Which contains more alcohol , a 12 oz. bottle of beer, a 4 oz glass of wine, or a 4 oz. of 100 proof liquor?

Can someone help me? It needs to have a diagram that has arrows.

daser333 [38]

Answer: The enthalpy change for formation of butane is -125 kJ/mol

Explanation:

The balanced chemical reaction is,

$C_4H_{10}(g)+\frac{13}{2}O_2(g)\rightarrow 4CO_2(g)+5H_2O(l)$

The expression for enthalpy change is,

$\Delta H=[n\times H_f{products}]-[n\times H_f{reactants}]$

Putting the values we get :

$\Delta H=[4\times H_f_{CO_2}+5\times H_f_{H_2O}]-[1\times H_f_{C_4H_{10}}+\frac{13}{2}\times H_f_{O_2}]$

$-2877=[(4\times -393)+(5\times -286)]-[1\times H_f_{C_4H_{10}}+\frac{13}{2}\times 0]$

$H_f_{C_4H_{10}=-125kJ/mol$

Thus enthalpy change for formation of butane is -125 kJ/mol

5 0

3 years ago

If the diet is deficient in one or more of the _____amino acids, the body must break down existing proteins to provide the amino

AysviL [449]

Essential amino acids
Amino acids are the molecules that make up proteins when they are linked up. There are certain essential amino acids, namely they are <span>histidine, isoleucine, leucine, lysine, methionine, phenylalanine, threonine, tryptophan, and valine.
All of these must be obtained from one's diet. If they are not, the body will use existing proteins, such as muscle mass, to obtain them.</span>

3 0

3 years ago

Read 2 more answers

The heat of vaporization for water is 40. 7 kJ/mol. How much heat energy must 150. 0 g of water absorb to boil away completely?

netineya [11]

The heat absorbed by the water sample for vaporization has been 6,105 kJ. Thus, option D is correct.

The heat of vaporization has been the amount of heat required to vaporize 1 gram of liquid.

The heat required for the vaporization has been given as:

$Q=m\Delta H_{vap}$

<h3 /><h3>Computation for the Heat of vaporization</h3>

The heat of vaporization of water has been given as, $\Delta H_{vap}=40.7\;\rm kJ/mol$

The mass of water sample has been, $m=150\;\rm g$

Substituting the values for the heat energy absorbed, <em>Q:</em>

<em />

<em /> $Q=150\;\times\;40.7\;\rm kJ\\
\textit Q=6,105\;kJ$

The heat absorbed by the water sample for vaporization has been 6,105 kJ. Thus, option D is correct.

Learn more about heat of vaporization, here:

brainly.com/question/2427061

3 0

3 years ago

Given that the freezing point depression constant for water is 1.86°c kg/mol, calculate the change in freezing point for a 0.907

melamori03 [73]

Answer : The correct answer for change in freezing point = 1.69 ° C

Freezing point depression :

It is defined as depression in freezing point of solvent when volatile or non volatile solute is added .

SO when any solute is added freezing point of solution is less than freezing point of pure solvent . This depression in freezing point is directly proportional to molal concentration of solute .

It can be expressed as :

ΔTf = Freezing point of pure solvent - freezing point of solution = i* kf * m

Where : ΔTf = change in freezing point (°C)

i = Von't Hoff factor

kf =molal freezing point depression constant of solvent. $\frac{^0 C}{m}$

m = molality of solute (m or $\frac{mol}{Kg}$ )

Given : kf = 1.86 $\frac{^0 C*Kg}{mol}$

m = 0.907 $\frac{mol}{Kg}$ )

Von't Hoff factor for non volatile solute is always = 1 .Since the sugar is non volatile solute , so i = 1

Plugging value in expression :

ΔTf = 1* 1.86 $\frac{^0 C*Kg}{mol}$ * 0.907 $\frac{mol}{Kg}$ )

ΔTf = 1.69 ° C

Hence change in freezing point = 1.69 °C

5 0

3 years ago

Which of the following happens to a molecule of an object when the object is cooled?

Romashka [77]

Loses kinetic energy

8 0

3 years ago