The molarity of the solution is 0.1625M.
<h3>What do you meant by Molarity</h3>
Molarity is the amount of a substance in a certain volume of solution. Molarity is defined as the moles of solute per litres of a solution .
Molarity → no of moles / Volume (L)
SI unit of Molarity is M or mol/ L
We have given here the mass of solution is 3.10×10²g .
molality of the solution is 0.125m
Molality → no of moles / mass in kg
→ 0.125×3.10×10²/ 1000
→ no.of moles = 0.0162
For molarity we can assume volume as 1000 ml .
Molarity = 0.0162×1000/ 100
Molarity →0.162 M.
So, the molarity of solution will be 0.162M.
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You will need a periodic table to help you answer this problem. The atomic numbers are arrange from lowest to highest in the periodic table. You can locate element number 55 to be Cesium with an atomic weight of 132.905 amu. So, you start from element 56. The following elements are:
56 Barium 137.328 amu
57 Lanthanium 138.905 amu
58 Cerium 140.116 amu
59 <span>Praseodymium 140.908 amu
60 Neodymium 144.243 amu
Neodymium is already greater than 144 amu. Therefore, these elements only include Barium, Lanthanium, Cerium and Praseodymium.</span>
Answer:
3.51× 10²³ formula units
Explanation:
Given data:
Mass of CaO = 32.7 g
Number of formula units = ?
Solution:
First of all we will calculate the number of moles.
Number of moles = mass/molar mass
Number of moles = 32.7 g/ 56.1 g/mol
Number of moles = 0.583 mol
Number of formula units:
1 mole = 6.022 × 10²³ formula units
0.583 mol × 6.022 × 10²³ formula units / 1 mol
3.51× 10²³ formula units
The number 6.022 × 10²³ is called Avogadro number.
True.
A catalyst is a substancr that increases the rate of a chemical reaction without itself undergoing any permanent chemical change.
A chemical formula identifies each constituent element by its chemical symbol and indicates the proportionate number of atoms of each element.
<em>For example, the empirical formula of ethanol may be written C2H6O because the molecules of ethanol all contain two carbon atoms, six hydrogen atoms, and one oxygen atom.</em>
