"An orange compound was added to the top of a chromatography column. Solvent was added immediately, with the result that the entire volume of solvent in the solvent reservoir turned orange. No separation could be obtained from the chromatography experiment. What went wrong?

The gas is accurately described by the ideal gas equation. Calculations reveal that the given conditions match Argon (Ar) most closely. Thus, the correct option is B) Ar.

To identify the gas accurately, we can use the ideal gas equation: PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.

First, let's calculate the number of moles (n) using the given mass (m) of the gas and its molar mass (M). The molar mass of the gas is necessary for this step.

[tex]\[ n = \frac{m}{M} \][/tex]

For this calculation, we need to know the molar mass of each gas option. The molar masses are approximately:

- CO2: 44.01 g/mol

- Ar: 39.95 g/mol

- Ne: 20.18 g/mol

- N2: 28.02 g/mol

Let's go through the calculations step by step.

Given values:

- Mass (m) = 27.1 g

- Volume (V) = 50.0 L

- Pressure (P) = 0.500 atm

- Temperature (T) = 449 K

1. Calculate moles (n) using the given mass:

[tex]\[ n = \frac{m}{M} \][/tex]

For CO2: [tex]\( n_{\text{CO2}} = \frac{27.1 \, \text{g}}{44.01 \, \text{g/mol}} \approx 0.616 \, \text{mol} \)[/tex]

For Ar: [tex]\( n_{\text{Ar}} = \frac{27.1 \, \text{g}}{39.95 \, \text{g/mol}} \approx 0.679 \, \text{mol} \)[/tex]

For Ne: [tex]\( n_{\text{Ne}} = \frac{27.1 \, \text{g}}{20.18 \, \text{g/mol}} \approx 1.342 \, \text{mol} \)[/tex]

For N2: [tex]\( n_{\text{N2}} = \frac{27.1 \, \text{g}}{28.02 \, \text{g/mol}} \approx 0.968 \, \text{mol} \)[/tex]

2. Use the ideal gas equation to calculate moles (n):

[tex]\[ n = \frac{PV}{RT} \][/tex]

Substitute values:

[tex]\[ n = \frac{(0.500 \, \text{atm} \times 50.0 \, \text{L})}{(0.0821 \, \text{L}\cdot\text{atm/mol}\cdot\text{K} \times 449 \, \text{K})} \][/tex]

[tex]\[ n \approx \frac{25}{36.8029} \approx 0.681 \, \text{mol} \][/tex]

3. Compare the calculated moles:

- [tex]\( n_{\text{CO2}} \approx 0.616 \, \text{mol} \)[/tex]

- [tex]\( n_{\text{Ar}} \approx 0.679 \, \text{mol} \)[/tex]

- [tex]\( n_{\text{Ne}} \approx 1.342 \, \text{mol} \)[/tex]

- [tex]\( n_{\text{N2}} \approx 0.968 \, \text{mol} \)[/tex]

The closest match is [tex]\( n_{\text{Ar}} \approx 0.679 \, \text{mol} \)[/tex], so the gas is likely Argon (Ar). Therefore, option B) Ar is the correct answer.

Complete question :- A real gas is in a regime in which it is accurately described by the ideal gas equation of state. 27.1 g of the gas occupies 50.0 L at a pressure of 0.500 atm and a temperature of 449 K. Identify the gas.

A) CO2

B) Ar

C) Ne

D) N2

Answer: one simple distillation column is required to separate the stream into five pure products. With four different flat bottom flask, for collection of the distilled products

Explanation: simple distillation works with the difference in boiling points of the liquid to be separated. For the separation of five different constituent to be possible, we have to know the boiling points of the constituents.

For your understanding, let's define constituents in the liquid to be A, B, C, D, E. And the boiling points increases respectively. Start by heating the liquid to the boiling point of A to extract A. After a while check if the constituents A is still dropping in the flat bottom flask, if it has stopped dropping, it simply means that we have extracted all A constituents in the liquid, label the Flask A. Get another flask to extract constituent B.

Heat the mixture to the boiling point of B, after a while check if constituent B is still dropping in the flat bottom flask, if it has stopped dropping,it means that we have extracted all B constituent in the liquid, label the Flask B. Get another flask for C.

Repeat the same process for C and D.

