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Lady bird [3.3K]
2 years ago
13

How many times higher is the concentration of H+ in the Hubbard Brook sample than in unpolluted rainwater?

Chemistry
1 answer:
Anna [14]2 years ago
8 0

Answer:

1. 7 (a neutral solution)

Answer: 10-7= 0.0000001 moles per liter

2. 5.6 (unpolluted rainwater)

Answer: 10-5.6 = 0.0000025 moles per liter

3. 3.7 (first acid rain sample in North America)

Answer: 10-3.7 = 0.00020 moles per liter

The concentration of H+ in the Hubbard Brook sample is 0.00020/0.0000025, which is 80 times higher than the H+ concentration in unpolluted rainwater.

Explanation:

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A radioactive material, with half-life of six months, has 100 thousand unstable nuclei.
goldfiish [28.3K]

Answer:

See Explanation

Explanation:

Given that;

N/No = (1/2)^t/t1/2

Where;

No = amount of radioactive isotope originally present

N = A mount of radioactive isotope present at time t

t = time taken

t1/2 = half life

N/1000=(1/2)^3/6

N/1000=(1/2)^0.5

N = (1/2)^0.5 * 1000

N= 707 unstable nuclei

Since the value of the initial activity of the radioactive material was not given, the activity of the radioactive material after three months is given by;

Decay constant = 0.693/t1/2 = 0.693/6 months = 0.1155 month^-1

Hence;

A=Aoe^-kt

Where;

A = Activity after a time t

Ao = initial activity

k = decay constant

t = time taken

A = Aoe^-3 *0.1155

A=Aoe^-0.3465

3 0
3 years ago
A monoprotic weak acid when dissolved in water is 0.66% dissociated and produces a solution with a pH of 3.04. Calculate the Ka
raketka [301]

Answer:

Ka = 6.02x10⁻⁶

Explanation:

The equilibrium that takes place is:

  • HA ⇄ H⁺ + A⁻
  • Ka = [H⁺][A⁻]/[HA]

We <u>calculate [H⁺] from the pH</u>:

  • pH = -log[H⁺]
  • [H⁺] = 10^{-pH}
  • [H⁺] = 9.12x10⁻⁴ M

Keep in mind that [H⁺]=[A⁻].

As for [HA], we know the acid is 0.66% dissociated, in other words:

  • [HA] * 0.66/100 = [H⁺]

We <u>calculate [HA]</u>:

  • [HA] = 0.138 M

Finally we <u>calculate the Ka</u>:

  • Ka = \frac{[9.12x10^{-4}]*[9.12x10^{-4}]}{[0.138]} = 6.02x10⁻⁶
3 0
2 years ago
Calculate the number of moles of magnesium, chlorine, and oxygen atoms in 7.80 moles of magnesium perchlorate, Mg(ClO4)2.
Dmitry [639]
 The number of moles of moles of  Magnesium,chlorine and oxygen atoms in 7.80 moles of Mg(ClO4)2 is calculated as below

find the total number of each atom in Mg(ClO4)2

that is mg = 1 atom
        Cl =  1x2 = 2 atoms
        O  = 4  x2 = 8 atoms

then multiply 7.80 moles with total number of each atom , to get the number moles of each atom
that is

Mg = 7.80 x1= 7.80  moles
cl =   7.80  x2=15.6  moles
O =  7.80 x8= 62.4 moles

5 0
3 years ago
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