In an extraction experiment similar to your expt. 4, you attempt to separate 2-naphthol from cyclohexane, both of which are dissolved in ether. You attempt the separation by extracting the organic layer with a Ì´5% NaOH (aq) solution (made by adding 5.00 g of NaOH to every 100 mL of water). Assuming that all of the base present in the aqueous solution reacts with the phenol during your extractions, what is the minimum volume of the aqueous NaOH solution needed to deprotonate 20.0 g of 2-naphthol in order to extract it into the water layer? Clearly show all calculations. After removal of the phenol from the ether layer, only cyclohexane would remain. Concisely explain how you could efficiently isolate relatively pure cyclohexane from the ether layer.

The compound has the molecular formula CH₄O. H : O = 1 : 4 : 1 As a result, the empirical formula is CH₄O.

The organic compound sample's mass is 0.5 g, or (0.188 + 0.062 + 0.25) The vapor density should not have a unit.

The organic compound has a molar mass = 16 × 2 g/mol

                                                                            = 32 g/mol

Moles of C atoms = (32 g) × (0.188/0.5) / (12 g/mol) = 1

Moles of H atoms = (32 g) × (0.062/0.5) / (1 g/mol) = 4

Moles of O atoms = (32 g) × (0.25/0.5) / (16 g/mol) = 1

Hence, molecular formula = CH₄O

Thus, the compound has the molecular formula CH₄O. H : O = 1 : 4 : 1 As a result, the empirical formula is CH₄O.

Organic molecules contain covalently bound carbon and hydrogen as well as frequently additional elements. Benzoic acid, diethyl malonate, propanoic acid, butanoic acid, malonic acid, amines, heterocyclic compounds, and benzoic aldehyde are all examples of organic compounds. Organic substances include the carbohydrates, fats (lipids), proteins, and nucleic acids that make up the components of life. Organic molecules include petroleum and natural gas, which constitute the majority of fossil fuels.

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Answer:

Explanation:

The first step in order to solve this particular question is to make sure that the two reactions given in the question is balanced. Therefore, we have;

Reaction 1: C6H12O6 -----------------------------> 2 C2H5OH + 2 CO2.

Reaction 2: C6H12O6 ------------------------------> 2 C2H3CO2H + 2 H2O.

Next, we determine the number of moles of water and that of glucose. Recall that we are given from the question that the open flow reactor =  4000 kg of a 12% glucose-water solution flows in that is to say the percentage for water is [100% - 12% = 88%]. Also, the molar mass of water = H₂O = 18 kg/kmol and that for glucose =180 kg/kmol.

Number of moles of water = (4000 kg × 88%) ÷ 18 = 195.6 kmol.

Number of moles of glucose= (4000kg × 12%) ÷ 180 = 2.67 kmol.

Next thing to do is to determine the number of moles in the unreacted glucose . Therefore, the 90 kg of unreacted glucose ÷ 180kg/kmol = 0.5  kmol.

So, we have that During fermentation, 120 kg of carbon dioxide is produced.  Thus, the number of kmol = 120kg÷ 44kg/mol = 2.73kmol.

For reaction 1, we have 2 moles of CO₂ that is to say the extent of the reaction = 2.73kmol / 2 moles of CO₂ = 1.365 kmol.

For reaction 2, we have 2 moles of CO₂ that is to say the extent of the reaction = 2.67kmol - 0.5  kmol - 1.365kmol = 0.805 kmol.

For both reaction, the kmol for outflow of glucose = 2.73 kmol.

Also, 2 × 0.805 + 1.365 × 0 = 1.61kmol.

Hence, 195.6 kmol + 1.61 =197.21 kmol.

The mass of ethanol = 46.1 kg/kmol × 2.73 kmol = 125.853 kg.

The weight percent of ethanol has been 29.72%, and the percent mass of propionic acid has been 47.79%.

