the equation is p1 x v1 divided by T1 = p1 x v2 = T2 but since the pressure is kept constant you do not even need it so the equation would now be v1 divided by t1 = v2 divided by t2
2135 cm3 divided by 127 degrees celcius = x divided by 206
answer: 3460 cm3
First we need to find the number of moles of both K and O reacted
K - 0.779 g / 39 g/mol
= 0.02 mol
the mass of O₂ reacted = 1.417 g - 0.779 g = 0.638 g
O₂ moles = 0.638 g / 32 g/mol
= 0.02 mol
the number of both K and O₂ moles reacted are equal
therefore stoichiometry of K to O₂ reacted are 1:1
then the formula of potassium superoxide is KO₂
Answer:
= 82%
Explanation:
Percentage purity is calculated by the formula;
% purity = (mass of pure chemical/total mass of sample) × 100
In this case;
1 mole of Ca(NO3)2 = 164 g
but; 164 g of Ca(NO3)2 = 40 g Ca
Therefore; mass of Ca(NO3)2 = 164 /40
= 4.1 g
Thus;
% purity of Ca(NO3)2 = (Mass of Ca(NO3)2/ mass of the sample)× 100
= (4.1 g/ 5 g) × 100
= 82%
Functional groups create reactive sites in molecules.
The polar part of a molecule that can hydrogen bond to water is said be hydrophilic.
Pi (π) bonds create active sites and will react with electron-deficient species.
A electronegative heteroatom like nitrogen, oxygen, or a halogen makes a carbon atom electrophilic.
(carbon will have less electronic density, which is attracted by the more electronegative heretoatoms, and it will tend to attract electron rich chemical species, and in this situation we say that the carbon atom is electrophilic).
The nonpolar part of a molecule that is not attracted to water is said to be hydrophobic.
A lone pair on a heteroatom makes it basic and nucleophilic.
(the heteroatom with the lone pair will tend to attract electron poor chemical species, and in this situation we say that the heteroatom is nucleophilic).
Answer:
Mass released = 8.6 g
Explanation:
Given data:
Initial number of moles nitrogen= 0.950 mol
Initial volume = 25.5 L
Final mass of nitrogen released = ?
Final volume = 17.3 L
Solution:
Formula:
V₁/n₁ = V₂/n₂
25.5 L / 0.950 mol = 17.3 L/n₂
n₂ = 17.3 L× 0.950 mol/25.5 L
n₂ = 16.435 L.mol /25.5 L
n₂ = 0.644 mol
Initial mass of nitrogen:
Mass = number of moles × molar mass
Mass = 0.950 mol × 28 g/mol
Mass = 26.6 g
Final mass of nitrogen:
Mass = number of moles × molar mass
Mass = 0.644 mol × 28 g/mol
Mass = 18.0 g
Mass released = initial mass - final mass
Mass released = 26.6 g - 18.0 g
Mass released = 8.6 g
