Answer:
3.481 x 10²⁴ atoms
Explanation:
Data Given:
Number of moles of NH₄NO₃ = 5.78 mol
Number of atoms of NH₄NO₃ = ?
Solution:
Formula used
no. of moles = no. of atoms / Avogadro's number
As we have to find no. of atoms
So, we have to rearrange the above equation
no. of atoms = no. of moles x Avogadro's number . . . . . (1)
Where
Avogadro's number = 6.022 x 10²³ atoms
Put values in above Equation 1
no. of atoms = 5.78 mol x 6.022 x 10²³ atoms/mol
no. of atoms = 3.481 x 10²⁴ mol
So,
no. of atoms in 5.78 mole NH₄NO₃ = 3.481 x 10²⁴ atoms
Just choose one of them. Getting one question wrong isn't the end of the world.
I suggest water.
Answer:
Explanation:
Your strategy here will be to
use the chemical formula of carbon dioxide to find the number of molecules of
CO
2
that would contain that many atoms of oxygen
use Avogadro's constant to convert the number of molecules to moles of carbon dioxide
use the molar mass of carbon dioxide to convert the moles to grams
So, you know that one molecule of carbon dioxide contains
one atom of carbon,
1
×
C
two atoms of oxygen,
2
×
O
This means that the given number of atoms of oxygen would correspond to
4.8
⋅
10
22
atoms O
⋅
1 molecule CO
2
2
atoms O
=
2.4
⋅
10
22
molecules CO
2
Now, one mole of any molecular substance contains exactly
6.022
⋅
10
22
molecules of that substance -- this is known as Avogadro's constant.
In your case, the sample of carbon dioxide molecules contains
2.4
⋅
10
22
molecules CO
2
⋅
1 mole CO
2
6.022
⋅
10
23
molecules CO
2
=
0.03985 moles CO
2
Finally, carbon dioxide has a molar mass of
44.01 g mol
−
1
, which means that your sample will have a mass of
0.03985
moles CO
2
⋅
44.01 g
1
mole CO
2
=
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
∣
∣
a
a
1.8 g
a
a
∣
∣
−−−−−−−−−
The answer is rounded to two sig figs, the number of sig figs you have for the number of atoms of oxygen present in the sample.
Your question looks a bit incomplete as you have the same contents in options a) and d). According to your list, I can't see the correct answer, but I can give you one.The difference between the potential energy of the products of the potential energy of the reactants is equal to the enthalpy of the reaction.
