The combustion of any hydrocarbon yields water and carbon dioxide. We will now construct a balanced equation:
C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
Each mole of propane requires 5 moles of oxygen.
The middle carbon is 4-degree since it is attached to 4 carbons. All other carbons are 1-degree since they are attached to only 1 carbon.
Hydrogens attached with 1-degree carbon are all same. Hydrogen are often refereed to as protons. No carbon is attached to 4-degree carbon. So all hydrogens in this structure are same.
This structure is called
NeoPentane
Remember: heat lost = heat gained
When calculating heat loss or gain, remember
mass*(spec heat cap)*(change in T)
The unknown loses heat- we don't know the spec heat cap, so we'll call it x.
The water gains. I've omitted the units, but always use when solving problems on your own.
75*x*(96.5-37.1) = 1150*4.184*(37.1-25)
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Now it's all set up- use algebra to get x, the spec heat cap of the unk in J/g*degC
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Answer:
Shifts the equilibrium to the left. reduces solubility.
Explanation:
- MgF2(s) ↔ Mg2+(aq) + 2F-(aq)
S S 2S
∴ Ksp = 6.4 E-9 = [ Mg2+ ] * [ F- ]² = S * (2S)²
⇒ 4S² * S = 6.4 E-9
⇒ 4S³ = 6.4 E-9
⇒ S³ = 1.6 E-9
⇒ S = 1.1696 E-3 M
- NaF(s) → Na+(aq) + F-(aq)
0.10M 0.10M 0.10M
- MgF2(s) ↔ Mg2+(aq) + 2F-(aq)
S' S' 2S' + 0.10
⇒ Ksp = 6.4 E-9 = (S')*(2S' + 0.10)²
If we compare the concentration (0.10 M) of the ion with Ksp ( 6.4 E-9 ); thne we can neglect S' as adding:
⇒ 6.4 E-9 = (S')*(0.10)² = 0.01S'
⇒ S' = 6.4 E-7 M
∴ % S' = ( 6.4 E-7 / 0.1 )*100 = 6.4 E-4% <<< 5%, we can make the assumption
We can observe that S >> S' ( 1.1696 E-3 M >> 6.4 E-7 M ), which shows that the solubility is reduced by the efect of the common ion from the salt, which causes the equilibrium to shift to the left, precipitating part of MgF2(s).
Molarity = number of moles of solute/liters of solution
number of moles of solute = molarity x liters of solution
Part (a): <span>30.00 ml of 0.100m Cacl2
number of moles of CaCl2 = 0.1 x 0.03 = 3x10^-3 moles
1 mole of CaCl2 contains 2 moles of chlorine, therefore 3x10^-3 moles of CaCl2 contains 6x10^-3 moles of chlorine
Part (b): </span><span>10.0 ml of 0.500m bacl2
number of moles of BaCl2 = 0.5 x 0.01 = 5x10^-3 moles
1 mole of BaCl2 contains 2 moles of chlorine, therefore 5x10^-3 moles of BaCl2 contains 10x10^-3 moles of chlorine
Part (c): </span><span>4.00 ml of 1.000m nacl
number of moles of NaCl = 1 x 0.004 = 0.004 moles
1 mole of NaCl contains 1 mole of chlorine, therefore 4x10^-3 moles of NaCl contains 4x10^-3 moles of chlorine
Part (d): </span><span>7.50 ml of 0.500m fecl3
number of moles of FeCl3 = 0.5 x 0.0075 = 3.75x10^-3 moles
1 mole of FeCl3 contains 3 moles of chlorine, therefore 3.75x10^-3 moles of FeCl3 contains 0.01125 moles of chlorine
Based on the above calculations, the correct answer is (d)</span>