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Alex787 [66]
3 years ago
12

What is the buffer component ratio, (bro-)/(hbro) of a bromate buffer that has a ph of 8.08. ka of hbro is 2.3 x 10-9?

Chemistry
1 answer:
Alla [95]3 years ago
3 0

pKa= -log(ka)

= -log(2.3*10^-9)

= 8.64

Now pH can be calculated using equation:

pH=pka+log(BrO-)/(HBrO)

8.08 =8.64+log(BrO-)/(HBrO)

log(BrO-)/(HBrO)=8.08-8.64

= -0.56

(BrO-)/(HBrO)= 10^-0.56

=0.275

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Please help on this one?
Bezzdna [24]

Answer:

\text{C. } _{36}^{85}\text{Kr}

Explanation:

Your nuclear equation is  

_{35}^{85}\text{Br} \longrightarrow \, _{-1}^{0}\text{e} +\, _{x}^{y}\text{X}

The main point to remember in balancing nuclear equations is that

  • the sum of the superscripts and must be the same on each side of the equation.
  • the sum of the subscripts must be the same on each side of the equation.  

Then  

85 = 0 + y, so y = 85 - 0 = 0  

35 = -1 + x, so x = 35 + 1 = 36

The nucleus with atomic number 36 and atomic mass 85 is krypton-85.  

The nuclear equation becomes  

_{35}^{85}\text{Br} \longrightarrow \, _{-1}^{0}\text{e} + \, _{36}^{85}\text{Kr}

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3 years ago
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When humans burn fossil fuels, most of the carbon quickly enters the_______
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Answer:

Atmosphere.

Explanation:

Carbon moves from fossil fuels to the atmosphere when fuels are burned. When humans burn fossil fuels to power factories, power plants, cars and trucks, most of the carbon quickly enters the atmosphere as carbon dioxide gas.

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What is the study of acid-base chemistry called in the environment
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Answer:

An acid is a substance that donates protons (in the Brønsted-Lowry definition) or accepts a pair of valence electrons to form a bond (in the Lewis definition). A base is a substance that can accept protons or donate a pair of valence electrons to form a bond. Bases can be thought of as the chemical opposite of acids.

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Which of the following is true of the Sun? *
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The Sun is the closest star to Earth.

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What mass of precipitate forms when 185.5 ml of 0.533 m naoh is added to 627 ml of a solution that contains 15.8 g of aluminum s
Law Incorporation [45]

Answer:

2,57 g of precipitate.

Explanation:

For the reaction:

6 NaOH + Al₂(SO₄)₃ → 2 Al(OH)₃ + 3 Na₂SO₄

The precipitate is Al(OH)₃.

185,5mL of 0,533M NaOH are:

0,1855L × 0,533M = <em>0,0989 moles NaOH</em>

Moles of Al₂(SO₄)₃ are:

15,8g × \frac{1mol}{342,15g} = <em>0,0462 moles Al₂(SO₄)₃</em>

For the total reaction of 0,0989 moles NaOH with Al₂(SO₄)₃ you need:

0,0989moles NaOH × \frac{1molAl_{2}(SO_{4})_{3}}{6 moles NaOH} = <em>0,0165 moles Al₂(SO₄)₃</em>

As you have <em>0,0462 moles Al₂(SO₄)₃ </em>the limiting reactant is NaOH.

0,0989 moles of NaOH produce:

0,0989moles NaOH × \frac{2molAl(OH)_{3}}{6 moles NaOH} = <em>0,0330 moles of Al(OH)₃</em>

These moles are:

0,0330 moles of Al(OH)₃ × (78 g/mol) = <em>2,57 g of Al(OH)₃ ≡ mass of precipitate</em>

<em></em>

I hope it helps!

<em> </em>

3 0
3 years ago
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