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disa [49]
3 years ago
11

You see a worm crawling along the ground and wonder if scientists consider it living or nonliving. Which item below would not be

true when classifying the worm as living or non‐living?
A worm can move so it must be living.
A worm can reproduce and make little baby worms (called wormlets) so it must be alive.
Worm’s pass dirt through their body pulling food out and discarding the stuff they don’t want so it must be alive.
If you cut a worm in half, it can re‐grow so it must be alive.
Chemistry
1 answer:
lisov135 [29]3 years ago
4 0

Answer: a. A worm can move so it must be living.

Explanation:

If a worm is moving it is classified as a moving worm also known to be living if moving.

A worm could reproduce therefore it is living.

Worms do pass dirt into and through there bodies to eat of the minerals and discarding things through and let out.

If you do cut a worm in half it may damage it but since it has basically two heads on girl one boy and cut in half it may split them up, and make damage but they will eventually grow back if the injuries are not severe.

        I Gave Some Facts and I Hope This Will Help!

HAPPY NEW YEAR!

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This stadium can hold 100,000, or 1 x 10^5, people. The number of atoms in a grain of iron is about 1 x 10^18. Would you need 1 
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<h3>Answer:</h3>

1 x 10^13 stadiums

<h3>Explanation:</h3>

From the question;

1 x 10^5 people can fill 1 stadium

We are given,  1 x 10^18 atoms of iron

We are required to determine the number of stadiums that  1 x 10^18 atoms of iron would occupy.

We are going to assume that a stadium would occupy a number of atoms equivalent to the number of people.

Therefore;

One stadium =  1 x 10^5 atoms

Then, to find the number of stadiums that will be occupied by  1 x 10^18 atoms;

No. of stadiums = Total number of atoms ÷ Atoms in a single stadium

                           =  1 x 10^18 atoms ÷  1 x 10^5 atoms

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Air at 25°C with a dew point of 10 °C enters a humidifier. The air leaving the unit has an absolute humidity of 0.07 kg of water
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          Dew point = 10^{o}C = (10 + 273) K = 283 K

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Using Antoine's equation we get the following.

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            ln (P^{s}_{a}) = 16.26205 - \frac{3799.887}{283 - 46.854}

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As total pressure (P_{t}) = atmospheric pressure = 760 mm Hg

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The absolute humidity of inlet air = \frac{P^{s}_{a}}{P_{t} - P^{s}_{a}} \times \frac{18 kg H_{2}O}{29 \text{kg dry air}}

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Thus, we can conclude that kg of water the must be evaporated into the air for every 100 kg of dry air entering the unit is 6.265 kg.

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