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alexgriva [62]
2 years ago
14

How many moles of hcl are required to make 500 ml of a 3.0 mol/l solution of hcl?

Chemistry
1 answer:
LiRa [457]2 years ago
8 0

Answer:

1.5 moles

Explanation:

To find the number of moles of HCl in 500 mL of a 3 M solution of HCl, we consider moles in 1 liter/ 1000 mL.

3 moles HCl is contained in 1000 mL

x moles is HCl is contained in 500 mL

=  \frac{3 \:  \times  \: 500}{1000}  \: moles \\  = 1.5 \: moles

Hence the number of moles of HCl in 500 mL is 1.5 moles.

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Hi!

Electrons are particles which basically 'orbit' around the nucleus. Protons and neutrons are condensed, in a fixed position inside the nucleus. 

With this in mind, the answer will be C.

Hopefully, this helps! =)
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4 years ago
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What volume in milliliters of 1.420 M sulfuric acid is needed to neutralize 3.209 g of aluminum hydroxide 3 H 2 SO 4 (aq)+2 Al(O
nadya68 [22]

Answer:

V=43.46mL

Explanation:

Hello!

In this case, since the reaction between sulfuric acid and aluminum hydroxide is:

3H_2SO_4+2Al(OH)_3\rightarrow Al_2(SO_4)_3+6H_2O

Whereas the ratio of sulfuric acid to aluminum hydroxide is 3:2; thus, we first compute the moles of sulfuric acid that complete react with 3.209 g of aluminum hydroxide:

n_{H_2SO_4}=3.209gAl(OH)_3*\frac{1molAl(OH)_3}{78.00gAl(OH)_3} *\frac{3molH_2SO_4}{2molAl(OH)_3} \\\\n_{H_2SO_4}=0.0617molH_2SO_4

Then, given the molarity, it is possible to obtain the milliliters as follows:

V=\frac{n}{M}=\frac{0.0617mol}{1.420mol/L}*\frac{1000mL}{1L}\\\\V=43.46mL

Best regards!

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