Two atoms that isotopes of one another must have number of what

What would be the major product obtained from hydroboration–oxidation of the following alkenes?

zmey [24]

Answer:

a. 3-methylbutan-2-ol

b. 2-methylcyclohexan-1-ol

Explanation:

For this reaction, we must remember that the hydroboration is an "anti-Markovnikov" reaction. This means that the "OH" will be added at the least substituted carbon of the double bond.

In the case of 2-methyl-2-butene, the double bond is between carbons 2 and 3. Carbon 2 has two bonds with two methyls and carbon 3 is attached to 1 carbon. Therefore the "OH" will be added to carbon three producing 3-methylbutan-2-ol.

For 1-methylcyclohexene, the double bond is between carbons 1 and 2. Carbon 1 is attached to two carbons (carbons 6 and 7) and carbon 2 is attached to one carbon (carbon 3). Therefore the "OH" will be added to carbon 2 producing 2-methylcyclohexan-1-ol.

See figure 1

I hope it helps!

8 0

2 years ago

What is a trade-off? Don't mind the highlighted yellow circle

kirill [66]

Answer:

a new scientific discovery that benefits the environment

5 0

1 year ago

A compound formed from titanium and oxygen has the formula TiO2. Which of the following is true about the ratio of number Ti and

Finger [1]

Answer:

2

Explanation:

There are some basic laws that guide the combination of elements chemically. These are the law of conservation of mass, law of definite proportion, law of multiple proportion and the law of reciprocal proportion.

For this question, the useful law to use is the law of definite proportion. Here, it is stated that no matter the method of preparation or source of preparation, the elements of a chemical compound are always present in a fixed ratio.

What this means that at any point in time, the compound titanium dioxide contains one atom of titanium and two atoms of oxygen. This means that both atoms are present at all times in a proportion of 1 to 2 .

4 0

3 years ago

The useful metal manganese can be extracted from the mineral rhodochrosite by a two-step process. In the first step, manganese(I

ale4655 [162]

Answer:

The answer is "6.52 kg and 13.1 kg"

Explanation:

For point a:

$Actual\ yield = 6.52 \ kg\\\\Percent \ yield= 66\%\\\\Percent \ yield = \frac{Actual \ yield}{Theoretical\ yield} \times 100 \%\\\\Theoretical\ yield \ of \ MnO_2 = \frac{Actual \ yield}{Percent \ yield} \times 100\%\\\\=\frac{6.52 \ kg}{66 \%} \times 100\% =9.88 \ kg\\\\$

Equation:

$3MnCO_3 +O_2 \longrightarrow 2MnO_2 + 2CO_2\\\\$

Calculating the amount of $MnCO_3$

$= 9.88 \ kg \times \frac{1000 \ g}{1 \ kg} \times \frac{1 \ mol \ MnO_2}{86.94 \ g} \times \frac{2 \ Mol \ MnCO_3}{2 \ mol \ MnO_2} \times \frac{114.95\ g}{1 \ mol \ MnCO_3 }\times \frac{1\ kg}{1000\ g}\\\\= 13.1 \ kg$

For point b:

$Actual\ yield = 4.0 \ kg\\\\Percent\ yield=97.0\%\\\\Percent \ yield = \frac{Actual \ yield}{Theoretical \ yield} \times 100 \% \\\\Theoretical \ yield\ of\ Mn = \frac{Actual \ yield}{Percent \ yield} \times 100\%\\\\$

$=\frac{4.0 \ kg}{97.0\%} \times 100\% =4.12 \ kg$

Equation:

$3MnO_2 +4AL \longrightarrow 3Mn + 2AL_2O_3\\\\$

Calculating the amount of $MnO_2:$

$= 4.12 \ kg \times \frac{1000 \ g}{1 \ kg} \times \frac{1 \ mol \ Mn}{54.94 \ g} \times \frac{3 \ Mol \ MnO_2}{3 \ mol \ Mn} \times \frac{86.94 \ g}{1 \ mol \ MnO_2 }\\\\= 6516 \ g \\\\=6.52 \ kg\\\\$

7 0

2 years ago

How much heat is added if .0948g of water is increased in temperature by .728 degrees C?

Verdich [7]

Answer:

0.289J of heat are added

Explanation:

We can relate the change in heat of a substance with its increasing in temperature using the equation:

q = m*ΔT*S

Where Q is change in heat

m is mass of substance (In this case, 0.0948g of water)

ΔT = 0.728°C

S is specific heat (For water, 4.184J/g°C)

Replacing:

q = 0.0948g*0.728°C*4.184J/g°C

q = 0.289J of heat are added

5 0

2 years ago