a. double replacement
b. AB + CD ⇒ AD +CB
c and d
H₃PO₄ : Phosphoric acid
KOH : Potassium hydroxide
K₃PO₄ : Tripotassium phosphate
H₂O : water
<h3>Further explanation</h3>
Given
Reaction
H₃PO₄ + 3 KOH → K₃PO₄ + 3 H₂O
Required
Type of reaction
General equation
Name of reactants and product
Solution
Double replacement reaction : Cations and anions of different compounds switch places
It can also be said as <em>a neutralization reaction</em> because it produces water(reaction between an acid and a base)
AB + CD ⇒ AD +CB
- Name of reactants and products
H₃PO₄ : Phosphoric acid
KOH : Potassium hydroxide
K₃PO₄ : Tripotassium phosphate
H₂O : water
Answer:
am going to go ahead and say balanced
Explanation:
in the picture
Answer: POH=8
Explanation:PH = -log( H+concentration)
PH = 6
POH = 14 - PH = 8
a. 1.05 x 10⁻³ mol
b. 0.155% w/v
<h3>Further explanation</h3>
Given
50 ml Ca(OH)₂
19.5 ml of 0.1059 N HCl
Required
Amount of Ca(OH)₂
The Ca(OH)₂ concentration in % w/v.
Solution
Titration formula :
M₁V₁n₁=M₂V₂n₂
or
N₁V₁=N₂V₂
n = acid base valence=amount of H⁺/OH⁻(Ca(OH)₂=2, HCl=1)
a.
Input the value(1=Ca(OH)₂, 2= HCl) :
M₂=N₂=0.1059 M
M₁. 50 ml . 2 = 0.1059 . 19.5 . 1
M₁ = 0.021
Amount of Ca(OH)₂ :
mol Ca(OH)₂ = 0.021 x 50 ml = 1.05 mlmol = 1.05 x 10⁻³ mol
b. mass of Ca(OH)₂
= mol x MW
= 1.05 x 10⁻³ mol x 74 g/mol
= 0.0777 g
%w/v = (g solute / volume of solution) x 100
%w/v =( 0.0777 g/ 50 ml) x 100 =0.155% w/v
Answer:
0.0269 M
Explanation:
There is some info missing. I think this is the original question.
<em>Suppose 0.816 g of potassium nitrate is dissolved in 300 mL of a 14.0 mM aqueous solution of sodium chromate. Calculate the final molarity of potassium cation in the solution. You can assume the volume of the solution doesn't change when the potassium nitrate is dissolved in it. Be sure your answer has the correct number of significant digits.</em>
<em />
The molecular equation corresponding to this reaction is:
2 KNO₃(aq) + Na₂CrO₄(aq) ⇄ K₂CrO₄(aq) + 2 NaNO₃(aq)
The full ionic equation is:
2 K⁺(aq) + 2 NO₃⁻(aq) + 2 Na⁺(aq) + CrO₄²⁻(aq) ⇄ 2 K⁺(aq) + CrO₄²⁻(aq) + 2 Na⁺(aq) + 2 NO₃⁻(aq)
As we can see, the moles of K⁺ are equal to the initial moles of KNO₃. The molar mass of KNO₃ is 101.10 g/mol. The moles of KNO₃ (and K⁺) are:
0.816 g × (1 mol/ 101.10 g) = 8.07 × 10⁻³ mol
The molarity of K⁺ is:
8.07 × 10⁻³ mol / 0.300 L = 0.0269 M