A 500 mL hypertonic saline solution is labeled as consisting of 5.21 % w/w NaCl. Given that the density of salt water is 1.02 g/mL, what is the molarity of the saline solution? Molar mass of NaCl = 58.44 g/mol.

Answer:

E1 = 9.83 x [tex]10^{20}[/tex] [tex]NC^{-1}[/tex]

E2 = 1.748 x [tex]10^{21}[/tex] [tex]NC^{-1}[/tex]

Explanation:

E1 = k Q/r2 = 8.99 x [tex]10^{9}[/tex] x 7 x [tex]10^{-6}[/tex] / 8 x [tex]10^{-9}[/tex] x 8 x [tex]10^{-9}[/tex] = 9.83 x [tex]10^{20}[/tex] [tex]NC^{-1}[/tex]

E2 = k Q/r2 = 8.99 x [tex]10^{9}[/tex] x 7 x  [tex]10^{-6}[/tex] / 6 x  [tex]10^{-9}[/tex] x 6 x [tex]10^{-9}[/tex] = 1.748 x [tex]10^{21}[/tex] [tex]NC^{-1}[/tex]

The direction of the electric field will be from E1 to E2...

Explanation:

Pseudo-noble gas configuration which can also be called pseudo inert configuration is when elements have fully filled d-orbital, along with s- and p- orbitals. Examples are ions of elements, this is because they lose their valence electrons to be fully filled in order to have a stable octet.

Example of Group 3A(13) ion is Aluminium ion; Al3+. Al3+ takes up the configuration of Neon when it loses its 3 valence electrons.

Electronic configuration of Al and Al3+

Al = [Ne] 3s2 3p1

Al3+ = 1s2 2s2 2p6

A pseudo-noble gas configuration occurs when an ion or atom has an electron configuration that resembles that of a noble gas. An example of an ion from Group 3A(13) that has a pseudo-noble gas configuration is aluminium (Al3+).

A pseudo-noble gas configuration refers to the electron configuration of an ion or atom that resembles that of a noble gas. This occurs when an ion gains or loses electrons to acquire the same electron configuration as a noble gas. An example of an ion from Group 3A(13) that has a pseudo-noble gas configuration is aluminium (Al3+).

Aluminum has an electron configuration of 1s2 2s2 2p6 3s2 3p1 in its ground state. When it loses its three valence electrons, its configuration becomes 1s2 2s2 2p6, which is the same as that of the noble gas, neon (Ne). Therefore, aluminium has a pseudo-noble gas configuration.

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Answer:

a) The flame test of Na, at 589nm will show a golden yellow color when the sodium ion is emitted. These is due to the electropositive nature of the alkali metals.

While the flame test for K at 404nm will ignite a violet color when the Potassium ion is emitted. This properties are due to the high electropositive nature of the group1 elements which also indicate their strong reducing agent.

b) The cobalt glass filter act as hindrance during the flame test to seperate or filter the golden yellow color caused as a result of the presence of the sodium, as it makes the violet color to be more visible.

c) These is due to the Oxidizing ability of KClO₄ or KClO₃ compared to the salts of sodium. Also is the low solubility of the two salts and their solubility constant (Ksp) compared to sodium salts.

Explanation:

a) The flame test of Na, at 589nm will show a golden yellow color when the sodium ion is emitted. These is due to the electropositivity nature of the alkali metals.

While the flame test for K at 404nm will ignite a violet color when the Potassium ion is emitted. This properties are due to the high electropositivity nature of the group1 elements which also indicate their strong reducing agent.

b) The cobalt glass filter act as hindrance during the flame test to seperate or filter the golden yellow color caused as a result of the presence of the sodium, as it makes the violet color to be more visible.

c) These is due to the Oxidizing ability of KClO₄ or KClO₃ compared to the salts of sodium. Also is the low solubility of the two salts and their solubility constant (Ksp) compared to sodium salts.

Answer:

5.00 µM

Explanation:

Given data

We can find the final concentration using the dilution rule.

C₁ × V₁ = C₂ × V₂

C₂ = C₁ × V₁ / V₂

C₂ = 250.0 µM × 1.00 mL / 50.0 mL

C₂ = 5.00 µM

The concentration of the diluted solution is 5.00 µM.

Answer:

The answer according to the given equation is S = 0.00077 cm

Explanation:

According to this equation

S = (4Dt)^1/2

S = (4* 1^e-8 * 15)^0.5

S = 0.00077 cm

According to the Approximation equation for diffusion time

t ≅ S^2 / 2D

S = 0.00055 cm  

Using the equation S = (4Dt)^1/2, the average distance olive oil can move in a membrane in 15 seconds is calculated to be 7.75 micrometers.

