Answer: D. 19.9 g hydrogen remains.
Explanation:
To calculate the moles, we use the equation:
a) moles of 
b) moles of 
According to stoichiometry :
1 mole of require 1 mole of
Thus 0.0787 moles of 
require=
of
Thus is the limiting reagent as it limits the formation of product and acts as the excess reagent. (10.0-0.0787)= 9.92 moles of are left unreacted.
Mass of 
Thus 19.9 g of remains unreacted.
Answer:
13.06800 nanometers
Explanation:
i hope you had understand !
Answer:
Explanation:
T1 = 150°C = (150 + 273.15)K = 423.15K
T2 = 45°C = (45 + 273.15)K = 318K
V1 = 693mL = 693cm³
Applying Charle's law, the volume of a given gas is directly proportional to is temperature provided that pressure remains constant.
V = kT
V1 / T1 = V2 / T2
693 / 423.15 = V2 / 318
V2 = (693 * 318) / 423.15 = 520.79cm³
The new volume of the gas is 520.79cm³
Answer:
There will be one Al3+ ion.
There will be 3 NO3- ions
Explanation:
Dissociation equation:
Al(NO₃)₃ → Al³⁺ + 3NO₃¹⁻
When aluminium nitrate dissociate it produces one silver ion (Al³⁺) and three (NO₃¹⁻) ions.
Properties of Al(NO₃)₃:
It is inorganic compound having molecular mass 169.87 g/mol.
It is white odor less compound.
Its density is 4.35 g/mL.
Its melting and boiling points are 120°C and 440°C.
It is soluble in water.
It is sued to treat infections.
It is used in the photographic films.
It s toxic and must be handled with great care.
