48*10⁷ g = 231.5 *10⁴ kg

NARA [144]3 years ago
3 0
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
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Answer:

vHe / vNe = 2.24

Explanation:

To obtain the velocity of an ideal gas you must use the formula:

v = √3RT / √M

Where R is gas constant (8.314 kgm²/s²molK); T is temperature and M is molar mass of the gas (4x10⁻³kg/mol for helium and 20,18x10⁻³ kg/mol for neon). Thus:

vHe = √3×8.314 kgm²/s²molK×T / √4x10⁻³kg/mol

vNe = √3×8.314 kgm²/s²molK×T / √20.18x10⁻³kg/mol

The ratio is:

vHe / vNe = √3×8.314 kgm²/s²molK×T / √4x10⁻³kg/mol / √3×8.314 kgm²/s²molK×T / √20.18x10⁻³kg/mol

vHe / vNe = √20.18x10⁻³kg/mol / √4x10⁻³kg/mol

<em>vHe / vNe = 2.24</em>

<em />

I hope it helps!

8 0
3 years ago
What are some common tools scientists use to measure length and mass?
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To measure length scientists may use rulers, meter sticks, etc. and to measure mass they may use a balance.
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Looters break a statue into pieces. How do you expect the weathering of pieces of rock to change?
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5 0
3 years ago
How do you convert from grabs to moles of a substance
Lorico [155]

The correct question is as follows:

How do you convert from grams to moles of a substance

1. Divide by the molar mass

2. Subtract the molar mass

3. Add the molar mass

4. Multiple by the molar mass

Answer: In order to convert from grams to moles of a substance divide by the molar mass.

Explanation:

The number of moles of a substance is the mass of substance in grams divided by its molar mass.

The formula to calculate moles is as follows.

Moles = \frac{mass}{molar mass}

This means that grams are converted to moles when grams is divided by molar mass.

Thus, we can conclude that in order to convert from grams to moles of a substance divide by the molar mass.

8 0
3 years ago
The decomposition of acetaldehyde is a second order reaction with a rate constant of 4.71 10−8 L/mol/s. What is the instantaneou
pochemuha

Answer: Rate of decomposition of acetaldehyde in a solution is 1.45\times 10^{-14}mol/Ls

Explanation:

Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

For a reaction : A\rightarrow products

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Thus rate of decomposition of acetaldehyde in a solution is 1.45\times 10^{-14}mol/Ls

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