Answer:
Mass = 29.23 g
Explanation:
Given data:
Volume of solution = 814.2 mL 814.2/1000 = 0.8142 L)
Molarity of solution = 0.227 M
Mass of solute in gram = ?
Solution:
Molarity = number of moles / volume in L
By putting values,
0.227 M = number of moles / 0.8142 L
Number of moles = 0.227 M × 0.8142 L
Number of moles = 0.184 mol
Mass in gram:
Mass = number of moles × molar mass
Molar mass of calcium acetate = 158.17 g/mol
Mass = 0.184 mol × 158.17 g/mol
Mass = 29.23 g
Answer:
See explanation
Explanation:
Before the advent of the wave-particle duality theory proposed by Louis de Broglie, there was a sharp distinction between mater and waves.
However, Louis de Broglie introduced the idea that mater could display wave-like properties. Erwin Schrödinger developed this idea into what is now known as the wave mechanical model of the atom.
In this model, electrons are regarded as waves. We can only determine the probability of finding the electron within certain high probability regions within the atom called orbitals.
This idea has been the longest surviving atomic model and has greatly increased our understanding of atoms.
Answer:
1.2×10² mmole of Na₂S₂O₃
Explanation:
From the question given above, the following data were obtained:
Volume = 0.6 L
Molarity = 0.2 mol/L
Mole of Na₂S₂O₃ =?
Molarity is simply defined as the mole of solute per unit litre of water. Mathematically, it is expressed as:
Molarity = mole /Volume
With the above formula, we can obtain the number of mole of Na₂S₂O₃ in the solution as illustrated below:
Volume = 0.6 L
Molarity = 0.2 mol/L
Mole of Na₂S₂O₃ =?
Molarity = mole /Volume
0.2 = Mole of Na₂S₂O₃ / 0.6
Cross multiply
Mole of Na₂S₂O₃ = 0.2 × 0.6
Mole of Na₂S₂O₃ = 0.12 mole
Finally, we shall convert 0.12 mole to millimole (mmol). This can be obtained as follow:
1 mole = 1000 mmol
Therefore,
0.12 mole = 0.12 mole × 1000 mmol / 1 mole
0.12 mole = 120 = 1.2×10² mmole
Thus, the chemist added 1.2×10² mmole of Na₂S₂O₃
This rock is balanced by roots on the ground that are very strong
Answer:
Explanation:
depending on the activity series there will probably be a single replacement reaction possibly heat or color change and the copper precipitate out of solution
