A. immigration, how is this chemistry?
The mass of hydrated salt - 2.123 g
mass of anhydrous salt - 1.861 g
mass that has been reduced is the mass of water that has been heated and lost from the compound thereby making the salt anhydrous.
therefore mass of water lost - 2.123 - 1.861 = 0.262 g
number of moles of water lost - 0.262 g / 18 g/mol = 0.0146 mol
number of moles of salt - 1.861 g / 380.6 g/mol = 0.00490 mol
molar ratio of moles of water to moles of salt
molar ratio = 0.146 mol / 0.00490 mol = 2.98 rounded off to 3
for every 1 mol of salt there are 3 moles of water
therefore empirical formula - Cu₃(PO₄)₂.3H₂O
6 sodium and 6 Bromine in 6NaBr
Answer:
for one mole of C2H6 there are 7/2 mole of O2 required. so for4. 50 moles you require 4.50 x 7/2 = 15.75 moles of O2.
Explanation:
i hope it's helpful
Answer:
Kc = 1.09x10⁻⁴
Explanation:
<em>HF = 1.62g</em>
<em>H₂O = 516g</em>
<em>F⁻ = 0.163g</em>
<em>H₃O⁺ = 0.110g</em>
<em />
To solve this question we need to find the moles of each reactant in order to solve the molar concentration of each reactan and replacing in the Kc expression. For the reaction, the Kc is:
Kc = [H₃O⁺] [F⁻] / [HF]
<em>Because Kc is defined as the ratio between concentrations of products over reactants powered to its reaction coefficient. Pure liquids as water are not taken into account in Kc expression:</em>
<em />
[H₃O⁺] = 0.110g * (1mol /19.01g) = 0.00579moles / 5.6L = 1.03x10⁻³M
[F⁻] = 0.163g * (1mol /19.0g) = 0.00858moles / 5.6L = 1.53x10⁻³M
[HF] = 1.62g * (1mol /20g) = 0.081moles / 5.6L = 0.0145M
Kc = [1.03x10⁻³M] [1.53x10⁻³M] / [0.0145M]
<h3>Kc = 1.09x10⁻⁴</h3>
