Answer:
4.03dm³
Explanation:
The reaction expression is given as:
3H₂ + N₂ → 2NH₃
Volume of hydrogen = 12dm³
AT rtp:
1 mole of gas occupies volume of 22.4dm³
x mole of hydrogen will occupy a volume of 12dm³
Number of moles of hydrogen = 
= 0.54mole
From the balanced reaction equation:
3 mole of hydrogen gas combines with 1 mole of Nitrogen gas
0.54 mole of hydrogen as will therefore combine with 
= 0.18moles of nitrogen gas
Since ;
1 mole of gas occupies a volume of 22.4dm³
0.18moles of Nitrogen gas will occupy 0.18 x 22.4 = 4.03dm³
The answer is: the mass of 6.02 x 1023 representative particles of the element.
The base SI unit for molar mass is kg/mol, but chemist more use g/mol (gram per mole).
For example, molar mas of ammonia is 17.031 g/mol.
M(NH₃) = Ar(N) + 3 · Ar(H) · g/mol.
M(NH₃) = 14.007 + 3 · 1.008 · g/mol.
M(NH₃) = 17.031 g/mol.
The molar mass (M) is the mass of a given substance (in this example ammonia) divided by the amount of substance.
Well i do think they're the same.
Answer:
(1) breaking a pencil (2) rusting of iron
Explanation:
breaking a pencil does not alter the chemical properties of the pencil, it merely breaks it into 2 while the rusting of iron is changing the properties chemically because the iron is oxidizing and reacting with the water and oxygen in the atmosphere
The rate constant of the reaction K we can get it from this formula:
K=㏑2/ t1/2 and when we have this given (missing in question):
that we have one jar is labeled t = 0 S and has 16 yellow spheres inside and the jar beside it labeled t= 10 and has 8 yellow spheres and 8 blue spheres and the yellow spheres represent the reactants A and the blue represent the products B
So when after 10 s and we were having 16 yellow spheres as reactants and becomes 8 yellow and 8 blue spheres as products so it decays to the half amount so we can consider T1/2 = 10 s
a) by substitution in K formula:
∴ K = ㏑2 / 10 = 0.069
The amount of A (the reactants) after N half lives = Ao / 2^n
b) so no.of yellow spheres after 20 s (2 half-lives) = 16/2^2 = 4
and the blue spheres = Ao - no.of yellow spheres left = 16 - 4 = 12
c) The no.of yellow spheres after 30 s (3 half-lives) = 16/2^3 = 2
and the blue spheres = 16 - 2 = 14