To calculate the distance that olive oil, a lipid, could move in a membrane in 15 seconds with a diffusion coefficient of 1 μm2/s, we use the equation S = (4Dt)1/2. Plugging in the values, we get S = (4 × 1 μm2/s × 15 s)1/2. We perform the calculation as follows:

The average distance that olive oil can move in the membrane in 15 seconds is 7.75 micrometers.

Answer: 4376.6g/mol

Explanation:Please see attachment for explanation

The molar mass of the protein is 4376.6 g/mol.

The molar mass of a chemical compound is defined as the mass of a sample divided by the amount of substance in that sample, measured in moles.

Solution:

Given, the mass of protein is 3.3×10-2 mg

Mass converted into gram = 3.3×10-2 mg = M 0.000033 g

Therefore, xg of the protein dissolve in 1 L

xg pf protein =[tex]\bold{\dfrac{0.000033}{0.00025} = 0.132\; g/L}[/tex]

Given, the osmotic pressure of 0.56 torr.

[tex]\bold{\pi = 0.56\;torr =\dfrac{0.56}{760} = 0.00737\; atm p}[/tex]

Thus, the volume of the solution is 0.007 L

Given, Temperature = 25 °C

Converting into kelvin = 25+273 = 298 k.

R = 0.082 atm.

Now, by the formula  [tex]\bold{M_2= \dfrac{W_2RT}{\pi V}}[/tex]

[tex]\bold{M_2= \dfrac{0.132\times0.082\times298}{0.00737\times0.007} =4376.6 \;g/mol}[/tex]

Thus, the molar mass of protein is 4376.6 g/mol.

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Answer:

The correct answer is option C) "equilibrium price of peanut butter definitely falls".

Explanation:

The equilibrium price of a product is the market price established at a point where the quantity of products is equal to the quantity demanded by the consumers. In this case, the market of peanut butter is facing the decrease in the price of two substitutes of peanut butter: deli turkey slices and peanut brittle. As a result, the equilibrium price of peanut butter definitely falls since the quantity demanded by the consumers will certainly fall.

Final answer:

A decrease in the price of deli turkey slices and peanut brittle affects the demand and supply for peanut butter differently, making the final impact on the market for peanut butter ambiguous without further information.

Explanation:

The question considers the effects of changes in the prices of substitute goods in consumption (deli turkey slices) and production (peanut brittle) on the market for peanut butter. A decrease in the price of deli turkey slices, a substitute in consumption, would lead to consumers substituting away from peanut butter towards deli turkey slices, thereby decreasing the demand for peanut butter. Conversely, a decrease in the price of peanut brittle, a substitute in production, might signal that manufacturers could switch to producing more peanut brittle as it becomes more profitable, potentially decreasing the supply of peanut butter if resources are diverted. Consequently, the equilibrium quantity of peanut butter might increase or decrease, depending on the relative magnitudes of the shifts in supply and demand, while the equilibrium price might rise or fall for similar reasons. Therefore, without additional information on the extents of these changes, the final impact on the equilibrium price and quantity of peanut butter cannot be definitively determined, making option E the correct answer.

Answer:

The total pressure at equilibrium is 0.07503 atm

Explanation:

The partial pressure of the product at equilibrium will be calculated as follows;

Kp = Kc[RT]³

given;

equilibrium constant Kc = 1.58 X 10⁻⁸

gas constant R = 0.0821 L.atm/mol.K

temperature T = (250 +273) = 523 k

Kp = 1.58 X 10⁻⁸ *(0.0821)³ *(523)³ = 1.251 X 10⁻³

NH₂COONH₄(s) ⇌ 2NH₃(g) + CO₂

NH₂COONH₄(s): Kp = 0, since it is in solid state

2NH₃(g) + CO₂: Kp = 1.251 X 10⁻³

I.C.E Analysis on the product

          2NH₃(g)       CO₂

I  :      0                  0

C :      2x                x

E :     (2x-0)          (x-0)

At equilibrium, E: (2x-0)(x-0) = 1.251 X 10⁻³

(2x)(x) = 1.251 X 10⁻³

2x² = 1.251 X 10⁻³

x²   = (1.251 X 10⁻³)/2

x²  = 6.255 X 10⁻⁴

x  = √(6.255 X 10⁻⁴)

x  = 0.02501 atm

Partial pressure of 2NH₃(g)  = 2x = 2(0.02501 atm) = 0.05002 atm

Partial pressure of  CO₂ = x = 0.02501 atm

Total pressure = P(NH₃(g)) +P(CO₂)

Total pressure = 0.05002 atm + 0.02501 atm = 0.07503 atm

Therefore, the total pressure at equilibrium is 0.07503 atm

From data provided, the total pressure at equilibrium is 0.203 atm.

The total pressure, Ptotal at equilibrium is calculated from the equation of the reaction given below:

From the equation of the reaction, If x moles of ammonium carbamate decomposes, it will produce 2x moles of NH3(g) and x moles of CO2(g).

Ammonium carbamate is a solid, and so it does not appear in the expression for Kc.

