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Over [174]
2 years 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]2 years 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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Convert 5.0x10^24 molecules to liters.
makvit [3.9K]
Number of moles = 5 x 10^24 / 6.02 x 10^23 = 8.305 moles. Volume= moles x 22.4 = 186.032 liters. Hope this helps!
5 0
2 years ago
Percentage yield of sodium peroxide if 5 g of sodium oxide produces 5.5 g of sodium peroxide
Rama09 [41]
<h3>Answer:</h3>

87.40 %

<h3>Explanation:</h3>

Concept being tested: Percent yield of a product

We are given;

Mass of Sodium oxide 5 g

Experimental or Actual yield of sodium peroxide IS 5.5 g

We are required to calculate the percent yield of sodium peroxide;

The equation for the reaction that forms sodium peroxide is

2Na₂O + O₂ → 2Na₂O₂

<h3>Step 1; moles of sodium oxide</h3>

Moles = mass ÷ molar mass

Molar mass of sodium oxide is 61.98 g/mol

Therefore;

Moles = 5 g ÷ 61.98 g/mol

          = 0.0807 moles

<h3>Step 2: Theoretical moles of sodium peroxide produced </h3>

From the equation, 2 moles of sodium oxide produces 1 mole of sodium peroxide.

Thus, moles of sodium peroxide used is 0.0807 moles

<h3>Step 3: Theoretical mass of sodium peroxide used</h3>

Mass = Number of moles × Molar mass

Molar mass of sodium peroxide = 77.98 g/mol

Therefore;

Theoretical mass = 0.0807 moles × 77.98 g/mol

                            = 6.293 g

Theoretical mass of Na₂O₂ is 6.293 g

<h3>Step 4: Percent yield of Na₂O₂</h3>
  • We know that percent yield is given by the ratio of actual yield to theoretical yield expressed as a percentage.

Percent yield=(\frac{Actual yield}{theoretical yield})100

Percent yield(Na_{2}O_{2})=(\frac{5.5g}{6.293g})100

                       = 87.40 %

Therefore, the percentage yield of sodium peroxide is 87.4%

8 0
2 years ago
100 points!!!!!!!!!!!!!!!!!! how does a volcanic eruption benefit the surrounding area? A. Lava and ash bury animal habitats. B.
Shkiper50 [21]
I think it would be C) The surrounding soil can become very fertile

4 0
2 years ago
Read 2 more answers
what property of the noble gases most likely prevented the gases from being readily/easily discovered?
Vlad [161]

These gases very rarely react, with others and also noble gases are odourless and colourless.

Explanation:

  • Noble gases will not react with anything so that is the reason why they are known as an inert gas.
  • Noble gases are present in group 18 on the periodic table and following the rule of the octet which is they completed their orbital by s2p6 which is the highest energy level.
  • Most elements are discovering through their reactivity with the other elements, commonly with oxygen. In the case of a noble gas, it is difficult for a scientist to work with the gases which have very less or no chemical property in terms of their reactivity.

7 0
2 years ago
he first-order rate constant for the gas-phase decomposition of dimethyl ether, (CH3)2O → CH4 + H2 + CO is 3.2 ✕ 10−4 s−1 at 450
seropon [69]

Answer:

0.290 atm is the pressure of the system after 7.7min

Explanation:

The general first-order rate constant is:

ln [A] = -kt + ln [A]₀

<em>Where [A] is concentration of A in time t,</em>

<em>K is rate constant, 3.2x10⁻⁴s⁻¹</em>

<em>[A]₀ is initial concentration = 0.336atm.</em>

<em />

7.7 min are:

7.7min * (60s / 1min) = 462s

Solving:

ln [A] = -kt + ln [A]₀

ln [A] = -<em>3.2x10⁻⁴s⁻¹*462s</em> + ln [0.336atm]

ln [A] = -1.238

[A] =

<h3>0.290 atm is the pressure of the system after 7.7min</h3>

<em />

6 0
2 years ago
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