After Extracting D we don't need to distillate E because we already have a pure form of E inside to the conical flask.

SEE PICTURE TO UNDERSTAND WHAT A SIMPLE DISTILLATION LOOKS LIKE

Answer: The mass of cryolite produced is 65.06 kg

Explanation:

To calculate the number of moles, we use the equation:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]      .....(1)

Given mass of aluminium oxide = 15.8 kg = 15800 g     (Conversion factor:  1 kg = 1000 g)

Molar mass of aluminium oxide = 102 g/mol

Putting values in equation 1, we get:

[tex]\text{Moles of aluminium oxide}=\frac{15800g}{102g/mol}=154.9mol[/tex]

For the given chemical reaction:

[tex]Al_2O_3(s)+6NaOH(l)+12HF(g)\rightarrow 2Na_3AlF_6+9H_2O(g)[/tex]

As all the reactants are getting completely utilized. So, the amount of product can be determined by any 1 of the reactant.

By Stoichiometry of the reaction:

1 mole of aluminium oxide produces 2 moles of cryolite

So, 154.9 moles of aluminium oxide produces = [tex]\frac{2}{1}\times 154.9=309.8mol[/tex] of cryolite

Now, calculating the mass of cryolite by using equation 1, we get:

Moles of cryolite = 309.8 moles

Molar mass of cryolite = 210 g/mol

Putting values in equation 1, we get:

[tex]309.8mol=\frac{\text{Mass of cryolite}}{210g/mol}\\\\\text{Mass of cryolite}=(309.8mol\times 210g/mol)=65058g=65.06kg[/tex]

Hence, the mass of cryolite produced is 65.06 kg

Answer: Histones, DNA methylation

Explanation:

The double stranded DNA are wrapped twice around a histone protein and depending on how loosely or tightly DNA are wounded around histion, DNA can be either read or not. The nature of histones have the ability to control which sections of DNA get copied and expressed. Histone synthesis stops when DNA synthesis ceases, histone are modified by acetylation, methylation ADP--ribosylation and phosphorylation. It has shown that environmental factors such as diet can have an effect on protein structure. Alteration of the histone protein structure can affect the expression of some genes.

DNA methylation can also inhibit the expression of some genes, which can be caused by environmental factit's.

Environmental factors can influence DNA and its packaging at the molecular level through epigenetic regulation and histone modification.

Environmental factors can have an influence on DNA and its packaging at the molecular level. One way is through epigenetic regulation, where environmental factors can cause chemical modifications to DNA and histone proteins. For example, certain environmental factors can lead to the addition of methyl groups to specific cytosine nucleotides in DNA, which can affect gene expression. Another way is through the modification of histone proteins, such as acetylation and deacetylation, which can influence the packaging state of DNA and its accessibility for transcription.

Answer:

Ca(OH)₂ and HCl(aq)

Explanation:

We have the products of a reaction, CaCl₂ and H₂O. Since the products are a salt and water, this is likely to be a neutralization reaction. In such a reaction, an acid and a base react to form a salt and water.

The complete molecular equation is:

Ca(OH)₂ + 2 HCl(aq) → CaCl₂ + 2 H₂O

The question is incomplete, here is the complete question:

A student sets up the following equation to convert a measurement. (The stands for a number the student is going to calculate.) Fill in the missing part of this equation.

[tex]23.Pa.cm^3=?kPa.m^3[/tex]

Answer: The measurement after converting is [tex]23\times 10^{-9}kPa.m^3[/tex]

Explanation:

We are given:

A quantity having value [tex]23.Pa.cm^3[/tex]

To convert this into [tex]kPa.m^3[/tex], we need to use the conversion factors:

1 kPa = 1000 Pa

[tex]1m^3=10^6cm^3[/tex]

Converting the quantity into [tex]kPa.m^3[/tex], we get:

[tex]\Rightarrow 23.Pa.cm^3\times (\frac{1kPa}{1000Pa})\times (\frac{1m^3}{10^6cm^3})\\\\\Rightarrow 23\times 10^{-9}kPa.m^3[/tex]

Hence, the measurement after converting is [tex]23\times 10^{-9}kPa.m^3[/tex]

Answer:

The concentration is 4.16 grams per liter, we call the solution saturated.

Explanation:

Step 1: Data given

The solubility of calcium chromate in water is 4.16 grams per liter.

