Answer:
16.5 dm³
Explanation:
Data Given:
no. moles of O₂ = 0.735 moles
volume of O₂ = ?
Solution:
Now
we have to find volume of O₂ gas
Formula used for this purpose
No. of moles = Volume / molar volume
where
molar volume at STP for Oxygen (O₂) = 22.4 dm³/mol
No. of moles O₂ = Volume of O₂ / 22.4 dm³/mol . . . . . .(1)
Put values in equation 1
0.735 = Volume of O₂ / 22.4 dm³/ mol
rearrange above equation
Volume of O₂ = 0.735 x 22.4 dm³/ mol
Volume of O₂ = 16.5 dm³
So,
the volume of O₂ at STP is 16.5 dm³
Answer:
The concentration of the solution is 5.8168 × 
mol.
Explanation:
Here, we want to calculate the concentration of the solution.
The unit of this is mol/dm^3
So the first thing to do here is to calculate the number of moles of the solute present, which is the number of moles of AlCO3
The number of moles = mass/molar mass
molar mass of AlCO3 = 27 + 12 + 3(16) = 27 + 12 + 48 = 87g/mol
Number of moles = 33.4/87 = 0.384 moles
This 0.384 moles is present in 660 L
x moles will be present in 1 dm^3
Recall 1 dm^3 = 1L
x * 660 = 0.384 * 1
x = 0.384/660 = 0.00058168 = 5.8168 * 10^-4 mol/dm^3
Answer:
4.12 mol
Explanation:
Given data:
Moles of LiOH required = ?
Volume of solution = 4.2 L
Molarity of solution = 0.98 M
Solution:
Molarity is used to describe the concentration of solution. It tells how many moles are dissolve in per litter of solution.
Formula:
Molarity = number of moles of solute / L of solution
we will calculate the moles from above given formula.
0.98 M = number of moles / 4.2 L
0.98 M × 4.2 L = number of moles
Number of moles = 0.98 M × 4.2 L
Number of moles = 4.12 mol (M = mol/L)
Answer: Electrovalent or Ionic Compounds
Explanation:
Electrovalent Compounds Form bonds that are characterised by transfer of electrons from metallic atoms to non-metal licenses atoms during a chemical reaction.
The metallic atom after donating their valence electrons, become positively charged, while the non-metal license atoms becomes negatively charged after acquiring extra electrons.
A typical example of electrovalent compounds can be found between the association of Group 1(Alkali Metals) elements and the Group 7(Halogen Family) elements.