Kc = 1.58 × 10^-8

Therefore:

Kc = [NH3(g)]^2[CO2(g)]

Kc = (2x)^2(x) = 4x^3

1.58 × 10-8 = 4x^3

Thus, x = 0.00158 M

Hence:

[NH3(g)] = 2 × 0.00158 = 0.00316 M

[CO2(g)] = 0.00158 M

From the ideal gas equation:

where

Also, concentration is given by:

Therefore, P = cRT.

Substituting and calculating for Ptotal:

Ptotal = (0.00316 M) × RT + (0.00158 M) × RT

= ((0.00316 + 0.00158) M) × (0.08206) × ((523) K)

Ptotal = 0.203 atm

Therefore, the total pressure at equilibrium is 0.203 atm.

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Answer:a large amount of solute

Explanation:

A solution is composed of a solute and a solvent. If the amount of solute is greater than that of solvent, the solution is concentrated. A concentrated solution contains quite a large amount of solute while a dilute solution contains less amount of solute. This is the difference between diluted and concentrated solutions.

Answer:

Answer:a large amount of solute

Explanation:

A solution is composed of a solute and a solvent. If the amount of solute is greater than that of solvent, the solution is concentrated. A concentrated solution contains quite a large amount of solute while a dilute solution contains less amount of solute. This is the difference between diluted and concentrated solutions.

Explanation:

The change in the energy of the water for this process is calculated as 11.3375 kJ, considering the heat transferred and work done.

To determine the change in energy of the water within the piston-cylinder setup, we need to account for the heat transferred and the work done. The given values are:

The net heat energy (Q) added to the system is:

For the work done by the system, we use:

where W is work, P is pressure (101.325 kPa), and ΔV is the volume change (0.5 m³). Converting kPa to kJ by recognizing 1 kPa×m³ = 1 kJ, we get:

Using the first law of thermodynamics, the change in internal energy (ΔU) is:

Substituting the values, we get:

Therefore, the change in the energy of the water for this process is 11.3375 kJ.

Answer:

Vapor pressure of solution is 66.7 mmHg

Explanation:

Colligative property about vapor pressure lowering. That's we must use to solve this problem.

Formula is: ΔP = P° . Xm

P° is Vapor pressure of pure solvent

ΔP = P° - P' (vapor pressure of solution)

Xm = mole fraction of solute (mol of solute / total moles)

Let's determine the total moles, firstly.

Total moles = moles of solute + moles of solvent

Moles of solute → 0.217 mol

Moles of solvent → 179 g / molar mass of benzene

179 g / 78 g/mol = 2.29  mol

2.29 mol + 0.217 mol = 2.507 moles

Xm for solute = 0.217 mol / 2.507 mol = 0.0865

Let's replace the data in the formula:

73 mmHg - P' = 73 mmHg  . 0.0865

P' = - (73 mmHg  . 0.0865  - 73mmHg)

P' = 66.7 mmHg

Answer:

The usual 'particles' that we see in the macroscopic world (let's call them objects), are big and massive. This usually means that they have high characteristic frequencies => low wavelengths.Explanation: umm on google i found

Answer:

109.09°C

Explanation:

Given that:

the capacity of the cooling car system = 5.6 gal

volume of solute = volume of the water; since a 50/50 blend of engine coolant and water (by volume) is used.

∴ [tex]\frac{5.60}{2}gallons = 2.80 gallons[/tex]

Afterwards, the mass of the solute and the mass of the water can be determined as shown below:

mass of solute = [tex](M__1}) = Density*Volume[/tex]

                          [tex]= 1.1g/mL *2.80*\frac{3785.41mL}{1gallon}[/tex]

                         [tex]= 11659.06grams[/tex]

On the other hand; the mass of water = [tex](M__2})= Density*Volume[/tex]

                         [tex]= 0.998g/mL *2.80*\frac{3785.41mL}{1gallon}[/tex]

                        [tex]= 10577.95 grams[/tex]

Molarity = [tex]\frac{massof solute*1000}{molarmassof solute*massofwater}[/tex]

              =  [tex]\frac{11659.06*1000}{62.07*10577.95}[/tex]

              = 17.757 m

              ≅ 17.76 m

∴  the boiling point of the solution is calculated using the  boiling‑point elevation constant for water and the Molarity.

[tex]\Delta T_{boiling} = k_{boiling}M[/tex]

where,

[tex]k_{boiling}[/tex] = 0.512 °C/m

[tex]\Delta T_{boiling}[/tex] =  100°C + 17.56 × 0.512

              = 109.09 °C

The equilibrium molarity of SO4 2- in a 0.049 M aqueous solution of H2SO4 will be approximately 0.049 M after the initial dissociation, as sulfuric acid is a strong acid and dissociates completely in its first step, and the second dissociation is generally weaker.

