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2 years ago

Draw the structure of (z)-3-methylhept-3-ene. be sure the stereochemistry is drawn clearly.

Chemistry

1 answer:

aivan3 [116]2 years ago

6 0

In alkene if two substituent and hydrogen are attached in the isomer may be cis or trans. When two or more substitution are attached to an alkene the isomer may be Z or E.All cis are Z isomer. The structure of (Z)-3-methy-3-heptene is as the following attachment

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For the atoms that do not follow the octet rule state how many electrons surround these atoms. Express your answers as integers

wolverine [178]

Answer:

The octet rule isn't all its cracked up to be. It works fine for some of the elements in the second period, but begins to come unglued for many others.

NNO does follow the octet rule, and has at least three resonance structures

N=N=O <−> N≡N−O <−> N−N≡O

Formal charges suggest that the one in the middle is most likely.

NBF3 -- Really? Show me, because I don't think there is such a compound.

NICl2^- or NICl^2- ..... which is it?

NICl2 (neutral) is an analogue of NCl3 where iodine has replaced one chlorine. It follows the octet rule.

Explanation:

7 0

1 year ago

From the following list of elements, those that will always form ionic compounds in a 1:2 ratio with zinc.

harina [27]

Answer:

A. iodine

C. fluorine

F. bromine

Explanation:

Ionic bonds occur mostly between metals and non-metals. Usually, a wide electronegativity difference is preferred between the two atoms. This makes one atom more desirous to gain electron and other more willing to donate electrons.

To have Zn forming a compound in the ratio of 1 to 2, the combining power must be similar to this.

The dominant oxidation state of Zn is the is the +2 state.

The other combining atoms must have the ability to recieve the two electrons.

The halogens fit perfectly into this picture. They need just an electron to attain nobility. They are also highly electronegative. If two halogens combines with the Zn, then the ionic bond will result.

The halogens are fluorine,chlorine, bromine, iodine and astatine.

They will form these compounds:

ZnF₂, ZnBr₂ and ZnI₂

3 0

2 years ago

A water treatment tablet contains 20.0 mg of tetraglycine hydroperiodide, 40.0% of which is available as soluble iodine. If two

Galina-37 [17]

Answer:

$16\ \text{ppm}$

Explanation:

Mass of one tablet = 20 mg

Mass of two tablets = $2\times 20=40\ \text{mg}$

Percent that is soluble in water = 40%

Mass of tablet that is soluble in water = $0.4\times 40=16\ \text{mg}$

So, mass of solute is $16\ \text{mg}$

Density of water = 1 kg/L

Volume of water = 1 L

So, mass of 1 L of water is $1\times 1=1\ \text{kg}=1000\ \text{g}$

PPM is given by

$\dfrac{\text{Mass of solute}}{\text{Mass of solvent}}\times 10^6=\dfrac{16\times 10^{-3}}{1000}\times 10^6\\ =16\ \text{ppm}$

Hence, the concentration of iodine in the treated water $16\ \text{ppm}$ .

8 0

1 year ago

Suppose you had used carbon tetrachloride, a liquid of density 2.20 g/mL, to determine the actual volume measured by your pipet.

VARVARA [1.3K]

Answer:

Carbon tetrachloride would be 2.2 fold heavier than water

Explanation:

Carbon tetrachloride (2.20g/mL) is denser than water (1.00g/mL)

4 0

1 year ago

A 0.72-mol sample of PCl5 is put into a 1.00-L vessel and heated. At equilibrium, the vessel contains 0.40 mol of PCl3(g) and 0.

Sonja [21]

Answer:

Equilibrium constant for $PCl_5$ is 0.5

Equilibrium constant for decomposition of $NO_2$ is $1.79 \times 10^{-14}$

Explanation:

$PCl_5$ dissociates as follows:

$PCl_5 \rightleftharpoons PCl_3+Cl_2$

initial 0.72 mol 0 0

at eq. 0.72 - 0.40 0.40 0.40

Expression for the equilibrium constant is as follows:

$k=\frac{[PCl_3][Cl_2]}{[PCl_5]}$

Substitute the values in the above formula to calculate equilibrium constant as follows:

$k=\frac{[0.40/1][0.40/1]}{0.32/1} \\=\frac{0.40 \times 0.40}{0.32} \\=0.5$

Therefore, equilibrium constant for $PCl_5$ is 0.5

Now calculate the equilibrium constant for decomposition of $NO_2$

It is given that $3.3 \times 10^{-3} \%$ is decomposed.

$NO_2$ decomposes as follows:

$2NO_2 \rightleftharpoons 2NO + O_2$

initial 1.0 M 0 0

at eq. concentration of $NO_2$ is:

$[NO_2]_{eq}=1-(0.000066) = 0.999934\ M$

$[NO]_{eq}=6.6 \times 10^{-5}\ M$

$[O_2]_{eq}=3.3\times 10^{-5} = 3.3\times 10^{-5}\ M$

Expression for equilibrium constant is as follows:

$K=\frac{[NO]^2[O_2]}{[NO_2]^2}$

Substitute the values in the above expression

$K=\frac{[6.6\times 10^{-5}]^2[3.3 \times 10^{-5}]}{[0.999934]^2} \\=1.79\times 10^{-14}$

Equilibrium constant for decomposition of $NO_2$ is $1.79 \times 10^{-14}$

8 0

2 years ago