Answer: IUPAC NOMENCLATURE
Explanation:
IUPAC stands for International Union of Pure and Applied Chemistry. They devised a systematic method for naming compounds in order to create a uniform global unambiguous system of nomenclature hence making it easier for researchers to share information more freely without the hindrance of reporting the same compound using different names in different parts of the world thus creating confusion in chemical literature.
Answer:
OH−(aq), and H+(aq)
Explanation:
Redox reactions may occur in acidic or basic environments. Usually, if a reaction occurs in an acidic environment, hydrogen ions are shown as being part of the reaction system. For instance, in the reduction of the permanganate ion;
MnO4^-(aq) + 8H^+(aq) +5e-------> Mn^2+(aq) + 4H2O(l)
The appearance of hydrogen ion in the reaction equation implies that the process takes place under acidic reaction conditions.
For reactions that take place under basic conditions, the hydroxide ion is part of the reaction equation.
Hence hydrogen ion and hydroxide ion are included in redox reaction half equations depending on the conditions of the reaction whether acidic or basic.
Hey there!:
8) ΔTb = i*Kb*m
m is molality
Since same number of mol is added to same amount of water in both cases
m will be same for both
is 1 for glucose since it is covalent compound
is 4 of Al(NO3)3 as it breaks into 1 Al₃⁺ and 3 NO₃⁻
So, ΔTb will be 4 times in aluminum nitrate case
So, boiling point will change by 4ºC
9) use Q = m* L
L = heat of vaporization so:
T1=T2=100ºC
5.40 * 1000 => 5400 cal/g
Q = 5400 / 540
Q = 10 grams
Hope that thlps!
If 1000 ml (1 L) of CH₃COOH contain 1.25 mol
let 250 ml of CH₃COOH contain x
⇒ x =
= 0.3125 mol
∴ moles of CH₃COOH in 250ml is 0.3125 mol
Now, Mass = mole × molar mass
= 0.3125 mol × [(12 × 2)+(16 × 2)+(1 × 4)] g/mol
= 18.75 g
∴ Mass of CH₃COOH present in a 250 mL cup of 1.25 mol/L solution of vinegar is <span>18.75 g</span>
Answer:- 544.5 mL of water need to be added.
Solution:- It is a dilution problem. The equation used for solving this type of problems is:

where,
is initial molarity and
is the molarity after dilution. Similarly,
is the volume before dilution and
is the volume after dilution.
Let's plug in the values in the equation:



Volume of water added = 907.5mL - 363mL = 544.5 mL
So, 544.5 mL of water are need to be added to the original solution for dilution.