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motikmotik
3 years ago
12

How many moles of aluminum are in this aluminum can? Answer: ____________ (25 pts) Calculate the number of aluminum atoms in thi

s can. Answer: _______________ (25 pts) The Amount of Aluminum is 12.69g
Chemistry
1 answer:
Sonja [21]3 years ago
5 0

Answer:a)No of moles of Aluminium=0.4703

b)number of aluminum atoms=2.832 x 10^23 atoms

Explanation:

a)Given that The Amount/ Mass of Aluminum is 12.69g

We know that

No of moles =Mass/ Molar mass

Molar mass of Aluminium, Al is 26.982 g/mol

No of moles =12.69g/26.982 g/mol

No of moles =0.4703

Also

b) 1  mole of aluminium must contain  6.022  x 10^23  atoms of aluminium  

which is  known as Avogadro's constant.

Therefore, 0.4703 moles will contain 0.4703 x  6.022  x 10^23  atoms =2.832 x 10^23 atoms

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3 years ago
For all of the following questions 20.00 mL of 0.200 M HBr is titrated with 0.200 M KOH.
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Answer :

The concentration of H^+ before any titrant added to our starting material is 0.200 M.

The pH based on this H^+ ion concentration is 0.698

Explanation :

First we have to calculate the concentration of H^+ before any titrant is added to our starting material.

As we are given:

Concentration of HBr = 0.200 M

As we know that the HBr is a strong acid that dissociates complete to give hydrogen ion H^+ and bromide ion Br^-.

As, 1 M of HBr dissociates to give 1 M of H^+

So, 0.200 M of HBr dissociates to give 0.200 M of H^+

Thus, the concentration of H^+ before any titrant added to our starting material is 0.200 M.

Now we have to calculate the pH based on this H^+ ion concentration.

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pH=-\log [H^+]

pH=-\log (0.200)

pH=0.698

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3 0
3 years ago
How many moles of sodium carbonate are contained by 57.3g of sodium carbonate
Lady_Fox [76]

Answer:

\boxed {\boxed {\sf 0.541 \  mol \ Na_2CO_3}}

Explanation:

We are asked to find how many moles of sodium carbonate are in 57.3 grams of the substance.

Carbonate is CO₃ and has an oxidation number of -2. Sodium is Na and has an oxidation number of +1. There must be 2 moles of sodium so the charge of the sodium balances the charge of the carbonate. The formula is Na₂CO₃.

We will convert grams to moles using the molar mass or the mass of 1 mole of a substance. They are found on the Periodic Table as the atomic masses, but the units are grams per mole instead of atomic mass units. Look up the molar masses of the individual elements.

  • Na:  22.9897693 g/mol
  • C: 12.011 g/mol
  • O: 15.999 g/mol

Remember the formula contains subscripts. There are multiple moles of some elements in 1 mole of the compound. We multiply the element's molar mass by the subscript after it, then add everything together.

  • Na₂ = 22.9897693 * 2= 45.9795386 g/mol
  • O₃ = 15.999 * 3= 47.997 g/mol
  • Na₂CO₃= 45.9795386 + 12.011 + 47.997 =105.9875386 g/mol

We will convert using dimensional analysis. Set up a ratio using the molar mass.

\frac {105.9875386  \ g \ Na_2CO_3}{1 \ mol \ Na_2CO_3}

We are converting 57.3 grams to moles, so we multiply by this value.

57.3 \ g \ Na_2CO_3} *\frac {105.9875386  \ g \ Na_2CO_3}{1 \ mol \ Na_2CO_3}

Flip the ratio so the units of grams of sodium carbonate cancel.

57.3 \ g \ Na_2CO_3} *\frac {1 \ mol \ Na_2CO_3}{105.9875386  \ g \ Na_2CO_3}

57.3 } *\frac {1 \ mol \ Na_2CO_3}{105.9875386 }

\frac {57.3 }{105.9875386 } \ mol \ Na_2CO_3

0.5406295944 \ mol \ Na_2CO_3

The original measurement of moles has 3 significant figures, so our answer must have the same. For the number we found that is the thousandth place. The 6 in the ten-thousandth place to the right tells us to round the 0 up to a 1.

0.541 \  mol \ Na_2CO_3

There are approximately <u>0.541 moles of sodium carbonate</u> in 57.3 grams.

6 0
3 years ago
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