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3241004551 [841]
2 years ago
7

Based on your observations in the lab, what do you think happens to the majority of water (not, ice) at the poles? at the equato

r?
Chemistry
1 answer:
ahrayia [7]2 years ago
5 0
I think the best way to  explain it is that the water at the poles and in the equators is that temperature of the water at the poles are mostly low because of the insufficient heat energy and the water in the poles has a higher temperature compare to the water in the poles.
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Identify and calculate the number of representative particles in 2.15 moles of gold.<br>​
LekaFEV [45]

Answer:

1.29 * 10^{24} particles of gold

Explanation:

To convert the number of moles of any substance, in this case gold, you need Avogadro's number.

Avogadro's number is always 6.022 × 10^{23}

2.15 moles Au × \frac{6.022*10^{23} particles}{1 mole Au} = 1.29 * 10^{24} particles of gold

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3 years ago
Is the formation of acid rain an exothermic or endothermic reaction?
Monica [59]
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3 years ago
Nault 25000L 250 our = 25000L 15040 8. How many grams of CaCl2 are needed to make 150.0 mL of a 0.500 M CF solution? (Note: CaCl
yuradex [85]

Answer:

You need 8,324 g of CaCl₂ yo make this solution

Explanation:

Molarity is a way to express concentration in a solution, in units of moles of solute per liter of solution.

To know the grams of CaCl₂ it is necessary to know, first, the moles of this substance with the desired volume and concentration , thus:

0,1500 L × \frac{0,500 mol}{L} = 0,075 CaCl₂ moles

Now, with the molar mass of CaCl₂ you will obtain the necessary grams, thus:

0,075 CaCl₂ moles  × \frac{110,98 g}{mol} = 8,324 g of CaCl₂

So, you need <em>8,324 g of CaCl₂</em> to make 150,0 mL of a 0,500M solution

I hope it helps!

3 0
3 years ago
BALANCING CHEMICAL EQUATIONS WORKSHEET
kherson [118]

Answer:

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Explanation:

4 0
2 years ago
The diagram above shows the atomic emission line spectrum of hydrogen. The wavelength of each line is shown at the top of the li
Anni [7]
Hello there,

Your correct answer would be "<span>the energy of a photon".

Hope this helps.

~Hottwizzlers


</span>
6 0
3 years ago
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