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3 years ago

Watch the animation and identify the correct conditions for forming a hydrogen bond.

Chemistry

1 answer:

torisob [31]3 years ago

8 0

Answer:

options B,C and E are correct conditions for forming hydrogen bonds.

Explanation:

In order for particles to stick together or shows certain characteristics, there must be a bond of attractions that is making them to be able to stay together. Some of these attractions are strong while some are weak. The kind of attraction that is being considered in this question is the hydrogen bonding which is considered as part of the weak forces of attractions.

In order for hydrogen bonds to occur, there must be hydrogen atoms [option B is correct] and a much more electronegative elements such as oxygen, Fluorine and Nitrogen. Hence, option C is correct.

Since the Fluorine or oxygen or nitrogen have much more electronegative properties than hydrogen, it causes a partial negative charge on Fluorine or oxygen or nitrogen and a partial positive charge on the Hydrogen. Thus, option E is correct.

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The ability to put in a light bulb with greater ease is because the bottom half is what type of simple machine?

Travka [436]

A screw simple machine

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3 years ago

Estimate the coordination number for the cations in each of these ceramic oxides and also the coordination numbers of the oxygen

Juli2301 [7.4K]

Answer:

CN cation, anion ] respectively are thus [6,4] , [2,3] ,[6,6] , [6,6] , [6,3] , [12,6,2] , [6], [6].

Explanation:

The coordination number CN is the number of ligand atoms bonded (coordinate bonds) directly to the central of the metal ion. It is not the same as the oxidation state of the metal ion or complex.

Coordination number – the number of anions surrounding the cation.

In solving for CN we need to understand Pauling's rules.

According to Linus Pauling, 1932

“Pauling’s rules” for crystal structures, makes assumptions for ionic bonding. It states that ionic structure is understood using electrostatic rules of attraction and repulsion.

Cations and anions surround each other to neutralize charge – and these one can rationalize crystal structure with coordination number.

Ratio of cationic/anionic radius

• The structure of D-Al2O3 results in coordination number of 6 and 4 for cation and anion respectively.

• The average oxygen coordination number in v-B2O3 is equal to the average cation coordination number × cation/anion ratio (2/3).

• Co-ordination number of Ca2+ ion is =6;

In CaO crystal, Ca2+ is a cation and O2- is an anion. Cationic (Ca2+) has radius 100 pm and anionic (O2-) has radius 145 pm.

Ratio of cationic/anionic radius is:

r⁺/r⁻ = 100 / 145

r⁺/r⁻ = 0.69

CaO will form FCC lattice.

Coordination number in FCC lattice is 6. Therefore CN of Ca2+ = 6.

For MgO:

r Mg2+/ r O2- = 86pm / 126 pm =0.683

The cordination number for the cation is 6. MgO with ions Mg+2 and O-2 will have a AX type stochiometry exhibiting the

crystal structure of sodium chloride.

For TiO2:

The CN of the titanium (IV) cation is 6, which is twice the CN of the oxide anion, which is 3.

This fits with the formula unit of TiO2, since there are twice as many O2− ions as Ti4+ ions.

Consequently the crystal structure of all ionic compounds reflects the formula unit.

For LaAlO3 a Cubic perovskites (ABX3)

In perovskite structures, B cations are coordinated by six X anions, while A

cations present CN = 12 (also coordinated by X anions). The X anions have CN = 2, being

coordinated by two A cations, since the distance A-O is about 40% larger than the B-O

bond distance. The correct ionic radii (rA, rB, rX), taken from one of Shannon’s work.

rA = 1.36 pm

rB = 0.535 pm

rX = 1.35 pm

Forsterite Mg2SiO4

We have mixed sites of Si4+ in tetrahedral site, Mg2+ in octahedral site, O atoms anions .

They all forms octahedral chains/strips. The CN is estimated to (6) octahedral, with an average ratio 0.414

Nickel Cobaltite Ni(Co2O4)3

bidentate ligand includes the Oxalate, three oxalate ligands form six-coordinate bonds around the Ni2+ ion.

