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alexdok [17]
4 years ago
15

A prescription medication requires 5.98 mg per kg of body weight.

Chemistry
1 answer:
valkas [14]4 years ago
5 0

<u>Answer:</u>

<u>For A:</u> The 0.0132 grams is required per pound of the body weight.

<u>For B:</u> The dost required for 191 lb patient is 5.96 grams.

<u>Explanation:</u>

  • <u>For A:</u>

We are given:

Medication requirement = 5.98 mg/kg

To convert the given amount in grams per pounds, we use the conversion factor:

1 kg = 2.20462 lb

1 g = 1000 mg

Converting the value into grams per pounds, we get:

\Rightarrow (\frac{5.98mg}{kg})\times (\frac{1g}{1000mg})\times (\frac{2.20462lb}{1kg})\\\\\Rightarrow 0.0132g/lb

Hence, the 0.0132 grams is required per pound of the body weight.

  • <u>For B:</u>

To calculate the amount of medication for 191 lb patient, we use unitary method:

For 1 lb body weight, the amount required is 0.0132 grams.

So, for 191 lb body weight, the amount required is \frac{0.0312g}{1lb}\times 191lb=5.96g

Hence, the dost required for 191 lb patient is 5.96 grams.

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Neither N₂ or O₂ are greenhouses gases because they do not cause the greenhouse effect.

<h3>What are greenhouses gases?</h3>

Greenhouse gases trap heat in the atmosphere causing the temperature of the planet to rise.

Greenhouse gases cause the greenhouse effect.

<h3>What is the greenhouse effect?</h3>

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The energy needed to remove an electron from an atom is called
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Some of the water you spilled on your shirt evaporates
Montano1993 [528]

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No

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What is the ph for hcl and ch3cooh?
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A 10.0 mL sample of 0.75 M CH3CH2COOH(aq) is titrated with 0.30 M NaOH(aq) (adding NaOH to CH3CH2COOH). Determine which region o
Llana [10]

Answer:

5.75

Explanation:

Weak acid and strong base,

Make it a two part problem

The first will be Partial neutralization of the weak acid, while the second is the equilibrium and final pH

1.Parameters

HC2H3O2 =10mL x 0.75M = 7.5 mmol

NaOH =22mL x 0.30M = 6.6 mmol

R HC2H3O2 + OH- --> C2H3O2- + H2O

I 7.5 mmol 6.6 mmol 0 mmol ignore

C -6.6 mmol -6.6 mmol +6.6 mmol

E 0.9 mmol 0.0 mmol 6.6 mmol

2.) HC2H3O2 --> H+ + C2H3O2-

Initial concentrations

[HC2H3O2]: 0.9 mmol / 32mL = 0.0281M

[H+]:

[C2H3O2-]: 6.6 mmol x 32mL = 0.206M

Concentrations at equilibrium

[HC2H3O2]: 0.0281M - x

[H+]: [C2H3O2-]: 0.206M + x

If x is small and can be ignored except [H+]

Substitute and solve

1.3 x 10-5 = (x)(0.206)

(0.0281)

X= 1.77 x 10-6M => pH = 5.75

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3 years ago
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