Answer:
the final mole of the flexible container = 12.92 moles
Explanation:
Given that :
initial volume of a flexible container = 6.13 L
initial mole of a flexible container = 6.51 mol
final volume of a flexible container = 18.3 L
final mole of a flexible container = ???
Assuming the pressure and temperature of the gas remain constant, calculate the number of moles of gas added to the container.
Therefore,
n = 19.43
19.43 = 6.51 + n₂
n₂ = 19.43 - 6.51
n₂ = 12.92 moles
Thus; the final mole of the flexible container = 12.92 moles
Answer:
15 grams of water
Explanation:
15 grams of water of water would lose heat the faster compared to higher masses of water.
Water generally is a poor conductor heat.
- To heat up a unit of water, significant amount of energy must be added to the body of water.
- With time, the body continues to increase in temperature.
- A 500g mass of water will take more time to lose heat.
Answer:
D. 15.8atm
Explanation:
Given parameters:
Initial pressure = 13atm
Initial temperature = 34°C = 34 + 273 = 307K
Final temperature = 100°C = 100 + 273 = 373K
Unknown:
Final pressure = ?
Solution:
To solve this problem, we apply a derivation of the combined gas law taking the volume as a constant.
The expression is shown mathematically below;
=
P and T pressure and temperature values
1 and 2 are initial and final states
Insert the parameters and solve for T₂;
=
P₂ = 15.8atm
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.