Answer:
4 moles of H₃PO₄
Explanation:
The reaction expression is given as;
3KOH + H₃PO₄ → K₃PO₄ + 3H₂O
Number of moles of water = 12moles
Unknown:
Number of moles of H₃PO₄ = ?
Solution:
From the balanced reaction expression we see that;
3 moles of water is produced from 1 mole of H₃PO₄
So; 12 moles of water would be produced from 
= 4 moles of H₃PO₄
Answer:
3. V = 0.2673 L
4. V = 2.4314 L
5. V = 0.262 L
6. V = 2.224 L
Explanation:
3. assuming ideal gas:
∴ R = 0.082 atm.L/K.mol
∴ V1 = 225 L
∴ T1 = 175 K
∴ P1 = 150 KPa = 1.48038 atm
⇒ n = RT/PV
⇒ n = ((0.082 atm.L/K.mol)(175 K))/((1.48038 atm)(225 L))
⇒ n = 0.043 mol
∴ T2 = 112 K
∴ P2 = P1 = 150 KPa = 1.48038 atm
⇒ V2 = RT2n/P2
⇒ V2 = ((0.082 atm.L/K.mol)(112 K)(0.043 mol))/(1.48038 atm)
⇒ V2 = 0.2673 L
4. gas is heated at a constant pressure
∴ T1 = 180 K
∴ P = 1 atm
∴ V1 = 44.8 L
⇒ n = RT/PV
⇒ n = ((0.082 atm.L/K.mol)(180 K))/((1 atm)(44.8 L))
⇒ n = 0.3295 mol
∴ T2 = 90 K
⇒ V2 = RT2n/P
⇒ V2 = ((0.082 atm.L/K.mol)(90 K)(0.3295 mol))/(1 atm)
⇒ V2 = 2.4314 L
5. V1 = 200 L
∴ P1 = 50 KPa = 0.4935 atm
∴ T1 = 271 K
⇒ n = RT/PV
⇒ n = ((0.082 atm.L/K.mol)(271 K))/((0.4935 atm)(200 L))
⇒ n = 0.2251 mol
∴ P2 = 100 Kpa = 0.9869 atm
∴ T2 = 14 K
⇒ V2 = RT2n/P2
⇒ V2 = ((0.082 atm.L/K.mol)(14 K)(0.2251 mol))/(0.9869 atm)
⇒ V2 = 0.262 L
6.a) ∴ V1 = 24.6 L
∴ P1 = 10 atm
∴ T1 = 25°C = 298 K
⇒ n = RT/PV
⇒ n = ((0.082 atm.L/K.mol)(298 K))/((10 atm)(24.6 L))
⇒ n = 0.0993 mol
∴ T2 = 273 K
∴ P2 = 101.3 KPa = 0.9997 atm
⇒ V2 = RT2n/P2
⇒ V2 = ((0.082 atm.L/K.mol)(273 K)(0.0993 mol))/(0.9997 atm)
⇒ V2 = 2.224 L
Answer: the answer should and most definitely be D.
Explanation: I mean think about it after a while only a few radioactive nuclei are left which means it will dye down after a while which also makes it very boring hope this helps :)
Conductivity, malleability, and high melting points. Hope this helps :)
Answer:
Option (B) 7
Explanation:
C3H6O2(l) + O2(g) → CO2(g) + H2O(l)
To know the coefficient of O2 in the above equation, let us balance the equation.
The above equation can be balance as follow:
C3H6O2(l) + O2(g) → CO2(g) + H2O(l)
There are 3 atoms of C on the left side and 1 atom on the right side. It can be balance by putting 3 in front of CO2 as shown below:
C3H6O2(l) + O2(g) → 3CO2(g) + H2O(l)
There are 6 atoms of H on the left side and 2 atoms on the right side. It can be balance by putting 3 in front of H2O as shown below:
C3H6O2(l) + O2(g) → 3CO2(g) + 3H2O(l)
There are a total of 4 atoms of O on the left side and a total of 9 atoms on the right side. It can be balance by putting 7/2 in front of O2 as show below:
C3H6O2(l) + 7/2O2(g) → 3CO2(g) + 3H2O(l)
Multiply through by 2
2C3H6O2(l) + 7O2(g) → 6CO2(g) + 6H2O(l)
Now, the equation is balanced.
From the balanced equation above, the coefficient of O2 is 7.
