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motikmotik
2 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]2 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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Explain the reason for low melting and boiling point of a covalent compound ​
mylen [45]

Answer:

Covalent compounds have weak forces of attraction between the binding molecules. Thus less energy is required to break the force of bonding. Therefore covalent compounds have low melting and boiling point.

Explanation:

3 0
3 years ago
Read 2 more answers
________________ is 'The movement of a liquid up a narrow tube due to adhesive affects.'
jolli1 [7]

Answer: capillary action

Explanation: it occurs when the adhesion forces (attraction between two surfaces or substances) in the liquid are stronger than the cohesion forces (attraction between the same molecule)

7 0
3 years ago
How many moles are in 2.8x10^23 atoms of Calcium?
Zielflug [23.3K]

Answer: 1 mole ➡️ 6.022×10²³ atoms of si.

X mole ➡️ 2.8×10²⁴ atoms of si.

X = 2.8×10×10²³/6.022×10²³

= 28/6.022

= 4.65 moles.

Explanation:

8 0
3 years ago
'Suppose you take 0.0332 kg of ammonium carbonate and dissolve it using 0.0395 kg of water. What would be the mass percent conce
VMariaS [17]

The solution is 45.7 % (NH₄)₂CO₃ by mass.

Mass of solution = 0.0332 kg + 0.0395 kg = 0.0727 kg

% (NH₄)₂CO₃ = Mass of (NH₄)₂CO₃/Total mass × 100 % = 0.0332 kg/0.0727 kg × 100 % = 45.7 %


8 0
3 years ago
0.28gram of NH3 on decomposition gave 0.25gram of nitrogen and hydrogen. Find the volume of hydrogen evolved at Ntp. (2gram hydr
hodyreva [135]

The volume of H₂ evolved at NTP=0.336 L

<h3>Further explanation</h3>

Reaction

Decomposition of NH₃

2NH₃ ⇒ N₂ + 3H₂

conservation mass : mass reactants=mass product

0.28 NH₃= 0.25 N₂ + 0.03 H₂

2 g H₂ = 22.4 L

so for 0.03 g :

\tt \dfrac{0.03}{2}\times 22.4=0.336

7 0
3 years ago
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