Hybridization, Specific heat & Change of Formation

2

sp

Bond angle: 180°

3

sp²

Bond angle: 120°

4

sp³

Bond angle: 109.5°

5

sp³d

90, 120, 180°

6

sp³d²

Bond angle: 90, 180°

If it doesn’t have hydrogen

Use octet rule

Ex. SeF6 → 34 e / 8 = 4.25

4 B.G. + 1 L.P. = 5 → sp³d

Remember **

Double bonds are still equal to one

Ex. Make sure there is an octet **

Diatomic elements

H2, O2, N2, F2, Cl2, Br2, I2

P4, S8 *

How much heat is absorbed by a copper penny with mass 3.10 g whose temperature rises from -8.0 °C to 37.0 °C?

Q = m x Cs x T

T is Final - Inital (37 - (-8.0))

3.10 g x Cu specific heat given x (37.0 - (-8.0))

Ending answer is in Joules (J) & in the correct amount of sig figs (3)

system produces 3.0 × 10² J of heat energy & surroundings do 7.6 J of work on the system

E = q + w

When the system produces or releases, its negative (-) & when the surroundings do work its (+)

Surroundings releases 3.0 kJ of heat & system produces 0.5 kJ of work

E = q + w

When surroundings release or produce its positive (+) and when systems does or produces work, its negative (-)

When all masses are the same, which of the following wild reach the highest temperature upon gaining 200.0 J of heat?

The one with the lowest specific heat because it requires the least energy to increase its temperature.

When all the masses are different, which would reach the highest temperature upon gaining 200.0 J of heat?

The highest specific capacity because it had the smallest mass & to allow for a temperature increase

How much heat is absorbed by a copper penny with mass 3.10 g whose temperature rises from -8.0 °C of 37.0 °C?

Q = m x Cs x T

T = final - inital

3.10 g x ____ specific heat capacity x T

= 53.7 J

Answer in joules and correct sig figs

Al2(SO4)3 (s)

2 Al + 3 S8 + 6 O2

3 Fe2O3 + CO → 2 Fe3O4 + CO2

Products - Reactants

2 (-1118) + (-394) — 3 (-824) + (-111)

Calculate separately & answer will be in kJ & usually a negative value

IF7 + I2 → IF5 + 2 IF

H°rxn = -89 kJ

Products - Reactants

-89 kJ = -840 + 2x — [(-941 + 0)]

KCF

-89kJ = -840 + 2x + 941

Calculate & the end goal should be in kJ (usually a negative number)

If the initial pressure of a system was 1.00 atm and the volume was tripled and the kelvin temperature was doubled, what is the final pressure?

A) 0.667 atm

B) 6.00 atm

C) 2.00 atm

D) 1.50 atm

P2 = 1.00 × 1/3 × 2 = 2/3 =0.667

A gas bottle contains 0.650 mol of gas at 730 mm Hg pressure. If the final pressure is 1.15 atm, how many moles of gas were added to the bottle?

A) 0.717 mol

B) 0.128 mol

C) 0.778 mol

D) 0.0680 mol

730 mm / 760 (standard) = 0.9605

0.650 × 1.15 / 0.9605 =0.778

0.778 - 0.650 again = 0.128 mol

A 45.0 L steel tank at 20.0°C contains C2H2 gas at a pressure of 1.39 atm. Assuming ideal behavior, how many grams of acetylene are in the tank?

A) 67.8 g

B) 990 g

C) 10.0 g

D) 2.60 g

1.39 × 45.0 / 0.08206 × 293.15 K =2.16

2.16 x molar mass of C2H2 (26.04) = 67.8 g

The density of ________ is 0.716 g/L at STP.

A) Ne

B) F2

C) CH4

D) CO

E) NO

Density = M * V

0.716 = m / 22.04 → 0.716 × 22.04 = 16.04 g/mol & it matches w CH4

The value of ΔH° for the reaction below is -482 kJ. Calculate the heat (kJ) released to the surroundings when 14.0 g of CO (g) reacts completely.

2CO (g) + O2(g) → 2CO2 (g)

A) 482 kJ

B) 240. kJ

C) -482 kJ

D) 120. kJ

E) 3370 kJ

CO = 28.01 g/mol

14.0g / 28.01 g/mol = 0.500 mol

Take -482 & divide by coefficient (2) for CO

Multiply by 0.500 & answer is +120 kJ

The value of ΔH° for the reaction below is -186 kJ.

H2 (g) + Cl 2 (g) → 2HCl (g)

The value of ΔH°f for HCl (g) is ________ kJ/mol.

A) -3.72 × 10^2

B) -1.27 × 10^2

C) -186

D) +186

E) -93.0

Products - Reactants

-186 = 2 HCl

-186/2 =-93 kJ/mol

when given a full equation 4NH3 (g) + 5O2 (g) → 4NO (g) + 6H2O (l)

It’s products-reactants

The total number of π bonds in the H—C≡C—CH=CH-C≡C—C≡N molecule is ________.

A) 3

B) 4

C) 5

D) 6

E) 7

Each triple bond has 2pi bonds

Each double bond has 1 pi bond

3 × 2 = 6

1 × 1 =1

= 7