Believe it or not it’s a pure substance, two hydrogen atoms combine with one oxygen atom chemically, forming a new substance that has properties different from hydrogen alone or oxygen alone.
Answer:
C = 107.97 mol/L
Explanation:
Given data:
Volume of solution = 1.38 mL (1.38 mL× 1 L /1000 mL = 0.00138 L)
Mass of ammonium sulfite = 17.36 g
Concentration of solution =?
Solution:
We will calculate the number of moles of ammonium sulfite.
Number of moles = mass/molar mass
Number of moles = 17.36 g / 116.15 g/mol
Number of moles = 0.149 mol
Concentration:
C = n/V
C = concentration
n = number of moles of solute
v = volume in L
C = 0.149 mol / 0.00138 L
C = 107.97 mol/L
The SI unit of specific heat is J per gram per degree
Celsius. Thus it follows that specific heat could be calculated in this way:
Specific Heat = Energy / (mass x change in temperature)
Thus,
Specific Heat = 3.912 cal / (9.84 oz x (191.2 ˚F – 73.2 ˚F))
Specific Heat = 3.369 x 10^-3 cal/oz-˚F
Yogurt, kefir, kombucha, sauerkraut, pickles, miso, tempeh, kimchi, sourdough bread and some cheeses.
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:
![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)

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>