For each of the following pairs of solutes and solvent, predict whether the solute would be soluble or insoluble. After making your predictions, you can check your answers by looking up the compounds in the Merck Index or the CRC Handbook of Chemistry and physics. Generally, the Merck Index I the easier reference book to use. If the substance has a solubility greater than 40mg/mL, you may conclude that it is soluble.

a.Malic acid in water: Soluble

b.Naphthalene in water: Insoluble

c.Amphetamine in ethyl alcohol: Insoluble

d.Aspirin in Water: Insoluble

e.Sucinic Acid in Hexane: Insoluble

f.Ibuprofen in Diethyl Ether: Insoluble

g.1-Decanol in Water: Slightly soluble because of OH

Answer:

Balanced molecular reaction

Na2S2O5 + H2O ----> 2NaHSO3

Net ionic reaction

(S2O5)^(2-) + H2O ----> 2((HSO3)^(1-))

Explanation:

The sodium bisulfite compound is produced in this manner because Sodium metabisulfite has better preservative properties. It is more stable compared to sodium bisulfite and readily dissolves in water to give the required sodium bisulfite.

Sodium bisulfite sold in the market contains sodium metabisulfite as it is the more stable one of the pair.

And of all the ways of preparing the compound, this is the cheapest and easiest one.

Answer: check explanation.

Explanation:

The balanced molecular equation for the reaction of sodium metabisulfite (Na2S2O5) with water to produce sodium bisulfite (NaHSO3) is given below;

Na2S2O5 + H2O --------> 2NaHSO3.

Therefore, the net ionic equation for this reaction is given below;

S2O5^2- + H2O ------> 2HSO3^-1.

S2O5^2- + H+ ------------> 2HSO3^-1

=====> Why is sodium bisulfite prepared using this method?

sodium bisulfite is prepared using this method because of the following reasons;

(1). It is the most easy way of synthesisizing/producing sodium bisulfite (NaHSO3).

(2). In order to produce sodium bisulfite (NaHSO3) through this method, it does not require high cost,that is to say that it is financial friendly.

(3). Because of high preservative properties of sodium metabisulfite (Na2S2O5).

(4). It reacts with water with ease to produce sodium bisulfite (NaHSO3).

Chloroform-Chloroform and Acetone-Acetone interactions are weaker than chloroform-acetone interactions.

Explanation:

Raoult's law states, In a solution, vapor pressure is equal to the sum of the vapor pressures of individual component and this implies when it is multiplied by the mole fraction of that component in the solution. Raoult's Law is expressed by the formula:

                         P solution = Χ solvent [tex]\times[/tex] P0 solvent.

Raoult's law assumes that the components in the mixture are ideal, it assumes that the Chloroform-Chloroform interactions, Chloroform-Acetone, and Acetone-Acetone are the same. A negative deviation is shown by the measurement of observed vapor pressure which is less than that of calculated. Hence the interaction of Chloroform-Chloroform is weaker than the Chloroform-Acetone interactions.  

A solution of chloroform and acetone exhibits a negative deviation from Raoult's law, which means that the interactions between chloroform and acetone are stronger than those of the individual components

A solution of chloroform (CHCl3) and acetone((CH3)2CO) exhibits a negative deviation from Raoult's law. This means the solution has a lower vapor pressure than predicted by Raoult's Law due to the strength of the new intermolecular interactions being stronger than what was present in the pure components. The correct answer is X. chloroform-chloroform and acetone-acetone interactions are weaker than chloroform-acetone interactions. This is because in a solution exhibiting negative deviation, the interactions between the different molecules in the mixture are stronger than those of the pure components. This stronger interaction between chloroform and acetone molecules reduces the overall vapor pressure, leading to a negative deviation from Raoult's Law.

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Answer:

The reaction requires anti-Markovnikov product without sigmatropic rearrangement.

Explanation:

The reaction is known to begin with the concerted syn addition of B and H across the double bond, with the boron adding to the less substituted carbon atom.

The second step of the reaction involves hydrogen peroxide and a base such as NaOH are added, NaOH deprotonates the hydrogen peroxide.

The resulting NaOOH then attacks the boron and sets up the key migration step, where the carbon-boron bond migrates to the oxygen bound to boron, breaking the weak oxygen-oxygen bond . Then the -OH expelled then returns to form a bond on the boron resulting in a deprotonated alcohol (alkoxide). The alkoxide is then protonated by water or some other comparably acidic species.