The question involves the dissociation of sulfuric acid (H2SO4) in water, which occurs in two steps. The first dissociation is strong, with the equation H2SO4 (aq) → 2H+ (aq) + SO42- (aq). Because sulfuric acid is a strong acid, the initial dissociation is essentially complete, and the equilibrium molarity of SO42- will be equal to the initial molarity of H2SO4, barring any further reactions.

Given a 0.049 M solution of H2SO4, after the first dissociation, we have 0.049 M of SO42-. The second dissociation of HSO4- to form SO42- is weak. However, without the acid dissociation constant (Ka) value or any other provided equilibrium concentrations, one cannot calculate the additional contribution of SO42- from the second dissociation. Therefore the most straightforward answer, assuming the second dissociation's contribution is negligible compared to the first, is that the equilibrium molarity of SO42- due to the first dissociation is approximately 0.049 M.

Answer: The limiting reagent is iron (II) chloride and the mass of sodium chloride formed is 21.2 grams

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 iron (II) chloride = 23 g

Molar mass of iron (II) chloride = 126.8 g/mol

Putting values in equation 1, we get:

[tex]\text{Moles of iron (II) chloride}=\frac{23g}{126.8g/mol}=0.181mol[/tex]

Given mass of sodium phosphate = 41 g

Molar mass of sodium phosphate = 164 g/mol

Putting values in equation 1, we get:

[tex]\text{Moles of sodium phosphate}=\frac{41g}{164g/mol}=0.25mol[/tex]

The chemical equation for the reaction of iron (II) chloride and sodium phosphate follows:

[tex]3FeCl_2+2Na_3PO_4\rightarrow 6NaCl+Fe_3(PO_4)_2[/tex]

By Stoichiometry of the reaction:

3 moles of iron (II) chloride reacts with 2 moles of sodium phosphate

So, 0.181 moles of iron (II) chloride will react with = [tex]\frac{2}{3}\times 0.181=0.1206mol[/tex] of sodium phosphate

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

Thus, iron (II) chloride is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

3 moles of iron (II) chloride produces 6 mole of sodium chloride.

So, 0.181 moles of iron (II) chloride will produce = [tex]\frac{6}{3}\times 0.181=0.362moles[/tex] of sodium chloride.

Now, calculating the mass of sodium chloride from equation 1, we get:

Molar mass of sodium chloride = 58.5 g/mol

Moles of sodium chloride = 0.362 moles

Putting values in equation 1, we get:

[tex]0.362mol=\frac{\text{Mass of sodium chloride}}{58.5g/mol}\\\\\text{Mass of sodium chloride}=(0.362mol\times 58.5g/mol)=21.2g[/tex]

Hence, the limiting reagent is iron (II) chloride and the mass of sodium chloride formed is 21.2 grams

To determine the limiting reagent between FeCl2 and Na3PO4, the masses of the reactants are converted to moles and compared according to the stoichiometry of the balanced chemical equation. The reactant that produces the least amount of product is the limiting reagent, and using this information, the mass of NaCl produced can be calculated.

The student is asking which reactant is the limiting reagent in the reaction between FeCl2 and Na3PO4, and the amount of NaCl that can be formed as a result. To solve this, we must first balance the chemical equation and then convert the masses of the reactants to moles. After that, we use stoichiometry to compare the mole ratios and identify the limiting reactant. Lastly, we calculate the mass of NaCl produced by the reaction, based on the limiting reactant's moles.

Let's balance the equation: FeCl2 + Na3PO4 → Fe3(PO4)2 + NaCl. We don't have the balanced equation here, but typically, you'd see NaCl being produced alongside iron(III) phosphate. To compute the moles of each reactant, we use their molar masses (FeCl2 = 126.75 g/mol, Na3PO4 = 163.94 g/mol) and determine the mole ratio according to the balanced equation. The smaller mole ratio indicates the limiting reagent. Using the stoichiometry of the balanced reaction, we then calculate the amount of NaCl that can be formed.

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Answer:

i asked to where my uncle shoes he told my  no

Explanation:

Answer:

(a) No overlap

(b) There is overlap

(c) Two

(d) See explanation below

Explanation:

1/λ = Rh (1/n₁² - 1/n₂² )

where λ is the wavelength of the transion, n₁ and n₂ are the principal energy levels ( n₁ < n₂ )

To solve this question, our strategy is to :

1. Calculate the longest wavelength for n₁ = 1,   which  corresponds to  the transition with n₂ = 2.

2. Calculate the  shortest wavelength for n₁ = 2, which  corresponds to n₂ = infinity.

3. Compare the values to check if there is overlap

Lets plug the numbers to visualize this better:

Rydberg´s equation : 1/λ = 1.097 x 10⁷ /m x  (1/n₁² - 1/n₂² )

For n₁ = 1, longest wavelength ( n₂ = 2 ) :

