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Anna [14]
3 years ago
15

As a chemist for an agricultural products company, you have just developed a new herbicide,"Herbigon," that you think has the po

tential to kill weeds effectively. A sparingly soluble salt, Herbigon is dissolved in 1 M acetic acid for technical reasons having to do with its production. You have determined that the solubility product Ksp of Herbigon is 9.00×10−6.
Although the formula of this new chemical is a trade secret, it can be revealed that the formula for Herbigon is X-acetate (XCH3COO, where "X" represents the top-secret cation of the salt). It is this cation that kills weeds. Since it is critical to have Herbigon dissolved (it won't kill weeds as a suspension), you are working on adjusting the pH so Herbigon will be soluble at the concentration needed to kill weeds. What pH must the solution have to yield a solution in which the concentration of X+ is 1.50×10−3 M ? The pKa of acetic acid is 4.76.
Chemistry
1 answer:
coldgirl [10]3 years ago
7 0

i hope i helped for this problem

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Answer:

42,3g of H₂O

Explanation:

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4NH₃(g) + 5O₂(g) ⟶ 4NO(g) + 6H₂O(g)

62,8 g of NH₃ are:

62,8g×(1mol/17,031g) = <em>3,69 moles of NH₃</em>

62,8 g of O₂ are:

62,8g×(1mol/32g) =<em> 1,96 moles of O₂</em>

For a complete reaction of 1,96 moles of O₂ you need:

1,96mol O₂×(4mol NH₃ / 5molO₂) = 1,57 moles NH₃. As you have 3,69 moles, limiting reactant is O₂.

Assuming a complete reaction, 1,96mol O₂ produce:

1,96mol O₂×(6mol H₂O / 5molO₂) = 2,35 moles of H₂O. In grams:

2,35 moles of H₂O×(18,01g/1mol) = <em>42,3g of H₂O</em>

I hope it helps!

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3 years ago
Why do astronauts have to wear pressurized suits in space? Need to know for homework!
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A pan containing 20 grams of water was allowed to cool from temperature of 95 degrees C. If the amount of heat released is 1200
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81 °C

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3 years ago
The ph of a 0.55 m aqueous solution of hypobromous acid, hbro, at 25.0°c is 4.48. what is the value of ka for hbro?
elena-14-01-66 [18.8K]

The solution would be like this for this specific problem:

Given:

 

pH of a 0.55 M hypobromous acid (HBrO) at 25.0 °C =  4.48

 

[H+] = 10^-4.48 = 3.31 x 10^-5 M = [BrO-] <span>

Ka = (3.31 x 10^-5)^2 / 0.55 = 2 x 10^-9</span>

 

To add, Hypobromous Acid does not require acid adjustment, which is necessary for chlorine-based product and is stable and effective in pH ranges of 5-9.<span>

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8 0
3 years ago
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