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Ad libitum [116K]
3 years ago
9

As an athlete exercises, sweat is produced and evaporated to help maintain a proper body temperature. On average, an athlete los

es approximately 307 g of sweat during an hour of exercise. How much energy is needed to evaporate the sweat that is produced? The heat of vaporization for water is 2257J/g. energy required:
Chemistry
1 answer:
dmitriy555 [2]3 years ago
7 0

<u>Answer:</u> For the given amount of sweat lost, the amount of energy required will be 692,899 Joules.

<u>Explanation:</u>

We are given:

Heat of vaporization for water = 2257 J/g

Amount of sweat lost = 307 grams

Applying unitary method:

For 1 g of sweat lost, the energy required is 2257 Joules

So, for 307 grams of sweat lost, the energy required will be = \frac{2257J}{1g}\times 307g=692,899J

Hence, for the given amount of sweat lost, the amount of energy required will be 692,899 Joules.

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A. 2.0x1024 atoms of S to moles (3.3)
Fynjy0 [20]

3.3 moles of sulfur

Explanation:

To find the number of moles knowing the number of atoms we use Avogadro's number to formulate the following reasoning:

if in         1 mole of sulfur (S) there are 6.022 × 10²³ atoms of sulfur (S)

then in   X moles of sulfur (S) there are 2 × 10²⁴ atoms of sulfur (S)

X = (1 × 2 × 10²⁴) / (6.022 × 10²³)

X = 3.3 moles of sulfur

Learn more about:

Avogadro's number

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5 0
3 years ago
<img src="https://tex.z-dn.net/?f=%5Ctext%7BThis%20is%20for%20a%20school%20project.%20I%20just%20need%20help%20on%20these%20two%
BigorU [14]

Answer:

Water molecules in the solid state, such as in ice and snow, form weak bonds (called hydrogen bonds) to one another. These ordered arrangements result in the basic symmetrical, hexagonal shape of the snowflake.

Explanation:

5 0
3 years ago
Silver has two naturally isotopes and has an atomic mass of 107.868 amu. One isotope is Ag-109 isotope (108.905 amu) and has a n
Triss [41]

Answer:  106.905

Explanation:  If there are only 2 isotopes, and 1 of them is 48.16%, the second must, by default, be (100 - 48.16%) = 51.84%  The final, averaged, atomic mass is 107.868.  This is made up of each isotope's atomic mass times the percentage of that isotope in the total sample.  The weighted value of the known isotope (109) plus that of the unknown must come to the observed value of 107.868 amu.  (107.868 - 52.45 = 55.42).  Divide that by the % for that isotope (55.42/0.5184) = 106.90 amu for the second isotope.

<u>Atomic Mass</u>  <u>% of Sample</u> <u>Weighted Value</u>

    108.905         48.16%              52.45

          X               51.84%              <u>55.42</u>

                                                     107.87

      X = (55.42/0.5184) = 106.90 amu

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