Hydroboration is a syn addition that gives an anti-Markovnikov product without sigmatropic rearrangement.

The reasons why hydroboration–oxidation reagents were chosen to effect the transformation will be:

It should be noted that hydroboration–oxidation is vital for the production of alcohol. It's needed for the transformation rather than using other reagents because hydroboration-oxidation reagents yield the anti-Markovnikov product of addition.

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Answer:

Primary alcohol.

Explanation:

Jones reagent is mixture of chromium trioxide (CrO3) and sulfuric acid (H2SO4) dissolved in a mixture of acetone and water. Alternatively, potassium dichromate (K2CrO7) can be used in place of chromium trioxide because of its carcinogenic nature.

This oxidation reaction is an organic reaction for the oxidation of primary alcohols to aldehydes then carboxylic acid and secondary alcohols to ketones.

Lucas reagent is a solution of anhydrous zinc chloride (ZnCl) in concentrated hydrochloric acid. The reaction involves substitution reaction in which the chloride replaces a hydroxyl group.

A positive test is indicated by a change in appearance of the solution, from clear and colourless to fog-like, which shows the formation of a chloroalkane. Accurate results for this test are observed in tertiary alcohols, as they form alkyl halides the fastest due to the stability of their intermediate tertiary carbocation.

Therefore, an alcohol in which there was a positive Jones test and a negative Lucas test indicates the presence of primary alcohol.

This is because:

A primary alcohol would test positive to Jones test but in Lucas test, the substitution reaction is the slowest as compared to the secondary and tertiary alcohols.

1° alcohols < 2° alcohols < 3° alcohols

So a primary alcohol will give a negative result to Lucas reagent.

Acetic acid, CH3CO2H, is composed of carbon, hydrogen, and oxygen atoms. The electron-dot structure is represented by placing valence electrons as dots around the symbol of each atom, showing their arrangement and the bonding between them.

Acetic acid, CH3CO2H, is composed of carbon, hydrogen, and oxygen atoms. The two carbon atoms are connected by a single bond, and both oxygen atoms are connected to the same carbon atom. To draw the electron-dot structure for acetic acid, we represent each atom using its symbol and show the valence electrons as dots surrounding the symbol.

In the case of acetic acid, the carbons are each connected to three hydrogens and one oxygen, and the oxygen is also connected to another carbon. Therefore, for each carbon atom, one dot is placed next to each hydrogen atom and three dots are placed next to the oxygen atom. Two additional dots are placed next to the oxygen atom connected to the other carbon. The arrangement of the dots represents the bonding between the atoms.

The resulting electron-dot structure for acetic acid would be represented as CH3CO2H.

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Answer:

Al 72.61%

Mg 27.39%

Explanation:

To obtain the mass percentages, we need to place the individual masses over the total mass and multiply by 100%.

If we observe clearly, we can see that the parameters given are the moles. We need to convert the moles to mass.

To do this ,we need to multiply the moles by the atomic masses. The atomic mass of aluminum is 27 while that of magnesium is 24.

Now, the mass of aluminum is thus = 27 * 0.0898 = 2.4246g

The mass of magnesium is 0.0381 * 24 = 0.9144g

We can now calculate the mass percentage.

The total mass is 0.9144 + 2.4246 = 3.339g

% mass of Al = 2.4246/3.339 * 100 = 72.61%

% mass of Mg = 0.9144/3.39 * 100 = 27.39%

Answer:

Mass % Al = 72.3 %

Mass % Mg = 27.7 %

Explanation:

Step 1: Data given

Number of moles Al = 0.0898 moles

Number of moles Mg = 0.0381 moles

Molar mass Al = 26.98 g/mol

Molar mass Mg = 24.3 g/mol

Step 2: Calculate mass Al

Mass Al = moles Al * molar mass Al

Mass Al = 0.0898 moles * 26.98 g/mol

Mass Al = 2.42 grams

Step 3: Calculate mass Mg

Mass Mg = 0.0381 moles * 24.3 g/mol

Mass Mg = 0.926 grams

Step 4: Calculate total mass

Total mass = mass Al + mass Mg

Total mass = 2.42 grams + 0.926 grams

Total mass = 3.346 grams

Step 5: Calculate mass %

Mass % Al = (mass Al/ total mass) * 100%

Mass % Al = (2.42 grams / 3.346 grams ) *100%

Mass % Al = 72.3 %

Mass % Mg = (mass Mg/ total mass)*100%

Mass % Mg = (0.926 / 3.346) *100 %

Mass % Mg = 27.7 %

Answer : The concentration of a solution with an absorbance of 0.420 is, 0.162 M

Explanation :

Using Beer-Lambert's law :

[tex]A=\epsilon \times C\times l[/tex]

As per question, at constant path-length there is a direct relation between absorbance and concentration.

