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Olin [163]
3 years ago
7

GIVING BRAINLY, FOLLOW, 5 STARS AND HEARTS

Chemistry
1 answer:
Alina [70]3 years ago
8 0

Answer:

The one on the top,'' As the water cool, it is returned to the boiler where it is heated,''

Explanation:

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A sample of water contains 5.24x10^22 molecules. How many moles are in this sample?
Helga [31]

Answer:

0.087 moles of water

Explanation:

Given data:

Number of molecules of water = 5.24×10²² molecules

Number of moles of water = ?

Solution:

The given problem will solve by using Avogadro number.

1 mole = 6.022 × 10²³ molecules of water

5.24×10²² molecules × 1 mol / 6.022 × 10²³ molecules

0.87×10⁻¹ mol

0.087 mol

Avogadro number:

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.  The number 6.022 × 10²³ is called Avogadro number.

8 0
3 years ago
Why is geography considered a science
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7 0
3 years ago
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Which of the following is a strong acid?<br><br> OH–<br><br><br> HCl<br><br> NaCl<br><br> NaOH
Paul [167]
The answer is HCL 
Hope this helps!

5 0
3 years ago
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Functions of the following ingredients in the production of dairy products.ii. Enzyme in cheese​
Serjik [45]

Answer:

sugar in melk

Explanation:

cow eat sugar then make sugar :)

5 0
3 years ago
The normal freezing point of water (H2O) is 0.00 oC and its Kf value is 1.86 oC/m. A nonvolatile, nonelectrolyte that dissolves
adelina 88 [10]

Answer:

3.74g of ethylene glycol must be added to decrease the freezing point by 0.400°C

Explanation:

One colligative property is the freezing point depression due the addition of a solute. The equation is:

ΔT=Kf*m*i

<em>Where ΔT is change in temperature = 0.400°C</em>

<em>Kf is freezing point constant of the solvent = 1.86°C/m</em>

<em>m is molality of the solution (Moles of solute / kg of solvent)</em>

<em>And i is Van't Hoff constant (1 for a nonelectrolyte)</em>

Replacing:

0.400°C =1.86°C/m*m*1

0.400°C / 1.86°C/m*1 = 0.215m

As mass of solvent is 280.0g = 0.2800kg, the moles of the solute are:

0.2800kg * (0.215moles / 1kg) = 0.0602 moles of solute must be added.

The mass of ethylene glycol must be added is:

0.0602 moles * (62.10g / mol) =

3.74g of ethylene glycol must be added to decrease the freezing point by 0.400°C

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6 0
3 years ago
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