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Sodium chloride because it contains the most reactive metal(sodium) and most reactive non-metal(chlorine).
Answer:
pH = 2.0
Explanation:
To find the pH of a solution, take the -log[H+]. In this case, the -log(9.4 x 10^-3) equals 2.02687 which makes 2.0 when accounting for significant figures.
Answer:
(NH₄)₃PO₄ (l) + Al(NO₃)₃(l) -----------→ AlPO₄(l) + 3NH₄NO₃(l)
Explanation:
Data Give:
Reaction between ammonium phosphate solution and solution of aluminum nitrate
- Write a balanced chemical equation
Details:
To write a balanced chemical equation we have to know formula units of compounds or molecules
Formula units
ammonium phosphate : (NH₄)₃PO₄
aluminum nitrate: Al(NO₃)₃
ammonium nitrate: NH₄NO₃
Now to write a chemical equation
- we have to write the chemical formulas or formula unit of each compound
- write the reactant on left side of the arrow
- write the product on the right side of the arrow
- put a plus sign in 2 reactants and products on each side of the arrow
- balance the equation by putting coefficient with compound formula
- write the phase symbols on the right corner of the compound formula in brackets
So the Reaction will be
(NH₄)₃PO₄ + Al(NO₃)₃ -----------→ AlPO₄ + NH₄NO₃
Now balance the Chemical equation
(NH₄)₃PO₄ + Al(NO₃)₃ -----------→ AlPO₄ + 3NH₄NO₃
Now write the phase Symbols
(NH₄)₃PO₄ (l) + Al(NO₃)₃(l) -----------→ AlPO₄(l) + 3NH₄NO₃(l)
all compounds in the reaction are in liquid form and soluble in water
*** Note:
There is no aluminum nitrite in chemicals formulas
Also ammonium nitrite can not be used in pure isolated form due to its highly instability
Answer:
c.
Explanation:
A serial dilution is a dilution that is made fractionated. The stock solution is diluted, then this now solution is diluted, and then successively. The final dilution is the multiplication of the steps dilutions.
The representation of the dilution is v/v (volume per volume) indicates how much of the stock solution is in the total volume of the solution. So 1/5 indicates 1 mL to 5 mL of the solution. If the final volume must be 1 mL, then the stock solution must be 0.2 mL (0.2/1 = 1/5), and the volume of the solvent is 1 mL - 0.2 = 0.8 mL.
The second solution is done with a dilution of 1/10 or 1 mL of the first solution in 10 mL of the total solution. Because the solution has 1 mL, then the volume of the first solution must be 0.1 mL (0.1/1 = 1/10), and the volume of the solvent that must be added is 1 mL - 0.1 mL = 0.9 mL.