[tex]\frac{A_1}{A_2}=\frac{C_1}{C_2}[/tex]

where,

A = absorbance of solution

C = concentration of solution

l = path length

[tex]A_1[/tex] = initial absorbance = 0.350

[tex]A_2[/tex] = final absorbance = 0.420

[tex]C_1[/tex] = initial concentration = 0.135 M

[tex]C_2[/tex] = final concentration = ?

Now put all the given value in the above relation, we get:

[tex]\frac{0.350}{0.420}=\frac{0.135}{C_2}[/tex]

[tex]C_2=0.162M[/tex]

Thus, the concentration of a solution with an absorbance of 0.420 is, 0.162 M

Answer:

              1,1,1-trichloroethane being a saturated halogenated alkane will undergo substitution reaction via free radical mechanism. The mechanism is divided in three steps,

Step 1: Initiation:

                          In this step the reaction is started by treating the chlorine gas either with UV light or by sunlight. This results in the formation of free radical.

Step 2: Propagation:

                          In this step the radical formed will react with the hydrogen atom resulting in formation of HCl and generating free radical of corresponding alkane. Hence, the radical will agian react with Cl2 molecule generating another Chlorine radical and corresponding halogenated compound i.e. 1,1,1,2-tetrachloroethane.

Step 3: Termination:

                           This is the last step. In this step the reaction is stopped/terminated. The free radicals react with each other forming a single bonds and stopping the formation of further radicals.

The mechanism is shown below,

Answer: Inittiation: Br2 -> Br. + Br.

Propagation: CH3CH(CH3)CH3+Br. -> CH3CH(CH3)CH2. +HBr

CH3CH(CH3)CH2. + Br. -> CH3CH(CH3)CH2Br

CH3CH(CH3)CH2Br +Br. -> CH3C.(CH3)CH2Br +HBr

CH3C.(CH3)CH2Br + Br. -> CH3CBr(CH3)CH2Br

Termination: Br. + Br. ->Br2

Answer:

C flyash = 10677789.55 μg/m³

Explanation:

A coal-fired power plant:

∴ rate = 50 ft³/s = (1.416 m³/s)(3600 s/h) = 5097.6 m³/h

hot gas-flyash:

∴ rf = 120 Lb/h = 54431.1 g/h = 54431100000 μg/h

⇒ C flyash = ? [=] μg/m³

⇒ C flyash = (54431100000 μg/h)/(h/5097.6 m³)

⇒ C flyash = 10677789.55 μg/m³

Answer:

35.26 rad

Explanation:

Let's assume the cube in the figure below. If it's in the first octant, then origin (0, 0, 0) is one of the vertices and it's also the common vertex of the diagonals (OB and OE).

The point B is at the y-axis, so since the length is 8, it is (8, 0, 8), and the point E is (8, 8, 8). The vectors of the diagonals are the subtraction of the coordinates of the two points, so OB = <8, 0, 8> and OE = <8, 8, 8>. The angle between two vectors in the tridimensional space is:

θ = cos⁻¹[(OB · OE)/(|OB|·|OE|)]

The module (| |) of a vector <x, y, z> is √(x² + y² + z²)

θ = cos⁻¹[(<8, 0, 8> · <8, 8, 8>)/(√(8² + 0² + 8²) · √(8² + 8² + 8²))]

θ = cos⁻¹[(8*8 + 8*0 + 8*8)/(√128 ·√192)]

θ = cos⁻¹[128/156.77]

θ = cos⁻¹[0.8165]

θ = 35.26 rad

Explanation:

Molarity of solution = 1.00 M = 1.00 mol/L

In 1 L of solution 1.00 moles of calcium chloride is present.

