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Over [174]
1 year ago
12

Butyl butyrate is an ester that is a naturally occurring oil used in the flavor industry for its fruity scent. If the steam dist

illation of butyl butyrate with water has partial pressures of 50 mm Hg and 710 mm Hg respectively, how much of the distillate is water
Chemistry
1 answer:
Snezhnost [94]1 year ago
8 0

Answer:

The correct answer is 62.5 %.

Explanation:

Based on the given information, the partial pressure of butyl butyrate is 50 mmHg and the partial pressure of water is 710 mmHg.  

Hence, the total pressure is 710+50 = 760 mmHg

According to Dalton's law of partial pressure,  

Partial pressure = mole fraction * total pressure

Mole fraction of water is,  

Partial pressure of water/Total pressure = 710/760 = 0.93

Similarly, the mole fraction of butyl butyrate is,  

Partial pressure of butyl-butyrate/Total pressure = 50/760 = 0.07

Therefore, mole% of water is 0.93 * 100 = 93 %

For calculating mass%,  

Mass of H2O = 0.93 * 18 = 16.8 grams (The molecular mass of water is 18 grams per mole)

The molecular mass of butyl-butyrate is 144 gram per mole

The mass of butyl-butyrate = 144 * 0.07 = 10.08 grams

The mass percent of water will be,  

Mass % of water/Total mass % * 100 = 16.8 / 10.08 + 16.8 * 100 = 62.5%.  

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Under standard conditions, a given reaction is endergonic (i.e., ΔG >0). Which of the following can render this reaction favo
sleet_krkn [62]

Answer:

Maintaining a high starting-material concentration can render this reaction favorable.

Explanation:

A reaction is <em>favorable</em> when <em>ΔG < 0</em> (<em>exergonic</em>). ΔG depends on the temperature and on the reaction of reactants and products as established in the following expression:

ΔG = ΔG° + R.T.lnQ

where,

ΔG° is the standard Gibbs free energy

R is the ideal gas constant

T is the absolute temperature

Q is the reaction quotient

To make ΔG < 0 when ΔG° > 0 we need to make the term R.T.lnQ < 0. Since T is always positive we need lnQ to be negative, what happens when Q < 1. Q < 1 implies the concentration of reactants being greater than the concentration of products, that is, maintaining a high starting-material concentration will make Q < 1.

5 0
1 year ago
Which statement about the electron-cloud model is true? It is the currently accepted atomic model. It can easily be replaced by
yan [13]

Answer:

3 choice

Explanation:

7 0
1 year ago
Read 2 more answers
Naomi is investigating the properties of a solid material. It takes 120 joules to raise the temperature of 10 grams of the mater
MrRissso [65]
When heat energy is supplied to a material it can raise the temperature of mass of the material.
Specific heat is the amount of energy required by 1 g of material to raise the temperature by 1 °C.
equation is 
H = mcΔt
H - heat energy 
m - mass of material 
c - specific heat of the material 
Δt - change in temperature
substituting the values in the equation 
120 J = 10 g x c x 5 °C
c = 2.4 Jg⁻¹°C⁻¹
3 0
2 years ago
A beaker has 0.2 M of Na2SO4. What will be the concentration of sodium and sulfate ions?
netineya [11]

The concentration of sodium and sulphate ions are [Na^+] = 0.4 M, [SO_4^2-] = 0.2 M

Explanation:

The molar concentration is defined as the number of moles of a molecule or an ion in 1 liter of a solution.

In the given solution, the concentration of the salt sodium sulphate is 0.2M. So, 0.2 moles of sodium sulphate is present in 1 liter of solution.

Assuming 100% dissociation,

1 molecule of sodium sulphate gives 2 ions of sodium and 1 ion of sulphate.

So 0.2 moles of sodium sulphate will give 0.4 moles of sodium ions and 0.2 moles of sulphate ions.

7 0
2 years ago
The isotope 64Cu has t1/2 of 12.7 hours. If the initial concentration of this isotope in an aqueous solution is 845 ppm, what wi
Kisachek [45]

Answer:

A = 679.2955 ppm

Explanation:

In this case, we already know that 64Cu has a half life of 12.7 hours. The expression to use to calculate the remaining solution is:

A = A₀ e^-kt

This is the expression to use. We have time, A₀, but we do not have k. This value is calculated with the following expression:

k = ln2 / t₁/₂

Replacing the given data we have:

k = ln2 / 12.7

k = 0.0546

Now, let's get the concentration of Cu:

A = 845 e^(-0.0546*4)

A = 845 e^(-0.2183)

A = 845 * 0.8039

A = 679.2955 ppm

This would be the concentration after 4 hours

7 0
1 year ago
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