When phosphorus (P) and bromine (Br) bond, what kind of bond would happen?

What is a mixture called when it has its different components mixed evenly within the substance? Heterogeneous mixture Homogeneo

larisa86 [58]

It would be considered a Homogeneous Mixture. A mixture with two or more components mixed evenly is a Homogeneous mixture.

3 0

3 years ago

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A runner completes a 400-meter race in 2.5 mins. What is the runner's average speed?

NNADVOKAT [17]

The average speed is 160 meters a minute.

8 0

3 years ago

What is water oxidation number

-Dominant- [34]

Answer:

The formula for water is . The oxidation number of hydrogen is +1. Since there are two of them, the hydrogen atoms contribute to a charge of +2. The water molecule is neutral; therefore, the oxygen must have an oxidation number of to balance the charge.

6 0

3 years ago

Give regents and mechanism

Kryger [21]

Answer:

Reagents: 1) $BH_3$ 2) $H_2_O2$ , $OH^-$

Mechanism: Hydroboration

Explanation:

In this case, we have a hydration of alkenes reaction. But, in this example, we have an anti-Markovnikov reaction. In other words, the "OH" is added in the least substituted carbon. Therefore we have to choose an anti-Markovnikov reaction: "hydroboration".

The first step of this reaction is the addition of borane ( $BH_3$ ) to the double bond. Then in the second step, we have the deprotonation of the hydrogen peroxide, to obtain the peroxide anion. In the third step, the peroxide anion attacks the molecule produced in the first step to produce a complex compound in which we have a bond " $BH_2-O-OH$ ". In step number 4 we have the migration of the C-B bond to oxygen. Then in step number 5, we have the attack of $OH^-$ on the $O-BH_2$ to produce an alkoxide. Finally, the water molecule produce in step 2 will protonate the molecule to produce the alcohol.

See figure 1

I hope it helps!

4 0

3 years ago

The activation energy of an uncatalyzed reaction is 95kJ/mol. The addition of a catalyst lowers the activation energy to 55kJ/mo

notka56 [123]

Answer:

a) at 25°C the rate of reaction increases by a factor of 1,027*10^7

b) at 25°C the rate of reaction increases by a factor of 1,777*10^5

Explanation:

using the Arrhenius equation

k= ko*e^(-Ea/RT)

where

k= reaction rate

ko= collision factor

Ea= activation energy

R= ideal gas constant= 8.314 J/mol*K

T= absolute temperature

for the uncatalysed reaction

k1= ko*e^(-Ea1/RT)

for the catalysed reaction

k2= ko*e^(-Ea2/RT)

dividing both equations

k2/k1= e^(-(Ea2-Ea1)/RT)

a) at 25°C

k2/k1 = e^(-(55kJ/mol-95kJ/mol)/(8.314J/mol*K*298K)* (1000J/kJ ) ) = 1,027*10^7

therefore at 25°C , k2/k1 = 1,027*10^6

b) at 125°C

k2/k1 = e^(-(55kJ/mol-95kJ/mol)/(8.314J/mol*K*298K)* (1000J/kJ ) ) = 1,777*10^5

therefore at 125°C , k2/k1 = 1,777*10^5

Note:

when the catalysts is incorporated, the catalysed reaction and the uncatalysed one run in parallel and therefore the real reaction rate is

k real = k1 + k2 = k2 (1+k1/k2)

since k2>>k1 → 1+k1/k2 ≈ 1 and thus k real ≈ k2

6 0

3 years ago