Mass of solute or calcium chloride = m

[tex]m = 1 mol\times 111 g/mol = 111 g[/tex]

Mass of solution = M

Volume of solution = V = 1L = 1000 mL

Density of solution , d= 1.07 g/mL

[tex]M=d\times V=1.07 g/mL\times 1000 mL=1,070 g[/tex]

1) The value of %(m/M):

[tex]\frac{m}{M}\times 100=\frac{111 g}{1,070 g}\times 100=10.37\%[/tex]

2) The value of %(m/V):

[tex]\frac{m}{V}\times 100=\frac{111 g}{1000 L}\times 100=11.1\%[/tex]

[tex]Molality = \frac{\text{Moles of compound }}{\text{mass of solvent in kg}}[/tex]

[tex]Normality=\frac{\text{Moles of compound }}{n\times \text{volume of solution in L}}[/tex]

n = Equivalent mass

n = [tex]\frac{\text{molar mass of ion}}{\text{charge on an ion}}[/tex]

3) Normality of calcium ions:

Moles of calcium ion = 1 mol (1 [tex]CaCl_2[/tex] mole has 1 mole of calcium ion)

[tex]n=\frac{40 g/mol}{2}=20 [/tex]

[tex]=\frac{1 mol}{20 g/mol\times 1L}=0.050 N[/tex]

4) Normality of chlorine ions:

Moles of chlorine ion = 2 mol (1 [tex]CaCl_2[/tex] mole has 2 mole of chlorine ion)

[tex]n=\frac{35.5 g/mol}{1}=35.5[/tex]

[tex]=\frac{2 mol}{35.5 g/mol\times 1L}=0.056 N[/tex]

Moles of calcium chloride = 1.00 mol

Mass of solvent =  Mass of solution - mass of solute

= 1,070 g - 111 g = 959  g = 0.959 kg ( 1 g =0.001 kg)

5) Molality of the solution :

[tex]\frac{1 mol}{0.959 kg}=1.043 mol/kg[/tex]

Moles of calcium chloride = [tex]n_1=1mol[/tex]

Mass of solvent = 959 g

Moles of water = [tex]n_2=\frac{959 g}{18 g/mol}=53.28 mol[/tex]

Mass of solvent = 959 g

6) Mole fraction of calcium chloride =

[tex]\chi_1=\frac{n_1}{n_1+n_2}=\frac{1mol}{1 mol+53.28 mol}=0.01842[/tex]

7) Mole fraction of water =

[tex]\chi_2=\frac{n_2}{n_1+n_2}=\frac{53.28 mol}{1mol+53.28 mol}=0.9816[/tex]

8) Mass of solution = m'

Volume of the solution= v = 100 mL

Density of solution = d = 1.07 g/mL

[tex]m'=d\times v=1.07 g/ml\times 100 g= 107 g[/tex]

Mass of 100 mL of this solution 107 grams of solution.

9) Volume of solution = V = 100 mL

Mass of solution = M'' = 107 g

Mass of solute = m

The value of %(m/V) of solution = 11.1%

[tex]11.1\%=\frac{m}{100 mL}\times 100[/tex]

m = 11.1 g

Mass of solvent = M''- m = 107 g -11.1 g = 95.9 g

95.9 grams of water was present in 100 mL of given solution.

Answer: The concentration of salt (potassium phosphate), phosphoric acid and KOH in the solution is 0.0342 M, 0.0533 M and 0 M respectively.

Explanation:

To calculate the number of moles for given molarity, we use the equation:

[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}[/tex]     .....(1)

For KOH:

Initial molarity of KOH solution = 0.205 M

Volume of solution = 25.0 mL = 0.025 L   (Conversion factor:   1 L = 1000 mL)

Putting values in equation 1, we get:

[tex]0.205M=\frac{\text{Moles of KOH}}{0.025L}\\\\\text{Moles of KOH}=(0.205mol/L\times 0.025L)=5.123\times 10^{-3}mol[/tex]

For phosphoric acid:

Initial molarity of phosphoric acid solution = 0.175 M

Volume of solution = 25.0 mL = 0.025 L

Putting values in equation 1, we get:

[tex]0.175M=\frac{\text{Moles of }H_3PO_4}{0.025L}\\\\\text{Moles of }H_3PO_4=(0.175mol/L\times 0.025L)=4.375\times 10^{-3}mol[/tex]

The chemical equation for the reaction of KOH and phosphoric acid follows:

[tex]3KOH+H_3PO_4\rightarrow K_3PO_4+3H_2O[/tex]

By Stoichiometry of the reaction:

3 moles of KOH reacts with 1 mole of phosphoric acid

So, [tex]5.123\times 10^{-3}[/tex] moles of KOH will react with = [tex]\frac{1}{3}\times 5.123\times 10^{-3}=1.708\times 10^{-3}mol[/tex] of phosphoric acid

As, given amount of phosphoric acid is more than the required amount. So, it is considered as an excess reagent.

