Explanation:
A. Hydrogen bonding is present in CS2 but not in CO2.
B. CS2 has greater dipole moment than CO2 and thus the dipole-dipole forces in CS2 are stronger.
C. CS2 partly dissociates to form ions and CO2 does not. Therefore, ion-dipole interactions are present in CS2 but not in CO2.
D. The dispersion forces are greater in CS2 than in CO2.
<u><em>PLS MARK BRAINLIEST :D</em></u>
Mass of H₂ needed to react with O₂ : 1.092 g
<h3>Further explanation</h3>
The concentration of a substance can be expressed in several quantities such as moles, percent (%) weight / volume,), molarity, molality, parts per million (ppm) or mole fraction. The concentration shows the amount of solute in a unit of the amount of solvent.
Reaction
O₂(g) + 2H₂(g) → 2H₂O(g)
mass of O₂ : 8.75 g
mol O₂(MW=32 g/mol) :

From the equation, mol ratio of O₂ : H₂ = 1 : 2, so mol H₂ :

Mass H₂ (MW=2 g/mol) :

Answer:
All of the above.
Explanation:
In positive deviation from Raoult's Law occur when the vapour pressure of components is greater than what is expected value in Raoult's law.
When a solution is non ideal then it shows positive or negative deviation.
Let two solutions A and B to form non- ideal solutions.let the vapour pressure of component A is
and vapour pressure of component B is
.
= Vapour pressure of component A in pure form
= Vapour pressure of component B in pure form
=Mole fraction of component A
=Mole fraction of component B
The interaction between A- B is less than the interaction A- A and B-B interaction.Therefore, the escaping tendency of liquid molecules in mixture is greater than the escaping tendency in pure form.Hence, the vapour pressure of a mixture is greater than the initial value of vapour pressure.
,
Therefore, 
Therefore, the enthalpy of mixing is greater than zero and change in volume is greater than zero.
Hence, option a,b,c and d are true.
Answer:
This question is incomplete but the correct option is B
Explanation:
This question is incomplete because of the absence of the "Reference Table S", however the question can still be answered in the absence of the table. The energy described in the question is the ionization energy (energy required to remove the most loosely bound electron in an atom). This question seeks to know the atom (from the options provided) with the least ionization energy.
Ionization energy increases from left to right across the period because it's easier to remove a single electron (valence electron) from the outermost shell than to remove two electrons from the same shell; thus the more the valence electrons (in a shell), the higher the ionization energy. Thus, bromine (Br) and tin (Sn) have high ionization energies because they have more number of electrons in there outermost shell.
<u>Berylium (Be) and strontium (Sr) are both in the group 2 of the periodic table because they both have 2 electrons in there outermost shell. Ionization energy decreases down a group. This is because the farther an electron is from the nucleus, the weaker the force of attraction between the nucleus and the electron. Thus, strontium (Sr) would have a lesser ionization energy between the two and would indeed have the least ionization among the options provided</u>. Hence, the correct option is B
<u>Answer:</u> The chemical equation is written below.
<u>Explanation:</u>
Every balanced chemical equation follows law of conservation of mass.
This law states that mass can neither be created nor be destroyed but it can only be transformed from one form to another form. This also means that total number of individual atoms on reactant side must be equal to the total number of individual atoms on the product side.
The chemical equation for the reaction of elemental boron and oxygen gas follows:

By Stoichiometry of the reaction:
4 moles of elemental boron reacts with 3 moles of oxygen gas to produce 2 moles of diboron trioxide.
The chemical equation for the reaction of diboron trioxide and water follows:

By Stoichiometry of the reaction:
1 mole of diboron trixoide reacts with 3 moles of water to produce 2 moles of boric acid.
Hence, the chemical equations are written above.