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.
Answer:
Explanation:
Structure of the 2,2,4,4-tetramethyl-3-pentanone is give in the attachment
In 2,2,4,4-tetramethyl-3-pentanone, no alpha hydrogen is present, therefore, enol form is not possible and hence, exist only in keto form.
Explanation for existence of cyclohexa-2,4-diene-1-one only in enol form:
keto form of cyclohexa-2,4-diene-1-one not aromatic and hence less stable.
Whereas enol form it is aromatic which makes it highly stable. that's why cyclohexa-2,4-diene-1-one exists only in enol form.
Answer:
0.37atm
Explanation:
Given parameters:
Initial pressure = 0.25atm
Initial temperature = 0°C = 273K
Final temperature = 125°C = 125 + 273 = 398K
Unknown:
Final pressure = ?
Solution:
To solve this problem, we use a derivative of the combined gas law;
= 
P and T are pressure and temperature
1 and 2 are initial and final values
= 
P2 = 0.37atm
Eat eat eat eat eat eat eat eat eat
The hydrate form of CuSO4 has 5 water molecules (CuSO4-5H20) copper (II) Sulfate pentahydrate or commonly known as blue vitriol.
To solve, the following molar masses are to be known.
CuSO4.5H2O (hydrate) - 249.7g/mole
CuSO4 (anhydrous) -159.6g/mole
Also there molar ratio of the hydrate and CuSO4 is 1.
the mass of the hydrate is to be divided by the molar mass of the hydrate then multiplied by the ratio (1) to get the moles of hydrate and multiplied by the molar mass of the anhydrous to get the mass in grams.
moles = (100g/249.7)*1 = 0.4 moles hydrate
grams = 0.4*159.6 = 64.9 grames hydrate
