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slega [8]
3 years ago
6

A patient is required to take an IV drug for cancer treatment. The required dosage is 5.0 mg drug her lb patient body weight eve

ry day. Each of the bags of the drug contains 250 mg drug. If you are treating a 150 lb patient,How many bags of the drug are needed each day.
Chemistry
1 answer:
Kruka [31]3 years ago
4 0

Answer:

3 bags are required.

Explanation:

Find the number of mg needed for 150 pounds

150 pounds * 5 mg/pound = 750 mg

1 bag = 250 mg

x bags = 750 mg               Cross multiply

250 * x= 750                      Divide by 250

x = 750/250

x = 3

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Write the recation of formation of ions by fluorine and potassium​
bekas [8.4K]

Answer: The reaction of formation of ions by fluorine and potassium​ is K^{+} + F^{-} \rightarrow KF.

Explanation:

Atomic number of potassium is 19 and its electronic distribution is 2, 8, 8, 1. So, in order to attain stability it needs to lose 1 electron due to which it forms K^{+} ion.

Atomic number of fluorine is 9 and its electronic distribution is 2, 7. In order to attain stability it needs to gain one electron and hence forms F^{-} ion.

Therefore, when both these ions come in contact with each other then it leads to the formation of KF compound.

The reaction equation for this is as follows.

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3 years ago
at a particular temperature, asample of pure water has a Kw of 1.7x10-12. what is the hydroxide concentration of this sample
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Answer:

Hydroxide concentration of the sample is 1.3x10⁻⁶M

Explanation:

The equilibrium constant of water, Kw, is:

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Kw is defined as:

Kw = 1.7x10⁻¹² = [H⁺] [OH⁻]

As the sample is of pure water, both H⁺ and OH⁻ ions have the same concentration because come from the same equilibrium, that is:

[H⁺] = [OH⁻]

We can write the Kw expression:

1.7x10⁻¹² = [OH⁻] [OH⁻]

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<h3>Hydroxide concentration of the sample is 1.3x10⁻⁶M</h3>
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