Thus, KOH is considered as a limiting reagent because it limits the formation of product.

Excess moles of phosphoric acid = [tex](4.375-1.708)\times 10^{-3}=2.667\times 10^{-3}mol[/tex]

By Stoichiometry of the reaction:

3 moles of KOH produces 1 mole of potassium phosphate

So, [tex]5.123\times 10^{-3}[/tex] moles of KOH will produce = [tex]\frac{1}{3}\times 5.123\times 10^{-3}=1.708\times 10^{-3}moles[/tex] of potassium phosphate

Moles of potassium phosphate = [tex]1.708\times 10^{-3}moles[/tex]

Volume of solution = [25.0 + 25.0] = 50.0 mL = 0.050 L

Putting values in equation 1, we get:

[tex]\text{Molarity of potassium phosphate}=\frac{1.708\times 10^{-3}}{0.050}=0.0342M[/tex]

Moles of excess phosphoric acid = [tex]2.667\times 10^{-3}moles[/tex]

Volume of solution = [25.0 + 25.0] = 50.0 mL = 0.050 L

Putting values in equation 1, we get:

[tex]\text{Molarity of phosphoric acid}=\frac{2.667\times 10^{-3}}{0.050}=0.0533M[/tex]

Moles of KOH remained = 0 moles

Volume of solution = [25.0 + 25.0] = 50.0 mL = 0.050 L

Putting values in equation 1, we get:

[tex]\text{Molarity of KOH}=\frac{0}{0.050}=0M[/tex]

Hence, the concentration of salt (potassium phosphate), phosphoric acid and KOH in the solution is 0.0342 M, 0.0533 M and 0 M respectively.

Answer:

18 Carats because of 75% Purity...

Explanation:

Red Gold Jewelry always contains a mixing of 25% copper by mass to make it durable and strong. That according to the simple parallel ratio rule, gives us a purity level of gold to be exactly 18 Carats. In other words the Red Gold is 75% pure...

The question pertains to the calculation of gold purity in carats. Without an exact gold percentage in the red gold jewelry, a precise purity or carat value cannot be identified.

The question relates to the understanding of carats and the purity of gold used in jewellery. Generally, 24 carats is defined as 100% pure gold. Hence, the number of carats indicates the proportion of gold in an alloy.

The question asks about a piece of red gold jewellery but lacks the necessary percentage to determine its carat value. If we had an exact percentage of gold in the item, we could calculate the carat value by multiplying the given percentage by 24.

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Answer: ΔG =23.169kJ/mol

Explanation:

Solution

To Calculate Gibbs free energy ΔG for the reaction above we use the equation ΔG=ΔH−TΔS.

Where

ΔH= 38.468 kJ/mol = 38468 J/mol

∆S = +51.4 J mol−1 K−1).

T = 25◦C =298k

ΔG= 38468J/mol−298k(51.4 J mol−1 K−1).

ΔG = 38468 J/mol - 15317.2J/mol

ΔG = 23168.8J/mol

ΔG =23.169kJ/mol

Answer: The pressure of carbon dioxide needed is 2.94 atm

Explanation:

To calculate the partial pressure of carbon dioxide, we use the equation given by Henry's law, which is:

[tex]C_{CO_2}=K_H\times p_{CO_2}[/tex]

where,

[tex]K_H[/tex] = Henry's constant = [tex]3.4\times 10^{-2}mol/L.atm[/tex]

[tex]C_{CO_2}[/tex] = molar solubility of carbon dioxide gas = [tex]0.10mol/L[/tex]

[tex]p_{CO_2}[/tex] = pressure of carbon dioxide = ?