1/λ = 1.097 x 10⁷ /m x  (1/1 ² - 1/2² ) = 8227.5/m

λ  = 1/8227.5/m = 121 x 10⁻⁴ m x 1 x 10⁹ nm/m = 1.22 x 10² nm

For n₁ = 2, shortest wavelength ( n₂ = infinity ) :

1/λ = 1.097 x 10⁷ /m x  (1/2 ² ) = 2.7 x 10⁶ /m

λ  = 1/2.7 x 10⁶/m = 3.7 x 10⁻⁷ m x 1 x 10⁹ nm/m = 3.70 x 10² nm

There is no overlap between the n₁ = 1 and n₁ = 2 series ( there is no overlap    1.22 x 10² nm vs 3.70 x 10² nm )

(b)  Repeat the same  procedure as in part (a)

For n₁ = 3, longest wavelength ( n₂ = 4 ) :

1/λ = 1.097 x 10⁷ /m x  (1/3 ² - 1/4² ) =5.33 x 10⁵/m

λ  = 1/5.33 x 10⁵/m =1.88 x 10⁻⁶ m x 1 x 10⁹ nm/m = 1.88 x 10³ nm

For n₁ = 4, shortest wavelength ( n₂ = infinity ) :

1/λ = 1.097 x 10⁷ /m x  (1/4 ² ) = 6.86 x 10⁵ /m

λ  = 1/6.86 x 10⁵/m = 1.46 x10⁻⁶ m x 1 x 10⁹ nm/m = 1.46 x 10³ nm

There will be overlap

(c) Proceed as in the calculations above but now not only calculate for n₂ = 5 for n₁ = 4 but also a couple more and  verify if there is overlap and count them.

For n₁ = 4  lets calculate n₂ = 5, 6, 7

1/λ =  1.097 x 10⁷ /m x  (1/4 ² - 1/5² ) = 2.47 x 10⁵/m

λ  = 1/2.47 x 10⁵/m = 4.05 x10⁻⁶ m x 1 x 10⁹ nm/m = 4.05 x 10³ nm

The same calculation is done for n₂ = 6 and 7, with the following results:

2.63 x 10³ nm, 2.17 x 10³ nm

Now the shortest wavelength in n₁ = 5 is:

1/λ = 1.097 x 10⁷ /m x  (1/5² ) = 4.39 x 10⁵ / m

λ  = 1/4.39 x 10⁵/m = 2.28 x10⁻⁶ m x 1 x 10⁹ nm/m = 2.28 x 10³ nm

There will be an overlap with 2 lines of n₁ = 4 (2.63 x 10³ nm, 2.17 x 10³ nm )

(d) The overlap tell  us that the energy gap between energy levels becomes smaller as we could see from the calculations above. The spectra becomes confusing  as there is more overlaps.

Final answer:

The Aufbau principle leads to the format of the periodic table by arranging elements based on their electron configurations and the number of valence electrons they have.

Explanation:

The Aufbau principle, in connection with the periodic law, leads to the format of the periodic table by arranging elements based on their electron configurations and the number of valence electrons they have. The periodic table is organized in a way that groups elements with similar properties in the same vertical columns, also known as groups. This arrangement follows the filling of subshells with electrons according to the Aufbau principle.

Answer:

Benzoic acid

Explanation:

The strength of an acid is principally a measure of its dissociative capabilities in aqueous solutions. While strong acids dissociate completely in solution, weak acids dissociates only partially.

The relative strength of an acid can be obtained from its pKa value. The pKa value is the negative logarithm of the concentration of the Ka value.

Stronger acids have a pKa value usually negative. This is a pointer to the fact that the lower the pKa value, the stronger the strength of the acid in question.

Relatively therefore, Benzoic acid is stronger than acetic acid because it has a lesser value for pKa

The strength of an acid is determined by its pKa value, with lower values indicating stronger acids. Therefore, between CH3CO2H (acetic acid, pKa = 4.8) and C6H5CO2H (benzoic acid, pKa = 4.2), benzoic acid is the stronger acid.

When comparing acids, the strength of an acid is determined by its pKa value. The lower the pKa value, the stronger the acid is. In this case, the two acids that are being compared are CH3CO2H (acetic acid, pKa = 4.8) and C6H5CO2H (benzoic acid, pKa = 4.2). Therefore, given that the pKa of benzoic acid is lower than that of acetic acid, we can conclude that benzoic acid is the stronger acid. To summarize, the best answer to the question 'Which acid is the stronger acid, CH3CO2H (acetic acid) or C6H5CO2H (benzoic acid)?' is option a. benzoic acid.

