Answer is: 4.45 grams of methane gas <span>need to be combusted</span>.
Balanced chemical reaction: CH₄ + 2O₂ → CO₂ + 2H₂O.
Ideal gas law: p·V =
n·R·T.<span>
p = 1.1 atm.
T = 301 K.
V(H</span>₂O) <span>= 12.5 L.
R = 0,08206 L·atm/mol·K.
</span>n(H₂O) = <span>1.1 atm ·
12.5 L ÷ 0,08206 L·atm/mol·K · 301 K.
</span>n(H₂O) = 0.556 mol.
From chemical reaction: n(H₂O) : n(CH₄) = 2 : 1.
n(CH₄) = 0.556 mol ÷ 2 = 0.278 mol.
m(CH₄) = 0.278 mol · 16 g/mol.
m(CH₄) = 4.448 g.
Answer:
The different structures are shown in the attachment.
I and II - structural isomers
I and III - Structural isomers
I and IV - structural isomers
II and III - structural isomers
II and IV - structural isomers
III and IV - stereoisomers
Explanation:
The knowledge of Isomerism is tested here; there are two types of isomerism ; structural and stereoisomerism.
- Structural Isomers have similar molecular and different double bond positioning, these occurs mostly in ALKENE FAMILY.
- Stereo-isomers have the same molecular formular and similar patterns but differ in their spatial arrangement. trans and cis are typical examples of stereo-isomers.
From the question; Relationship between I and II is that they are structural isomers since they have the same molecular formula, but different bond atom arrangement and infact they are the same compound.
- Relationship between I and III is that they are structural isomers with similar molecular formular but differ in the double bond position.
- Relationship between I and IV is that they are structural isomers with similar molecular formula but different double bond arrangement.
- Relationship between II and III is that they are structural isomers with similar molecular formular but different double bond position
- Relationship between II and IV is that they are also structural isomers with the same molecular formular but different double bond position.
- Relationship between III and IV is that they are stereo-isomers with same molecular formula but different spatial arrangement, hence cis and trans.
The answer is B. Increase, Decrease
Answer:
-1.37 kJ/mol
Explanation:
The expression for the calculation of the enthalpy of dissolution of [tex[NH_4NO_3[/tex] is shown below as:-
Where,
is the enthalpy of dissolution of [tex[NH_4NO_3[/tex]
m is the mass
C is the specific heat capacity
is the temperature change
Thus, given that:-
Mass of ammonium nitrate = 5.60 g
Specific heat = 4.18 J/g°C
So,
Negative sign signifies loss of heat.
Also, 1 J = 0.001 kJ
So,

Also,
Molar mass of [tex[NH_4NO_3[/tex] = 80.043 g/mol
The formula for the calculation of moles is shown below:
Thus,

Thus, 