Putting values in above equation, we get:

[tex]0.10mol/L=3.4\times 10^{-2}mol/L.atm\times p_{CO_2}\\\\p_{CO_2}=\frac{0.10mol/L}{3.4\times 10^{-2}mol/L.atm}=2.94atm[/tex]

Hence, the pressure of carbon dioxide needed is 2.94 atm

Answer:

6.25×10¹⁹ e⁻

Explanation:

Let's apply a rule of three:

1 e⁻ has 1.60×10⁻¹⁹ C

There, we can think:

1.60×10⁻¹⁹ C of charge are made by 1 e⁻

If we want to produce 1 C, we would need ( 1 . 1) / 1.60×10⁻¹⁹

= 6.25×10¹⁹ e⁻

Answer:

HCOOH + H2O <===> HCOO- + H3O+

Explanation:

Answer:

a) Germanium = 5.76 x 〖10〗^11 〖cm〗^(-3) , Semiconductor is n-type.

b) Silicon = 2.25 x 〖10〗^5 〖cm〗^(-3) , Semiconductor is n-type.

For clear view of the answers: Please refer to calculation 5 in the attachments section.

Explanation:

So, in order to find out the concentration of holes and electrons in a sample of germanium and silicon which have the concentration of donor atoms equals to 〖10〗^15 〖cm〗^(-3). We first need to find out the intrinsic carrier concentration of silicon and germanium at room temperature (T= 300K).

Here is the formula to calculate intrinsic carrier concentration: For calculation please refer to calculation 1:

So, till now we have calculated the intrinsic carrier concentration for germanium and silicon. Now, in this question we have been given donor concentration (N_d) (N subscript d), but if donor concentration is much greater than the intrinsic concentration then we can write: Please refer to calculation 2.

So, now we have got the concentration of electrons in both germanium and silicon. Now, we have to find out the concentration of holes in germanium and silicon (p_o).  (p subscript o)

Equation to find out hole concentration: Please refer to calculation 3. and Calculation 4. in the attachment section.

Good Luck Everyone! Hope you will understand.  

(a) The germanium with 10^15 cm^-3 donor atoms: Concentration of electrons = 10^15 cm^-3, Concentration of holes ≈ 0, Semiconductor is n-type. (b) For silicon with 10^15 cm^-3 donor atoms: Concentration of electrons = 10^15 cm^-3, Concentration of holes ≈ 0, Semiconductor is n-type.

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Explanation:

Ionic equation

NaCl(aq) --> Na+(aq) + Cl-(aq)

Na2SO4(aq) --> 2Na+(aq) + SO4^2-(aq)

In NaCl solution, 1 mole of Na+ is dissociated in 1 liter of solution while in Na2SO4, 2 moles of Na+ is dissociated in 1 liter of solution.

Molecular weight of NA2SO4 = (23*2) + 32 + (16*4)

= 142 g/mol

Molecular weight of NaCl = 23 + 35.5

= 58.5 g/mol

Masses

% Mass of NA+ in Na2SO4 = mass of Na+/total mass of Na2SO4 * 100

= 46/142 * 100

= 32.4%

% Mass of NA+ in NaCl = mass of Na+/total mass of NaCl * 100

= 23/58.5 * 100

= 39.3%

Therefore, the % mass of Na+ in NaCl and Na2SO4 are different so it cannot be used.

You cannot substitute Na2SO4 directly for NaCl based on mass since they have different molar masses. The same mass of Na2SO4 will provide more Na+ ions than NaCl, leading to a change in the Na+ ion concentration.

No, you cannot substitute the same number of grams of Na2SO4 for the NaCl in a solution. This is because NaCl and Na2SO4 have different molar masses and therefore different numbers of moles per gram. The concentration of a solution is determined by the number of moles of solute per unit volume of solvent, not the mass. Hence, using the same mass of a different compound would alter the concentration of Na+ ions in the solution.

For instance, if one mole of NaCl gives us one mole of Na+, one mole of Na2SO4 will provide two moles of Na+. In other words, the same mass of Na2SO4 contains more Na+ ions than the same mass of NaCl. So using the same mass of Na2SO4 in place of NaCl will result in a solution with a higher Na+ ion concentration.

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Answer:

0.681 atm

Explanation:

To solve this problem, we make use of the General gas equation.

Given:

P1 = 785 torr

V1 = 2L

T1 = 37= 37 + 273.15 = 310.15K

P2 = ?

V2 = 3.24L

T2 = 58 = 58+273.15 = 331.15K

P1V1/T1 = P2V2/T2

Now, making P2 the subject of the formula,

P2 = P1V1T2/T1V2

P2 = [785 * 2 * 331.15]/[310.15 * 3.24]

P2 = 515.715 Torr

We convert this to atm: 1 torr = 0.00132 atm

515.715 Torr = 515.715 * 0.00132 = 0.681 atm

Answer:

6.6 is the [tex]pK_a[/tex] of the weak acid.

