Which set of reagents will best convert 2,2-dimethylpropan-1-ol (neopentyl alcohol) to 4,4-dimethylpentan-2-ol

i think its volume but im not 100% sure

Answer:

D.

Explanation:

cuz liter is volume

Answer:

The answer is "35 atoms".

Explanation:

The Formula for isopropyl alcohol is: [tex]C_3H_8O[/tex]

In 1 molecule of [tex]C_3H_8O[/tex] = 8 atoms of hydrogen  

In 4 molecule of = atoms of hydrogen

According to the given formula [tex]C_3H_7O[/tex]

In 5 molecule of [tex]C_3H_7O[/tex] [tex]= 5 \times 7 = 35[/tex] atoms of hydrogen.

Explanation:

I mol of water = 22.4 lts of O2

thus, 4.21 moles of water = 4.21× 22.4= 94.304 lts.

Answer:

This region of thick, hot crust and mantle extends 1,300 kilometers north and south of the Iceland hotspot and some 1,000 kilometers north and south of the Azores hotspot. Together, the two hotspots appear to be feeding a huge supply of magma to nearly the entire northern segment of the Mid-Atlantic Ridge.

Explanation:

please give me a heart

Answer:

Easter island

Answer:

A control is important for an experiment because it allows the experiment to minimize the changes in all other variables except the one being tested.

Explanation:

do u mean that kind of control

Answer:

true

Explanation:

plucking and abrasion

abrasion is when a glacier scrapes rocks below and plucking is when sediments are picked up by a glacier

Answer:

See explanation

Explanation:

An initiation step is the first step in a free radical mechanism. In this step, free radicals are generated from a stable compound. The initiation step tends to increase the concentration of radicals present in the system.  It leads to a net increase in the number of free radicals in the system.

For the chlorination of 2,2-dimethylpropane, the initiation step is;

Cl2 + hv --------->2Cl.

Note that hv is used to represent light

Answer: ice is less dense than liquid water. If ice was more dense, Earth would freeze.

Explanation: There are many reasons why life on Earth depends on the characteristics of water. One could discuss hydrogen bonds and its role as a solvent, but the unusual property of water is is the change in density with change in temperature. Water is densest at 4 degC, which is why ice floats - it is less dense than cold water (it melts quickly in warm water, so density isn’t impotant at higher temperatures). Most liquids are less dense than the solid, frozen form. If this was the case with water, any ice that formed would sink, and sease would freeze from the bottom up. Furthermore, the lowest layers would be insulated and would not all melt in summer. Thus over time, the seas would become a thin layer of liquid water at best, over solid ice. Life could not develop without liquid seas. In addition, ice is reflective, reducing the amount of sunlight absorbed, further reducing temperatures. Without ocean circulation, polar areas would be even colder, and there would be no rain.

Answer:

Cccccccccccccccccccccccccccccccccccccccccccc

Answer:

[tex]r=0.0341\frac{M}{s}[/tex]

Explanation:

Hello!

In this case, considering the given table, we are able to represent the symbolic rate law as shown below:

[tex]r=k[BF_3]^m[NH_3]^n[/tex]

Thus, by using the following steps, we can find both m (BF3 order of reaction) and n (BF3 order of reaction):

- Experiment 1 and 2 for the calculation of n:

[tex]\frac{r_1}{r_2}=\frac{k[BF_3]_1^m[NH_3]_1^n}{k[BF_3]_2^m[NH_3]_2^n}[/tex]

So we plug in to obtain:

[tex]\frac{0.2130}{0.1065}=\frac{k*0.250^m*0.250^n}{k*0.250^m*0.125^n}\\\\2=\frac{0.250^n}{0.125^n}\\\\2=2^n\\\\log(2)=n*log(2)\\\\n=1[/tex]

So the order of reaction with respect to NH3 is 1.

- Experiment 3 and 4 for the calculation of m

[tex]\frac{r_3}{r_4}=\frac{k[BF_3]_3^m[NH_3]_3^n}{k[BF_3]_4^m[NH_3]_4^n}[/tex]

So we plug in to obtain:

[tex]\frac{0.0682}{0.1193}=\frac{k*0.200^m*0.100^n}{k*0.350^m*0.100^n}\\\\0.57=0.57^m\\\\log(0.57)=m*log(0.57)\\\\m=1[/tex]

So the order of reaction with respect to BF3 is also 1.

Now, we can compute the rate constant by solving for it on any of the experiments there, say experiment 1:

[tex]k=\frac{r_1}{[BF_3][NH_3]} =\frac{0.2130M/s}{0.250M*0.250M}\\\\k=3.41M^{-1}s^{-1}[/tex]

Thus, the initial reaction rate for the 0.50M BF3 and 0.020M NH3 is:

[tex]r=3.41M^{-1}s^{-1}*0.50M*0.020M\\\\r=0.0341\frac{M}{s}[/tex]

Best regards!

