It's a because if you add them together you till get 1.40
Answer:
The concentration of a saturated solution of CuF₂ in aqueous 0.20 M NaF is 4.0×10⁻⁵ M.
Explanation:
Consider the ICE take for the solubility of the solid, CuF₂ as:
CuF₂ ⇄ Cu²⁺ + 2F⁻
At t=0 x - -
At t =equilibrium (x-s) s 2s
The expression for Solubility product for CuF₂ is:
![K_{sp}=\left [ Cu^{2+} \right ]\left [ F^- \right ]^2](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D%5Cleft%20%5B%20Cu%5E%7B2%2B%7D%20%5Cright%20%5D%5Cleft%20%5B%20F%5E-%20%5Cright%20%5D%5E2)
Given s = 7.4×10⁻³ M
So, Ksp is:
Ksp = 1.6209×10⁻⁶
Now, we have to calculate the solubility of CuF₂ in NaF.
Thus, NaF already contain 0.20 M F⁻ ions
Consider the ICE take for the solubility of the solid, CuF₂ in NaFas:
CuF₂ ⇄ Cu²⁺ + 2F⁻
At t=0 x - 0.20
At t =equilibrium (x-s') s' 0.20+2s'
The expression for Solubility product for CuF₂ is:
Solving for s', we get
<u>s' = 4.0×10⁻⁵ M</u>
<u>The concentration of a saturated solution of CuF₂ in aqueous 0.20 M NaF is 4.0×10⁻⁵ M.</u>
Molarity = number of moles of solute/liters of solution
number of moles of solute = molarity x liters of solution
Part (a): <span>30.00 ml of 0.100m Cacl2
number of moles of CaCl2 = 0.1 x 0.03 = 3x10^-3 moles
1 mole of CaCl2 contains 2 moles of chlorine, therefore 3x10^-3 moles of CaCl2 contains 6x10^-3 moles of chlorine
Part (b): </span><span>10.0 ml of 0.500m bacl2
number of moles of BaCl2 = 0.5 x 0.01 = 5x10^-3 moles
1 mole of BaCl2 contains 2 moles of chlorine, therefore 5x10^-3 moles of BaCl2 contains 10x10^-3 moles of chlorine
Part (c): </span><span>4.00 ml of 1.000m nacl
number of moles of NaCl = 1 x 0.004 = 0.004 moles
1 mole of NaCl contains 1 mole of chlorine, therefore 4x10^-3 moles of NaCl contains 4x10^-3 moles of chlorine
Part (d): </span><span>7.50 ml of 0.500m fecl3
number of moles of FeCl3 = 0.5 x 0.0075 = 3.75x10^-3 moles
1 mole of FeCl3 contains 3 moles of chlorine, therefore 3.75x10^-3 moles of FeCl3 contains 0.01125 moles of chlorine
Based on the above calculations, the correct answer is (d)</span>
It's 2, glass. Water, nitrogen, and sucrose don;t have a crystalline structure.
We can solve this without a concrete formula through dimensional analysis. This works by manipulating the units such that you end up with the unit of the final answer. Manipulate them by cancelling units that appear both in the numerator and denominator side. As a result, we must be left with the units of g. The current in A or amperes is equivalent to amount of Coulombs per second. Since this involves Coulombs, we will use the Faraday's constant which is 96,500 C/mol electron. The reaction is:
Cr³⁺(aq) + 3e⁻ --> Cr(s)
This means that for every 3 moles of electron transferred, 1 mole of Chromium metal is plated. The molar mass of Cr: 52 g/mol. The solution is as follows:
Mass of Chromium metal = (8 C/s)(60 s/1 min)(160 min)(1 mol e⁻/96,500 C)(1 mol Cr/3 mol e)(52 g/mol)
<em>Mass of Chromium metal = 13.79 g</em>
