The molar mass of Sb2S3 is approximately equal to 339.7 g/mol. We calculate the number of moles of Sb2S3 by dividing the given mass by the molar mass.
n = 23.5 g / (339.7 g/mol)
n = 0.0692 mols
To calculate for the number of formula units, we multiply the number of mols by the Avogadro's number,
number of formula units = (0.0692 mols)(6.022 x 10^3)
= 4.167 x 10^22 formula units
<span>The alkali metals and hydrogen are reactive because they have only one electron to give in order to complete their valence shell. It is easier to give that one electron so when given the opportunity they will. This means they will react with anything polar or willing to take an electron.</span>
Answer:
The mass of 3.491 × 10¹⁹ molecules of Cl₂ of Cl₂ is 4.11 × 10⁻³ grams
Explanation:
The number of particles in one mole of a substance id=s given by the Avogadro's number which is approximately 6.023 × 10²³ particles
Therefore, we have;
One mole of Cl₂ gas, which is a compound, contains 6.023 × 10²³ individual molecules of Cl₂
3.491 × 10¹⁹ molecules of Cl₂ is equivalent to (3.491 × 10¹⁹)/(6.023 × 10²³) = 5.796 × 10⁻⁵ moles of Cl₂
The mass of one mole of Cl₂ = 70.906 g/mol
The mass of 5.796 × 10⁻⁵ moles of Cl₂ = 70.906 × 5.796 × 10^(-5) = 4.11 × 10⁻³ grams
Therefore;
The mass of 3.491 × 10¹⁹ molecules of Cl₂ of Cl₂ = 4.11 × 10⁻³ grams.
1- change of state (boiling)
2-increase in acidity
3-methylxanthines (caffeine) theobromine and theophylline
4- lemon juice
Answer:
5 L
Explanation:
We'll begin by calculating the molarity of the CaCl₂ solution. This can be obtained as follow:
Mole of CaCl₂ = 0.5 mole
Volume = 2 L
Molarity =?
Molarity = mole /Volume
Molarity = 0.5 / 2
Molarity = 0.25 M
Finally, we shall determine the volume of the diluted solution. This can be obtained as follow:
Molarity of stock solution (M₁) = 0.25 M
Volume of stock solution (V₁) = 2 L
Molarity of diluted solution (M₂) = 0.1 M
Volume of diluted solution (V₂) =?
M₁V₁ = M₂V₂
0.25 × 2 = 0.1 × V₂
0.5 = 0.1 × V₂
Divide both side by 0.1
V₂ = 0.5 / 0.1
V₂ = 5 L
Thus the volume of the diluted solution is 5 L
