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Leya [2.2K]
3 years ago
14

5) Which of these are intensive properties?

Chemistry
1 answer:
evablogger [386]3 years ago
6 0

Answer:

For number 5 it is the 2nd one some stants has a mass of 1.25 g And the 2nd 1 number 6It is the 1st 1A subject has a melting point of 40

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Is the method for converting molecules to moles the same as that for atoms?
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Answer:

One mole is the Avogadro's number of particles (atoms, molecules, ions or electrons) in a substance. Converting to moles is fairly easy because the conversion is always the same.

8 0
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When 6.0 mol Al react with 13 mol HCl, what is the limiting reactant, and how many moles of H2 can be formed?
Ad libitum [116K]

You're looking for the number of moles of H2, and you have 6.0 mol Al and 13 mol HCL.

For the first part, you have to make your way from 6.0 mol of Al to mol of H2, right? For that to happen, you need to make a conversion factor that will cancel the mol Al, in such case use the 2 moles of Al from your equation to cancel them out. At the top of the equation, you can use the number of moles of H2 from the equation and find the moles that will be produced for the H2.

6.0mol Al x 3 mol H2/2 mol Al = 9 mol H2

For the second part, you have to make the same procedure, make a conversion factor that will cancel the mol of HCL and for that you need to use the 6 mol HCL from your equation, and at the numerator you can put the 3 mol of H2 from the equation so that you can find the number of moles of H2 that will be produced.

13 mol HCL x 3 mol H2/6 mol HCL = 6.5 mol H2

As it can be seen, HCL produces the less amount of H2 moles. Therefore, the reaction CANNOT produce more than 6.5 mol H2, in that case 6.5 mol will be the maximum number of moles that will be produced at the end because HCL does not have enough to produce more than 6.5 mol.

In that case HCL is the limiting reactant because it limits that will be produced, and so the answer is B!

6 0
3 years ago
Show
SIZIF [17.4K]

Answer: 0.9375 g

Explanation:

To calculate the number of moles for given molarity, we use the equation:

\text{Moles of solute}={\text{Molarity of the solution}}\times{\text{Volume of solution (in L)}}     .....(1)

Molarity of HCl solution = 0.75 M

Volume of HCl solution = 25.0 mL = 0.025 L

Putting values in equation 1, we get:

\text{Moles of} HCl={0.75}\times{0.025}=0.01875moles  

CaCO_3(s)+2HCl(aq)\rightarrow CaCl_2(s)+CO_2(g)+H_2O(l)  

According to stoichiometry :

2 moles of HCl require = 1 mole of CaCO_3

Thus 0.01875 moles of HCl will require=\frac{1}{2}\times 0.01875=0.009375moles  of CaCO_3

Mass of CaCO_3=moles\times {\text {Molar mass}}=0.009375moles\times 100g/mol=0.9375g

Thus 0.9375 g of CaCO_3 is required to react with 25.0 ml of 0.75 M HCl

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