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2 years ago

How does the skin protect your internal organs from the outside world?

1 answer:

8 0

Your skin protects your internal organs in so many different ways. I will name a few for you! :)

Protects from damage - Your skin protects your internal organs from damage such as cuts and bruising. Our skin is not incredibly strong, but it does make a huge difference compared to our organs being bare.

Protects from bacteria - Your skin blocks most bacteria from directly having contact with your internal organs. Most people get sick because bacteria is let into contact with internal organs through our mouths, noses, etc. Our skin does not let bacteria in.

Keeps internal organs warm - Skin is not fur, it does not provide significant resistance to cold. However, it does make a big difference compared to bare organs. It protects our internal organs from heat by sweating, as well as cold by preventing too much cold from reaching our organs.

I hope this helped. Have a great day!

You might be interested in

Consider the reaction of solid aluminum iodide and potassium metal to form solid potassium iodide and aluminum metal.The balance

ratelena [41]

Answer:

674.26 g of AlI₃

Explanation:

We'll begin by calculating the theoretical yield of aluminum (Al). This can be obtained as follow:

Percentage yield of Al = 67.8%

Actual yield of Al = 30.25 g

Theoretical yield of Al =?

Percentage yield = Actual yield /Theoretical yield × 100/

67.8% = 30.25 / Theoretical yield

67.8 / 100 = 30.25 / Theoretical yield

0.678 = 30.25 / Theoretical yield

Cross multiply

0.678 × Theoretical yield = 30.25

Divide both side by 0.678

Theoretical yield = 30.25 / 0.678

Theoretical yield of Al = 44.62 g

Next, we shall determine the mass of AlI₃ that reacted and the mass of Al produced from the balanced equation. This can be obtained as follow:

AlI₃(s) + 3K(s) → 3KI(s) + Al(s)

Molar mass of AlI₃ = 27 + (3×127)

= 27 + 381 = 408 g/mol

Mass of AlI₃ from the balanced equation = 1 × 408 = 408 g

Molar mass of Al = 27 g/mol

Mass of Al from the balanced equation = 1 × 27 = 27 g

Summary:

From the balanced equation above,

408 g of AlI₃ reacted to produce 27 g of Al.

Finally, we shall determine the mass of

AlI₃ required to produce 44.62 g of Al. This can be obtained as follow:

From the balanced equation above,

408 g of AlI₃ reacted to produce 27 g of Al.

Therefore, Xg of AlI₃ will react to produce 44.62 g of Al i.e

Xg of AlI₃ = (408 × 44.62)/27

Xg of AlI₃ = 674.26 g

Thus, 674.26 g of AlI₃ is needed for the reaction.

8 0

3 years ago

How many grams of oxygen would be needed to completely react with 18 grams of methane (CH4)?

S_A_V [24]

Answer:

35.91 grams of oxygen

Explanation:

6 0

3 years ago

PLEASE HELP!! GIVING BRAINLIEST

RSB [31]

Answer:

The answer is "0.00172172603".

Explanation:

Given:

$Mass (M) = 1.60 \times 10^{-3} \ g\\\\Density (D) = 9.293 \times 10^{-1} \ \frac{g}{cm^3}\\\\Volume (V) = ?$

Formula:

$\to \bold{V = \frac{M}{D}}$

$= \frac{1.60 \times 10^{-3}}{9.293 \times 10^{-1}} \\\\ = \frac{1.60 \times 10^{1}}{9.293 \times 10^{3}} \\\\ = \frac{1.60 \times 10}{9.293 \times 1000} \\\\ = \frac{1.60 }{9.293 \times 100} \\\\ = \frac{1.60 }{929.3 } \\\\= 0.00172172603$