Molar C6H12O6 = 180.15 g/mol
180.15 g -------------- 6.02x10²³ molecules
?? g ------------------- 3.5x10²² molecules
( 3.5x10²²) x 180.15 / 6.02x10²³ =
=> 10.47 g
Answer:
The reactivity of Group 7 elements decreases down the group. ... The electrons in the outer shell move further away from the nucleus as we go down the group and the attraction force between the electrons and the nucleus become weaker and weaker. This weaker attraction in the larger atoms makes it harder to gain electron.
The balanced chemical reaction is:
Na2SO4 + Ba(OH)2 = 2NaOH +BaSO4
We are given the amount of the reactants to be used. These values will be the starting point of our calculations.
0.0820 mol/L Ba(OH)2 (2.27 L solution) = 0.18614 mol Ba(OH)2
0.0664 mol/L Na2SO4 (3.06 L solution) = 0.20318 mol Na2SO4
The limiting reactant is the Ba(OH)2. The amount for this compound will be used.
0.18614 mol Ba(OH)2 (1 mol BaSO4 / 1 mol Ba(OH)2 ) (233.43 g BaSO4 / 1 mol BaSO4)= 43.4507 g BaSO4
Answer:
the partial pressure of the O2 is 0.167 atm
Explanation:
The computation of the partial pressure of the O2 is shown below:
As we know that
P = P_N2 + P_O2 + P_CO2
P_O2 = P - P_N2 - P_CO2
= (1.007 - 0.79 - 0.05)
= 0.167 atm
Hence, the partial pressure of the O2 is 0.167 atm
we simply applied the above formula
Let's assume that change of pressure of PCl₅ is X
According to the ICE table,
PCl₅(g) ⇄ PCl₃(g) + Cl₂(g)
Initial Pressure 0.123 - -
Change -X +X +X
Equilibrium 0.123 - X X X
Kp = Ppcl₃(g) x Pcl₂(g) / Ppcl₅(g)
by substitution,
0.0121 = X × X / (0.123 - X)
0.0121 × (0.123 - X) = X²
1.4883 × 10⁻³ - 0.0121X = X²
X² + 0.0121X - 1.4883 × 10⁻³ = 0
X₁ = 0.03735 or X₂ = -0.03865
X cannot be a negative value since X is a pressure.
Hence,
X = 0.03735 atm
Hence, partial pressure of PCl₃ is 0.03735 atm
