The net ionic equation formed is
Ag^+(aq)+Cl^−(aq)→AgCl(s)
Chromium(III) nitrate and silver(I) chloride are the products of the balanced molecular equation for the reaction between chromium(III) chloride and silver(I) nitrate. An (s) next to the chemical formula for silver(I) chloride designates it as an insoluble salt.
CrCl3(aq)+3AgNO3(aq)→Cr(NO3)3(aq)+3AgCl(s)
Silver and the chloride ions are the two ions that must interact to create silver(I) chloride. By designating ions as the reactants and silver(I) chloride as the product, the net ionic equation is formed.
Ag^+(aq)+Cl^−(aq)→AgCl(s)
Ionic Equation:
In general, anions and cations react to generate a compound in a dissolved media, which is known as an ionic reaction. Water-insoluble salts are created when the ions of water-soluble salts interact with one another in an aqueous media.
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Answer:
%N = 25.94%
%O = 74.06%
Explanation:
Step 1: Calculate the mass of nitrogen in 1 mole of N₂O₅
We will multiply the molar mass of N by the number of N atoms in the formula of N₂O₅.
m(N): 2 × 14.01 g = 28.02 g
Step 2: Calculate the mass of oxygen in 1 mole of N₂O₅
We will multiply the molar mass of O by the number of O atoms in the formula of N₂O₅.
m(O): 5 × 16.00 g = 80.00 g
Step 3: Calculate the mass of 1 mole of N₂O₅
We will sum the masses of N and O.
m(N₂O₅) = m(N) + m(O) = 28.02 g + 80.00 g = 108.02 g
Step 4: Calculate the percent composition of N₂O₅
We will use the following expression.
%Element = m(Element)/m(Compound) × 100%
%N = m(N)/m(N₂O₅) × 100% = 28.02 g/108.02 g × 100% = 25.94%
%O = m(O)/m(N₂O₅) × 100% = 80.00 g/108.02 g × 100% = 74.06%
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Buoyant
Buoyancy is the ability to float in fluids.
Gravity is the force of all objects attracted to one.
Friction is the force caused when two objects rub against each other causing them to slow
<span>Archimedes principle explains that the magnitude of that force is proportional to the difference in the pressure between the top and the bottom of the column, and</span> is also equivalent to the weight of the fluid that would otherwise occupy the column, i.e. the displaced fluid.
<span>Assume you have 1 L of solution.
Moles F- = M F- x L F- = (0.0610)(1) = 0.0610 moles F-
0.0610 moles F- x(19.0 g F-/1mole F-) = 1.159 g F- in 1 L of solution
1 L solution x (1000 mL / 1 L) x (1.00 g / mL) = 1000 g of solution
mass % F- = (g F- / g solution) x 100 = (1.159 / 1000) x 100
= 0.1159%
parts per million F- = mg F- /L = 1159 / 1 = 1159 ppm F-</span>