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Answer:

The empirical formula is C4H8O

The molecular formula is C8H16O2

Explanation:

Step 1: Data given

Mass of the unknown sample = 5.90 grams

Temperature = 20.0 °C

PRessure = 1 bar

Volume CO2 = 7.98 L

Volume of H2O = 5.91 mL = 0.00591 L

Density of water at 20.0 °C = 0.998 g/mL

Step 2: Calculate moles CO2

p*V= n*R*T

⇒ with p = the pressure = 1 bar = 0.986923 atm

⇒ with V = the volume of CO2 = 7.98 L

⇒ with n = the number of moles CO2 = TO BE DETERMINED

⇒ with R = The gas constant = 0.08206 L*atm/K*mol

⇒ with T = the temperature = 20.0 °C = 293 Kelvin

n = (p*V)/(R*T)

n = (0.986923*7.98)/(0.08206*293)

n = 0.3276 moles

Step 3: Calculate mass water

Mass water = volume * density

Mass water = 5.91 mL * 0.998 g/mL

Mass water = 5.89818 grams

Step 4: Calculate moles H2O

Moles H2O = 5.89818 grams / 18.02 g/mol

Moles H2O = 0.3273 moles

Step 5: Calculate moles of hydrogen

For 1 mol H2O we have 2 moles of hydrogen

For 0.3273 moles H2O we have 2*0.3273 moles = 0.6546 moles

Step 6: Calculate moles of carbon

1 mol CO2 has 1 mol C

0.3276 moles moles CO2 has 0.3276 moles C

Step 7: Calculate mass C

Mass C = 0.3276 moles * 12.0 g/mol

Mass C = 3.93 grams

Step 8: calculate mass of oxygen

Mass of O = mass of sample - (mass of C + mass of H)

Mass O = 5.90 grams - (3.93 +0.661 )

Mass O = 1.309 grams

Step 9: Calculate moles O

Moles O = 1.309 grams / 16.0 g/mol

Moles O = 0.0818 moles

Step 10: Calculate mol ratio

We divide by the smallest amount of moles

C: 0.3276 / 0.0818 = 4

H: 0.6546 / 0.0818 = 8

O: 0.0818 / 0.0818 = 1

The empirical formula is C4H8O

If the molar mass is 144.2

The molecular formula is C8H16O2

g calcium nitrate = 21.75 g, g ammonium fluoride = 22.66 g

g calcium fluoride = 10.34 g, g dinitrogen monoxide = 5.82 g, and

g water = 2.38 g.

The balanced chemical equation for the reaction is:

[tex]\rm 3 Ca(NO_3)2 + 6 NH_4F \rightarrow 6 NH_4NO_3 + CaF_2 + N_2O + 3 H_2O[/tex]

Given the molar masses:

[tex]\rm Ca(NO_3)2[/tex] = 164.09 g/mol

[tex]\rm NH_4F[/tex] = 37.04 g/mol

[tex]\rm CaF_2[/tex] = 78.08 g/mol

[tex]\rm N_2O[/tex] = 44.02 g/mol

[tex]\rm H_2O[/tex] = 18.02 g/mol

First, calculate the moles of each reactant:

Moles of [tex]\rm Ca(NO_3)2[/tex] = 21.75 g / 164.09 g/mol ≈ 0.1323 mol

Moles of [tex]\rm NH_4F[/tex] = 22.66 g / 37.04 g/mol ≈ 0.6113 mol

Based on the balanced equation, the limiting reactant is [tex]\rm Ca(NO_3)2[/tex], which reacts with 0.1323 moles.

Calculate the masses of the products:

g calcium fluoride = 0.1323 mol * 78.08 g/mol ≈ 10.34 g

g dinitrogen monoxide = 0.1323 mol * 44.02 g/mol ≈ 5.82 g

g water = 0.1323 mol * 18.02 g/mol ≈ 2.38 g

Since [tex]\rm NH_4F[/tex] is in excess, some of it remains unreacted:

Unreacted [tex]\rm NH_4F[/tex] = (0.6113 mol - 0.1323 mol) * 37.04 g/mol ≈ 21.90 g

In summary:

g calcium nitrate = 21.75 g

g ammonium fluoride = 22.66 g

g calcium fluoride = 10.34 g

g dinitrogen monoxide = 5.82 g

g water = 2.38 g

The reaction consumes calcium nitrate and ammonium fluoride to produce calcium fluoride, dinitrogen monoxide, and water vapour.

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Answer:

t = 4.62 sec        

Explanation:

For every first order reaction the rate constant K is given as

           [tex]k =(\frac{2.303}{t} )log\frac{[A_{o} ]}{[A]}[/tex]

        [tex][A_{o} ] = initial concentration = 0.52\frac{mol}{L}[/tex]

       [tex][A] =final concentration = 0.358 \frac{mol}{L}[/tex]

         [tex]K = 0.0808 sec^{-1}[/tex]

      [tex]t = (\frac{2.303}{K} ) log (\frac{[A_{o} ]}{[A]} )[/tex]

         [tex]= (\frac{2.303}{0.0808} )log (\frac{0.52}{0.358} )[/tex]

        t = 4.62 sec

Answer:0.0771mins

Explanation:The first order rate law eqn =Ca=Caoe^-kt

Ca=final mass remaining

Cap=initial mass

K=rate or decay constant

t=time

e=exponential

ca=0.358

cao=0.52

K=0.0808/sec

Substituting,we have

0.358=0.52e^-0.0808t

0.358/0.52=e^-0.0808t

0.688=e^-0.0808t

Taking naturaing logarithm of both sides(ln of both sides)

ln(0.688)=-0.0808t

-0.3739=-0.0808t

t=0.3739/0.0808

t=4.628secs

In mins,4.628/60=0.0771mins.

