Answer:
SN1 = a stepwise loss of a leaving group to form a carbocation followed by nucleophilic attack
Explanation:
Since 2-methyl-2-butanol is tertiary alcohol, the first step will be loss of leaving group to form a 3° carbocation which is very stable and favours SN1, followed by attack of nucleophile
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<em>~Hope this helped! :)</em></span>
This is a common laboratory experiment called calorimetry which determines the specific heat capacity of the sample metal.
By the Conservation of Energy,
Energy of Metal = Energy of Water
mCmetalΔT = mCwaterΔT, wherein Cwater = 4.187 J/g·°C
(26.5 g)(Cmetal)(98 - 32.5°C) = (150 g)(4.187 J/g·°C)(32.5 - 20°C)
Solving for Cmetal,
<em>Cmetal = 4.523 J/g·°C</em>
Answer:
0.643 mol.
Explanation:
- We can use the general law of ideal gas: <em>PV = nRT.</em>
where, P is the pressure of the gas in atm (P = 4.0 atm).
V is the volume of the gas in L (V = 4000 mL = 4.0 L).
n is the no. of moles of the gas in mol (n = ??? mol).
R is the general gas constant (R = 0.0821 L.atm/mol.K),
T is the temperature of the gas in K (T = 30ºC + 273 = 303 K).
<em>∴ n = PV/RT </em>= (4.0 atm)(4.0 L)/(0.0821 L.atm/mol.K)(303 K) = <em>0.643 mol.</em>
Balance your chemical reaction first. Start by balancing carbon so you have 2 carbons on both sides.
C2H6 + O2 —> 2CO2 + H2O
Now balance your hydrogen so you have 6 hydrogens on both sides.
C2H6 + O2 —> 2CO2 + 3H2O
Now balance your oxygens. You have 7 oxygens on the right and 2 on the left, so multiply O2 by 3.5.
C2H6 + 3.5O2 —> 2CO2 + 3H2O
However, you can’t have a decimal in a coefficient. So, multiply everything by two.
2C2H6 + 7O2 —> 4CO2 + 6H2O
Now use your mole ratio of 6 mol H2O for every 2 mol of C2H6 to solve.
1.4 mol C2H6 • 6 mol H2O / 2 mol C2H6 = 4.2 mol H2O
D) 4.2 moles
