Answer:
a)
,
, b)
, 
Explanation:
a) The ideal gas is experimenting an isocoric process and the following relationship is used:

Final temperature is cleared from this expression:


The number of moles of the ideal gas is:



The final temperature is:


The final pressure is:



b) The ideal gas is experimenting an isobaric process and the following relationship is used:

Final temperature is cleared from this expression:




The final volume is:



Answer : The given compound belongs to ether and alcohol.
Explanation :
The chemical formula of the given compound is, 
First we have to calculate the degree of unsaturation.
Formula used:
Degree of unsaturation = 
where,
C = number of carbon
H = number of hydrogen
N = number of nitrogen
X = number of halogen
Degree of unsaturation = 
The degree of unsaturation is, 0 that means there is no double or triple bond in the compound only single bond is present between the atoms.
Thus, the given compound belongs to ether and alcohol.
Answer:
1) Covalent bond, in chemistry, the interatomic linkage that results from the sharing of an electron pair between two atoms. The binding arises from the electrostatic attraction of their nuclei for the same electrons.
2)A molecule is the smallest particle in a chemical element or compound that has the chemical properties of that element or compound. Molecules are made up of atoms that are held together by chemical bonds. These bonds form as a result of the sharing or exchange of electrons among atoms.
3)Hydrogen (H2), oxygen (O2), and chlorine (Cl2) molecules, for example, each contains two atoms. Another form of oxygen, ozone (O3), has three atoms, and sulfur (S8) has eight atoms. All elemental molecules are made of atoms of a single element.
Explanation:
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Answer:
8.44 atm
Explanation:
From the question given above, the following data were obtained:
Initial volume (V₁) = 2.25 L
Initial temperature (T₁) = 350 K
Initial pressure (P₁) = 1.75 atm
Final volume (V₂) = 1 L
Final temperature (T₂) = 750 K
Final pressure (P₂) =?
The final pressure of the gas can be obtained as illustrated below:
P₁V₁/T₁ = P₂V₂/T₂
1.75 × 2.25 / 350 = P₂ × 1 / 750
3.9375 / 350 = P₂ / 750
Cross multiply
350 × P₂ = 3.9375 × 750
350 × P₂ = 2953.125
Divide both side by 350
P₂ = 2953.125 / 350
P₂ = 8.44 atm
Thus, the final pressure of the gas is 8.44 atm.
<span>.750 moles X (6.02 x10^23 atoms/1 mol)= 4.52 X 10^23 atoms is the answer </span>