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3 years ago

What is the iupac name for this compound ch3-c(ch3) 2-ch2-chcl-ch3?

Chemistry

1 answer:

myrzilka [38]3 years ago

4 0

The IUPAC name of the above mentioned compound is "2-Chloro-4,4-dimethylpentane"

Exaplanation:

Since the above organic compound is an compound with only one saturated bond, It can be considered as a single bond compound, and hence we can conclude that as alkane.
It also has 5 carbon atoms, so it is termed as pentane.
From right to left we have to number the atoms, and 2 nd carbon atom contain Cl atom so it is termed as 2-Chloro and in the 4th position carbon atom contains 2 methyl groups, so it is termed as, 2-Chloro-4,4-dimethylpentane.

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2 years ago

Hello, everyone!

marusya05 [52]

Answer: 27.09 ppm and 0.003 %.

First, for air pollutants, ppm refers to parts of steam or gas per million parts of contaminated air, which can be expressed as cm³ / m³. Therefore, we must find the volume of CO that represents 35 mg of this gas at a temperature of -30 ° C and a pressure of 0.92 atm.

Note: we consider 35 mg since this is the acceptable hourly average concentration of CO per cubic meter m³ of contaminated air established in the "National Ambient Air Quality Objectives". The volume of these 35 mg of gas will change according to the atmospheric conditions in which they are.

So, according to the law of ideal gases,

PV = nRT

where P, V, n and T are the pressure, volume, moles and temperature of the gas in question while R is the constant gas (0.082057 atm L / mol K)

The moles of CO will be,

n = 35 mg x $\frac{1 g}{1000 mg}$ x $\frac{1 mol}{28.01 g}$

→ n = 0.00125 mol

We clear V from the equation and substitute P = 0.92 atm and

T = -30 ° C + 273.15 K = 243.15 K

V = $\frac{0.00125 mol x 0.082057 \frac{atm L}{mol K} x 243 K}{0.92 atm}$

→ V = 0.0271 L

As 1000 cm³ = 1 L then,

V = 0.0271 L x $\frac{1000 cm^{3} }{1 L}$ = 27.09 cm³

Then the acceptable concentration c of CO in ppm is,

c = 27 cm³ / m³ = 27 ppm

To express this concentration in percent by volume we must consider that 1 000 000 cm³ = 1 m³ to convert 27.09 cm³ in m³ and multiply the result by 100%:

c = 27.09 $\frac{cm^{3} }{m^{3} }$ x $\frac{1 m^{3} }{1 000 000 cm^{3} }$ x 100%

c = 0.003 %

So, the acceptable concentration of CO if the temperature is -30 °C and pressure is 0.92 atm in ppm and as a percent by volume is 27.09 ppm and 0.003 %.

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3 years ago

Given 0.02 of KF, how many liters of solution are needed to make a 7.2 solution?

Dmitry_Shevchenko [17]

Answer:

1/360

Explanation:

let x = liters

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3 years ago

Values for the molar mass of nitrogen, oxygen, and nitrogen dioxide

CaHeK987 [17]

Answer:

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Explanation:

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3 years ago

What is the iupac name for this compound ch3-c(ch3) 2-ch2-chcl-ch3? ​

What is the iupac name for this compound ch3-c(ch3) 2-ch2-chcl-ch3?