Explanation:

To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:

[tex]pH=pK_a+\log(\frac{[salt]}{[acid]})[/tex]

We are given:

[tex]pK_a[/tex] = negative logarithm of acid dissociation constant =?

The ratio of conjugate base to acid is = [tex]\frac{[salt]}{acid}=4[/tex]

pH = 7.2

Putting values in above equation, we get:

[tex]7.2=pK_a+\log(4)[/tex]

[tex]pK_a=7.2-\log(4)=6.598\approx 6.6[/tex]

6.6 is the [tex]pK_a[/tex] of the weak acid.

Answer:

0.00077 qt

Explanation:

Density -

Density of a substance is given by the mass of the substance divided by the volume of the substance .

Hence , d = m / V

V = volume

m = mass ,

d = density ,

From the question ,

The mass mercury = 100 g

Density of mercury = 13.6 g/cm³ .

Hence , by using the above formula ,and putting the corresponding values , the volume of mercury is calculated as -

d = m / V

13.6 g/cm³ = 100 g  / V

V = 7.35 cm³

1 cm³ = 0.001 L

V = 7.35 * 0.001 L = 0.0073 L

Since ,

1 L = 1.06 qt

V = 0.0073* 1.06 qt = 0.0077 qt

A triglyceride formed from glycerol and myristic acid involves esterifying three myristic acid molecules with a glycerol backbone through a dehydration reaction, releasing three water molecules and forming a simple triglyceride.

The triglyceride formed from the esterification of glycerol and three molecules of myristic acid is constructed by attaching each of the three fatty acid molecules to the glycerol backbone through a dehydration reaction. In this process, each fatty acid's carboxyl group (COOH) reacts with one of the hydroxyl groups (OH) on the glycerol molecule, resulting in the formation of an ester bond and the release of water.

The chemical structure of glycerol (H₂C-OH | HC-OH | H₂C-OH) bonded with three myristic acid molecules (which have the formula CH₃(CH₂)₌₁₂COOH) will show three ester linkages replacing the hydroxyl groups of glycerol with the alkyl chains of myristic acid.

Each dehydration synthesis reaction between the carboxyl group of a fatty acid and a hydroxyl group of glycerol results in the release of one molecule of water, leading to a total of three water molecules released when forming the triacylglycerol. If all three OH groups on the glycerol molecule are esterified with myristic acid, the molecule formed is a simple triglyceride because it contains only one type of fatty acid.

Answer:

Option D. 4.02 kJ

Explanation:

A simple calorimetry problem

Q = m . C . ΔT

ΔT = Final T° - Initial T°

C = Specific heat capacity

m = mass

Let's replace the data

Q = 125 g . 2.42 J/g∘C . (34.8°C -21.5 °C)

Q= 4023.25 J

We must convert the answer to kJ

4023.25 J . 1kJ /1000 =4.02kJ

Answer:

(a)    ⁴¹₁₉K

(b)  ¹⁰⁶₄₆Pd

(c)  ¹²⁵₅₂Te

(d)   ⁸⁸₃₈Sr

Explanation:

The identity of an element  is its atomic number, by convention we write the atomic mass as superscript and the and the atomic number as subscript to the left of the element .

(a) Z = 19   A = 41         symbol:      ⁴¹₁₉K

(b) Z = 46  A = 106       symbol:   ¹⁰⁶₄₆Pd

(c) Z = 52   A = 125       symbol:  ¹²⁵₅₂Te

(d) Z = 38   A  = 88       symbol:    ⁸⁸₃₈Sr

The isotopes correspond to Potassium-41 (41K), Palladium-106 (106Pd), Tellurium-125 (125Te), and Strontium-88 (88Sr). Z and A denote atomic and mass numbers, which help in identifying particular isotopes.

The appropriate symbol for each of these isotopes would be:

Z and A represent atomic and mass numbers respectively. These isotopes are written with a mass number (A) preceding the symbol and atomic number (Z) is typically understood from the symbol itself.

