The compounds that are produced upon this combustion reaction would be Carbon Dioxide and water.
CO2 = Carbon and Oxygen
H2O = Hydrogen and Oxygen.
The exact molecular amounts or moles can be determined by balancing this combustion reaction.
Answer: The value of 
for chloroform is 
when 0.793 moles of solute in 0.758 kg changes the boiling point by 3.80 °C.
Explanation:
Given: Moles of solute = 0.793 mol
Mass of solvent = 0.758
As molality is the number of moles of solute present in kg of solvent. Hence, molality of given solution is calculated as follows.
Now, the values of 
is calculated as follows.
where,
i = Van't Hoff factor = 1 (for chloroform)
m = molality
= molal boiling point elevation constant
Substitute the values into above formula as follows.
Thus, we can conclude that the value of for chloroform is when 0.793 moles of solute in 0.758 kg changes the boiling point by 3.80 °C.
Answer:
A fluid is a medium that has a defined mass and volume, but no fixed shape, at a constant temperature and pressure. This may include gases, liquids, plasmas, and to some extent plastic solids. A fluid can flow and deform, preventing it from carrying loads in a static equilibrium. A fluid is always compressible and internal frictional forces always occur due to the viscosity of the fluid.
Answer:
KBr is limiting reactant.
Explanation:
Given data:
Mass of KBr =4g
Mass of Cl₂ = 6 g
Limiting reactant = ?
Solution:
Chemical equation:
2KBr + Cl₂ → 2KCl + Br₂
Number of moles of KBr:
Number of moles = mass/molar mass
Number of moles = 4 g/ 119 gmol
Number of moles = 0.03 mol
Number of moles of Cl₂:
Number of moles = mass/molar mass
Number of moles = 6 g/ 70 gmol
Number of moles = 0.09 mol
Now we will compare the moles of reactant with product.
KBr : KCl
2 : 2
0.03 : 0.03
KBr : Br₂
2 : 1
0.03 : 1/2×0.03= 0.015
Cl₂ : KCl
1 : 2
0.09 : 2/1×0.09 = 0.18
Cl₂ : Br₂
1 : 1
0.09 : 0.09
Less number of moles of product are formed by the KBr thus it will act as limiting reactant while Cl₂ is present in excess.
Answer:
a) Se²⁻> S²⁻ > O²
b) Te²⁻ > I- >Cs+
c) Cs+ > Ba²⁺ > Sr²⁺
Explanation:
(a) Se²⁻, S²⁻, O²⁻
In general, ionic radius decreases with increasing positive charge.
As the charge on the ion becomes more positive, there are fewer electrons.
The ion has a smaller radius. In general, ionic radius increases with increasing negative charge.
For ions of the same charge (e.g. in the same group) the size increases as we go down a group in the periodic table
Se²⁻> S²⁻ > O²
(b) Te²⁻, Cs⁺, I⁻
Te²⁻ > I- >Cs+
Te2- hast the biggest size, because of the double negative charge.
Cs+ has the smallest size since it has the most positive charge, compared to Te2- and I-.
(c) Sr²⁺, Ba²⁺, Cs⁺
Cs+ > Ba²⁺ > Sr²⁺
Cs+ has the biggest size, because its more downward (compared to Sr2+) and more to the left (compared) ot Ba2+.
Sr2+ has the smallest size because it's more upwords (compared to Cs+ and Ba2+)
