Can anyone heelp me pls

Give a reason why hydroxide ions to an acid solution increase in ph

yanalaym [24]

Answer:

creating new water molecules and lowering the amount of free H+ in the system, thereby raising the overall pH.

Explanation:

5 0

3 years ago

Radioactivity is a random Process, Why?

Pie

Explanation:

Radioactive decay is the set of various processes by which unstable atomic nuclei (nuclides) emit subatomic particles (radiation). Decay is said to occur in the parent nucleus and produces a daughter nucleus. This is a random process, i.e. it is impossible to predict the decay of individual atoms.

HOPE THSI HELPS HAVE A NICE DAY!

4 0

3 years ago

Can someone pls give me the answer to this question?

Svetllana [295]

Answer:

D

Explanation:

5 0

3 years ago

What is the symbol for the ion that contains 12 protons, 10 electrons, and 12 neutrons?

elena-s [515]

Magnesium ion or Mg 2+

4 0

3 years ago

The Handbook of Chemistry and Physics gives solubilities of the following compounds in grams per 100 mL water. Because these com

Elanso [62]

Answer:

(a) $Ksp=4.50x10^{-7}$

(b) $Ksp=1.55x10^{-6}$

(c) $Ksp=2.27x10^{-12}$

(d) $Ksp=1.05x10^{-22}$

Explanation:

Hello,

In this case, given the solubility of each salt, we can compute their molar solubilities by using the molar masses. Afterwards, by using the mole ratio between ions, we can compute the concentration of each dissolved and therefore the solubility product:

(a) $BaSeO_4(s)\rightleftharpoons Ba^{2+}(aq)+SeO_4^{2-}(aq)$

$Molar\ solubility=\frac{0.0188g}{100mL} *\frac{1mol}{280.3g}*\frac{1000mL}{1L}=6.7x10^{-4}\frac{mol}{L}$

In such a way, as barium and selenate ions are in 1:1 molar ratio, they have the same concentration, for which the solubility product turns out:

$Ksp=[Ba^{2+}][SeO_4^{2-}]=(6.7x10^{-4}\frac{mol}{L} )^2\\\\Ksp=4.50x10^{-7}$

(B) $Ba(BrO_3)_2(s)\rightleftharpoons Ba^{2+}(aq)+2BrO_3^{-}(aq)$

$Molar\ solubility=\frac{0.30g}{100mL} *\frac{1mol}{411.15g}*\frac{1000mL}{1L}=7.30x10^{-3}\frac{mol}{L}$

In such a way, as barium and bromate ions are in 1:2 molar ratio, bromate ions have twice the concentration of barium ions, for which the solubility product turns out:

$Ksp=[Ba^{2+}][BrO_3^-]^2=(7.30x10^{-3}\frac{mol}{L})(3.65x10^{-3}\frac{mol}{L})^2\\\\Ksp=1.55x10^{-6}$

(C) $NH_4MgAsO_4(s)\rightleftharpoons NH_4^+(aq)+Mg^{2+}(aq)+AsO_4^{3-}(aq)$

$Molar\ solubility=\frac{0.038g}{100mL} *\frac{1mol}{289.35g}*\frac{1000mL}{1L}=1.31x10^{-4}\frac{mol}{L}$

In such a way, as ammonium, magnesium and arsenate ions are in 1:1:1 molar ratio, they have the same concentrations, for which the solubility product turns out:

$Ksp=[NH_4^+][Mg^{2+}][AsO_4^{3-}]^2=(1.31x10^{-4}\frac{mol}{L})^3\\\\Ksp=2.27x10^{-12}$

(D) $La_2(MoOs)_3(s)\rightleftharpoons 2La^{3+}(aq)+3MoOs^{2-}(aq)$

$Molar\ solubility=\frac{0.00179g}{100mL} *\frac{1mol}{1136.38g}*\frac{1000mL}{1L}=1.58x10^{-5}\frac{mol}{L}$

In such a way, as the involved ions are in 2:3 molar ratio, La ion is twice the molar solubility and MoOs ion is three times it, for which the solubility product turns out:

$Ksp=[La^{3+}]^2[MoOs^{-2}]^3=(2*1.58x10^{-5}\frac{mol}{L})^2(3*1.58x10^{-5}\frac{mol}{L})^3\\\\Ksp=1.05x10^{-22}$

Best regards.

7 0

4 years ago