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xz_007 [3.2K]
3 years ago
10

Increased industrialization has caused a rise in harmful acid rain precipitation that affects plant and marine life. A sample of

acid rain has a proton concentration 10,000 times greater than pure water and more than 100,000 times greater than seawater. What is the approximate pH of this sample?
Chemistry
1 answer:
wolverine [178]3 years ago
3 0

Answer:

pH of the sample of acid rain is 3.

Explanation:

Pure water has a theoretical pH of 7.00. As pH = -log [H⁺], [H⁺] = 1x10⁻⁷M

Now, the sample of acid rain has a proton concentration 10,000 times greater than pure water. That is:

[H⁺] = 10,000 * 1x10⁻⁷M = 1x10⁻³M

The pH of this sample is:

pH = -log 1x10⁻³M

<h3>pH = 3</h3>
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How much energy is needed to raise the temperature of a 200 g piece of lead by 250 o C?
devlian [24]

<u>Answer:</u> The amount of energy required to raise the given amount of temperature is 6400 J

<u>Explanation:</u>

To calculate the heat absorbed by the lead, we use the equation:

q=mc\Delta T

where,

q = heat absorbed

m = mass of lead = 200 g

c = heat capacity of lead = 0.128 J/g°C

\Delta T = change in temperature = 250°C

Putting values in above equation, we get:

q=200g\times 0.128J/g^oC\times 250^oC=6400J

Hence, the amount of energy required to raise the given amount of temperature is 6400 J

4 0
3 years ago
A sample of oxygen gas in one container has a volume of 20.0mililiter at 297 K and 101.3 kPa. The entire sample is transferred t
VashaNatasha [74]

Answer: V_2=\frac{101.3kPa\times 20.0ml\times 283K}{297K\times 94.6kPa}

Explanation:

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,:

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

V_2=\frac{P_1V_1T_2}{T_1P_2}

where,

P_1 = initial pressure of gas = 101.3 kPa

P_2 = final pressure of gas = 94.6 kPa

V_1 = initial volume of gas = 20.0 ml

V_2 = final volume of gas = ?

T_1 = initial temperature of gas = 297K

T_2 = final temperature of gas = 283K

Now put all the given values in the above equation, we get the final volume of gas.

V_2=\frac{101.3kPa\times 20.0ml\times 283K}{297K\times 94.6kPa}

V_2=20.4ml

Thus the correct numerical setup for calculating the new volume is \frac{101.3kPa\times 20.0ml\times 283K}{297K\times 94.6kPa}

3 0
3 years ago
A compound distributes between benzene (solvent 1) and water (solvent 2) with a distribution coefficient, K = 2.7. If 1.0g of th
mote1985 [20]

Explanation:

The given data is as follows.

Solvent 1 = benzene,          Solvent 2 = water

 K_{p} = 2.7,         V_{S_{2}} = 100 mL

V_{S_{1}} = 10 mL,       weight of compound = 1 g

       Extract = 3

Therefore, calculate the fraction remaining as follows.

                  f_{n} = [1 + K_{p}(\frac{V_{S_{2}}}{V_{S_{1}}})]^{-n}

                                  = [1 + 2.7(\frac{100}{10})]^{-3}

                                  = (28)^{-3}

                                  = 4.55 \times 10^{-5}

Hence, weight of compound to be extracted = weight of compound - fraction remaining

                                  = 1 - 4.55 \times 10^{-5}

                                  = 0.00001

or,                               = 1 \times 10^{-5}

Thus, we can conclude that weight of compound that could be extracted is 1 \times 10^{-5}.

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3 years ago
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If 1.00 g of a hydrocarbon is combusted and found to produce 3.14 g of co2, what is the empirical formula of the hydrocarbon?
photoshop1234 [79]
The combustion reaction is as expressed,

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The mass fraction of carbon in CO2 is 3/11. Hence,
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Given that we have 1 g of the hydrocarbon, the mass of H is equal to 0.14 g. 

     moles of C = 0.86 g C / 12 g = 0.0713
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The empirical formula for the hydrocarbon is therefore, CH₂.
7 0
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