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2 years ago

A triple bond is made up of sigma bond(s) and 1,2 pi bond(s). 2,1

Chemistry

2 answers:

wolverine [178]2 years ago

8 0

Answer:

1 sigma bond and 2 pi bonds

Explanation:

A triple bond is made up of 1 sigma bond and 2 pi bonds.

Given example is that of ethyne

valina [46]2 years ago

3 0

Answer:

1 $\sigma$ and 2π bonds

Explanation:

Always remember that

First bond among any two atoms is always Sigma bond .then rest forming bonds are pi bonds
Breaking a Signa bond is far difficult than breaking pi bonds
That's why tripple bond is unstable

The general example is

N=N bond

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Silver is a white metal that is an excellent conductor. Silver tarnishes when exposed to air and light. The density of silver is

Stells [14]

Answer:

c tarnishes in air

Explanation:

After silver has been exposed to air that contains sulphur gases, discoloration would occur. there would be darkening that is caused by the reaction with gases.When any silver object tarnishes, it brings about a disfiguring of that object. Hydrogen sulphide would be needed for this to happen. silver sulphide is black and a if a thin layer should form on any surface, it ill darken it. This is what we refer to as tarnishing.

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3 years ago

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A certain compound is made up of two chlorine atoms, one carbon atom, and one oxygen atom. what is the chemical formula of this

Ostrovityanka [42]

COCl₂ (phosgene)
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3 0

3 years ago

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Consider the reaction below. Initially the concentration of SO2Cl2 is 0.1000 M. Solve for the equilibrium concentration of SO2Cl

tensa zangetsu [6.8K]

Answer:

$[SO_2Cl_2] = 0.09983 M$

Explanation:

Write the balance chemical equation ,

$SO_2Cl_2((g) = SO_2(g) + Cl_2(g)$

initial concenration of $SO_2Cl_2((g) =0.1M$

lets assume that degree of dissociation= $\alpha$

concenration of each component at equilibrium:

$[SO_2Cl_2] = 0.1-0.1\alpha$

$[SO_2] = 0.1\alpha$

$[Cl_2] = 0.1\alpha$

$Kc =\frac{0.1\alpha \times 0.1\alpha}{0.1-0.1\alpha}$

$Kc =\frac{0.1\alpha \times \alpha}{1-\alpha}$

as $\alpha$ is very small then we can neglect $1-\alpha$

therefore ,

$Kc ={0.1\alpha \times \alpha}$

$\alpha =\sqrt{\frac{Kc}{0.1}}$

$\alpha = 1.73 \times 10^{-3}$

Eqilibrium concenration of $[SO_2Cl_2] = 0.1-0.1\alpha = 0.1-0.1\times 0.00173$

$[SO_2Cl_2] = 0.09983 M$

4 0

3 years ago

In a 0.730 M solution, a weak acid is 12.5% dissociated. Calculate Ka of the acid.

Mamont248 [21]

Answer:

Approximately $1.30 \times 10^{-2}$ , assuming that this acid is monoprotic.

Explanation:

Assume that this acid is monoprotic. Let $\rm HA$ denote this acid.

$\rm HA \rightleftharpoons H^{+} + A^{-}$ .

Initial concentration of $\rm HA$ without any dissociation:

$[{\rm HA}] = 0.730\; \rm mol \cdot L^{-1}$ .

After $12.5\%$ of that was dissociated, the concentration of both $\rm H^{+}$ and $\rm A^{-}$ (conjugate base of this acid) would become:

$12.5\% \times 0.730\; \rm mol \cdot L^{-1} = 0.09125\; \rm mol \cdot L^{-1}$ .

Concentration of $\rm HA$ in the solution after dissociation:

$(1 - 12.5\%) \times 0.730\; \rm mol \cdot L^{-1} = 0.63875\; \rm mol\cdot L^{-1}$ .

Let $[{\rm HA}]$ , $[{\rm H}^{+}]$ , and $[{\rm A}^{-}]$ denote the concentration (in $\rm mol \cdot L^{-1}$ or $\rm M$ ) of the corresponding species at equilibrium. Calculate the acid dissociation constant $K_{\rm a}$ for $\rm HA$ , under the assumption that this acid is monoprotic:

$\begin{aligned}K_{\rm a} &= \frac{[{\rm H}^{+}] \cdot [{\rm A}^{-}]}{[{\rm HA}]} \\ &= \frac{(0.09125\; \rm mol \cdot L^{-1}) \times (0.09125\; \rm mol \cdot L^{-1})}{0.63875\; \rm mol \cdot L^{-1}}\\[0.5em]&\approx 1.30 \times 10^{-2} \end{aligned}$ .

5 0

3 years ago

Select the procedure that will NOT help saturate a thin‑layer chromatography (TLC) developing chamber with developing solvent va

fenix001 [56]

Answer:

Explanation:

(A) Cover the chamber. => It helps to keep the solvent in his gas state

(B) Gently swirl the solvent in the chamber prior to placing the TLC plate inside => The mechanical energy can promote the conversion from the liquid state to the gas state and help to the saturation process.

(C) Add ninhydrin (a visualization technique) to the developing solvent =>There is no change in the saturation process

(D) Place a paper wick (like a piece of filter paper) inside the chamber=>Increases the area for evaporation. The solvent can go up across the paper by capillarity a then can be evaporated increasing the saturation in the chamber.

3 0

3 years ago