Question in the image

Why does every chemical reaction require a certain amount of activation energy

attashe74 [19]

Every chemical reaction requires a minimum amount of energy to undergo reaction.

The total energy required is Threshold energy. The chemicals have some internal energy and they require some extra energy which is known as activation energy.

The reaction occurs when two molecules or atoms come closer to each other and for this they need kinetic energy.

Now for an effective collision the two molecules or atoms should collide with a minimum amount of energy so that they may undergo reaction so again they need some energy

These kinetic energy and energy for effective collision is actually called as activation energy.

5 0

3 years ago

A sample of He gas (3.0 L) at 5.6 atm and 25°C was combined with 4.5 L of Ne gas at 3.6 atm and 25°C at constant temperature in

In-s [12.5K]

Answer:

3.6667

Explanation:

For helium gas:

Using Boyle's law

${P_1}\times {V_1}={P_2}\times {V_2}$

Given ,

V₁ = 3.0 L

V₂ = 9.0 L

P₁ = 5.6 atm

P₂ = ?

Using above equation as:

${P_1}\times {V_1}={P_2}\times {V_2}$

${5.6}\times {3.0}={P_2}\times {9.0} atm$

${P_2}=\frac {{5.6}\times {3.0}}{9.0} atm$

${P_1}=1.8667\ atm$

The pressure exerted by the helium gas in 9.0 L flask is 1.8667 atm

For Neon gas:

Using Boyle's law

${P_1}\times {V_1}={P_2}\times {V_2}$

Given ,

V₁ = 4.5 L

V₂ = 9.0 L

P₁ = 3.6 atm

P₂ = ?

Using above equation as:

${P_1}\times {V_1}={P_2}\times {V_2}$

${3.6}\times {4.5}={P_2}\times {9.0} atm$

${P_2}=\frac {{3.6}\times {4.5}}{9.0} atm$

${P_1}=1.8\ atm$

The pressure exerted by the neon gas in 9.0 L flask is 1.8 atm

Thus total pressure = 1.8667 + 1.8 atm = 3.6667 atm.

6 0

3 years ago

How many moles are in 38 grams of Argon? Round to 2 decimal places. Use "e" for exponents

yuradex [85]

Answer: 39.948

Explanation:

How many grams Argon in 1 mol? The answer is 39.948. We assume you are converting between grams Argon and mole. You can view more details on each measurement unit: molecular weight of Argon or mol The molecular formula for Argon is Ar.

3 0

3 years ago

Suppose you start with a solution of red dye #40 that is 2.3 ✕ 10−5 M. If you do three successive volumetric dilutions pipetting

larisa [96]

Answer: Therefore, the concentration of final solution is $4.0\times 10^{-8}M$