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kaheart [24]
3 years ago
9

A 28-kg child can consume a maximum of 23 children's acetaminophen tablets in an 8-h period without exceeding the safety-limit m

aximum allowable dose. Given that each children's tablet contains 80 mg of acetaminophen, determine the maximum allowable dose in mg per pound of body weight for one day. (a) 80 mg/lb (b) 90 mg/lb (c) 430 mg/lb (d) 720 mg/lb (e) 3.7 mg/lb
Chemistry
1 answer:
Nesterboy [21]3 years ago
6 0

Answer:

(b)90mg/lb

Explanation:

The child can consume a maximum of 23 tablets with 80mg of acetaminophen, the total acetaminophen ingest is:

23*80=1840mg

This value is for 8 hours, 8 hours is 1/3 of one day. This means that the dose that can be consumed in 24 hours is 3 times 1840mg.

1840*3=5520mg

To obtain the maximum allowable dose in mg per pound of body weight, divide this last value by the weight of the child in pounds.

Conversion of units: 28Kg(\frac{2.20lb}{1Kg} )=61.6lb

\frac{5520}{61.6} =89.61\approx90mg/lb

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Which of the following are examples of chemical changes? Select all that apply.
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b. milk spoiling and c. firecrackers exploding

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2 years ago
Predict the electronegativity of the yet undiscovered element with Z = 119.
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We know that the element Z = 119 would be placed right below the Fr, in the column of the alcaline metals.

We also know that the trend in the electronegativity is to decrease when you go up-down ia group.

The known electronegativities of the elements of this group are:

Li: 0.98
Na: 0.93
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4 0
3 years ago
How many moles of Carbon are in 3.06 g of Carbon
natta225 [31]

Answer:

\boxed {\boxed {\sf 0.255 \ mol \ C }}

Explanation:

If we want to convert from grams to moles, the molar mass is used. This is the mass of 1 mole. They are found on the Periodic Table as the atomic masses, but the units are grams per mole (g/mol) instead of atomic mass units (amu).

Look up the molar mass of carbon.

  • Carbon (C): 12.011 g/mol

Set up a ratio using the molar mass.

\frac {12.011 \ g \ C}{ 1 \ mol \ C}

Since we are converting 3.06 grams to moles, we multiply by that value.

3.06 \ g \ C*\frac {12.011 \ g \ C}{ 1 \ mol \ C}

Flip the ratio. This way, the ratio is still equivalent, but the units of grams of carbon cancel.

3.06 \ g \ C* \frac{1 \ mol \ C}{12.011 \ g\ C}                      

3.06 * \frac{1 \ mol \ C}{12.011 }    

\frac {3.06}{12.011 } \ mol \ C                                

0.25476646 \ mol \ C

The original measurement of grams (3.06) has 3 significant figures, so our answer must have the same. For the number we calculated, that is the thousandth place.

  • 0.25476646

The 7 in the ten-thousandth place tells us to round the 4 up to a 5.

0.255 \ mol \ C

3.06 grams of carbon is approximately <u>0.255 moles of carbon.</u>

3 0
2 years ago
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