Answer:
Soluble salts can be made by reacting acids with soluble or insoluble reactants. Titration must be used if the reactants are soluble. Insoluble salts are made by precipitation reactions.
Making insoluble salts
An insoluble salt can be prepared by reacting two suitable solutions together to form a precipitate.
Determining suitable solutions
All nitrates and all sodium salts are soluble. This means a given precipitate XY can be produced by mixing together solutions of:
X nitrate
sodium Y
For example, to prepare a precipitate of calcium carbonate:
X = calcium and Y = carbonate
mix calcium nitrate solution and sodium carbonate solution together
calcium nitrate + sodium carbonate → sodium nitrate + calcium carbonate
Ca(NO3)2(aq) + Na2CO3(aq) → 2NaNO3(aq) + CaCO3(s)
It also works if potassium carbonate solution or ammonium carbonate solution is used instead of sodium carbonate solution. Remember that all common potassium and ammonium salts are soluble.
please mark as brainliest
Explanation:
Since gravitational force is inversely proportional to the square of the separation distance between the two interacting objects, more separation distance will result in weaker gravitational forces. So as two objects are separated from each other, the force of gravitational attraction between them also decreases.
-google
its probably 40
ELETRICK probably wondering why metal shocks (psst why does wires charge phones)
1) Start by standardizing the solution of NaOH by using the solution of H2SO4 whose concentration is known.
2) Equation:
2Na OH + H2SO4 --> Na2 SO4 + 2H2O
3) molar ratios
2 mol NaOH : 1 mol H2SO4
4) Number of moles of H2SO4 in 50.0 ml of 0.0782 M solution
M = n / V => n = M*V = 0.0782 M * 0.050 l = 0.00391 mol H2SO4
5) Number of moles of NaOH
2 moles NaOH / 1mol H2SO4 * 0.00391 mol H2SO4 = 0.00782 mol NaOH
6) Concentration of the solution of NaOH
M = n / V = 0.00782 mol / 0.0184 ml = 0.425 M
7) Standardize the solution of HCl
Chemical reaction:
NaOH + HCl --> NaCl + H2O
8) Molar ratios
1 mol NaOH : 1 mol HCl
9) Number of moles of NaOH in 27.5 ml
M = n / V => n = M * V = 0.425 M * 0.0275 l = 0.01169 moles NaOH
10) Number of moles of HCl
1 mol HCl / 1mol NaOH * 0.01169 mol NaOH = 0.01169 mol HCl
11) Concentration of the solution of HCl
M = n / V = 0.01169 mol / 0.100 l = 0.1169 M
Rounded to 3 significant figures = 0.117 M
Answers:
[NaOH] = 0.425 M
[HCl] = 0.117 M
