First let us compute for the number of moles of butane
(molar mass = 58.12 g/mol)
number of moles = 145 g / (58.12 g/mol) = 2.49 mol
<span>We use the ideal
gas equation to calculate the volume:</span>
<span> V = n R T / P</span>
V = 2.49 mol * 62.36367 L torr / mol K * 308.15 K / 745
torr
<span>V = 64.35 L</span>
Answer:
1x10⁻¹²
Explanation:
- Cu₂S(s) ⇌ 2Cu⁺(aq) + S²⁻(aq)
At equilibrium:
The equilibrium constant for the the reaction can be written as:
[Cu⁺] is squared because it has a stoichiometric coefficient of 2 in the reaction. <em>Cu₂S has no effect on the constant because it is a solid</em>.
Now we can <u>calculate the equilibrium constant</u>:
- Keq = (1.0x10⁻⁵)² * 1.0x10⁻² = 1x10⁻¹²
Group 2A of the Periodic table contains the Alkaline Earth matals. Those elements are: Beryllium ( Be ), Magnesium ( Mg ), Calcium ( Ca ), Strontium ( Sr ), Barium ( Ba ) and Radium ( Ra ). Beryllium is the lightest member of the group 2A. Ionic radius increases moving down the group ans decreases from left to right across the period of elements. In the group 2A ionic radius increases from Beryllium ( 59 pm - picometers ) to Radium ( 163 pm ). Answer: Beryllium<span> has the lowers ionic radius. </span>
Explanation:
Alkali metals forms the group 1 of the modern periodic table. They include Lithium, sodium, potassium, rubidium, cesium, and francium.
The general electronic configuration of the alkali metals -
. They have 1 valence electron.
Ionization energy is the minimum amount of energy which is required to knock out the loosely bound valence electron from the isolated gaseous atom.
<u>Thus, removal of the valence electron leads to the alkali metals having noble gas electronic configuration and which is stable. Thus, the removal of the second electron is very tough and thus, there a large increase between the first and second ionization energies of the alkali metals.</u>