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Answer:

The wavelength in nm = 5120

The wavelength in A°= 51,200

The frequency of the absorbed radiation is [tex]5.859\times 10^{13} Hertz[/tex]

Explanation:

1) The wave number of the CO bond = [tex]\bar v=1953 cm^{-1}[/tex]

The wavelength corresponding to this wave number =[tex]\lambda [/tex]

[tex]\lambda =\frac{1}{\bar v}=\frac{1}{1953 cm^{-1}}=5.12\times 10^{-4} cm[/tex]

[tex]1 cm = 10^7 nm[/tex]

[tex]5.12\times 10^{-4}\times 10^7 nm=5120 nm[/tex]

[tex]1 cm = 10^8 \AA[/tex]

[tex]5.12\times 10^{-4}\times 10^8 \AA=51,200 \AA[/tex]

The wavelength in nm = 5120

The wavelength in A°= 51,200

2)

Wavelength of the wave = [tex]\lambda =5120 nm = 5120\times 10^{-9} nm[/tex]

[tex]1 nm = 10^{-9} m[/tex]

Frequecy of the wave = [tex]\nu [/tex]

[tex]\nu=\frac{c}{\lambda }[/tex]

c = Speed of light = [tex]3\times 10^8 m/s[/tex]

[tex]\nu=\frac{3\times 10^8 m/s}{5120\times 10^{-9} m}=5.859\times 10^{13} s^{-1}[/tex]

The frequency of the absorbed radiation is [tex]5.859\times 10^{13} Hertz[/tex]

The bond between hemoglobin and CO absorbs radiation of 1953 cm⁻¹. The corresponding wavelength is 5120 nm and 51200 Å. The corresponding frequency is 5.855 × 10¹³ Hz.

The bond between hemoglobin and CO absorbs radiation of 1953 cm⁻¹, that is, the wavenumber (ν) is 1953 cm⁻¹.

We can calculate the wavelength (λ) using the following expression.

[tex]\lambda = \frac{1}{\nu } = \frac{1}{1953cm^{-1} } = 5.120 \times 10^{-4} cm[/tex]

We will convert 5.120 × 10⁻⁴ cm to nm using the following conversion factors.

[tex]5.120 \times 10^{-4} cm \times \frac{1m}{100cm} \times \frac{10^{9}nm }{1m} = 5120 nm[/tex]

We will convert 5.120 × 10⁻⁴ cm to Å using the following conversion factors.

[tex]5.120 \times 10^{-4} cm \times \frac{1m}{100cm} \times \frac{10^{10}A }{1m} = 51200 nm[/tex]

We can calculate the frequency (f) of the absorbed radiation using the following expression.

[tex]f = \frac{c}{\lambda } = \frac{2.998 \times 10^{8}m/s }{5.120\times 10^{-6}m } = 5.855 \times 10^{13} Hz[/tex]

where,

The bond between hemoglobin and CO absorbs radiation of 1953 cm⁻¹. The corresponding wavelength is 5120 nm and 51200 Å. The corresponding frequency is 5.855 × 10¹³ Hz.

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Answer:

Options A and B are correct.

Sodium salts and Carbon based fuels satisfy the criteria.

Explanation:

Sodium salts and Carbon Based fuels produce bright yellow emissions that can mask other emission colors.

Magnesium Salts produce no emissions, although, Magnesium metal burns brightly white while Aluminium salts produce white emissions.

Hope this Helps!!!

Options A and B are correct.

Sodium salts and Carbon based fuels satisfy the criteria.

The following information should be considered:

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The formula weight of the unknown copper compound is approximately 13.90145 amu.

To find the formula weight (F.W.) of the unknown copper compound, we need to consider the percent copper composition and the atomic weight of copper. The atomic weight of copper is approximately 63.55 amu.

We can use the percent composition to calculate the mass of copper in one mole of the compound:

(0.2190) * F.W. = (0.2190) * (63.55 amu) = 13.90145 amu

Since one mole of the compound contains one mole of copper, we can equate the mass of copper to the formula weight:

13.90145 amu = F.W.

Therefore, the formula weight of the unknown copper compound is approximately 13.90145 amu.

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Answer:

2 moles of Methane gas will occupy 44.8 L

Explanation:

If one mole of any gas occupies 22.4 L under certain conditions of temperature and pressure, and those conditions are assumed in this question, then we comfortably solve this problem as follows;

1 mole of Methane gas ---------------> 22.4 L

2 moles of Methane gas -------------->?

Cross and multiply, 2 moles of Methane gas = 2 X 22.4 L = 44.8L

Therefore, 2 moles of Methane gas will occupy 44.8 L, if the conditions of temperature and pressure are maintained.

OPTION A IS THE RIGHT SOLUTION.

Answer:

The answer is A

Explanation: