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3 years ago

Identify the hybridization of the n atom in nf3.

2 answers:

7 0

The hybridization would be sp2. it goes from s, sp, sp2, sp3, sp3d, and so on. each bond gets one more level. since there are three bonds connecting to nitrogen it would be sp2.

guapka [62]3 years ago

7 0

Answer:

sp3 hybridization

Explanation:

Hello,

Based on the Lewis structure Nitrogen has 3 valence electrons with a lone pair of non-bonded electrons, thus, the hybridization calculation rule,

$Hybridization=\#Attached.Atoms+\# lone.pairs\\$

Allows us to identify that the $NF_3$ has three attached atoms and one lone pair. Now, by considering the convention: sp3=4, sp2=3 and sp=2 , one finds that:

$Hybridization=3+1=4$

Therefore, $NF_3$ has sp3 hybridization.

Best regards.

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Ultraviolet radiation and radiation of shorter wavelengths can damage biological molecules because they carry enough energy to b

Alja [10]

Answer:

439.7nm

Explanation:

Energy of a quantum can be calculated using below formula

E=hv...........eqn(1)

But v=λ/ c .........eqn(2)

If we substitute eqn(2) into eqn(1) we have

E= hc/(λ)

Where E= energy

h= Plank's constant= 6.62607004 × 10-34 m2 kg / s

c= speed of light

c= 2.998 × 10^8 m/s

λ= wavelength= ?

But the energy was given in Kj , it must be converted to Kj/ photon for unit consistency.

Energy E= 272 kJ/mol × 1mol/6.02× 10^23

Energy= 451.83× 10^-24 Kj/ photon

E= hc/(λ)...........eqn(1)

If we make λ subject of the formula

λ= hc/E

Then substitute the values we have

λ= [(6.626 × 10^-34) × (2.998 × 10^8)]/451.83× 10^-24

λ=(0.00043965) × (1Kj/1000J) × (10^9nm/1m)

λ=439.7nm

Hence, the longest wavelength of radiation with enough energy to break carbon-sulfur bonds is 439.7nm

4 0

3 years ago

Calculate the amount of heat that must be absorbed by 10.0 g of ice at –20°C to convert it to liquid water at 60.0°C. Given: spe

Murljashka [212]

Answer:

The amount of heat to absorb is 6,261 J

Explanation:

Calorimetry is in charge of measuring the amount of heat generated or lost in certain physical or chemical processes.

The total energy required is the sum of the energy to heat the ice from -20 ° C to ice of 0 ° C, melting the ice of 0 ° C in 0 ° C water and finally heating the water to 60 ° C.

So:

Heat required to raise the temperature of ice from -20 °C to 0 °C

Being the sensible heat of a body the amount of heat received or transferred by a body when it undergoes a temperature variation (Δt) without there being a change of physical state (solid, liquid or gaseous), the expression is used:

Q = c * m * ΔT

Where Q is the heat exchanged by a body of mass m, made up of a specific heat substance c and where ΔT is the temperature variation (ΔT=Tfinal - Tinitial).

In this case, m= 10 g, specific heat of the ice= 2.1 $\frac{J}{g*C}$ and ΔT=0 C - (-20 C)= 20 C

Replacing: Q= 10 g*2.1 $\frac{J}{g*C}$ *20 C and solving: Q=420 J

Heat required to convert 0 °C ice to 0 °C water

The heat Q necessary to melt a substance depends on its mass m and on the called latent heat of fusion of each substance:

Q= m* ΔHfusion

In this case, being 1 mol of water= 18 grams: Q= 10 g* $6.0 \frac{kJ}{mol} *\frac{1 mol of water}{18 g}$ = 3.333 kJ= 3,333 J (being kJ=1,000 J)

Heat required to raise the temperature of water from 0 °C to 60 °C

In this case the expression used in the first step is used, but being: m= 10 g, specific heat of the water= 4.18 $\frac{J}{g*C}$ and ΔT=60 C - (0 C)= 60 C

Replacing: Q= 10 g*4.18 $\frac{J}{g*C}$ *60 C and solving: Q=2,508 J

Finally, Qtotal= 420 J + 3,333 J + 2,508 J

Qtotal= 6,261 J

The amount of heat to absorb is 6,261 J

3 0

3 years ago

Mg3N2(s)+6H2O(l)→3Mg(OH)2(s)+2NH3(g) When 36.0 g of H2O react, how many grams of NH3 are produced? When 36.0 g of H2O react, how

Afina-wow [57]

Answer:

11.3 g of $NH_{3}$ are produced from 36.0 g of $H_{2}O$

Explanation:

1. The balanced chemical equation is the following:

$Mg_{3}N_{2}(s)+6H_{2}O(l)=3Mg(OH)_{2}(s)+2NH_{3}(g)$

2. Use the molar mass of the $H_{2}O$ , the molar mass of the $NH_{3}$ and the stoichiometry of the balanced chemical reaction to find how many grams of $NH_{3}$ are produced:

Molar mass $H_{2}O$ = 18 $\frac{g}{mol}$