The balanced chemical equations of the reactions have been:

\(\rm C_6H_1_2O_6\;\rightarrow\;2\;C_2H_5OH\;+\;2\;CO_2\)

\(\rm C_6H_1_2O_6\;\rightarrow\;2\;C_2H_3CO_2H\;+\;2\;H_2O\)

The solution has consisted of 12% glucose.

Water in the solution = 88%

The mass of the solution = 4000 kg.

The moles of glucose = 12% of 4000 kg

Moles of glucose = \(\rm \dfrac{12}{100}\;\times\;4000\;\times\;\dfrac{1}{180\;g/mol}\)

Moles of glucose = 2.67 kmol.

Moles of water =  88 % of 4000 kg

Moles of water = \(\rm \dfrac{88}{100}\;\times\;4000\;\times\;\dfrac{1}{18\;g/mol}\)

Moles of water = 195.6 kmol.

The unreacted glucose in the mixture = 90 kg

Moles of unreacted glucose:

Moles = \(\rm \dfrac{weight}{molecular\;weight}\)

Moles of unreacted glucose = \(\rm \dfrac{90\;\times\;1000\;g}{180\;g/mol}\)

Moles of unreacted glucose = 0.5 kmol

The mass of carbon dioxide produced = 120 kg.

Moles of carbon dioxide produced = \(\rm \dfrac{120\;\times\;1000\;g}{44\;g/mol}\)

Moles of carbon dioxide produced = 2.73 kmol

Since 1 mole of glucose produces 2 moles of carbon dioxide.

2.73 kmol of carbon dioxide has been produced from 1.365 kmol of glucose.

The moles of ethanol produced by reaction 1 = 2 moles/ mole glucose.

The glucose present has been 2.67 kmol.

The ethanol produced = 5.34 kmol.

Moles of propionic acid produced = 5.34 kmol.

The mass of 5.34 kmol ethanol = Moles × molecular weight

The mass of ethanol produced = 5.34 × 1000 × 46.07g/mol

The mass of ethanol produced = 246.0138 kg.

The mass of propionic acid produced = 5.34  × 1000 × 74.08 g/mol

The mass of propionic acid produced = 395.5872 kg.

The mass of water produced = 5.34 × 1000 × 18 g/mol

The mass of water produced = 96.12 kg.

The remained glucose = 90 kg

The total mass in the reactor:

= Mass of glucose + water + propionic acid + ethanol

= 90 + 96.12 + 395.5872 + 246.0138 kg

= 827.721 kg.

% Mass of ethanol = \(\rm \dfrac{246.0138}{827.721}\;\times\;100\)

% Mass of ethanol = 29.72%

% Mass of Propionic acid = \(\rm \dfrac{395.5872}{827.721}\;\times\;100\)

% Mass of Propionic acid = 47.79 %.

The weight percent of ethanol has been 29.72%, and the percent mass of propionic acid has been 47.79%.

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Answer:

the answer is chemical element because when a symbol consists of two letters the first letter is always capitalized

The growth of the new plant depicts the asexual reproduction characteristic. The characteristic that describes the growth of the new plant in Stephan's mother cutting a twig from a rose bush and planting it in the soil is asexual reproduction.

Asexual reproduction is the mode of reproduction by which organisms generate offspring that are identical to the parent's without the fusion of gametes. Asexual reproduction is a type of reproduction in which the offspring is produced from a single parent.

The offspring created are clones of the parent plant, meaning they are identical to the parent.The new plant in Stephan’s mother cutting a twig from a rose bush and planting it in the soil depicts the process of asexual reproduction, which is the ability of a plant to reproduce without seeds. In asexual reproduction, plants can reproduce vegetatively by cloning themselves using their roots, bulbs, or stems.

