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Mars2501 [29]
2 years ago
6

How many moles are in 3.01 x 10^23 atoms of zinc?

Chemistry
1 answer:
Lerok [7]2 years ago
6 0

Answer:

<h2>0.5 moles</h2>

Explanation:

To find the number of moles in a substance given it's number of entities we use the formula

n =  \frac{N}{L} \\

where n is the number of moles

N is the number of entities

L is the Avogadro's constant which is

6.02 × 10²³ entities

From the question we have

n =  \frac{3.01 \times  {10}^{23} }{6.02 \times  {10}^{23} }  =  \frac{3.01}{6.02}  \\  = 0.5

We have the final answer as

<h3>0.5 moles</h3>

Hope this helps you

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A chemist prepares a solution of silver perchlorate by measuring out of silver perchlorate into a volumetric flask and filling t
djverab [1.8K]

Complete Question:

A chemist prepares a solution of silver (I) perchlorate (AgCIO4) by measuring out 134.g of silver (I) perchlorate into a 50.ml volumetric flask and filling the flask to the mark with water. Calculate the concentration in mol/L of the silver (I) perchlorate solution. Round your answer to 2 significant digits.

Answer:

13 mol/L

Explanation:

The concentration in mol/L is the molarity of the solution and indicates how much moles have in 1 L of it. So, the molarity (M) is the number of moles (n) divided by the volume (V) in L:

M = n/V

The number of moles is the mass (m) divided by the molar mass (MM). The molar mass of silver(I) perchlorate is 207.319 g/mol, so:

n = 134/207.319

n = 0.646 mol

So, for a volume of 50 mL (0.05 L), the concentration is:

M = 0.646/0.05

M = 12.92 mol/L

Rounded to 2 significant digits, M = 13 mol/L

7 0
3 years ago
I don't know how to do this, pls help!
Aliun [14]
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1,036                                 <span>853
</span><span>The lattice energy increases as cations get smaller, as shown by LiF and KF.
</span><span>I think this one should be correct answer, because the compared substances have also the same anion, and we can compare cations in them.

2) The same cation Li , so wrong statement.

3)</span>The same cation Na , so wrong statement.

4) NaCl smaller cation then KF
  786                                   853
7 0
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Answer:

KE = 1/2*m*v^2

KE = 1/2*150kg*(20 m/s)^2

KE = 75kg * 400m²/s²

KE = 30,000 kg*m²/s²

KE = 30,000 N*m

KE = 30,000 J

Explanation:

Hope this helped.

A brainliest is always appreciated.

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2 years ago
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How is percent by mass and concentration related?
Julli [10]
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