Group I and Group II have the same number of outermost electrons as you go down each group but the shells increase therefore as you go downward it becomes much easier to remove electrons because of its wide radius however group 7 and 6 have seven and six electrons in their outermost shell respectively. Therefore down the group it is much easier to attract electrons and across the period it is much harder because the number of electrons on the outermost shell increase
When The rate of effusion is inversely proportional to the √molar mass of the substance.
and we have R(He) = 1L / 4.5 min so,
R(He)/R(Cl2) = (molar mass of Cl2/ molar mass of He)^0.5
and when we have the molar mass of Cl2 = 70.9 & the molar mass of He = 4
so by substitution:
(1L/4.5 min)/ R(Cl2) = (70.9 / 4)^0.5
(1L/4.5 min) / R(Cl2) = 4.21
∴R(Cl2) = (1L/4.5 min) / 4.21 = 1L/ (4.5*4.21)min = 1 L / 18.945 min
∴Cl2 will take 18.945 min for 1 L to effuse under identical conditions
Use Socratic or quiz let it help
Answer: 0.4M
Explanation:
Given that,
Amount of moles of NaOH (n) = ?
Mass of NaOH in grams = 40.0g
For molar mass of NaOH, use the atomic masses: Na = 23g; O = 16g; H = 1g
NaOH = (23g + 16g + 1g)
= 40g/mol
Since, n = mass in grams / molar mass
n = 40.0g / 40.0g/mol
n = 1 mole
Volume of NaOH solution (v) = 2.5 L
Concentration of NaOH solution (c) = ?
Since concentration (c) is obtained by dividing the amount of solute dissolved by the volume of solvent, hence
c = n / v
c = 1 mole / 2.5 L
c = 0.4 mol/L (Concentration in mol/L is the same as Molarity, M)
Thus, the concentration of a solution of a 40.0 g of NaOH in 2.5 L of solution is 0.4 mol/L or 0.4M
Answer: 8 moles
Explanation:
C3H8(g) + 5O2(g) ---> 3CO2(g) + 4H2O(g)
The balanced equation above shows the combustion of propane, C3H8 (for x is 3; y is 8) in oxygen, O2 to yield carbon dioxide, CO2 and water, H2O
The mole ratio on combustion are as follows:
5 moles of O2 react to produce 3 moles of CO2
Z mole of O2 react to produce 4.8 moles of CO2
To get the value of Z, cross multiply
4.8 moles x 5 mole = 3 mole x Z
24 = 3Z
Divide both sides by 3
24/3 = 3Z/3
8 = Z
Thus, in the combustion of propane, 8 moles of oxygen (O2) are needed to react to produce 4.8 moles of carbon dioxide (CO2)
