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Semmy [17]
2 years ago
7

the reaction of aluminum with chlorine gas is shown 2Al + 3Cl2 -> 2AlCl3 based on this equation how many molecules of chlorin

e gas are needed to react with 30 aluminum atoms?
Chemistry
1 answer:
Vanyuwa [196]2 years ago
3 0

45 molecules of chlorine gas (Cl₂) are needed to react with 30 atoms of aluminum (Al)

The balanced equation for the reaction is given below:

2Al + 3Cl₂ —> 2AlCl₃

From the balanced equation above,

2 atoms of Al required 3 molecules of Cl₂.

With the above information, we can determine the number of molecules of Cl₂ needed to react with 30 atoms of Al. This can be obtained as follow:

From the balanced equation above,

2 atoms of Al required 3 molecules of Cl₂.

Therefore,

30 atoms of Al will require = \frac{30 * 3}{2}\\ = 45 molecules of Cl₂.

Thus, 45 molecules of chlorine gas (Cl₂) are needed to react with 30 atoms of aluminum (Al)

Learn more: brainly.com/question/24918379

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What is he empirical formula for the compound that is of 1.85 moles of nitrogen and 4.63 miles of oxygen
BartSMP [9]

The empirical formula is N₂O₅.

The empirical formula is the <em>simplest whole-number ratio of atoms</em> in a compound.  

The ratio of atoms is the same as the ratio of moles, so our job is to calculate the <em>molar ratio of N:O</em>.  

I like to summarize the calculations in a table.  

<u>Element</u> <u>Moles</u>  <u>Ratio¹ </u>  <u> ×2²  </u>  <u>Integers</u>³

     N        1.85    1             2             2

     O        4.63    2.503   5.005     5

¹To get the molar ratio, you divide each number of moles by the smallest number (1.85).

²Multiply these values by a number (2) that makes the numbers in the ratio close to integers.

³Round off the number in the ratio to integers (2 and 5).

The empirical formula is N₂O₅.

4 0
3 years ago
If the concentration of the stock (provided) Cu(NH3)42 was 0.041 M, what concentration will the Cu2 be in beaker?
kodGreya [7K]

Answer:

[Cu^{2+}]=0.041 M

Explanation:

Hello!

In this case, since the molarity of a solution is defined in terms of the moles of the solute and the volume of solution, given that the concentration of Cu(NH₃)₄²⁺ is 0.041 M, and there is only one copper atom per Cu(NH₃)₄²⁺ ion, we can compute the concentration of Cu²⁺ as shown below:

[Cu^{2+}]=0.041\frac{molCu(NH_3)_4^{2+}}{L}*\frac{1molCu^{2+}}{1molCu(NH_3)_4^{2+}} =0.041 \frac{molCu(NH_3)_4^{2+}}{L}

[Cu^{2+}]=0.041 M

Best regards!

6 0
2 years ago
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