Its A Natural resources are found in abundance in the US.
Answer: 899 g
Explanation:
Since the mass of a mole of diatomic chlorine is about 70.9 g/mol, this means that 875 grams is about 875/70.9 = 12.3 mol.
Hence, we know that chlorine is the limiting reactant, and that about 12.3 moles of chlorine will be consumed.
This means that 12.3(2)=24.6 grams of HCl will be produced, which will have a mass of about (24.6)(36.46)=899 g
Answer:
S = 0.788 g/L
Explanation:
The solubility product (Kps) is an equilibrium solubization constant, which can be calculated by the equation:
![Kps = \frac{[product]^x}{[reagent]^y}](https://tex.z-dn.net/?f=Kps%20%3D%20%5Cfrac%7B%5Bproduct%5D%5Ex%7D%7B%5Breagent%5D%5Ey%7D)
Where x and y are the stoichiometric coefficients of the product and the reagent, respectively. Because of the aggregation form, the concentration of solids is always equal to 1 for use in this equation.
Analyzing the equation, we see that for 1 mol of
is necessary 2 mols of
, so if we call "x" the molar concentration of
, for
we will have 2x, so:
![Kps = [Fe^{+2}].[F^-]^2\\\\2.36x10^{-6} = x(2x)^2\\\\2.36x10^{-6} = 4x^3\\\\x^3 = 5.9x10^{-7}\\\\x = \sqrt[3]{5.9x10^{-7}} \\\\x = 8.4x10^{-3} mol/L](https://tex.z-dn.net/?f=Kps%20%3D%20%5BFe%5E%7B%2B2%7D%5D.%5BF%5E-%5D%5E2%5C%5C%5C%5C2.36x10%5E%7B-6%7D%20%3D%20x%282x%29%5E2%5C%5C%5C%5C2.36x10%5E%7B-6%7D%20%3D%204x%5E3%5C%5C%5C%5Cx%5E3%20%3D%205.9x10%5E%7B-7%7D%5C%5C%5C%5Cx%20%3D%20%5Csqrt%5B3%5D%7B5.9x10%5E%7B-7%7D%7D%20%5C%5C%5C%5Cx%20%3D%208.4x10%5E%7B-3%7D%20mol%2FL)
So, to calculate the solubility (S) of FeF2, which is in g/L, we multiply this concentration by the molar mass of FeF2, which is:
Fe = 55.8 g/mol
F = 19 g/mol
FeF2 = Fe + 2xF = 55.8 + 2x19 = 93.8 g/mol
So,
[tex]S = 8.4x10^{-3}x93.8
S = 0.788 g/L
Answer:
0.4444 g/cm³ ≅ 0.44 g/cm³ (2 significant figures).
Explanation:
<em>d = m/V,</em>
where, d is the density of the material (g/cm³).
m is the mass of the material (m = 28 g).
V is the volume of the material (V = 63.0 cm³).
<em>∴ d = m/V </em>= (28 g)/(63.0 cm³) = <em>0.4444 g/cm³ ≅ 0.44 g/cm³ (2 significant figures).</em>
C - straight line
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