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photoshop1234 [79]
2 years ago
12

copper hydroxide and potassium sulfate are produced when potassium hydroxide reacts with copper sulfate balanced equation

Chemistry
1 answer:
a_sh-v [17]2 years ago
8 0

This problem is requiring the balanced chemical equation that takes place when copper hydroxide and potassium sulfate are produced when reacting potassium hydroxide with copper sulfate.

<h3>Balancing chemical equations:</h3>

In chemistry, balancing chemical equations is based on the law of conservation of mass, which demands us to have equal number of atoms on both sides of the chemical equation. This can be accomplished by inserting coefficients in front of the chemical species.

For this particular case, we have potassium hydroxide with copper sulfate on the reactants side, however, copper can be copper (I) or copper (II) as it has 1+ and 2+ as its possible oxidation numbers. In addition, copper hydroxide and potassium sulfate as the products. Hence, we can assume this is all about copper (II) so we can write:

KOH+CuSO_4\rightarrow K_2SO_4+Cu(OH)_2

As we can see, potassium, hydrogen and oxygen have two atoms each on the products side, but just one on the reactants side; drawback we can overcome by putting a 2 in front of KOH so as to balance it:

2KOH+CuSO_4\rightarrow K_2SO_4+Cu(OH)_2

Learn more about balancing chemical equations: brainly.com/question/8062886

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Answer:

cartilage or soft tissue? you didn't give options :/

Explanation:

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A substance that produces hydroxide ions when placed in water is a(n) _____.
bekas [8.4K]

Answer: A substance that produces hydroxide ions when placed in water is base.

Explanation:

Bases are the substance:

  • Which gives negatively charged hydroxide(OH^-) ions in aqueous solution.
  • Which have pH value ranging from 7 to 14.
  • BOH(aq)\rightarrow B^+(aq)+OH^-(aq)

Where as acid gives positively charged hydronium ion(H^+) in aqueous solution.

7 0
3 years ago
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(3) A 10.00-mL sample of 0.1000 M KH2PO4 was titrated with 0.1000 M HCl Ka for phosphoric acid (H3PO4): Ka1= 7.50x10-3; Ka2=6.20
shtirl [24]

Answer:

The pH of this solution is 1,350

Explanation:

The phosphoric acid (H₃PO₄) has three acid dissociation constants:

HPO₄²⁻ ⇄ PO4³⁻ + H⁺        Kₐ₃ = 4,20x10⁻¹⁰  (1)

H₂PO₄⁻ ⇄ HPO4²⁻ + H⁺    Kₐ₂ = 6,20x10⁻⁸   (2)

H₃PO₄ ⇄ H₂PO4⁻ + H⁺       Kₐ₁ = 7,50x10⁻³   (3)

The problem says that you have 10,00 mL of KH₂PO₄ (It means H₂PO₄⁻) 0,1000 M and you add 10,00 mL of HCl (Source of H⁺) 0,1000 M. So you can see that we have the reactives of the equation (3).

We need to know what is the concentration of H⁺ for calculate the pH.

The moles of H₂PO₄⁻ are:

10,00 mL × ( 1x10⁻⁴ mol / mL) = 1x10⁻³ mol

The moles of H⁺ are, in the same way:

10,00 mL × ( 1x10⁻⁴ mol / mL) = 1x10⁻³ mol

So:

H₃PO₄   ⇄      H₂PO4⁻         +        H⁺           Kₐ₁ = 7,50x10⁻³   (3)

X mol     ⇄  (1x10⁻³-X) mol  + (1x10⁻³-X) mol                            (4)

The chemical equilibrium equation is:

Kₐ₁ = ([H₂PO4⁻] × [H⁺] / [H₃PO₄]

So:

7,50x10⁻³ = (1x10⁻³-X)² / X

Solving the equation you will obtain:

X² - 9,5x10⁻³ X + 1x10⁻⁶ = 0

Solving the quadratic formula you obtain two roots:

X = 9,393x10⁻³ ⇒ This one has no chemical logic because solving (4) you will obtain negative H₂PO4⁻ and H⁺ moles

X = 1,065x10⁻⁴

So the moles of H⁺ are : 1x10⁻³- 1,065x10⁻⁴ : 8,935x10⁻⁴ mol

The reaction volume are 20,00 mL (10,00 from both KH₂PO₄ and HCL)

Thus, the molarity of H⁺ ([H⁺]) is: 8,935x10⁻⁴ mol / 0,02000 L = 4,468x10⁻² M

pH is -log [H⁺]. So the obtained pH is 1,350

I hope it helps!

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Most modern medications are given in doses of milligrams. Thyroid medications, however, are typically given in doses of microgra
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Answer:

0.125 mg

Explanation:

<em>The correct answer would be 0.125 mg</em>

<u>According to the conversion factor, one milligram of a sample is equivalent to one thousand micrograms of the same sample.</u>

milligram = 10^{-3}

microgram = 10^{-6}

Hence,

1 milligram = 1000 micrograms or 1 microgram = 10^{-3} milligram

Therefore, 125 micrograms will be:

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