All categories

3 years ago

How many 2º alkyl bromides, neglecting stereoisomers, exist with the formula c6h13br?

Chemistry

1 answer:

deff fn [24]3 years ago

5 0

Secondary alkyl halide will be which have carbon attached to two other carbon atoms and the bromine is attached to such carbon.

So in case of bromo-hexane the following will be 2 degree or secondary alkyl halide

You might be interested in

What is the highest occupied molecular orbital in the O2 molecule?

Aleonysh [2.5K]

H2 is known to exist. For dihydrogen, H2, we can identify the frontier molecular orbitals (FMOs). The highest occupied molecular orbital (or HOMO) is the σ (sigma) 1s MO. The lowest unoccupied MO (LUMO) is the σ* (sigma star) 1s MO which is antibonding.

3 0

3 years ago

An aqueous solution of sodium thiosulfate releases heat when it crystallizes. What would happen when sodium thiosulfate crystals

rusak2 [61]

Answer:

An excellent experiment is to heat sodium thiosulfate in a water bath. The solid crystals will dissolve into the water in the hydrated crystals forming a supersaturated solution. ... Placing a small crystal in the supersaturated solution will cause the liquid to turn solid.

7 0

3 years ago

What is the bond order of li2−? express the bond order numerically?

jeka94

Atomic Number of Lithium is 3, so it has 3 electrons in its neutral state. Also, Li₂ will have 6 electrons. But the chemical formula we are given has a negative charge on it (i.e Li₂⁻) so there is an additional electron (RED) present on this compound. So, the total number of electrons are 7. The MOT diagram for this compound is shown below. According to diagram we are having 4 electrons in Bonding Molecular Orbitals (BMO) and 3 electrons in Anti-Bonding Molecular Orbitals (ABMO). Bond Order is calculated as,

Bond Order = (# of e⁻s in BMO - # of e⁻s in ABMO) ÷ 2

Bond Order = (4 - 3) ÷ 2

Bond Order = 1 ÷ 2
Or,
Bond Order = 1/2
Or,
Bond Order = 0.5

4 0

3 years ago

Arterial blood contains about 0.25 g of oxygen per liter at 37°C and standard atmospheric pressure. Under these conditions, the

Olin [163]

Firstly we need to determine the partial pressure of O2:

$\begin{gathered} P_{O_2}=X\times P_T \\ P_{O_2}:partial\text{ }pressure \\ X:mole\text{ }fraction \\ P_T:total\text{ }pressure \\ \\ P_{O_2}=0.209\times0.35\text{ }atm \\ P_{O_2}=0.073\text{ }atm \end{gathered}$

We will now use the Henry's Law equation to determine the solubility of the gas:

$\begin{gathered} c=K_H\times P_{O_2} \\ c:solubility\text{ }or\text{ }concentration\text{ }of\text{ }the\text{ }gas(M) \\ K_H:Henry^{\prime}sLawconstant=3.7\times10^{-2}M\text{ }atm^{-1} \\ P_{O_2}:partial\text{ }pressure\text{ }of\text{ }the\text{ }gas=0.073atm \\ \\ c=3.7\times10^{-2}M\text{ }atm^{-1}\times0.073 \\ c=2.7\times10^{-3}M \end{gathered}$

Answer: Solubility is 2.7x10^-3 M

6 0

11 months ago

A sealed helium balloon with an internal pressure of 1.00 atm and a volume of 4.50 L at 293 K is moved. What volume will the bal

bixtya [17]

Answer:

6.48 L

Explanation:

From the question,

Applying

PV/T = P'V'/T'......................... Equation 1

P = initial pressure of the helium balloon, V = Initial volume of the balloon, T = Initial temperature of the balloon, P' = Final pressure of the balloon, T' = Final temperature of the balloon, V' = Final volume of the balloon.

make V' the subject of the equation

V' = PVT'/P'T......................... Equation 2

Given: P = 1 atm, V = 4.5 L, T' = 253 K, T= 293 K, P' = 0.6 atm

Substitute these values into equation 2

V' = (4.5×1×253)/(0.6×293)

V' = 1138.5/175.8

V' = 6.48 L

8 0

3 years ago