Consider the intermediate equations:

3 equations. 1: upper C solid plus upper o subscript 2 gas right arrow upper C upper O subscript 2 gas Delta H 1 equals negative 393.5 kilojoules. 2: 2 upper C upper O gas plus upper O subscript 2 gas right arrow 2 upper C upper O subscript 2 gas Delta H 2 equals negative 566.0 kilojoules. 3: 2 upper H subscript 2 upper o gas right arrow 2 upper H subscript 2 gas plus upper O subscript 2 gas delta H 3 equals 483.6 kilojoules.

With the overall reaction:

Upper C (s) plus upper H subscript 2 upper O (g) right arrow upper C upper O (g) plus upper H subscript 2 (g).

What must be done to calculate the enthalpy of reaction? Check all that apply.

The first equation must be halved.

The first equation must be reversed.

The second equation must be halved.

The second equation must be reversed.

The third equation must be halved.

The third equation must be reversed.
What is the overall enthalpy of reaction?
=
kJ

first let us lay out the formula for potential

and kinetic energy:

pe = mgh

ke = (mv^2)/2

where: m=mass, v=velocity, g=gravitational

acceleration, h=height

calculating for mass using the known values of

pe:

340j = m (9.81 m/s^2) (50ft) (1m/3.28ft)

m = 2.27 kg

the law of conservation of energy states that

energy is neither created nor destroyed. therefore, the change in kinetic

energy from 50 ft to 35 ft would be equal to the change in potential energy

from 50 ft to 35 ft. however it would be opposite in signs since one is losing

while other is gaining.

ke(35ft) – ke(50ft) = - [ pe(35ft) – pe(50ft]

ke(50ft) = 0 since the brick is initially at rest

ke(35ft) = 340j - 2.27kg (9.81 m/s^2) (35ft)

(1m/3.28 ft)

ke(35ft)

= 102.38 j

as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2