Ca(s)+ H2O (l)----->H2 +Ca(oH)2
Ca(s)+2H2O(l)----->H2(g)+Ca(oH)2
I hope this helps
sorry if it's wrong
Answer:
0.5M
Explanation:
The equation for molarity is:
- M =
; where the "M" stands for molarity, the "mol" stands for moles of solute and the "liters" means the volume in liters of solution.
We are given that there are:
- 1.80 moles of NaCl (the moles of solute)
- 3.60 Liters of solution (the volume in liters of solution)
Now we just plug those numbers into the formula and get our answer:
- M=
= 0.5M
After doing the math and dividing the moles of solute by the liters of solution, we get that the molarity of the solution is 0.5M.
Answer:
Ksp = [ Cu+² ] [ OH-] ²
molar mass Cu(oH )2 ==> M= 63.546 (1) + 16 (2) + 1 (2) = 97.546 g/mol
Ksp = [ Cu+² ] [ OH-] ²
Ksp [ cu (OH)2 ] = 2.2 × 10-²⁰
|__________|___<u>Cu</u><u>+</u><u>²</u><u> </u>__|_<u>2</u><u>OH</u><u>-</u>____|
|<u>Initial concentration(M</u>)|___<u>0</u>__|_<u>0</u>______|
<u>|Change in concentration(M)</u>|_<u>+S</u><u> </u>|__<u>+2S</u>__|
|<u>Equilibrium concentration(M)|</u><u>_S</u><u> </u><u>_</u><u>|</u><u>2S___</u><u>|</u>
Ksp = [ Cu+² ] [ OH-] ²
2.2 ×10-²⁰ = (S)(2S)²= 4S³
![s = \sqrt[3]{ \frac{2.2 \times {10}^{ - 20} }{4} } = 1.8 \times {10}^{ - 7}](https://tex.z-dn.net/?f=s%20%3D%20%20%5Csqrt%5B3%5D%7B%20%5Cfrac%7B2.2%20%5Ctimes%20%20%7B10%7D%5E%7B%20-%2020%7D%20%7D%7B4%7D%20%7D%20%20%3D%201.8%20%5Ctimes%20%20%7B10%7D%5E%7B%20-%207%7D%20)
S = 1.8 × 10-⁷ M
The molar solubility of Cu(OH)2 is 1.8 × 10-⁷ M
Solubility of Cu (OH)2 =

<h3>
Solubility of Cu (OH)2 = 1.75428 × 10 -⁵ g/ L</h3>
I hope I helped you^_^
Answer:
density=1.43 g/L
Explanation:
Since the density formula is density = mass / volume, we need to find out the mass of the gas and the volume is that of the container.
The mass of the gas is 130.0318 g-129.6375 g=0.3943 g
The gas volume is 276mL*(1L/1000mL) 0.276 L
density = mass / volume=0.3943g/0.276L
density =1.43g/L