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3 years ago

A level chem naming isomers

Chemistry

1 answer:

anyanavicka [17]3 years ago

7 0

Answer:

Butan-1-ol or Butanol

2-methylpropan-2-ol

2-methylpropan-1-ol

Butan-2-ol

Explanation:

Data Give:

Three Isomers are given

names of These =?

Details:

Isomers:

Isomers are those chemical compounds that have same molecular formula but different structures.

In the given question the compound have same molecular formula that is C₄H₉OH but the structures are different.

So due to different structures they are named differently.

The 1st and last one are straight Chains and the other 2 are branched. all the structure have four carbon atoms, one OH group and 9 hydrogen atoms.

Names are given in attachment

butan-1-ol or Butanol

As it contain 4 carbon atoms an -OH group is attached at position 1.

________________

2-methylpropan-2-ol

In this -OH and -CH₃ is attached on carbon 2

_____________

2-methylpropan-1-ol

In this -CH₃ attached at position 2 and -OH is attached on Carbon 1.

______________

Butan-2-ol

In this 4 carbons are in straight chain and a -OH group is attached on position 2 of carbon

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Molar mass c3h8

maksim [4K]

Answer:

1. 44.11 g

2. 36.03 g

3. 8.08 g

4. 81.7%

5. 18.3%

Explanation:

1. 12.01+12.01+12.01+1.01+1.01+1.01+1.01+1.01+1.01+1.01+1.01=44.11

2. 12.01×3= 36.03

3. 1.01×8= 8.08

4.(36.03/44.11)×100= 81.7%

5. (8.08/44.11)×100= 18.3%

8 0

2 years ago

What can be said about a reaction with H = 62.4 kJ/mol and S = 0.145 kJ/(mol·K)?

Zigmanuir [339]

Answer:

At 430.34 K the reaction will be at equilibrium, at T > 430.34 the reaction will be spontaneous, and at T < 430.4K the reaction will not occur spontaneously.

Explanation:

1) Variables:

G = Gibbs energy
H = enthalpy
S = entropy

2) Formula (definition)

G = H + TS

=> ΔG = ΔH - TΔS

3) conditions

ΔG < 0 => spontaneous reaction
ΔG = 0 => equilibrium
ΔG > 0 non espontaneous reaction

4) Assuming the data given correspond to ΔH and ΔS

ΔG = ΔH - T ΔS = 62.4 kJ/mol + T 0.145 kJ / mol * K

=> T = [ΔH - ΔG] / ΔS

ΔG = 0 => T = [ 62.4 kJ/mol - 0 ] / 0.145 kJ/mol*K = 430.34K

This is, at 430.34 K the reaction will be at equilibrium, at T > 430.34 the reaction will be spontaneous, and at T < 430.4K the reaction will not occur spontaneously.

3 0

3 years ago

Read 2 more answers

Which of these factors influences air temperature?

Ksju [112]

Answer:

Explanation:

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3 years ago

What is the mass, in pounds, of 389 mL of a gas that has a density of 1.29 g/L?

devlian [24]

Answer: Mass of gas is 0.001 pounds.

Explanation:

Density is defined as the mass contained per unit volume.

$Density=\frac{mass}{Volume}$

Given : Mass of gas = ?

Density of gas = $1.29g/L$

Volume of gas = 389 ml = 0.389 L (1L=1000ml)

Putting in the values we get:

$1.29g/L=\frac{mass}{0.389L}$

$mass=0.5grams$

$mass=0.5\times 0.002lb=0.001lb$ (1g =0.002 lb)

Thus the mass of gas is 0.001 pounds.

7 0

2 years ago

How many milliliters of a 3.4 M NaCl solution would be needed to prepare each solution?

Ksenya-84 [330]

Answer:

a. Approximately $1.3\; \rm mL$ .

b. Approximately $7.2\; \rm mL$ .

Explanation:

The unit of concentration " $\rm M$ " is equivalent to " $\rm mol \cdot L^{-1}$ ", which means "moles per liter."

However, the volume of both solutions were given in mililiters $\rm mL$ . Convert these volumes to liters:

$\displaystyle 45\; \rm mL = 45\; \rm mL \times \frac{1\; \rm L}{1000\; \rm mL} = 0.045\; \rm L$ .

$\displaystyle 330\; \rm mL = 330\; \rm mL \times \frac{1\; \rm L}{1000\; \rm mL} = 0.330\; \rm L$ .

In a solution of volume $V$ where the concentration of a solute is $c$ , there would be $c \cdot V$ (moles of) formula units of this solute.

Calculate the number of moles of $\rm NaCl$ formula units in each of the two solutions:

Solution in a.:

$n = c \cdot V = 0.045\; \rm L \times 0.10\; \rm mol \cdot L^{-1} = 0.0045\; \rm mol$ .

Solution in b.:

$n = c \cdot V = 0.330\; \rm L \times 0.074\; \rm mol \cdot L^{-1} = 0.02442\; \rm mol$ .

What volume of that $3.4\; \rm M$ (same as $3.4 \; \rm mol \cdot L^{-1}$ ) $\rm NaCl$ solution would contain that many

For the solution in a.:

$\displaystyle V = \frac{n}{c} = \frac{0.0045\; \rm mol}{3.4\; \rm mol \cdot L^{-1}} \approx 0.0013\; \rm L$ .

Convert the unit of that volume to milliliters:

$\displaystyle 0.0013\; \rm L = 0.0013\; \rm L \times \frac{1000\; \rm mL}{1\; \rm L} = 1.3\; \rm mL$ .

Similarly, for the solution in b.:

$\displaystyle V = \frac{n}{c} = \frac{0.02442\; \rm mol}{3.4\; \rm mol \cdot L^{-1}} \approx 0.0072\; \rm L$ .

Convert the unit of that volume to milliliters:

$\displaystyle 0.0072\; \rm L = 0.0072\; \rm L \times \frac{1000\; \rm mL}{1\; \rm L} = 7.2\; \rm mL$ .

8 0

2 years ago

A level chem naming isomers​

A level chem naming isomers