Saturated Solution : A solution with solute that dissolves until it is unable to dissolve anymore, leaving the undissolved substances at the bottom.

Unsaturated Solution: A solution (with less solute than the saturated solution) that completely dissolves, leaving no remaining substances.

Supersaturated Solution: A solution (with more solute than the saturated solution) that contains more undissolved solute than the saturated solution because of its tendency to crystallize and precipitate.

Step 2:

If the calcium chromate solution has a concentration of 4.16 grams per liter.

If the concentration is less than 4.16 grams per liter, we call it unsaturated.

If the concentration is more than 4.16 grams per liter, we call it supersaturated.

The concentration is 4.16 grams per liter, we call the solution saturated.

Final answer:

A saturated solution contains the maximum amount of solute that can dissolve in a solvent at a specific temperature.

Explanation:

The solution is saturated because the concentration of calcium chromate in the solution is equal to its solubility in water, which is 4.16 grams per liter.

A saturated solution contains the maximum amount of solute that can dissolve in a given solvent at a specific temperature.

Saturated solutions are in a dynamic equilibrium where the rate of dissolution is equal to the rate of precipitation.

Answer:

71.2 %

Explanation:

This is a problem we can solve by making use of the stoichiometry of the balanced  ( very important !) chemical reaction:

CaCO3(s) + 2HCl(aq) → CO2(g) + CaCl2(aq) + H2O(l)

Now  determine  the moles of CaCO3 and CO2 and perform the calculations

# mol CaCO3 = 10.51 g / 100.09 g/mol = 0.105 mol

From the stoichiometry of the reaction, we know 1 mol CO2 is produced per mol of CaCO3, thus whe should expect

0.105 mol CaCO3 x 1 mol CO2/ mol CaCO3 = 0.105 mol

lets compare this theoretical value with the one obtained:

mol CO2 obtained = 3.29 g / 44 g/mol = 0.075

Therefore our yield is

(0.075mol / 0.105 mol) x 100 = 71.2 %

Final answer:

To calculate the moles of ammonium perchlorate needed to produce a certain amount of water, use the balanced chemical equation and the mole ratio between ammonium perchlorate and water.

Explanation:

To calculate the moles of ammonium perchlorate needed to produce a certain amount of water, we need to use the balanced chemical equation. From the equation 10Al(s) + 6NH4C1O4(s) → 4Al2O3 (s) + 2AlCl3 (s) + 12H₂O(g) + 3N2 (8), we can see that for every mole of NH4C1O4, 12 moles of H₂O are produced. Therefore, the moles of ammonium perchlorate needed can be calculated by multiplying the given moles of water by the ratio of moles of NH4C1O4 to moles of water, which in this case is 6.

The concentration of the aluminum sulfate solution is calculated to be 25.8 g/dL by dividing the mass of the aluminum sulfate (116.0 g) by the volume of the solution (4.5 dL).

The concentration of a solution is given by the ratio of the mass of the solute to the volume of the solution. In this case, we have 116.0 g of aluminum sulfate dissolved in a solution whose total volume is 450.0 mL. To convert this volume to deciliters (dL), we remember that 1 L = 10 dL and 1 L = 1000 mL. Therefore, the volume of the solution is 450.0 mL * (1 L / 1000 mL) * (10 dL / 1 L) = 4.5 dL. The concentration of aluminum sulfate in the solution is thus 116.0 g / 4.5 dL = 25.8 g/dL.

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Answer: the empirical formula is CH3O and the subscripts are 1, 3, 1

Explanation:Please see attachment for explanation

Answer: the atomic Mass of Z is 46.26

Explanation:Please see attachment for explanation

Answer:

The syringe should be at 2.50

Final answer:

To administer a 2.46 mL dose, draw up slightly more than this amount into the syringe and then adjust to the precise volume by expelling any excess. Ensure the plunger aligns with the 2.46 mL marking on the syringe for an accurate dose.

Explanation:

When administering medication in a clinical setting, it's important to measure the dosage accurately. If you've calculated that the correct dose is 2.46 mL, you should first draw up the liquid into the syringe to a volume slightly greater than needed, then carefully adjust down to the exact amount. For instance, you might withdraw 3-4 mL and then expel the excess until you have the precise 2.46 mL required for administration.

