The net equations are obtained from the double displacement of the cations and anions, then balance.
NH3(aq) + HC2H3O2 (aq) = NH4+(aq) + C2H3O2-(aq<span>)
</span><span>H+(aq) + C2H3O2-(aq) + NH3(aq) -> NH4+(aq) + C2H3O2-(aq)</span><span>
</span><span>2NaOH(aq) + H2SO4 (aq) = Na2SO4 (s)+ 2H2O (aq)
</span>H2S (aq) + Ba(OH)2 (aq) = BaS (s)+ 2H2O (aq)
Explanation:
Calcium is the element that has 2 valence electrons.
Answer:
a) yes, it was an hydrate
b) the number of waters of hydration, x = 6
Explanation:
a) yes it was an hydrate because the mass decreased after the process of dehydration which means removal of water thus some water molecules were present in the sample.
b) NiCl2. xH2O
mass if dehydrated NiCl2 = 2.3921 grams
mass of water in the hydrated sample = mass of hydrated - mass of dehydrated = 4.3872 - 2.3921 = 1.9951 g which represent the mass of water that was present in the hydrated sample.
NiCl2.xH2O
mole of dehydrated NiCl2 = m/Mm = 2.3921/129.5994 = 0.01846 mole
mole of water = m/Mm = 1.9951/18.02 = 0.11072 mole
Divide both by the smallest number of mole (which is for NiCl2) to find the coefficient of each
for NiCl2 = 0.01846/0.01846 = 1
for H2O = 0.11072/0.01846 = 5.9976 = 6
thus the hydrated sample was NiCl2. 6H2O
Grinding solid crystals increase the rate of dissolving for a solid solute in water because smaller crystals have more surface area. The solubility of a substance depends on the physical and chemical properties of the solute and solvent as well as the temperature, pressure and the pH of the solution. For example increase in temperature increases the rate at which a solute dissolves in a solvent.
Answer:
11.4
Explanation:
Step 1: Given data
- Concentration of the base (Cb): 0.300 M
- Basic dissociation constant (Kb): 1.8 × 10⁻⁵
Step 2: Write the dissociation equation
NH₃(aq) + H₂O(l) ⇄ NH₄⁺(aq) + OH⁻(aq)
Step 3: Calculate the concentration of OH⁻
We will use the following expression.
![[OH^{-} ]=\sqrt{Kb \times Cb } = \sqrt{1.8 \times 10^{-5} \times 0.300 } = 2.3 \times 10^{-3} M](https://tex.z-dn.net/?f=%5BOH%5E%7B-%7D%20%5D%3D%5Csqrt%7BKb%20%5Ctimes%20Cb%20%7D%20%3D%20%5Csqrt%7B1.8%20%20%5Ctimes%2010%5E%7B-5%7D%20%5Ctimes%200.300%20%7D%20%3D%202.3%20%5Ctimes%2010%5E%7B-3%7D%20M)
Step 4: Calculate the pOH
We will use the following expression.
![pOH =-log[OH^{-} ]= -log(2.3 \times 10^{-3} M) = 2.6](https://tex.z-dn.net/?f=pOH%20%3D-log%5BOH%5E%7B-%7D%20%5D%3D%20-log%282.3%20%5Ctimes%2010%5E%7B-3%7D%20M%29%20%3D%202.6)
Step 5: Calculate the pH
We will use the following expression.