Answer:

Explanation:

Airs and trees

Answer:

chemical

Explanation:

Explanation:

At convergent plate boundaries two plates are moving towards one another and different scenarios are in play.

Most importantly, at these margins the crust is being destroyed.

Some of the features found at this margin include, oceanic trenches, accretionary wedges, volcanic islands, earthquakes

Answer:

Hate 2 break it 2 u foo but there's no option for me 2 choose....

Answer:

Percent error for beaker A is 0.26%

Percent error for beaker B is 0.25%

Explanation:

The remaining part of the question is shown in the attachment below.

For beaker A

Mass of water = 9.997 g

Volume = 10mL

First, we will determine the experimental density

From

Density = Mass / Volume

∴ Density = 9.997g / 10mL

Density = 0.9997 g/mL

To calculate the percent error

Percent error = (Error/True value) × 100%

Error = Experimental value - True value

From the question, True value = 0.99707 g/mL

Experimental value = 0.9997 g/mL

∴ Error = 0.9997 g/mL - 0.99707 g/mL

Error = 0.00263 g/mL

Then,

Percent error = (0.00263/0.99707) × 100%

Percent error = 0.26377%

Percent errror ≅ 0.26%

Hence, percent error for beaker A is 0.26%

For beaker B

Mass of water = 9.9956 g

Volume = 10mL

Also, we will first determine the experimental density

From

Density = Mass / Volume

∴ Density = 9.9956g / 10mL

Density = 0.99956 g/mL

To calculate the percent error

Percent error = (Error/True value) × 100%

Error = Experimental value - True value

True value = 0.99707 g/mL

Experimental value = 0.99956 g/mL

∴ Error = 0.99956 g/mL - 0.99707 g/mL

Error = 0.00249 g/mL

Then,

Percent error = (0.00249/0.99707) × 100%

Percent error = 0.24973%

Percent errror ≅ 0.25%

Hence, percent error for beaker B is 0.25%

Answer:

150.1 mL

Explanation:

Step 1: Given data

Step 2: Calculate the volume of the sample of benzene

Density is an intrinsic property. It is equal to the quotient between the mass and the volume of the sample of benzene.

ρ = m/V

V = m/ρ

V = 131.9 g/(0.879 g/mL)

V = 150.1 mL

4 cm³

Math

Pre-Algebra

Order of Operations: BPEMDAS

Equality Properties

Chemistry

Gas Laws

Density = Mass over Volume

Step 1: Define

D = 25 g/cm³

m = 100 g

Step 2: Solve for V

Answer:

The answer is C or 3 which ever but it that one

Explanation:

Answer:

1.44

Explanation:

Science happened in my brain

Answer:

ocean waves are converted to electricity with wave energy converter, or WEC, devices. ... The surface portion moves faster than the submerged part, and the WEC converts that relative motion into electricity

The energy in ocean waves that should be transformed into electrical energy is because of Buoyant objects.

Buoys and turbines should be considered as the two examples of devices that can be used to capture the energy from the ocean's movement.

Buoys are considered to be structures that float on the sea and move up and down with the motion of the waves.

Learn more about energy here; https://brainly.com/question/17172186

Answer:

3.125%

Explanation:

After 1 half life it remains 50%

After 2 half lines it remains 25%

After 3 half lines it remains 12.5 %

After 4th half life it remains 6.25%

After 5th half life it remains 3.125%

290 g Na₂O

Math

Pre-Algebra

Order of Operations: BPEMDAS

Chemistry

Atomic Structure

Step 1: Define

4.7 mol Na₂O

Step 2: Identify Conversions

Molar Mass of Na - 22.99 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of Na₂O - 2(22.99) + 16.00 = 61.98 g/mol

Step 3: Convert

Step 4: Check

Follow sig fig rules and round. We are given 2 sig figs.

291.306 g Na₂O ≈ 290 g Na₂O

Answer:

no they do not

Explanation:

Answer:

It's just...Hair grown there...i guess it just grows as a male humans would

Explanation:

hope this helps.

Answer: When the spring is recoiling

Explanation:

Answer:

When the spring is recoiling

Explanation:

Answer:

[tex]\mathbf{H_2SeO_{4(aq)} + 3SO_{2(aq)} + 2H_2O_{(l) } \to Se_{(s)} + 3H_2SO_{4(aq)} }[/tex]

Explanation:

We are to write a balanced equation from the information given above by including the states of their matter according to SATP conditions;

[tex]\mathbf{H_2SeO_{4(aq)} + 3SO_{2(aq)} + 2H_2O_{(l) } \to Se_{(s)} + 3H_2SO_{4(aq)} }[/tex]

From above;

The oxidizing agent is [tex]\mathbf{H_2SeO_{4(aq)} }[/tex]

The reducing agent is [tex]\mathbf{SO_{2(aq)} }[/tex]