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BigorU [14]
3 years ago
6

If the caffeine concentration in a particular brand of soda is 2.85 mg/oz, drinking how many cans of soda would be lethal? Assum

e that 10.0 g of caffeine is a lethal dose, and there are 12 oz in a can.
Chemistry
1 answer:
alina1380 [7]3 years ago
7 0

Answer:

2,375 cans

Explanation:

The strategy here is to use the information given to calculate the lethal dosage contained in the number of cans we will compute.

We know the lethal dosage is

Ld = 10.0 g caffeine

and we also know that the oncentration of caffeine is:

2.85 mg/ oz

So our problem simplifies to calculate how many oz will contain the lethal dose, and then given the ounces per can determine how many cans are required.

First convert the lethal dose in grams to mg:

Ld =( 10 g x 1000 mg ) = 10,000 mg caffeine

10,000 mg x ( 1 Oz / 2.85 mg ) = 28,500 oz

28500 oz x ( 1 can/12 oz ) = 2,375 cans

We could also have calculated it in one step  using conversion factors:

Number of cans = 10000 mg x 1 oz/ 2.85 mg x 1 can / oz = 2,375 cans

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What is the pH and pOH of a 2.2 x 10^-3 HBr solution?
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Answer:

The pH and pOH of a 2.2*10⁻³ HBr solution is 2.66 and 11.34  respectively.

Explanation:

pH - short for hydrogen potential - is a measure of the acidity or alkalinity of a solution. So the pH is a parameter that indicates the concentration of hydrogen ions [H]⁺ that exist in a solution.

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pH is measured on a scale of 0 to 14. On this scale, a pH value of 7 is neutral, which means that the substance or solution is neither acidic nor alkaline. A pH value of less than 7 means that it is more acidic, and a pH value of more than 7 means that it is more alkaline.

HBr is a strong acid. Then, in aqueous solution it will be  totally dissociated. So the proton concentration is equal to the initial concentration of  acid:

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On the other hand, pOH is a measure of the concentration of hydroxyl ions in a solution. The sum of pH and pOH equals 14:

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pOH= 11.34

<u><em>The pH and pOH of a 2.2*10⁻³ HBr solution is 2.66 and 11.34  respectively.</em></u>

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