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aleksklad [387]
3 years ago
13

(CH3)2-CH-CH2-O(CH3)3IUPAC NAME

Chemistry
1 answer:
bazaltina [42]3 years ago
3 0

Answer:

1-(tert-butoxy)-2-methylpropane

Note: there is a mistake in formula, the correct formula is (CH₃)₂-CH-CH₂-O-C(CH₃)₃ not (CH₃)₂-CH-CH₂-O(CH₃)₃, because oxygen is a divalent compound.

Explanation:

<em>Structural formula is attached</em>

IUPAC naming rules

1. start numbering the chain from the functional group. In this compound we        start from oxygen side.

2. Here we can see that at position 1 there is an oxy group along with a tertiary carbon having three methyl groups. So we write it as 1-tert-butoxy. Which means that there is a methoxy group at position 1 along with a tertiary carbon.

3. At position 2 we can see that there is a methyl group attached to the main chain, so we write it as 2-methyl.

4. Now we count the total number of carbons in the main chain. As we can see that there are 3 carbons in the remaining or parent chain, so we write it as propane

5. So the IUPAC name of the compound will be 1-(tert-butoxy)-2-methylpropane

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Charles's law is an experimental gas law that shows the relationship between the temperature of a gas and its corresponding volu
iren [92.7K]

Answer:

The correct answer is (b)

Explanation:

Charles law describes the behavior of gases when heated. Charles law states that the volume of a given mass of gas would increase as its Kelvin temperature increases provided the pressure is held constant. That is the volume of a given mass of gas is directly proportional to its Kelvin temperature at constant pressure

3 0
3 years ago
In a labratory activity the density of a sample of vanadium is determined to be 6.9g/cm^3 at room temperature. what is the perce
Snowcat [4.5K]

Answer:

c : 13%

Explanation:

Data Give:

Experimental density of vanadium = 6.9 g/cm³

percent error = ?

Solution:

Formula used to calculate % error

  % error = [experimental value -accepted value/accepted value] x 100

The reported accepted density value for vanadium = 6.11 g/cm³

Put value in the above equation

                  % error = [ 6.9 - 6.11 / 6.11 ] x 100

                  % error = [ 0.79 / 6.11 ] x 100

                  % error = [ 0.129] x 100

                  % error = 12.9

Round to the 2 significant figure

% error = 13 %

So, option c is correct            

7 0
3 years ago
who else has found that something they did or readed or whatever it is they had completely forgot about and just remembered abou
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Answer:

Yes I do this a lot :)

Explanation:

6 0
3 years ago
Calculate the partial pressure of each gas and the total pressure if the temperature of the gas is 21 ∘C∘C. Express the pressure
Volgvan

The question is incomplete, complete question is ;

A deep-sea diver uses a gas cylinder with a volume of 10.0 L and a content of 51.8 g of O_2 and 33.1 g of He. Calculate the partial pressure of each gas and the total pressure if the temperature of the gas is 21°C.Express the pressures in atmospheres to three significant digits separated by commas.

Answer:

Partial pressure of the oxygen gas is 3.91 atm.

Partial pressure of the helium gas is 20.0 atm

Total pressure of the gases is 24.0 atm

Explanation:

Moles of oxygen gas = n_1=\frac{51.8}{32 g/mol}=1.619 mol

Moles of helium gas = n_2=\frac{33.1 g}{4 g/mol}=8.275 mol

Total moles of gas = n_1+n_2=(1.619 +8.275 ) mole=9.894 mol

Volume of the cylinder = V = 10.0 L

Total pressure in the cylinder = P = ?

Temperature of the gas in cylinder = T = 21°C = 21 + 273 K = 294 K

PV = nRT ( ideal gas equation )

P=\frac{nRT}{V}

=\frac{9.894 mol\times 0.0821 atm L/mol K\times 294 K}{10.0 L}

P = 23.88 atm ≈ 23.9

Partial pressure of the individual gas will be determined by the help of Dalton's law:

partial pressure = Total pressure × mole fraction of gas

Partial pressure of the oxygen gas

p_{1}=P\times \chi_{1}=P\times \frac{n_1}{n_1+n_2}

p_1=23.88 atm\times \frac{1.619 mol}{9.894 mol}=3.91 atm

Partial pressure of the helium gas

p_{2}=P\times \chi_{2}=P\times \frac{n_2}{n_1+n_2}

p_2=23.88 atm\times \frac{8.275 mol}{9.894 mol}=19.97 atm\approx 20.0 atm

6 0
3 years ago
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lara31 [8.8K]

well , it's true because they are of sp3d type occur on sets of four

3 0
3 years ago
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