M=mol/L, 0.323M=mol/0.01325. Rework to solve for mol and bam! (I.e. times the two numbers)
Answer:
pH = 12.7
Explanation:
First, we have to calculate the [Ca²⁺] in a solution of about 250 ppm CaCO₃.
Now, let's consider the dissolution of Ca(OH)₂ in water.
Ca(OH)₂(s) ⇄ Ca²⁺(aq) + 2 OH⁻(aq)
The solubility product Ksp is:
Ksp = [Ca²⁺] × [OH⁻]²
[OH⁻] = √(Ksp/[Ca²⁺]) = √(6.5 × 10⁻⁶/2.5 × 10⁻³) = 5.1 × 10⁻² M
Finally, we can calculate pOH and pH.
pOH = -log [OH⁻] = -log (5.1 × 10⁻²) = 1.3
pH + pOH = 14 ⇒ pH = 14 - pOH = 14 - 1.3 = 12.7
B) both poles
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<span>100 g of KClO3 @ 122.55 g/mol = 0.816 moles of KClO3
by the reaction
2 KClO3 --> 2 KCl & 3 O2
0.816 moles of KClO3 @ 3 moles O2 / 2 moles KClO3 = 1.224 moles of O2 can be made
using molar mass
1.224 moles of O2 @ 32.0 g/mol =
39.2 grams of O2 can be made</span>
Answer:
490.83 oK
Explanation:
First of all, we better agree on the meaning of the word <em>adiabatically</em>. It means without the loss or gain of heat. So nothing is given up to or taken from the environment.
This also assumes that no change has occurred in the pressure.
T1 = 310 oK
T2 = ?
V1 = 12 L
V2 = 19 L
T1/V1 = T2/V2
310 / 12 = x/19 Multiply both sides by 19
310*19 / 12 = T2 Multiply 310 * 19 on the left.
5890 / 12 = T2 Divide by 12
490.83 = T2