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jonny [76]
3 years ago
14

Choose the correct total number of electron domains (bonding and nonbonding) about a central atom if the angle(s) between the el

ectron domains are all about 120 degrees. a) The central atom has 2 electron domains. b) The central atom has 5 electron domains. c) The central atom has 6 electron domains. d) The central atom has 3 electron domains. e) The central atom has 4 electron domains.
Chemistry
1 answer:
slamgirl [31]3 years ago
5 0

Answer:

The central atom has 3 electron domains.

Explanation:

According to the Valence Shell electron pair repulsion theory (VSEPR) put forward by Gillespie and Nyholm in 1957, the shape of a molecule is determined by repulsion between all the electron pairs (electron domains) present in the valence shell.

The electron pairs or electron domains are known to position themselves as far apart in space as possible in order to minimize repulsions.

Hence, when the central atom of a molecule contains three electron domains, they are positioned at an angle of 120° from each other to minimize repulsions. Hence the answer.

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Answer:

diamns

Explanation:

6 0
2 years ago
The lead-containing reactant(s) consumed during recharging of a lead-acid battery is/are ________. The lead-containing reactant(
irga5000 [103]

Answer:  The  lead-containing reactant(s) consumed during recharging of a lead-acid battery is PbSO_4(s)

Explanation:

In lead acid battery, the anode is made up of lead and undergoes oxidation during discharging and cathode is made up of lead oxide and acts as cathode during discharging. The electrolyte used is dilute H_2SO_4.

Charging:

Cathode : reduction : PbSO_4(s)+2e^{-}\rightarrow Pb+SO_4^{2-}

Anode: oxidation : PbSO_4+2H_2O\rightarrow  PbO_2+SO_4^{2-}+4H^++2e^-

Overall reaction : 2PbSO_4+2H_2O\rightarrow Pb+PbO_2+2H_2SO_4

The  lead-containing reactant(s) consumed during recharging of a lead-acid battery is PbSO_4(s)

4 0
2 years ago
Read 2 more answers
Calculate the mass in grams of 8.35 × 1022 molecules of CBr4.
antiseptic1488 [7]

Answer: 45.983 g CBr₄

Explanation:

To convert from moles to grams, you know that we will need molar mass and Avogadro's number.

Avogadro's number: 6.022×10²³ molecules/mol

Molar mass: 331.627 grams/mol

Now that we have what we need, you can use these to solve for grams. 8.35*10^2^2molecules CBr_{4} *\frac{1mol}{6.022*10^2^3molecules} *\frac{331.627 g}{1 mol} =45.983 g

Our final answer is 45.983 g CBr₄.

4 0
3 years ago
Scoring Scheme: 3-3-2-1 Use one of your experimentally determined values of k, the activation energy you determined, and the Arr
Lena [83]

Answer:

K = 2.7x10⁻⁵ at 25ºC

Explanation:

A way to write Arrhenius equation is:

ln K = - Ea/R × (1/T) + lnA

If you graph ln K as Y and 1/T as X (Absolute temperature in K), the equation you will obtain is:

Y = -13815X +35.817

R² = 0.9927

(Taking the last k point as 0.0386) (ln 0.0386), <em>0.1386 has no sense</em>)

Your slope is -13815

-13815K = - Ea/R

-13815K×8.314J/molK = 114858J/mol = Ea

And your intercept =

lnA = 35.817

A = 3.59x10¹⁵

Now, you want to know rate constant at 25ºC = 298.15K. Replacing in the equation (Where Y is ln (activation energy) and X is 1/T):

Y = -13815X +35.817

Y = -13815(1/298.15K) +35.817

Y = -10.5187

lnK = -10.5187

<h3>K = 2.7x10⁻⁵ at 25ºC</h3>

8 0
3 years ago
If the seawater carbonate ion (CO3 2-) concentration is 270 µmol/kg, a) what is the approximate rate of calcification, and b) ap
STALIN [3.7K]

Explanation:

Below are attachments containing the graph and solution.

6 0
2 years ago
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