The technique of drawing up more than needed and then adjusting allows for removing air bubbles and ensuring an accurate dose. In practice, syringes often have volume marks in increments of 0.1 mL or 0.01 mL, depending on the syringe size. It is critical to use these markings and ensure that the top of the plunger aligns exactly with the marking for 2.46 mL when viewed at eye level.

Answer:

89.52 gm

Explanation:

35.7 gm of NaCl , dissolves in 100ml of water.

Then amount of NaCl dissolved in 1 ml water = 35.7/100= 3.75 gm

therefore, sodium chloride dissolved in 250 ml of water

= 3.75×250= 89.25 gm

The maximum amount of sodium chloride that will dissolve in 250 ml of water is 89.25 g.

The solubility of sodium chloride in water is 35.7 g per 100 ml at 0°C. To find the maximum amount of sodium chloride that will dissolve in 250 ml of water, we can set up a proportion using the solubility values:

35.7 g / 100 ml = x g / 250 ml

Cross-multiplying and solving for x, we get:

x = (35.7 g * 250 ml) / 100 ml = 89.25 g

Therefore, the maximum amount of sodium chloride that will dissolve in 250 ml of water is 89.25 g.

Answer:

0.246 kg

Explanation:

There is some info missing. I think this is the original question.

A chemist adds 370.0mL of a 2.25 M iron(III) bromide (FeBr₃) solution to a reaction flask. Calculate the mass in kilograms of iron(III) bromide the chemist has added to the flask. Be sure your answer has the correct number of significant digits.

We have 370.0 mL of 2.25 M iron(III) bromide (FeBr₃) solution. The moles of FeBr₃ are:

0.3700 L × 2.25 mol/L = 0.833 mol

The molar mass of iron(III) bromide is 295.56 g/mol. The mass corresponding to 0.833 moles is:

0.833 mol × 295.56 g/mol = 246 g

1 kilogram is equal to 1000 grams. Then,

246 g × (1 kg/1000 g) = 0.246 kg

Answer:

B = CHCl2 + Cl2 --> CHCl3 + Cl

Explanation:

Free radical halogenation is a chlorination reaction on Alkane hydrocarbons. This involves the splitting of molecules into radicals/ unstable molecules in the presence of sunlight/ U.V light which ensures bonding of the molecules.

Free radical chlorination is divided into 3 steps which are:

The initiation step

The propagation step

The termination step

So in reference to the question, propagation step involves two steps.

The first step is where the molecule in this case the methylene chloride(CH2Cl2) loses a hydrogen atom and then bond with a chlorine atom radical to give a nethylwnw chloride radical and HCl.

The second step involves the reaction of this methylene chloride got in the first step with chlorine molecule to form trichloride methane and a chlorine radical.

You would find in the attachment the 2 step mechanism.

The propagation steps in the monochlorination of methylene chloride (CH2Cl2) are option b and d. Propagation steps are characterized by the reaction of two radicals to form a non-radical species and a new radical, which will then enter the chain cycle. These steps occur after initiation and before termination in a radical chain reaction.

The question asks which options represent a propagation step in the monochlorination of methylene chloride (CH2Cl2). The correct answer is option b and d. These steps depict the reaction of CHCl2 with Cl2 with chlorine radical (.) or molecular chlorine (Cl2) to generate either CHCl3 and Chlorine radical (.) in option b or CHCl3 in option d, which are required for the propagation process in a halogenation reaction. It is essential to distinguish between initiation, propagation and termination steps in radical chain reactions, like this one, where a hydrogen atom on the methylene chloride is replaced by a chlorine atom.

Propagation steps are the 'middle' steps of these reactions. They are characterized by two radicals reacting to form a non-radical species and a new radical, which re-enters the propagation cycle. Both initiation and termination steps involve either the creation or removal of radicals, respectively. Radical chain reactions, halogenation, and methylene chloride are key concepts for understanding this question.

