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blondinia [14]
3 years ago
15

Which type of drug commonly marketed to consumers does not require a prescription?

Chemistry
1 answer:
Nimfa-mama [501]3 years ago
7 0
C. pain reliever
ex. tylenol, aleve, etc.
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The changing of a substance from a liquid into a vapor or gas is called
sergejj [24]

Answer: Boiling and Evaporation: Evaporation is the change of a substance from a liquid to a gas. Boiling is the change of a liquid to a vapor, or gas, throughout the liquid.

Explanation: Boiling and Evaporation: Evaporation is the change of a substance from a liquid to a gas. Boiling is the change of a liquid to a vapor, or gas, throughout the liquid.

6 0
1 year ago
Help me I hate science so much​
mestny [16]

Answer:

it would be sciemce

Explanation:

yeah because its  hargf

8 0
3 years ago
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In this reaction what is the correct coefficient for sodium chloride? pb(no3)2 ? nacl → pbcl2 ? nano3
Dovator [93]
<span>pb(no3)2 + 2 nacl → pbcl2 + 2nano3</span>
4 0
3 years ago
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A positively charged substance is brought near the ion. What will most likely happen and why
d1i1m1o1n [39]

<u>Answer:</u> It repels positive ion and attracts negative ion.

<u>Explanation:</u>

There are 2 types of ions:

1. Cations: These ions are positively charged ions which are formed when a substance looses electrons.

2. Anions: These ions are negatively charged ions which are formed when a substance gains electrons.

It is known that, like charges repel each other and unlike charges attract each other.

As, it is given that the substance is positively charged, so it will attract anion and repel cation.

4 0
4 years ago
On the basis of the information above, a buffer with a pH = 9 can best be made by using
telo118 [61]

Answer:

D H2PO4– + HPO42–

Explanation:

The acid dissociation constant for \mathbf{H_3PO_4 , H_2PO^{-}_4 ,  HPO_4^{2-}} are \mathbf{7\times 10^{-3}, \ \ 8\times 10^{-8} ,\ \  5\times 10^{-13}} respectively.

\mathbf{pka (H_3PO_4) = -log (7\times 10^{-3} )=2.2}

\mathbf{pka (H_2PO_4^-) = -log (8\times 10^{-8} )=7.1}

\mathbf{pka (HPO_4^{2-}) = -log (5\times 10^{-13} )=12.3}

The reason while option D is the best answer is that, the value of pKa for both

\mathbf{H_2PO^{-}_4 ,\  \& \  HPO_4^{2-}} lies on either side of the desired pH of the buffer. This implies that one is slightly over and the other is slightly under.

Using Henderson-Hasselbach equation:

\mathbf{pH = pKa + log \Big( \dfrac{HPO_4^{2-}}{H_2PO_4^-} \Big)}

3 0
3 years ago
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