Answer:
Mass = 1.33 g
Explanation:
Given data:
Mass of argon required = ?
Volume of bulb = 0.745 L
Temperature and pressure = standard
Solution:
We will calculate the number of moles of argon first.
Formula:
PV = nRT
R = general gas constant = 0.0821 atm.L/mol.K
By putting values,
1 atm ×0.745 L = n × 0.0821 atm.L/mol.K× 273.15 K
0.745 atm. L = n × 22.43 atm.L/mol
n = 0.745 atm. L / 22.43 atm.L/mol
n = 0.0332 mol
Mass of argon:
Mass = number of moles × molar mass
Mass = 0.0332 mol × 39.95 g/mol
Mass = 1.33 g
Answer:
28 g CO
Explanation:
First convert grams to moles.
1 mole C = 12.011 g (I'm just going to round to 12 for the sake of this problem)
12 g C • 
= 1 mol C
1 mol O = 15.996 g (I'm just going to round to 16)
16 g O • 
= 1 mol O
So the unbalanced equation is:
-> 
(the oxygen has a 2 subscript because it is part of HONClBrIF meaning when not in a compound these elements appear in pairs - called diatomic elements)
The balanced equation is:
-> 
However, carbon is the limiting reactant in this equation and two moles cannot react because only 12 g (1 mole) are present. Therefore, use the equation
-> .
1 mole of CO is formed, therefore 12 g + 16 g = 28 g CO.
Group 1 contains metals while group 18 contains noble gases.
So group 1 is different from group 18 as they both contains different types of atoms as group 1 contains metals while group 18 contains noble gases.
There is 1 valence shell electron in group 1 , they are highly reactive while valence shell of group 18 is fully filled and they are least reactive .
Answer:
hi
Explanation:
i am smart but i need this app cuz some are realy hard