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Answer:

12

Explanation:

12

Answer:

The solubility is  \(S = 0.00912 \ g/mL\)

Explanation:

From the question we are told that

     The original volume of sample is  \(V_o = 34.0 mL = 34 *10^{-3} \ L\)

      The temperature is  \(T = 22.0 ^oC\)

      The new volume of sample  is  \(V_n = 750.0 mL = 750 *10^{-3} \ L\)

       The weight of the crystal is  \(X = 0.31 \ g = 0.31 *10^{-3} \ kg\)

Now looking at the question we see that 34.0 mL of the sample is saturated with  0.31g of the crystal X

Generally the solubility of X in the water sample at  \(22.0 ^oC\) can be mathematically evaluate as

                \(S = \frac{0.31 }{34.0}\)

                \(S = 0.00912 \ g/mL\)

Calculating the equilibrium constant (K) for a reaction requires knowing the balanced equation of the reaction. I won't be able to give the exact value of K without the exact reaction equation. However, I can explain how to use the concentrations provided to determine K.

An equilibrium constant statement for the reaction has the general form:

\(K = [C]^c[D]^d / [A]^a[B]^b\)

In this case, let's assume the reaction is:

aA + bB ⇌ cC + dD

Given the initial and equilibrium concentrations of CO and Cl 2, the following values ​​can be applied to the concentrations:

[A] = [CO] (initial concentration)

[B] = [Cl2] (initial concentration)

[C] = [CO] (equilibrium concentration)

[D] = [Cl2] (equilibrium concentration)

Now that the concentration is determined, we can determine the equilibrium constant (K):

\(K = ([C]^c [D]^d) / ([A]^a [B]^b)K = ([CO]^c [Cl2]^d) / ([CO]^a [Cl2]^b)K = ([0.25]^c [0.15]^d) / ([0.79]^a [0.69]^b)\)

The equilibrium constant (K) for the reaction can be obtained by substituting the values ​​of a, b, c and d based on the balanced equation and the specified concentration.

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Answer:

I believe that in H-C≡N, Nitrogen has 4 bonding electrons and 2 non-bonding electrons.

(:

Explanation:

Answer:

106.905 amu is the mass of the other isotope

Explanation:

The atomic mass of an element is the sum of the masses of the isotopes multiplied by its abundance. The atomic mass of an element X with 2 isotopes is:

X = X-109*i + X-107*i

Where X is the atomic mass = 107.868 amu

X-109 = 108.905amu, i = 48.16% = 0.4816

X-107 = ?, i = 1-0.4816 = 0.5184

Replacing:

107.868amu = 108.905amu*0.4816 + X-107*0.5184

55.4194 = X-107*0.5184

106.905 = X-107

Answer: It is indeed A.Standard Solution.

Hope this helps you!

Answer:

they physical quantities that are uneffected by changes to a vector space basis

The solid starts melting at 0°C. The heating curve for water shows the temperature changes as heat is added to the substance.

The horizontal flat line on the graph represents the phase change from solid to liquid. In the case of water, this occurs at 0°C. This is known as the melting point, or the temperature at which a substance changes from a solid to a liquid at atmospheric pressure.

It is an important characteristic of a substance and can be used to identify it. It is also important in many industrial processes, such as melting metals for casting. Therefore, understanding the melting point of a substance is crucial in various fields of science and technology.

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1) Exact numbers are the result of a count. In contrast, measured numbers have a finite number of significant figures.

The volume of a piece of aluminum. This comes from a measure.

The number of centimeters in an inch. This is a defined number. It is exact.

Today's temperature. This comes from a measure.

The number of second in a year. This is a defined number.

Plate Tectonics is the study of the movement and changes in the rocks that make up the earth's crust. Isostasy is the equilibrium in the earth's crust maintained by a flow of rock material in the asthenosphere.

Hence this led to scientists understanding of the earth's crust has changed.

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Answer: 20.9 billion years.

Explanation: To estimate the age of the rock, we can use the fact that the half-life of uranium-238 is 4.47 billion years. This means that half of the uranium-238 in the rock would have decayed into lead-206 after 4.47 billion years, and half of what remains would decay in the next 4.47 billion years, and so on.