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0.87 grams of an 8% w/w progesterone gel must be mixed with 1.45 g of a 4% w/w progesterone gel to prepare a 2.32 g of 5.5% w/w gel

Let x represent the number of grams of an 8% w/w progesterone gel must be mixed with 1.45 g of a 4% w/w progesterone gel to prepare a (x + 1.45)g of 5.5% w/w gel

Hence:

(x * 8%) + (1.45 * 4%) = (x + 1.45) * 5.5%

0.08x + 0.058 = 0.055x + 0.07975

0.025x = 0.02175

x = 0.87 g

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To prepare a 5.5% w/w progesterone gel, you should mix 4.383 g of an 8% progesterone gel with 1.45 g of a 4% progesterone gel.

A weight/weight percent (w/w%) refers to the amount of a substance (in this case, progesterone) contained in a total solution. The equation we want to use here is 'mass of solute/mass of solution = concentration'. We know the concentrations of the individual gels and that of the final gel we want to achieve. We need to find out the amount of 8% gel to mix with 1.45 g of 4% gel to get a 5.5% gel.

This would be solved as follows:

Let's denote the weight of the 8% gel as x. The contribution of progesterone from the 8% gel is 0.08x, and from the 4% gel is 0.04*1.45 = 0.058 g. The total weight of the final gel solution is x+1.45 g. The total amount of progesterone is 0.08x + 0.058 g. To get a 5.5% solution, we set up the equation 0.08x + 0.058 / 1.45 + x = 0.055. Solving for x gives us 4.383 g.

Therefore, the answer is that 4.383 g of an 8% progesterone gel should be mixed with 1.45 g of the 4% gel to obtain a 5.5% gel.

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Final answer:

To calculate the equilibrium constant, K, for the reaction A(aq) + 2B(aq) ⇒ 2C(aq), we can use the concentrations of A, B, and C at equilibrium. The equilibrium constant (K) for this reaction is approximately 5.07 x 10^-3.

Explanation:

To calculate the equilibrium constant, K, for the reaction A(aq) + 2B(aq) ⇒ 2C(aq), we need to use the concentrations of A, B, and C at equilibrium. Given that the equilibrium concentration of C is 0.0454 M, and the initial concentrations of A and B are 0.150 M and 0.200 M respectively, we can set up an expression for K.

The mathematical expression for the equilibrium constant, Kc, is given by:

Kc = [C]² / ([A] * [B]²)

Substituting the given equilibrium concentration and initial concentrations into the expression, we get:

Kc = (0.0454)² / (0.150 * (0.200)²) = 5.06666667 x 10^-3

Therefore, the equilibrium constant (K) for this reaction is approximately 5.07 x 10^-3.

Question: A optically active compound, C5H10O, exhibits IR absorption at 1730 cm-1.

Its carbon NMR shifts are given below. The number of hydrogen's at each carbon, determined by DEPT, is given in parentheses after the chemical shift.

13C NMR: δ 22.6 (3), 23.6 (1), 52.8 (2), 202.4 (1)

Draw the structure of this compound in the window below

Explanation:

3-methylbutanal is a butanal substituted by the methyl group at the 3rd position. It is a volatile constituent in the olive. Also, it is used as a flavoring agent and a plant metabolite, it is also a Saccharomyces cerevisiae metabolite. It is also called as the Isovaleraldehyde organic compound. The liquid is colorless at STP, and also found in very low concentrations. It is also seen to be produced commercially for different use. Mostly used compound reagent in the preparation of pharmaceuticals and pesticides.

Answer:

1-bromo-1-tert-butylcyclohexane

Explanation:

The parent compound comprises of a cyclohexane to with a tertiary butyl carbon attached. We have been told that the reaction occurs by radical mechanism hence we must recall the order of stability of radicals: tertiary>a secondary> a primary. This implies that the reaction will occur at carbon 1 of the cyclohexane which is a tertiary carbon atom. This leads to the formation of a radical at the 1-position and bromination at that position hence the answer chosen above.

The radical bromination of t-butylcyclohexane favors the formation of 1-bromo-1-tert-butylcyclohexane due to the stability of the tertiary radical intermediate formed during the reaction.

The major product obtained upon radical bromination of t-butylcyclohexane will most likely be 1-bromo-1-tert-butylcyclohexane. This is because radical bromination is highly selective for the most stable radical intermediate, which forms at the position with the greatest number of accessible hydrogens. In the case of t-butylcyclohexane, the tertiary carbon adjacent to the t-butyl group will form the most stable tertiary radical upon abstraction of a hydrogen, which leads to substitution by a bromine atom to form the final product.

