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Mandarinka [93]
3 years ago
10

Dicarbon monoxide, C2O, is found in dust clouds in space. Analysis of it shows that the sequence of atoms in this molecule is C–

C–O. All bonds are double bonds and there are no unpaired electrons. How many lone pairs of electrons are present in a molecule of C2O?
Chemistry
2 answers:
lorasvet [3.4K]3 years ago
8 0

Hello there,


Ione pairs of electrons that are present in a molecule of C2O are...

4

Each Oxygen forms two bonds with Carbon


Hope I Helped!

-Char

katovenus [111]3 years ago
5 0

Answer:

Three

Explanation:

Each C atom supplies four valence electrons and each O atom supplies six, for a total of 14 valence electrons.

If there are two double bonds (eight bonding electrons), there must be six nonbonding electrons (i.e., three lone pairs).

The Lewis structure of C₂O is :C=C=Ö:

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Margaret [11]

Answer:

ugh it’s not coming out

Explanation:

5 0
2 years ago
You have 50 ml of a complex mixture of weak acids that contains some HF (pKa = 3.18) and some HCN (pKa = 9.21). Which is larger,
bonufazy [111]

Answer:

\frac{[F^{-}]}{[HF]} is larger

Explanation:

pK_{a}=-logK_{a} , where K_{a} is the acid dissociation constant.

For a monoprotic acid e.g. HA, K_{a}=\frac{[H^{+}][A^{-}]}{[HA]} and \frac{[A^{-}]}{[HA]}=\frac{K_{a}}{[H^{+}]}

So, clearly, higher the K_{a} value , lower will the the pK_{a}

In this mixture, at equilibrium, [H^{+}] will be constant.

K_{a} of HF is grater than K_{a} of HCN

Hence, (\frac{F^{-}}{[HF]}=\frac{K_{a}(HF)}{[H^{+}]})>(\frac{CN^{-}}{[HCN]}=\frac{K_{a}(HCN)}{[H^{+}]})

So, \frac{[F^{-}]}{[HF]} is larger

5 0
3 years ago
What is the kinetic energy of the emitted electrons when cesium is exposed to UV rays of frequency 1.3×1015Hz?
Sonja [21]

Answer:

We know that

ħf = ф + Ekmax  

where  

ħ = planks constant = 6.626x10^-34 J s  

f = frequency of incident light = 1.3x10^15 /s (1 Hz =

1/s)

ф = work function of the cesium = 2.14 eV  

Ekmax = max  kinetic energy of the emmitted electron.  

We distinguish that:

1 eV = 1.602x10^-19 J

So:

2.14 eV x (1.602x10^-19 J / 1 eV) = 3.428x10^-19 J

So,

Ekmax = (6.626x10^-34 J s) x (1.3x10^15 / s) - 3.428x10^-19 J

= 8.6138x10^-19 J - 3.428x10^-19 J = 5.1858x10^-19 J  

Answer:

5.19x10^-19 J  

Kinetic energy:

In physics, the kinetic energy of an object is the energy that it owns due to its motion. It is defined as the work required accelerating a body of a given mass from rest to its specified velocity. Having expanded this energy during its acceleration, the body upholds this kinetic energy lest its speed changes.

Answer details:

Subject: Chemistry

Level: College

Keywords:

• Energy  

• Kinetic energy

• Kinetic energy of emitted electrons

Learn more to evaluate:

brainly.com/question/4997492

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6 0
3 years ago
Read 2 more answers
An increase in temperature affects the reaction rate by >decreasing the velocities of particles that collide in the reaction.
Bond [772]

Answer:

increasing the number of molecules that have sufficient kinetic energy to react.

Explanation:

An increase in temperature affects the reaction rate by increasing the number of molecules that have sufficient kinetic energy to react.

or we say; temperature increase, leads to an increase in the amount of collisions between molecules.

4 0
3 years ago
Read 2 more answers
44.8% of a 250. mL acid solution is used in an experiment, what volume of acid (in mL) was used?
postnew [5]

Answer:

112 mL  

Explanation:

The formula for percent by volume is

\text{Percent by volume} = \dfrac{\text{Volume of solute}}{\text{Volume of solution}}\times 100 \, \%

If you have 250 mL of a solution that is 44.8 % v/v,

\begin{array}{rcl}44.8\, \% & = & \dfrac{\text{Volume of solute}}{\text{250 mL}}\times 100 \, \%\\\\44.8 \times \text{ 250 mL} & = & \text{Volume of solute} \times 100\\\text{Volume of solute} & = & \dfrac{44.8 \times 250\text{ mL}}{100}\\\\ & = & \textbf{112 mL}\\\end{array}

7 0
3 years ago
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