Which of the following elements has the largest atomic radius? (2 points)

Compound a reacts with one equivalent of h2 in the presence of a catalyst to give methylcyclohexane. Compound a can be formed up

V125BC [204]

Answer:

The compound a is 1-methyl cyclohexene (see attachment for structure).

Explanation:

The reaction of 1-Bromo-1-methylcyclohexane with sodium methoxide is a second-order reaction since the methoxide ion is a strong base and also a strong nucleophile. This ion attacks the alkyl halide faster than the alkyl halide can ionize to produce a first-order reaction. However, we can not see the product of nucleophilic substitution. The SN₂ mechanism is blocked due to the impediment of the 1-Bromo-1-methylcyclohexane. The main product, according to the Zaitsev rule, is the 1-methyl cyclohexene, thus forming a double bond.

Then, this cyclohexene is hydrogenated to form the cyclohexane.

6 0

3 years ago

Solid sodium azide (NaN3) produces solid sodium and nitrogen gas. How many grams of sodium azide are needed to yield a volume of

ch4aika [34]

Answer:

52.008 grams of sodium azide are needed to yield a volume of 26.5 L of nitrogen gas at a temperature of 295 K and a pressure of 1.10 atmospheres.

Explanation:

An ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law, an equation that relates the three variables if the amount of substance, number of moles n, remains constant and where R is the molar constant of the gases:

P*V = n*R*T

In this case, the balanced reaction is:

2 NaN₃ → 2 Na + 3 N₂

You know the following about N₂:

P= 1.10 atm
V= 26.5 L
n=?

$R=0.082057 \frac{atm*L}{mol*K}$

T= 295 K

Replacing in the equation for ideal gas:

1.10 atm* 26.5 L= n* $0.082057 \frac{atm*L}{mol*K}$ *295 K

Solving:

$n=\frac{1.10 atm*26.5 L}{0.082057 \frac{atm*L}{mol*K} *295K}$

n= 1.2 moles

Now, the following rule of three can be applied: if 3 moles of N₂ are produced by stoichiometry of the reaction from 2 moles of NaN₃, 1.2 moles of N₂ are produced from how many moles of NaN₃?

$moles of NaN_{3}=\frac{1.2 molesofN_{2} *2 molesofNaN_{3} }{3 molesofN_{2} }$

moles of NaN₃= 0.8

Since the molar mass of sodium azide is 65.01 g / mol, then one last rule of three applies: if 1 mol has 65.01 grams of NaN₃, 0.8 mol how much mass does it have?

$mass of NaN_{3} =\frac{0.8 mol*65.01 grams}{1 mol}$

mass of NaN₃=52.008 grams

52.008 grams of sodium azide are needed to yield a volume of 26.5 L of nitrogen gas at a temperature of 295 K and a pressure of 1.10 atmospheres.

6 0

3 years ago

Read 2 more answers

Why is chromatography particularly well suited to the needs of a drug analyst

Zarrin [17]

Chromatography is used in purification. Drugs analysts may use the technique to separate the active molecule in a drug molecule, for efficacy or toxicity analysis, from the other drug components.

Explanation:

Chromatography is used to separate a mixture of different components based on the size of their molecules. In liquid chromatography, the mixture is dissolved in a solvent that acts as the mobile phase and then passed along a stationary phase with different kinds of pores, As the mixture passes through the pores, their different components are separated because they take different times to pass through the stationary phase because of their different rates in passing through the pores.

In gas chromatography, a gas is used as a mobile phase while a liquid is used as the stationary phase.

Learn More:

For more on chromatography check out;

brainly.com/question/13232854

#LearnWithBrainly

6 0

3 years ago

A piece of lithium is placed on the surface of some water in a beaker. hydrogen is given off. Lithium hydroxide is formed. What

Gala2k [10]

A piece of lithium is placed on the surface of some water in a beaker. Hydrogen is given off and Lithium hydroxide is formed. The word equation would be as follows:


lithium + water→ lithium hydroxide + hydrogen

Hope this answers the question. Have a nice day.

8 0

3 years ago

What is the longest wavelength in the Balmer series? (Hint: the Rydberg constant for Hydrogen is 1.096776×107 1/m, and the Balme

boyakko [2]

Answer: The longest wavelength of light is 656.5 nm

Explanation:

For the longest wavelength, the transition should be from n to n+1, where: n = lower energy level

To calculate the wavelength of light, we use Rydberg's Equation:

$\frac{1}{\lambda}=R_H\left(\frac{1}{n_i^2}-\frac{1}{n_f^2} \right )$

Where,

$\lambda$ = Wavelength of radiation

$R_H$ = Rydberg's Constant = $1.096776\times 10^7m^{-1}$

$n_f$ = Higher energy level = $n_i+1=(2+1)=3$

$n_i$ = Lower energy level = 2 (Balmer series)

Putting the values in above equation, we get:

$\frac{1}{\lambda }=1.096776\times 10^7m^{-1}\left(\frac{1}{2^2}-\frac{1}{3^2} \right )\\\\\lambda =\frac{1}{1.5233\times 10^6m^{-1}}=6.565\times 10^{-7}m$

Converting this into nanometers, we use the conversion factor:

$1m=10^9nm$

So, $6.565\times 10^{-7}m\times (\frac{10^9nm}{1m})=656.5nm$

Hence, the longest wavelength of light is 656.5 nm

4 0

3 years ago