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:
energy required=qnet=87.75kJ
Explanation:
we will do it in three seperate step and then add up those value.
first step is to heat the sample of water upto 100C i.e upto boiling pont. because just after this sample of water started vaporization.
q 1= m c (T2-T1)
q1 = 36.0 g (4.18 J/gC) (100 - 65 C)
q1 = 5267 J
=5.267kJ
next is to vaporize the sample at 100C
q2 = 36.0 g / 18.0 g/mol X 40.7 kJ/mol
q2= 81.4 kJ
Finally, heat the steam upto 115C
q3 = m c (T2-T1)
q 3= 36.0 g (2.01 J/gC)(115-100C)
q3 = 1085 J
=1.085kJ
qnet=q1 +q2 +q3
energy required=qnet=87.75kJ
Answer:
V₂ = 1.5 L
Explanation:
Given data:
Initial volume of balloon = 1.76 L
Initial temperature = 295 K
Final temperature = 253.15 K
Final volume = ?
Solution:
According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.
Mathematical expression:
V₁/T₁ = V₂/T₂
V₁ = Initial volume
T₁ = Initial temperature
V₂ = Final volume
T₂ = Final temperature
Now we will put the values in formula.
V₁/T₁ = V₂/T₂
V₂ = V₁T₂/T₁
V₂ = 1.76 L ×253.15 K / 295 K
V₂ = 445.54 L.K /295 K
V₂ = 1.5 L
Answer:
a) Schmidt number
Explanation:
Prandtl number in heat transfer is analogues to Schmidt number in mass transfer.
Prandtl number in heat transfer is the ration of momentum diffusivity to the heat diffusivity.
Whereas, Schmidt number in mass transfer is the ratio of momentum diffusivity to the mass diffusivity.
Answer: Gas pressure is employed in a variety of industrial operations, and it was the catalyst for the industrial revolution. Many industrial machines were powered by steam in the early twentieth century. To heat the water molecules inside massive boilers, wood was burned.