Answer:

0.000273 M

Explanation:

Henry's states that at constant temperature the amount of a gas that dissolves in a liquid is directly proportional to the partial pressure in of that gas in equilibrium with that liquid.

Pressure of Oxygen = mole fraction of Oxygen × 1.00 atm

Mole fraction Oxygen = 21/100 × 1.00atm = 0.21 atm

Molar solubility of Oxygen = KH × PO2 = 0.0013 × 0.21 = 0.000273 M

The amount of a gas that dissolves in a liquid is proportional to the partial pressure of the gas above the liquid. Solubility of oxygen in water exposed to air at 1.00 atm is 0.000273 M.  

Henry's Law:

It states that at constant temperature the amount of a gas that dissolves in a liquid is proportional to the partial pressure of the gas above the liquid.

C = k P

Where,

C = concentration of a dissolved gas

k = Henry's Law constant = 0.0013 M/atm.

P = partial pressure of the gas = [tex]\bold {\dfrac {21}{100} \times 1.00\ atm = 0.21\ atm}[/tex]

Put the values in the formula,

[tex]\bold {C = 0.0013 \times 0.21 = 0.000273\ M }[/tex]

Therefore, solubility of oxygen in water exposed to air at 1.00 atm is 0.000273 M.

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Answer: C

Explanation:

N-methylpropanamide and methylamine are the compounds that posses hydrogen bonds.

Hydrogen bonds are formed when hydrogen is covalently bonded to a highly electronegative atom. Hydrogen bonds are weaker than covalent bonds but they significantly impact on the chemistry of the molecules in which they occur.

Primary, secondary and tertiary amines all form hydrogen bonds. The compounds that has hydrogen bonds are;

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Answer:

Hi

Acebutolol hydrochloride is the form of the hydrochloride salt of acebutolol, a synthetic derivative of butyranide with a hypotensive and antiarrhythmic activity. Acebutolol acts as a cardioselective beta-adrenergic antagonist with very little effect on bronchial receptors, having intrinsic sympathomimetic properties. Acebutolol is used in ventricular arrhythmias. Other indications may include hypertension, alone or in combination with other drugs. The salt scheme is found in the attached file.

Explanation:

Explanation:

According to the collision theory the rate of chemical reaction directly proportional to number of reaction between the reactant molecules. More the number of collision between the reactant molecules, more would  be the changes of reaction to occur that reactant converting into product.

Moreover, with an increase in concentration of reactant molecules, the chances of collision between the molecules increase and hence the rate of reaction also increase. Thus, we can say higher the concentration higher will be the rate of reaction.

Thus the above response about the theory is correct.

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Answer:

Explanation:

Melting range is one of the properties of solid that can help to identify its purity. It is necessary to note the whole range of temperature through which the transition takes place. The change in this range of melting temperature is indicative of impurity in the sample.

The lower end is the starting temperature at which solid starts to melt while the upper end is the temperature at which the last trace of solid converts to liquid.

Answer:

A. False.

Every substance contains the same number of molecules i.e 6.02x10^23 molecules

B. False.

Mass conc. = number mole x molar Mass

Mass conc. of 1mole of N2 = 1 x 28 = 28g

Mass conc. of 1mol of Ar = 1 x 40 = 40g

The mass of 1mole of Ar is greater than the mass of 1mole of N2

C. False.

Molar Mass of N2 = 2x14 = 28g/mol

Molar Mass of Ar = 40g/mol

The molar mass of Ar is greater than that of N2.

Explanation:

Answer:

The rank is: C₆H₁₂O₆ < NaBr < CaCl₂ < Cr(NO₃)₃

Explanation:

To take into account, the greater the number of ions that are produced in a solution, the entropy will be greater. In this case, we have that the number of ions of each compound is:

Cr(NO₃)₃ → Cr³⁺ + 3NO⁻₃ (it has 4 ions)

CaCl₂ → Ca²⁺ + 2Cl⁻ (it has 3 ions)

NaBr → Na⁺ + Br⁻ (it has 2 ions)

C₆H₁₂O₆ (does not ionize)

Answer:

The answer is granitic or felsic rocks

Explanation:

Felsic is a term used in geology applied to silicate minerals, magmas and rocks, rich in light elements such as silicon, oxygen, aluminum, sodium and potassium (describs igneous rocks that are relatively rich in elements that form feldspar and quartz). This term is a combination of the words "feldspar" and "silica". Felsic minerals are generally light in color and have a specific gravity of less than 3. The most common felsic minerals are quartz, muscovite, alkaline feldspars (eg orthoclase) and feldspars from the plagioclase series. The most common felsic rock is granite. At the opposite end of the rock spectrum are mafic (rich in iron) and ultramafic (rich in magnesium) rocks and minerals.

