What? That docent make any scene?
Missing question:
Nitrogen: <span>2.0 L; </span>1.0 atm; 25°<span>C.
Oxygen: 3</span>.0 L; 2.0 atm; 25°C.
<span>When the valve between the two containers is opened, nitrogen gas moves from one container to another container and gases are mixed together, total volume of nitrogen is than:
V(nitrogen) = 2,0 L + 3,0 L = 5,0 L.</span>
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Answer:
-973 KJ
Explanation:
The balanced reaction equation is;
N2H4(aq) + 2Cl2(g) + 4OH^-(aq)---------> 4Cl-(aq) + 4H ^+(aq) + 4OH^-(aq) + N2(g)
Reduction potential of hydrazine = -1.16 V
Reduction potential of chlorine = 1.36 V
From;
E°cell= E°cathode - E°anode
E°cell= 1.36 - (-1.16)
E°cell= 2.52 V
∆G°=- nFE°cell
n= number of moles of electrons = 4
F= Faraday's constant = 96500 C
E°cell = 2.52 V
∆G°=- (4 × 96500 × 2.52)
∆G°= -972720 J
∆G°= -972.72 KJ
Answer:
If 13.4 grams of nitrogen gas reacts we'll produce 16.3 grams of ammonia
Explanation:
Step 1: Data given
Mass of nitrogen gas (N2) = 13.4 grams
Molar mass of N2 = 28 g/mol
Molar mass of NH3 = 17.03 g/mol
Step 2: The balanced equation
N2 + 3H2 → 2NH3
Step 3: Calculate moles of N2
Moles N2 = Mass N2 / molar mass N2
Moles N2 = 13.4 grams / 28.00 g/mol
Moles N2 = 0.479 moles
Step 4: Calculate moles of NH3
For 1 mol N2 we need 3 moles H2 to produce 2 moles NH3
For 0.479 moles N2 we'll produce 2*0.479 = 0.958 moles
Step 5: Calculate mass of NH3
Mass of NH3 = moles NH3 * molar mass NH3
Mass NH3 = 0.958 moles * 17.03 g/mol
Mass NH3 = 16.3 grams
If 13.4 grams of nitrogen gas reacts we'll produce 16.3 grams of ammonia