Answer:

Mole fraction H₂ = 0.29

Partial pressure of H₂ → 88.5 kPa

Explanation:

You need to know this relation to solve this:

Moles of a gas / Total moles = Partial pressure of the gas / Total pressure

Total moles = 3 mol + 7.3 mol → 10.3 moles

Mole fraction H₂ → 3 moles / 10.3 moles = 0.29

Mole fraction = Partial pressure of the gas / Total pressure

0.29 . 304 kPa = Partial pressure of H₂ → 88.5 kPa

Final answer:

The mole fraction of H₂ in the gas mixture is 0.291 and the partial pressure of H₂ is 88.28 kPa.

Explanation:

In order to find the mole fraction of H₂, we need to first calculate the total number of moles in the mixture. The total moles will be the sum of the moles of hydrogen and nitrogen: 3.0 mol + 7.3 mol = 10.3 mol.

The mole fraction of H₂ is calculated by dividing the moles of H₂ by the total moles in the mixture: 3.0 mol / 10.3 mol = 0.291. This means that the mole fraction of H₂ is 0.291.

The partial pressure of H₂ can be calculated using Dalton's Law of Partial Pressures. The total pressure of the mixture is given as 304 kPa, which is equal to the sum of the partial pressures of the gases: PH₂ + PN₂ = 304 kPa. We want to find the partial pressure of H₂, so we can rearrange the equation: PH₂ = 304 kPa - PN₂.

Now we need to find the partial pressure of N₂. The mole fraction of N₂ can be calculated as 1 - mole fraction of H₂: XN₂ = 1 - 0.291 = 0.709. We can then use this mole fraction and the total pressure to find the partial pressure of N₂: PN₂ = XN₂ * total pressure = 0.709 * 304 kPa = 215.72 kPa. With the partial pressure of N₂ known, we can plug it back into the rearranged equation for PH₂: PH₂ = 304 kPa - 215.72 kPa = 88.28 kPa.

Answer:

A. Methanol

B. 2-chloropropan-1-ol

C. 2,2-dichloroethanol

D. 2,2-difluoropropan-1-ol

Explanation:

Primary alcohols are stronger acids than secondary alcohols which are stronger than tertiary alcohols.

This trend is so because of the stability of the alkoxide ion formed(stabilising the base, increases the acidity). A more stabilised alkoxide ion is a weaker conjugate base (dissociation of an acid in water).

By electronic factors, When there are alkyl groups donating electrons, the density of electrons on th O- will increase a d thereby make it less stable.

By stearic factors, More alkyl group bonded to the -OH would mean the bulkier the alkoxide ion which would be harder to stabilise.

Down the group of the periodic table, basicity (metallic character) decreases as we go from F– to Cl– to Br– to I– because that negative charge is being spread out over a larger volume that is electronegativity decreases down the group.

Electronegative atoms give rise to inductive effect and a decrease in indutive effects leads to a decrease in acidity. Therefore an Increasing distance from the -OH group lsads to a decrease in acidity.

From above,

A. Methanol

B. 2-chloropropan-1-ol

C. 2,2-dichloroethanol

D. 2,2-difluoropropan-1-ol

Methanol is more acidic than tert-butyl alcohol because it forms a more stable alkoxide ion on deprotonation. Likewise, 2-chloropropan-1-ol is more acidic than propan-1-ol due to the stabilizing effect of its chlorine atom on the alkoxide ion.

The acidity of an alcohol is determined by the stability of the resulting alkoxide ion on deprotonation. In other words, an alcohol that can form a more stable alkoxide ion will be more acidic.

(a) Between methanol and tert-butyl alcohol, methanol would be more acidic. This is because the alkoxide ion formed when methanol loses a proton is more stable. In tert-butyl alcohol, the large, bulky tert-butyl group hinders the solvation of the alkoxide ion, making it less stable and the alcohol less acidic.

(b) 2-chloropropan-1-ol would be more acidic than propan-1-ol because the electron-withdrawing chlorine atom stabilizes the alkoxide ion, making the alcohol more acidic.

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The best explanation for this phenomenon is that the heat absorbed by the mixture is lost to the surroundings very quickly. Hence option B is correct.

Mixture is defined as a substance consisting of two or more unrelated chemicals that are not chemically linked. The two categories of mixtures are heterogeneous and homogeneous mixtures. While heterogeneous mixtures include distinct components, homogenous mixtures appear consistent throughout.