Final answer:

An igneous rock with quartz and potassium feldspar has a felsic mineralogic content, rich in silica and light-colored minerals, and typically has a lighter color due to low ferromagnesian content.

Explanation:

An igneous rock containing quartz and potassium feldspar would most likely be classified within the range of felsic igneous rocks. Felsic rocks are rich in silica and characteristically contain light-colored minerals such as quartz and feldspar. Specifically, the presence of quartz, a pure silica mineral, and potassium feldspar, which includes silica along with aluminum and potassium, indicates a felsic composition.

These rocks generally have high silica content (65-75%), and they also commonly include minor amounts of mafic minerals such as biotite mica and amphibole.

Felsic igneous rocks can be further understood by their mineral proportions which can include around 25% K-feldspar and 30% quartz, with additional albitic plagioclase and biotite or amphibole. Their color is generally lighter due to the predominance of light-colored minerals and their low content of ferromagnesian components like iron and magnesium.

Answer:

See answer for details

Explanation:

In this case, we have the following substract:

Ph-CH2 - O - CH2CH3

That's the benzyl ethyl ether.

When this compound reacts with a concentrated Solution of HI, as we have an acid medium, we will have an SN1 reaction, where the Hydrogen will attach to the oxygen and the iodine will go to the most stable carbon, in this case, the carbon next to the ring is the most stable (because of the double bond of the ring, charges can be stabilized better this way), so product A will be the benzyl with the iodine and product B, will be the alcohol:

A: Ph - CH2 - I

B: CH3CH2OH

When B reacts again with HI, it's promoting another SN1 reaction, where the OH substract the H from the HI, the OH2+ will go out the molecule, leaving a secondary carbocation (CH2+) and then, the Iodine (I-) can go there via SN1 and the final product would be:

C: CH3CH2I

See picture for mechanism:

Final answer:

The student's question pertains to the reaction of benzyl ethyl ether with concentrated aqueous HI, following an SN1 mechanism. The reaction generates a halogenated product (A) and an alcohol (B), with the alcohol further reacting to form a reduction product (C). Specific structures cannot be provided without more information on the starting compound.

Explanation:

The student is inquiring about the reaction of benzyl ethyl ether with concentrated aqueous HI and the subsequent products. The reaction is characteristic of the SN1 mechanism of ether cleavage, particularly with ethers that have benzylic, allylic, or tertiary alkyl groups. In such reactions, the benzyl or allylic group forms a relatively stable carbocation, leading to the formation of a halogen product (product A), while the ether's other alkyl substituent becomes an alcohol (product B).

Further reaction of product B with HI would lead to the formation of product C, which is typically the reduction of the alcohol to an alkane or a similar process depending on the specific structure of product B. Without the detailed structure of benzyl ethyl ether, we cannot provide the exact structures of the final products A, B, and C.

Answer:

We need 19.8 grams of butane

Explanation:

Step 1: Data given

Mass of CO2 produced = 59.8 grams

Molar mass CO2 = 44.01 g/mol

Molar mass butane =     58.12 g/mol

Step 2: The balanced equation

2 C4H10 + 13 O2 → 8 CO2 + 10 H2O

Step 3: Caclulate moles CO2

Moles CO2 = mass CO2 / Molar mass CO2

Moles CO2 = 59.8 grams / 44.01 g/mol

Moles CO2 = 1.36 moles

Step 4: Calculate moles butane

For 2 moles butane we need 13 moles O2 to produce 8 moles CO2 and 10 moles H2O

For 1.36 moles CO2 we need 1.36/4 = 0.34 moles butane

Step 5: Calculate mass butane

Mass butane = moles butane * molar mass butane

Mass butane = 0.34 moles * 58.12 g/mol

Mass butane = 19.76 grams ≈ 19.8 grams

We need 19.8 grams of butane