Answer:
See below
Explanation:
PV = n RT R = .082057 L - Atm / (K-Mole)
.97 V = .118 (.082057)(305)
V = 3.04 liters
Will not change for argon.....will be the same for all ideal gases
Answer:
2.82 g
Explanation:
Step 1: Write the balanced precipitation reaction
3 Ba(NO₃)₂ (aq) + Al₂(SO₄)₃ (aq) ⇒ 3 BaSO₄(s) + 2 Al(NO₃)₃(aq)
Step 2: Calculate the reacting moles of Ba(NO₃)₂
45.0 mL (0.0450 L) of 0.548 M Ba(NO₃)₂ react.
0.0450 L × 0.548 mol/L = 0.0247 mol
Step 3: Calculate the moles of Al₂(SO₄)₃ that react with 0.0247 moles of Ba(NO₃)₂
The molar ratio of Ba(NO₃)₂ to Al₂(SO₄)₃ is 3:1. The reacting moles of Al₂(SO₄)₃ are 1/3 × 0.0247 mol = 8.23 × 10⁻³ mol
Step 4: Calculate the mass corresponding to 8.23 × 10⁻³ moles of Al₂(SO₄)₃
The molar mass of Al₂(SO₄)₃ is 342.2 g/mol.
8.23 × 10⁻³ mol × 342.2 g/mol = 2.82 g
7.8049856 g of aluminium phosphate is produced from 7.5 g of lithium phosphate in this balanced equation.
<h3>What are moles?</h3>
A mole is defined as 6.02214076 × 
of some chemical unit, be it atoms, molecules, ions, or others. The mole is a convenient unit to use because of the great number of atoms, molecules, or others in any substance.
Given data:
→ 
Moles of 7.5 g of lithium phosphate.
The molar mass of lithium phosphate is 115.79 g/mol.
Moles = 
Moles =
Moles =
Moles = 0.06477243285
Now we will compare the moles of 
with 
.
:
2 : 2
0.385 : 2÷2× 0.064 = 0.064 mol
Mass of :
Mass of = moles × molar mass
Mass of =0.064 mol × 121.9529 g/mol
Mass of = 7.8049856 g
Hence, 7.8049856 g of aluminium phosphate is produced from 7.5 g of lithium phosphate in this balanced equation.
Learn more about moles here:
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Answer:
it is iron and flourine gas
Explanation:
