Life science explores ____

Which of the following would result in butyl phenyl ether as the major product? a. Reaction of phenoxide with 1-bromobutane b. R

AlladinOne [14]

Answer:

Reaction of 1-butanol with bromobenzene

Explanation:

The reaction would yield ether as the major product is the reaction of the 1-butanol with bromobenzene. This is because the reaction does not have the large percentage of the undesired side product. In fact, the major product is about 85 % in composition, compared to the 15 % of the minor product. Hence, the reaction is efficient.

4 0

3 years ago

N2 (g) + __ H2 (g) --> _NH3 When the reaction above is completely

Advocard [28]

Answer:

3

Explanation:

left side has 2 N so right side must have a 2 which means 6 H on right side so to get 6 on left you have a coef. of 3 to make 6 H

7 0

2 years ago

Calculate the amount of heat that must be absorbed by 10.0 g of ice at –20°C to convert it to liquid water at 60.0°C. Given: spe

Murljashka [212]

Answer:

The amount of heat to absorb is 6,261 J

Explanation:

Calorimetry is in charge of measuring the amount of heat generated or lost in certain physical or chemical processes.

The total energy required is the sum of the energy to heat the ice from -20 ° C to ice of 0 ° C, melting the ice of 0 ° C in 0 ° C water and finally heating the water to 60 ° C.

So:

Heat required to raise the temperature of ice from -20 °C to 0 °C

Being the sensible heat of a body the amount of heat received or transferred by a body when it undergoes a temperature variation (Δt) without there being a change of physical state (solid, liquid or gaseous), the expression is used:

Q = c * m * ΔT

Where Q is the heat exchanged by a body of mass m, made up of a specific heat substance c and where ΔT is the temperature variation (ΔT=Tfinal - Tinitial).

In this case, m= 10 g, specific heat of the ice= 2.1 $\frac{J}{g*C}$ and ΔT=0 C - (-20 C)= 20 C

Replacing: Q= 10 g*2.1 $\frac{J}{g*C}$ *20 C and solving: Q=420 J

Heat required to convert 0 °C ice to 0 °C water

The heat Q necessary to melt a substance depends on its mass m and on the called latent heat of fusion of each substance:

Q= m* ΔHfusion

In this case, being 1 mol of water= 18 grams: Q= 10 g* $6.0 \frac{kJ}{mol} *\frac{1 mol of water}{18 g}$ = 3.333 kJ= 3,333 J (being kJ=1,000 J)

Heat required to raise the temperature of water from 0 °C to 60 °C

In this case the expression used in the first step is used, but being: m= 10 g, specific heat of the water= 4.18 $\frac{J}{g*C}$ and ΔT=60 C - (0 C)= 60 C

Replacing: Q= 10 g*4.18 $\frac{J}{g*C}$ *60 C and solving: Q=2,508 J

Finally, Qtotal= 420 J + 3,333 J + 2,508 J

Qtotal= 6,261 J

The amount of heat to absorb is 6,261 J

3 0

3 years ago

The cylinder of aluminum in class was approximately 13mm in circumference and 42mm in length. What would be it's approximate mas

Roman55 [17]

Answer:

Mass, m = 1.51 grams

Explanation:

It is given that,