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KatRina [158]
3 years ago
8

Calculate the mass in grams for each of the following liquids.

Chemistry
1 answer:
WITCHER [35]3 years ago
3 0
The density is calculated as mass per volume, so if we want to solve for mass, we would multiply density by volume.
For Part A: if we have a density of 0.69 g/mL, and a volume of 280 mL, multiplying these will give a mass of: (0.69 g/mL)(280 mL) = 193.2 g. Rounded to 2 significant figures, this is 190 g gasoline.
For Part B: if we have a density of 0.79 g/mL, and a volume of 190 mL, multiplying these will give a mass of: (0.79 g/mL)(190 mL) = 150.1 g. Rounded to 2 significant figures, this is equal to 150 g ethanol.
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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
  • include phase symbols

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

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