C. Which molecule is not linear?<br> a H₂O<br> b. CO₂<br> c. N₂<br> d. Cl₂

Which metal atoms can form ionic bonds by losing electrons from both the outermost and next to outermost principal energy levels

RoseWind [281]

Full question options;

(Fe, Pb, Mg, or Ca)

Answer:

Iron - Fe

Explanation:

We understand tht metals pretty much form bonds by losing their valence (outermost electrons). But this question specifically asks for metals that lose beyond their outermost electrons; next to outermost principal energy levels.

Pb, Mg, and Ca only lose their outermost electrons to form the following ions;

Pb2+, Mg2+, and Ca2+.

This is because their ions have achieved a stable octet configuration - the dreamland of atoms where they are satisfied and don't need to go into reactions again.

Iron on the other hand has the following electronic configurations;

Fe: [Ar]4s2 3d6

Fe2+: [Ar]4s0 3d6

Fe3+: [Ar]4s0 3d5

This means ion can lose both the ooutermost electrons (4s) and next to outermost principal energy levels (3d). So correct option is Iron.

5 0

3 years ago

Who is the editor in chief of The Washington Post?.

Stels [109]

Steven Ginsberg named managing editor of The Washington Post, rounding out senior management team

6 0

2 years ago

Be sure to answer all parts. Consider the reaction N2(g) + 3H2(g) → 2NH3(g)ΔH o rxn = −92.6 kJ/mol If 4.0 moles of N2 react with

Sphinxa [80]

Answer : The work done is, $1.98\times 10^4J$

Explanation :

The given balanced chemical reaction is:

$N_2(g)+3H_2(g)\rightarrow 2NH_3(g)$

When 4 moles of $N_2$ react with 12 moles of $H_2$ then it gives 8 moles of $NH_3$

First we have to calculate the change in moles of gas.

Moles on reactant side = Moles of $N_2$ + Moles of $H_2$

Moles on reactant side = 4 + 12 = 16 moles

Moles on product side = Moles of $NH_3$

Moles on reactant side = 8 moles

Change in moles of gas = 16 - 8 = 8 moles

Now we have to calculate the change in volume of gas.

Using ideal gas equation:

$PV=nRT$

where,

P = Pressure of gas = 1.0 atm

V = Volume of gas = ?

n = number of moles of gas = 8 mole

R = Gas constant = $0.0821L.atm/mol.K$

T = Temperature of gas = $25^oC=273+25=298K$

Putting values in above equation, we get:

$1.0atm\times V=8mole\times (0.0821L.atm/mol.K)\times 298K$

$V=195.7L$

As the number of moles of gas decreased. So, the volume also deceased. Thus, the volume of gas will be, -195.7 L

Now we have to calculate the work done.

Formula used :

$w=-p\Delta V$

where,

w = work done

p = pressure of the gas = 1.0 atm

$\Delta V$ = change in volume = -195.7 L

Now put all the given values in the above formula, we get:

$w=-p\Delta V$

$w=-(1.0atm)\times (-195.7L)$

$w=195.7L.artm=195.7\times 101.3J=19824.41J=1.98\times 10^4J$

conversion used : (1 L.atm = 101.3 J)

Thus, the work done is, $1.98\times 10^4J$

6 0

2 years ago

Based on the equation, how many grams of Br2 are required to react completely with 29.2 grams of AlCl3?

goblinko [34]

Answer:

answer is 34.996 or 35 grams

Explanation:

Br2 + AlCl3

mass 34.996 29.2

RFM 160 133.5

moles 0.2187 0.2187

3 0

3 years ago

Describe the crystallization process as applied in salt preparation

shtirl [24]

Explanation:

The principle used in the preparation of these salts is to dissolve

the cystine in an alcoholic alkali solution to which just sufficient

water to effect solution has been added, and, after filtering from

excess cystine, to precipitate the salt by addition of a suitable

indifferent solvent. While various solvents, such as acetone,

ether, or large amounts of alcohol caused precipitations, these were

either oily or amorphous. Only acetonitrile was found to possess

the power of initiating regular crystallization in the salt solutions.

While the solutions of the different salts require different amounts

of the solvent for complete precipitation, a partial substitution of

absolute ether for acetonitrile was found expedient in the case

of the most soluble of the salts, the K salt.

6 0

2 years ago

Read 2 more answers

C. Which molecule is not linear? a H₂O b. CO₂ c. N₂ d. Cl₂