So for the first one It's 784.56 J (GPE)
For the second one it's KE is equal to 100 J. If you can give me a few more minutes I'll try to get the GPE
Answer:
a) 0.525 mol
b) 0.525 mol
c) 0.236 mol
Explanation:
The combustion reactions (partial and total) will be:
C₇H₁₆ + (15/2)O₂ → 7CO + 8H₂O
C₇H₁₆ + 11O₂ → 7CO₂ + 8H₂O
---------------------------------------------------
2C₇H₁₆ + (37/2)O₂ → 7CO + 7CO₂ + 16H₂O
It means that the reaction will form 50% of each gas.
a) 0.525 mol of CO
b) 0.525 mol of CO₂
c) The molar mass of heptane is: 7*12 g/mol of C + 16*1 g/mol of H = 100 g/mol
So, the number of moles is the mass divided by the molar mass:
n = 11.5/100 = 0.115 mol
For the stoichiometry:
2 mol of C₇H₁₆ -------------- (37/2) mol of O₂
0.115 mol of C₇H₁₆ --------- x
By a simple direct three rule:
2x = 2.1275
x = 1.064 mol of O₂
Which is the moles of oxygen that reacts, so are leftover:
1.3 - 1.064 = 0.236 mol of O₂
Lower temperature
Let's verify
- Pressure=P
- volume=V
- Temperature=T
As per Boyles law
As per Charles law
So
At higher altitudes lower the pressure so lower the temperature
Answer:
Explanation:
Mineral
The naturally occurring mineral anglesite, PbSO4, occurs as an oxidation product of primary lead sulfide ore, galena.
Basic and hydrogen lead sulfates
A number of lead basic sulfates are known: PbSO4·PbO; PbSO4·2PbO; PbSO4·3PbO; PbSO4·4PbO. They are used in manufacturing of active paste for lead acid batteries. A related mineral is leadhillite, 2PbCO3·PbSO4·Pb(OH)2.
At high concentration of sulfuric acid (>80%), lead hydrogensulfate, Pb(HSO4)2, forms.[4]
Chemical properties
Lead(II) sulfate can be dissolved in concentrated HNO3, HCl, H2SO4 producing acidic salts or complex compounds, and in concentrated alkali giving soluble tetrahydroxidoplumbate(II) [Pb(OH)4]2− complexes.
PbSO4(s) + H2SO4(l) ⇌ Pb(HSO4)2(aq)
PbSO4(s) + 4NaOH(aq) → Na2[Pb(OH)4](aq) + Na2SO4(aq)
Lead(II) sulfate decomposes when heated above 1000 °C:
PbSO4(s) → PbO(s) + SO3(g)
Answer:
=> 0.10 moles
Explanation:
The question is asking us to find the moles of NaCl cid formed.
We have been provided with;
0.10 L of 1.0 M NaOH.
We know that mole ratio of the reaction is 1:1 and also the molarity of a solution is contained in 1 L or 1000 mL or 1000 cm³.
This means that;
1.0 M is contained in 1.0 L.
X mol is contained in 0.10 L
= 0.10 moles of NaCl formed
