The quantity of heat required to change the temperature of 1 g of a substance by 1 C is defined as

n-Butane (C4H10) is burned with stoichiometric amount of oxygen. Determine the mole fraction of carbon dioxide and water in the

Fudgin [204]

Answer:

See details below

Explanation:

The balanced reaction equation is given below:

$2C_{4} H_{10}$ + $13O_{2}$ → $8CO_{2}$ + $10H_{2} O$

Mole fraction of CO2 to H20

= 8/10 = $\frac{4}{5}$

Mole ratio of C4H10 to CO2 is 2:8 = 1:4

1 mole of n-butane - 38.12 g

4 moles - ?

= 152.48g fuel consumed.

8 0

2 years ago

The same reaction is begun with an initial concentration of 0.05 M O3 and 0.02 M NO. Under these conditions, the reaction reache

Ivahew [28]

The balanced equation of the reaction is:

O3(g) + NO (g) → O2 (g) + NO2 (g)

Then the ratios of reaction is 1 mol O3 : 1 mol NO : 1 mol O2 : 1 mol NO2

If you have initially 0.05 M of O3 and 0.02 M of NO, the reaction will end when all the NO is consumed.

The by the stoichiometry 0.02 mol of O3 will be consumed in 8 seconds.

And the rate of reaction is change in concetration divided by the time.

The change in concentration in O3 is 0.02 M

Then, the rate respect O3 is 0.02 M / 8 seconds = 0.0025 M/s

8 0

2 years ago

Sodium tert-butoxide (NaOC(CH3)3) is classified as bulky and acts as Bronsted Lowry base in the reaction. It is reacted with 2-c

IceJOKER [234]

Answer:but-1-ene

Explanation:This is an E2 elimination reaction .

Kindly refer the attachment for complete reaction and products.

Sodium tert-butoxide is a bulky base and hence cannot approach the substrate 2-chlorobutane from the more substituted end and hence major product formed here would not be following zaitsev rule of elimination reaction.

Sodium tert-butoxide would approach from the less hindered side that is through the primary centre and hence would lead to the formation of 1-butene .The major product formed in this reaction would be 1-butene .

As the mechanism of the reaction is E-2 so it will be a concerted mechanism and as sodium tert-butoxide will start abstracting the primary hydrogen through the less hindered side simultaneously chlorine will start leaving. As the steric repulsion in this case is less hence the transition state is relatively stabilised and leads to the formation of a kinetic product 1-butene.

Kinetic product are formed when reactions are dependent upon rate and not on thermodynamical stability.

2-butene is more thermodynamically6 stable as compared to 1-butene

The major product formed does not follow the zaitsev rule of forming a more substituted alkene as sodium tert-butoxide cannot approach to abstract the secondary proton due to steric hindrance.