If 20% of the uranium-238 in the rock has decayed into lead-206, then 80% of the original uranium-238 is still present in the rock. This means that the rock has gone through one half-life of uranium-238 decay, and we can estimate its age by using the graph on page 253.

According to the graph, the ratio of lead-206 to uranium-238 after one half-life is approximately 0.027. This means that the rock contains 0.027 times as much uranium-238 as it did originally, and the remaining 80% of the uranium-238 corresponds to 0.8 times the original amount.

Therefore, we can estimate the original amount of uranium-238 in the rock as follows:

Original amount of uranium-238 = (0.8) / 0.027 = 29.63 times the current amount

Next, we can use the fact that each half-life corresponds to a reduction in the amount of uranium-238 by a factor of 2, to estimate the number of half-lives that have passed since the rock formed:

Number of half-lives = log2(29.63) = 4.88

Finally, we can estimate the age of the rock as follows:

Age of the rock = (4.88) x (4.47 billion years per half-life) = 20.9 billion years

Therefore, we can estimate that the age of the rock is approximately 20.9 billion years.

The atomic number of bromine is 35. Thus the total number of electrons is 35. The number of core electrons is 28.

Bromine is 35th element in periodic table. It is a nonmetal and is placed in the 17th group called halogen group. Bromine relatively electronegative in nature and form ionic compounds.

The atomic number of an atom is the number of its total electrons in the neutral state. The electrons which are occupied in the higher energy levels that is in the outer orbitals far from the nucleus is called valence electrons. All other electrons are called inner electrons or core electrons.

Bromine contains total 35 electrons. Out of which 7 are valence electrons. Thus, the number of core electrons in the ground state of bromine is 35 -7 = 28 electrons.

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Answer:

To resolve this, we need to place the coefficient “2” in front of the sodium in the reactant, to give the equation shown below. 2 Na (s) + Cl 2 (g) → 2 NaCl (s) In this equation, there are two sodiums in the reactants, two sodiums in the products, two chlorines in the reactants and two chlorines in the products; the equation is now balanced.

Explanation:

The amount, in kJ, of heat needed to completely vaporize 50.0g of water at 100°C is 118.8 kJ.

The heat needed to completely vaporize 50.0g of water at 100°C can be calculated using the following formula:

q = m x Hv

where:

First, we need to convert 50.0g to moles by dividing by the molar mass of water which is approximately 18.015 g/mol3:

moles of water = 50.0 g / 18.015 g/mol moles of water = 2.776 mol

Thus:

q = (2.776 mol) x (40.65 kJ/mol) q = 112.8 kJ

In other words, 112.8 kJ of heat is needed to completely vaporize 50.0g of water at 100°C.

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Answer:

86.85°C

Explanation:

K = °C + 273.15

360K − 273.15 = 86.85°C ≈ 87°C

False

They are light indeed... but they do have mass

Hope this helped you- have a good day bro cya)

The symbol is shown as Cr³+ because it loses electrons. Option A

When we talk about the process of oxidation, what is going on is the loss of electrons. Thus it is possible to say that ocidation is electron loss. The electrons that are lost would lead to the formation of a specie that has a positive ion.

The magnitude of the positive charge that we see in the compound is based on the number of electrons that it has lost in the process of the oxidation of the compound. There are three electrons that have been lost for chromium as shown.

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Answer:

A combustion reaction is a reaction in which a substance reacts with oxygen gas, releasing energy in the form of light and heat.

Explanation:

1.The gas law that we will use to solve this problem is the combined gas law.

2.We will choose the combined gas law because there are three variables that are changing in this problem: volume, pressure, and temperature.

3The volume is decreasing in this problem, from 1.74 L to 1.56 L.

The unknown that we are solving for is T2, the temperature at which the balloon will have a volume of 1.56 L.

4We will use the temperature in Kelvin for our calculations. To convert from Celsius to Kelvin, we add 273.15 to the Celsius temperature. Therefore, the temperature we will use in our calculations is 293.35 K (20.2°C + 273.15).

