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:
The epic emotional journey of a suburban African American family as they navigate love, forgiveness and coming together in the wake of a tragic loss.
Explanation:
hope thsi helps if so please brainliest
Merry Christmas
The molecular mass of the immunoglobulin G, given the data from the question is 1.53×10⁵ g/mole
<h3>How to determine the molarity</h3>
We'll begin by calculating the molarity of the immunoglobulin G. This is illustrated below:
- Volume = 0.106 L
- Temperature (T) = 25 °C = 25 + 273 = 298 K
- Osmotic pressure (π) = 0.733 mbar = 0.733 × 0.000987 = 0.00072 atm
- Gas constant (R) = 0.0821 atm.L/Kmol
- Van't Hoff factor (i) = 1
- Molarity (M)
π = iMRT
M = π / iRT
M = 0.00072 / (1 × 0.0821 × 298)
M = 0.000029 M
<h3>How to determine the mole of immunoglobulin G</h3>
- Molarity = 0.000029 M
- Volume = 0.106 L
- Mole =?
Mole = Molarity × volume
Mole = 0.000029 × 0.106
Mole = 3.074×10⁻⁶ mole
<h3>How to determine the molar mass of mmunoglobulin G</h3>
- Mole = 3.074×10⁻⁶ mole
- Mass = 0.470 g
- Molar mass =?
Molar mass = mass / mole
Molar mass = 0.47 / 3.074×10⁻⁶
Molar mass = 1.53×10⁵ g/mole
Learn more about Osmotic pressure:
brainly.com/question/5925156
#SPJ1