Co-ordinate number of Nickel in [Ni(C2O4)3] 4− is 3×2=6.

5 0

3 years ago

What is the element present for copper

amm1812

Copper is an brown-orange color which it's atomic number is 29. With high thermal and electricity conductivity with it's smooth surface.

8 0

3 years ago

Natural gas is stored in a spherical tank at a temperature of 13°C. At a given initial time, the pressure in the tank is 117 kPa

drek231 [11]

Answer:

1. the absolute pressure in the tank before filling = 217 kPa

2. the absolute pressure in the tank after filling = 312 kPa

3. the ratio of the mass after filling M2 to that before filling M1 = 1.44

The correct relation is option c ( $\frac{M_{2} }{M_{1} } = \frac{P_{2} T_{1} }{P_{1} T_{2} }$ )

Explanation:

To find -

1. What is the absolute pressure in the tank before filling?

2. What is the absolute pressure in the tank after filling?

3. What is the ratio of the mass after filling M2 to that before filling M1 for this situation?

As we know that ,

Absolute pressure = Atmospheric pressure + Gage pressure

So,

Before filling the tank :

Given - Atmospheric pressure = 100 kPa , Gage pressure = 117 kPa

⇒Absolute pressure ( p1 ) = 100 + 117 = 217 kPa

Now,

After filling the tank :

Given - Atmospheric pressure = 100 kPa , Gage pressure = 212 kPa

⇒Absolute pressure (p2) = 100 + 212= 312 kPa

Now,

As given, volume is the same before and after filling,

i.e. $V_{1}$ = $V_{2}$

As we know that, P ∝ M

⇒ $\frac{p_{1} }{p_{2} } = \frac{m_{1} }{m_{2} }$

⇒ $\frac{m_{2} }{m_{1} } = \frac{p_{2} }{p_{1} }$

⇒ $\frac{m_{2} }{m_{1} } = \frac{312 }{217 } = 1.4378$ ≈ 1.44

Now, as we know that PV = nRT

As V is constant

⇒ P ∝ MT

⇒ $\frac{P}{T}$ ∝ M

⇒ $\frac{M_{2} }{M_{1} } = \frac{P_{2} T_{1} }{P_{1} T_{2} }$

So, The correct relation is c option.

6 0

3 years ago

9. Using the balanced equation from Question #8, how many grams of lead will be produced if 2.54 grams of PbS is burned with 1.8

MissTica

Answer: 2.24 grams of Pb

Explanation:

Step 1

Balanced chemical reaction;

2PbS + 3O2 → 2Pb + 2SO3

Step 2

Moles of both PbS and O2

Moles = mass / molar mass

Moles of PbS = 2.54 g / 239.3 g/mol = 0.0108 moles

Moles of O2 = 1.88 / 32 g/mol = 0.0588 moles

Step 3

Finding the limiting reactant.

Limiting reactant, is that reactant which is completely used in the reaction;

If we assume that PbS is the limiting reactant;

We have 0.0588 moles of O2. This needs ( 0.0588 * 2) / 3 = 0.0392 moles of PbS to fully react. But we have only 0.0108 moles of PbS available. That means that the PbS will be completely consumed hence the limiting reactant

If we assume O2 is the limiting reactant;

We have 0.0108 moles of PbS. That needs ( 0.0108 * 3) / 2 = 0.0162 moles of O2. But we have 0.0588 moles of O2 which is in excess further confirming that PbS is the limiting reactant since it will be depleted in the reaction.

Step 4

Moles of lead

For this step we apply the mole ratios with the limiting reactant;

Mole ratio of PbS : Pb = 2 : 2 = 1 : 1

Therefore;

Moles of Pb = (0.0108 moles * 1 ) 1

Moles of Pb =0.0108 moles

Step 5

Mass of Pb

Mass = moles * molar mass

Mass of Pb =0.0108 moles * 207.2 g/mol

Mass of Pb = 2.24 grams

5 0

2 years ago