5To find the final temperature, we can use the combined gas law equation:

(P1 x V1)/T1 = (P2 x V2)/T2

where P1 = 1.02 atm, V1 = 1.74 L, T1 = 293.35 K, P2 = 0.980 atm, and V2 = 1.56 L.

Substituting the values into the equation and solving for T2, we get:

T2 = (P2 x V2 x T1)/(P1 x V1)

= (0.980 atm x 1.56 L x 293.35 K)/(1.02 atm x 1.74 L)

= 268.06 K

Therefore, the final temperature after the volume and pressure decrease is 268.06 K, which is approximately 253 K (option A).

Answer:

0.30625 Ci

Explanation:

From the question given above, the following data were obtained:

Half-life (t½) = 5.73E3 years

Original activity (Nₒ) = 0.350 Ci

Time (t) = 1.719E4 years

Difference in activities =?

Next, we shall determine number of half-lives that has elapse. This can be obtained as follow:

Half-life (t½) = 5.73E3 years = 5.73×10³ years

Time (t) = 1.719E4 years = 1.719×10⁴ years

Number of half-lives (n) =?

n = t / t(½)

n = 1.719×10⁴ / 5.73×10³

n = 3

Thus, 3 half-lives has elapsed.

Next, we shall determine the activity after 1.719E4 years. This can be obtained as follow:

Original activity (Nₒ) = 0.350 Ci

Number of half-lives (n) = 3

Activity remaining (N) =?

N = 1/2ⁿ × Nₒ

N = 1/2³ × 0.350

N = 1/8 × 0.350

N = 0.125 × 0.350

N = 0.04375 Ci

Finally, we shall determine the difference in activities. This can be obtained as follow:

Original activity (Nₒ) = 0.350 Ci

Activity remaining (N) = 0.04375 Ci

Difference in activities =?

Difference in activities = Nₒ – N

Difference in activities = 0.350 – 0.04375

Difference in activities = 0.30625 Ci.

Answer:

Explanation:

General Remarks

The balance numbers are the same. (The balance numbers are 2 and 2)

The balance number in front of the O2 is 1.

So for every mole of O2 needed 2 moles of NO2 should be produced.

If 10.0 moles of oxygen is used, then you should get 20 mols of NO2 should be produced.

Formula

% yield = (what was produced / what should have been produced) *100

Givens

What was produced = 7 moles

What should have been produced = 20 moles

Solution

Substitute the givens into the equation

% yield = 8/20 * 100

% yield = 0.40 * 100

Answer: 40%

Answer:

Store the genetic instructions needed to specify traits.

In DNA, the primary function of chromosome is store the genetic instructions needed to specify traits.

DNA is a hereditary material  which is present in human beings as well as all other living organisms.  Every cell which is present in an organism's body has DNA  which is the same. Most of the DNA is situated in the cell's nucleus and small amount of it can be found in the cell's mitochondria as well.

Information which is stored in DNA is stored as codes made up of four chemical bases namely, adenine, thymine , cytosine and guanine.Human DNA consists of 3 billion bases .The order of the bases determines information which is required for building and maintaining an organism.

DNA bases are capable of pairing up with each other. Adenine pairs with thymine and guanine pairs up with cytosine .Each base is also attached to a sugar molecule  and a phosphate group. A base, phosphate  sugar are together called as nucleotides.

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The representation is showing potassium atom (K) and Chlorine atom (Cl) when it ought to show their ions i.e potassium ion (K⁺) and Chlorine ion (Cl¯)

When potassium chloride, KCl dissolves in water, it dissociate to produce potassium ion (K⁺) and Chlorine ion (Cl¯) as shown below:

KCl(aq) —> K⁺(aq) + Cl¯(aq)

The representation given in the question is only showing potassium atom (K) and Chlorine atom (Cl). This makes it wrong as dissolution of ionic compounds in water will results in the corresponding ions of the element that makes up the compound

Please see attached photo

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