Name the following ketone:

g In the elimination reaction of t-butanol and t-butyl bromide. A. They have a common intermediate and require a strong base B.

11111nata11111 [884]

Answer:

Both reactions share a common intermediate and differ only in the leaving group

Explanation:

The elimination reaction of tertiary alkyl halides usually occur by E1 mechanism. In E1 mechanism, the substrate undergoes ionization leading to the loss of a leaving group and formation of a carbocation.

Loss of a proton from the carbocation completes the reaction mechanism yielding the desired alkene.

In the cases of t-butanol and t-butyl bromide, the mechanism is the same. The both reactions proceed by E1 mechanism. The leaving groups in each case are water and chloride ion respectively.

6 0

3 years ago

How do you round 44.981 using three significant figures.

sveta [45]

Three significant figures would be 45.0 since it would end with the nine so you look at the next digit to see if you round up or it remains the same. It is greater than five so round up to zero. This makes the ones place also go up so the answer is 45.0 The zero after the decimal counts as a significant figure.

3 0

3 years ago

Determine the overall reaction and its standard cell potential (in V) at 25°C for the reaction involving the galvanic cell made

mel-nik [20]

Answer:

Standard cell potential is 0.56V

Explanation:

Standard reduction potential ( $E^{0}$ ) are given below-

$E_{Cu^{+}\mid Cu}^{0}=0.52V$

$E_{Fe^{3+}\mid Fe}^{0}=-0.04V$

As $E_{Cu^{+}\mid Cu}^{0}$ is greater than $E_{Fe^{3+}\mid Fe}^{0}$ therefore Fe will be oxidized to $Fe^{3+}$ and $Cu^{+}$ will be reduced to Cu

Oxidation: $Fe(s)-3e^{-}\rightarrow Fe^{3+}(aq.)$

Reduction: $Cu^{+}(aq.)+e^{-}\rightarrow Cu(s)$

------------------------------------------------------------------------------------------------------

Overall: $Fe(s)+3Cu^{+}(aq.)\rightarrow Fe^{3+}(aq.)+3Cu(s)$

Standard cell potential, $E_{cell}^{0}=E_{Cu^{+}\mid Cu}^{0}-E_{Fe^{3+}\mid Fe}^{0}=(0.52+0.04)V=0.56V$

3 0

4 years ago

Convert 1 year to minutes

EleoNora [17]

The answer is 525600 minutes

7 0

3 years ago

Calcium metal reacts with water to form calcium hydroxide and hydrogen gas. How much hydrogen is formed when 0.50 g of calcium a

olga55 [171]

Answer:

0.025g

Explanation:

Firstly, we need to write a balanced chemical equation.

Ca + 2H2O —-> Ca(OH)2 + H2

From the chemical equation, we can see that one mole of calcium metal yielded one mole of hydrogen gas.

Now, we need to know the actual number of moles of hydrogen gas given off.

The number of moles is the mass divided by the atomic mass. The atomic mass of calcium is 40g/mol. The number of moles is thus 0.5/40 = 0.0125 mole

Since the mole ratio is 1:1, the number of moles of hydrogen gas produced too is 0.0125 moles.

We now Proceed to get the mass of the hydrogen gas produced. This is the number of moles of hydrogen multiplied by the molar mass of the diatomic hydrogen gas. The molar mass of the diatomic hydrogen gas is 2g/mol.

The amount given off is thus 2 * 0.0125 = 0.025g

6 0

4 years ago