This is how I got to that answer. Since we don't know how many atoms there are in a mole, we use the number 6.02 x 10^-23. Now, just plug in what you have in the equation:
<span>1.75 moles ChCl3 x (6.02 x 10 ^-23) / 1 mole = 1.0535 x 10^-22 atoms. </span>
The second one is more concentrated as they both times with the same thing but the second one (1.5) is bigger
CaCO₃ partially dissociates in water as Ca²⁺ and CO₃²⁻. The balanced equation is,
CaCO₃(s) ⇄ Ca²⁺(aq) + CO₃²⁻(aq)
Initial Y - -
Change -X +X +X
Equilibrium Y-X X X
Ksp for the CaCO₃(s) is 3.36 x 10⁻⁹ M²
Ksp = [Ca²⁺(aq)][CO₃²⁻(aq)]
3.36 x 10⁻⁹ M² = X * X
3.36 x 10⁻⁹ M² = X²
X = 5.79 x 10⁻⁵ M
Hence the solubility of CaCO₃(s) = 5.79 x 10⁻⁵ M
= 5.79 x 10⁻⁵ mol/L
Molar mass of CaCO₃ = 100 g mol⁻¹
Hence the solubility of CaCO₃ = 5.79 x 10⁻⁵ mol/L x 100 g mol⁻¹
= 5.79 x 10⁻³ g/L
Another product: CO₂
<h3>Further explanation</h3>
Given
Reaction
2C₄H₁₀ + 13O₂⇒ 8__+ 10H₂O
Required
product compound
Solution
In the combustion of hydrocarbons there can be 2 kinds of products
If there is excess Oxygen, you will get Carbon dioxide(CO₂) and water in the product
If Oxygen is low, you'll get Carbon monoxide(CO) and water
Or in other ways, we can use the principle of the law of conservation of mass which is also related to the number of atoms in the reactants and in the products
if we look at the reaction above, there are C atoms on the left (reactants), so that in the product there will also be C atoms with the same number of C atoms on the left
2C₄H₁₀ + 13O₂⇒ 8CO₂+ 10H₂O
Protons and neutrons are in the center of the atom, making up the nucleus. The charge on the proton and electron are exactly the same size but opposite. Neutrons have no charge.