Although the mixture of solid and liquid is heated, the temperature remains constant as the solid melts. The best explanation for this phenomena is that the heat that the mixture absorbs is swiftly transferred to the environment and used to overcome the intermolecular forces.

Thus, the best explanation for this phenomenon is that the heat absorbed by the mixture is lost to the surroundings very quickly. Hence option B is correct.

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Final answer:

The heat absorbed by the solid-liquid mixture at its melting point is used in overcoming the intermolecular attractions in the solid, allowing it to transition into the liquid phase without an increase in temperature.

Explanation:

When a solid coexists with its liquid phase at its melting point, the heat absorbed by the mixture is used in overcoming the intermolecular attractions in the solid. As a solid is heated, the particles vibrate more rapidly and gain enough energy to overcome the attractive forces within the solid. This allows the solid to transition into the liquid phase without an increase in temperature. The heat added during melting is not lost to the surroundings quickly or used in breaking the bonds within the molecules of the solid, but rather in overcoming the intermolecular attractions in the solid.

Answer:

3.90 L

Explanation:

The reaction between NH₄I and Pb(NO₃)₂ is a double replacement reaction, so, the ions will dissociate and change in the two substances. The ions are NH₄⁺, I⁻, Pb⁺², and NO₃⁻, so the reaction is:

2NH₄I + Pb(NO₃)₂ → PbI₂ + 2NH₄NO₃

Thus, by the stoichiometry of the reactions, 2 moles of NH₄I are necessary to react with 1 mol of Pb(NO₃)₂. According to Proust's law, the proportion of the reaction must be kept so:

2 moles/1 mol = n NH₄I/n Pb(NO₃)₂

The number of moles of Pb(NO₃)₂ that will react is the concentration multiplied by the volume in L, so:

n Pb(NO₃)₂ = 0.280 * 0.905 = 0.2534 mol

2/1 = n NH₄I/0.2534

n NH₄I = 0.5068 mol

The volume of NH₄I is:

n NH₄I = 0.130 *V

0.5068 = 0.130V

V = 3.90 L

Answer:

D

Explanation:

When a dilution is made, a volume of the stock solution is collected and then is mixed to the solvent. The total amount of the solute (number of moles or mass), must be equal in the volume of the sample and at the final volume, because of the Lavoiser's law (the matter can't be created nor destructed).

The mass or the number of moles is the concentration (C) multiplied by the volume (V), so, if 1 is the sample of the stock solution, and 2 the diluted solution:

C1*V1 = C2*V2

The final volume of the solution is 10 mL. So, let's identify the volume needed for each stock solution.

Acetaminophen

C1 = 500 ug/L

C2 = 50 ug/L

500*V1 = 50*10

V1 = 1 mL

Dextromethorphan

C1 = 1 mM = 1000 uM

C2 = 20 uM

1000*V1 = 20*10

V1 = 0.2 mL

So, the volume of water needed is the total less the volume of the stocks solutions less the volume of the embryo water:

V = 10 - 1 - 1 - 0.2 = 7.8 mL

Thus, to to the solution, it's necessary to add at 1 mL of the embryo water 1 ml of 500 ug/I acetaminophen and 0.2 ml of 1 mM dextromethorphan, and 7.8 ml of water.

The complete question is

Comparison of a homeodomain protein from yeast and Drosophila shows that only 17 of its 60 amino acids are identical. How is it possible for a protein to change over 70% of its amino acids and still fold in the same way?

Answer:

Many different strings of the amino acid can give rise to the identical protein folds. The amino acid differences between the Drosophila and homeodomain proteins from yeast are the functional proteins which do not change their structure and function.

There is a several amount of folding required which can cause change in the protein structure.

There is a basic functional proteins common in both the organism which do not change its structure and function.

Answer: False

Explanation: Electrolyte are good conductors of electricity because they contain ions which are able to move about

Answer:

There is 0.25 grams of C11H12O3 produced

Explanation:

Step1: Data given

vanillin = C8H8O3

Mass of vanillin = 0.2 grams

Molar mass of vanillin = 152.15 g/mol

Acetone =  58.08 g/mol

Step 2: The balanced equation

C8H8O3 + C3H6O → H2O + C11H12O3

Step 3: Calculate moles of C8H8O3

Moles C8H8O3 = mass / molar mass

Moles C8H8O3 = 0.2 grams / 152.15 g/mol

Moles C8H8O3 = 0.0013 moles

Step 4: Calculate moles of C11H12O3

For 1 mol vanillin we need 1 mol acetone to produce 1 mol C11H12O3

Step 5: Calculate mass of C11H12O3

Mass C11H12O3 = moles * molar mass

Mass C11H12O3 = 0.0013 moles * 192.21 g/mol

Mass C11H12O3 = 0.25 grams

There is 0.25 grams of C11H12O3 produced