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rosijanka [135]
4 years ago
6

Use the Gizmo to find the freezing, melting, and boiling points of water at 5,000 meters (16,404 feet). Write these values below

. Freezing point: _______ Melting point: _______ Boiling point: _______
Chemistry
1 answer:
Alex73 [517]4 years ago
3 0

Answer:

Freezing point: <u>32 ºF (0ºC)</u>

Melting point: <u>32 ºF (0ºC)</u>

Boiling point: <u>203°F (95°C)</u>

Explanation:

At the boiling point, <u>the vapor pressure of a liquid  equals the external pressure</u>. The normal melting point and boiling point of water at 1 atm are 0°C and 100°C, respectively. Decreasing the pressure under 1 atm (what happens when we are at high altitudes) will lower the boiling point since the external pressure will be lower, and it will become equal with the vapor pressure at a lower temperature.

However, the melting point and freezing point will stay unaffected since they don't depend on air-pressure; so at 0 or 5000 meters they will still be 0°C (remember that, <u>since water is a pure substance, the freezing and melting points will be the same</u>).

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What are 3 examples of making diluted solutions in "regular" life?<br> what are examples of diluted?
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6 0
3 years ago
How many atoms are present in 4.0 mol of sodium
gladu [14]
Avogadro's constant is the number of atoms of carbon-12 in 12g of
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atoms present equals avogadro's constant times number of moles for that substance. Type of substance does not change the number of atoms if you are given the quantity of moles, therefore the fact that it is sodium does not change the value.
<span>atoms present equals avogadro's constant times number of moles
</span>=6.023x10^23 x 4.0
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4 0
4 years ago
a student used 10 mL water instead of 30 mL for extraction of salt from mixture. How may this change the percentage of NaCl extr
Anit [1.1K]
It will be extracted only 1/3 of NaCl less in 10 mL of water than in 30 mL of water.

If it is known that solubility of NaCl is 360 g/L, let's find out how many NaCl is in 30 mL of water:

360 g : 1 L = x g : 30 mL

Since 1 L = 1,000 mL, then:
360 g : 1,000 mL = <span>x g : 30 mL

Now, crossing the products:
x </span>· 1,000 mL = 360 g · 30 mL
x · 1,000 mL = 10,800 g mL
x = 10,800 g ÷ 1,000 
x = 10.8 g

So, from 30 mL mixture, 10.8 g of NaCl could be extracted.

Let's calculate the same for 10 mL water instead of 30 mL.

360 g : 1 L = x g : 10 mL

Since 1 L = 1,000 mL, then:
360 g : 1,000 mL = <span>x g : 10 mL

Now, crossing the products:
x </span>· 1,000 mL = 360 g · 10 mL
x · 1,000 mL = 3,600 g mL
x = 3,600 g ÷ 1,000 
<span>x = 3.6 g
</span>
<span>So, from 10 mL mixture, 3.6 g of NaCl could be extracted.
</span>
Now, let's compare:
If from 30 mL mixture, 10.8 g of NaCl could be extracted and <span>from 10 mL mixture, 3.6 g of NaCl could be extracted, the ratio is:
</span>3.6/10.8 = 1/3

Therefore, i<span>t will be extracted only 1/3 of NaCl less in 10 mL of water than in 30 mL of water. </span>
5 0
3 years ago
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