A 1.0 oz piece of chocolate contains 15mg of caffeine, and a 6.0 oz cup of regular coffee contains 105mg of caffeine. How much chocolate in g, Ib, amu, and metric ton would you have to consume to get as much caffeine as you would from 2.0 cups of coffee?

Answer 1

Here's what I got.

I'll expand on the responses given here:

A one-ounce piece of chocolate contains fifteen milligrams of caffeine, and a six-ounce cup of

So, you know that you must consume #"14 oz"# of chocolate to get as much caffeine as you would from #"2.0"# cups of coffee.

Use the following conversion factors to get this value in grams, pounds, metric tons, and unified atomic mass units.

#"1 oz " = " 28.3495231 g"#
#"1 lb " = " 453.59237 g"#
#"1 u " = 1.66 * 10^(-24)"g"#
#"1 g " = 10^(-6)"t"#

Let's begin the conversion now.

#14color(red)(cancel(color(black)("oz"))) * "28.3495231 g"/(1color(red)(cancel(color(black)("oz")))) = 4.0 * 10^2"g"#
#4.0 * 10^2color(red)(cancel(color(black)("g"))) * "1 lb"/(453.59237color(red)(cancel(color(black)("g")))) = "0.88 lb"#
#4.0 * 10^2color(red)(cancel(color(black)("g"))) * "1 u"/(1.66 * 10^(0-24)color(red)(cancel(color(black)("g")))) = 2.4 * 10^(26)"u"#
#4.0 * 10^2color(red)(cancel(color(black)("g"))) * (10^(-6)"t")/(1color(red)(cancel(color(black)("g")))) = 4.0 * 10^(-4)"t"#
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Answer 2

To get as much caffeine as you would from 2.0 cups of coffee, you would need to consume:

  1. Chocolate in grams: 2 cups of coffee contain 210mg of caffeine. So, to get the same amount from chocolate, you would need to consume ( \frac{210, \text{mg}}{15, \text{mg/oz}} = 14 ) oz of chocolate. Converting ounces to grams: 1 oz = 28.35 grams, so 14 oz = 396.9 grams of chocolate.

  2. Chocolate in pounds: 1 lb = 16 oz, so ( \frac{396.9, \text{grams}}{453.6, \text{g/lb}} = 0.874 ) lbs of chocolate.

  3. Chocolate in atomic mass units (amu): Caffeine's molecular formula is C8H10N4O2. The molar mass of caffeine is approximately 194.19 g/mol. To find the amount of caffeine in amu, we divide the amount of caffeine in grams by the molar mass: ( \frac{396.9, \text{grams}}{194.19, \text{g/mol}} = 2.04 ) mol of caffeine. One mol of caffeine contains ( 6.022 \times 10^{23} ) molecules, so 2.04 mol contains ( 2.04 \times 6.022 \times 10^{23} ) molecules.

  4. Chocolate in metric tons: 1 metric ton = 1,000,000 grams. So, ( \frac{396.9, \text{grams}}{1,000,000, \text{g/metric ton}} = 0.0003969 ) metric tons of chocolate.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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