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krok68 [10]
3 years ago
5

What is the chemical formula for decomposition?

Chemistry
1 answer:
shutvik [7]3 years ago
6 0

Answer: AB A + B

Explanation:

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If the dosage for a medication is 225mg/lb, how many grams should a 25-lb child be given?
Oksanka [162]

Answer:

5.625 grams

Explanation:

Start your equation with what you have been given.  Place the units you need in your answer on the right side of the equal sign.

225mg

-----------   X  -----------   X ------------- =   ?   g

   lb

Now start to fill in your equation and use a conversions to get rid of the units you don't want.  Convert mg into grams first.  The child's weight (25 lb) is placed over 1 just to get the equation lined up properly so you can see how the units cancel out.

225 mg                 1 g                  25 lb            5.625 g

---------------   X   ---------------  X   -------------  =   ---------------

   lb                    1000 mg                 1                  1

The lb on the top and bottom cancel each other out and you are left with just grams.  Even though it is over one, that is the same at just 5.625 grams.

3 0
3 years ago
___________ are bonds in which valence electrons are shared.
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Valence electrons are shared in covalent bonds
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Which statements regarding the Henderson-Hasselbalch equation are true? If the pH of the solution is known as is the pKa for the
Sonbull [250]

Answer:

1, 2, and 3 are true.

Explanation:

The Henderson-Hasselbalch equation is:

pH = pka + log₁₀ \frac{[A^-]}{[HA]}

  • If the pH of the solution is known as is the pKa for the acid, the ratio of conjugate base to acid can be determined. <em>TRUE</em>

pH = pka + log₁₀ \frac{[A^-]}{[HA]}

If you know pH and pka:

10^(pH-pka) = \frac{[A^-]}{[HA]}

The ratio will be: 10^(pH-pka)

  • At pH = pKa for an acid, [conjugate base] = [acid] in solution. <em>TRUE</em>

pH = pka + log₁₀ \frac{[A^-]}{[HA]}

0 = log₁₀ \frac{[A^-]}{[HA]}

10^0 = \frac{[A^-]}{[HA]}

1 = \frac{[A^-]}{[HA]}

As ratio is 1,  [conjugate base] = [acid] in solution.

  • At pH >> pKa for an acid, the acid will be mostly ionized. <em>TRUE</em>

pH = pka + log₁₀ \frac{[A^-]}{[HA]}

If pH >> pKa,  10^(pH-pka) will be >> 1, that means that you have more [A⁻] than [HA]

  • At pH << pKa for an acid, the acid will be mostly ionized. <em>FALSE</em>

pH = pka + log₁₀ \frac{[A^-]}{[HA]}

If pH << pKa,  10^(pH-pka) will be << 1, that means that you have more [HA] than [A⁻]

I hope it helps!

6 0
3 years ago
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