Given that a chlorine-oxygen bond in ClO2(g) has an enthalpy of 243 kJ/molkJ/mol , an oxygen-oxygen bond has an enthalpy of 498 kJ/molkJ/mol , and the standard enthalpy of formation of ClO2(g) is? ΔH∘f=102.5kJ/molΔHf∘=102.5kJ/mol , use Hess's law to calculate the value for the enthalpy of formation per mole of ClO(g).

Explanation:

1 Cal of energy is equivalent to 4.184 Joules.

A.

770Cal of energy = 3221.680 joules

Therefore, 780kCal of energy = 3221680 Joules.

B.

3200000 Joules (in 2 s.f).

C.

I000 joules = 1 kJoule

3221680 joules = 3221.689 kJoules.

D.

3200 kJ (in 2 s.f).

1) The energy in joules is equal to 3221680 Joules.

2) The energy in joules to two significant figures is 3200000.

3) The energy in kilojoules is 3221.689 KJ.

4) The energy in kilojoules to two significant figures is 3200 KJ.

Significant figures can be used to generate numbers that can be written in the form of digits. We can find the number of significant digits by counting the values beginning from the first non-zero digit placed on the left.

The significant figures of a given number can be described as those significant digits, which deliver the meaning with respect to its accuracy.

Given the energy used for the bike = 770 kcal.

We know that 1 cal = 4.184 J

The energy in Joules = 770 × 1000 × 4.184 = 3221680 J

The energy in joules to two significant figures is equal to 3200000.

The energy in kilojoules =  3221689/1000 = 3221.689 KJ

The energy in kilojoules to two significant figures is equal to 3200 KJ.

Learn more about significant figures, here:

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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:

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:

Experimental concentration is 0.04M

Number density is 2.4092×10^25 molecules/m^3

Explanation:

Experimental concentration = 0.001773g/cm^3 × 1mole/44g × 1000cm^3/1L = 0.04mole/L = 0.04M

Number density = 0.04mole/L × 6.023×10^23molecules/1mol × 1000L/1m^3 = 2.4092×10^25molecules/m^3

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:

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.

The magnitude of ΔHsolute is often larger than the magnitude of ΔHhydration in general.

The magnitudes of ΔHsolute and ΔHhydration can vary depending on the solute and the solvent being used. However, in general, the magnitude of ΔHsolute is often larger than the magnitude of ΔHhydration.

ΔHsolute represents the energy required to separate the solute particles, which typically involves breaking intermolecular forces. This process is usually endothermic and requires more energy compared to the process of dissolving the solute in water. On the other hand, ΔHhydration represents the energy released when solute particles are surrounded by water molecules during dissolution, which is often exothermic but smaller in magnitude compared to ΔHsolute.

For example, when table salt (NaCl) dissolves in water, the magnitude of ΔHsolute required to break the ionic bonds between Na+ and Cl- ions is significantly larger than the magnitude of ΔHhydration as water molecules surround the separated ions.

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Answer: False

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

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

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:

(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.

Explanation:

Answer:

Vapor pressure of solution = 111.98 Torr

Explanation:

Colligative property to apply: Lowering vapor pressure

P° - P' = P° . Xm . i

P°, vapor pressure of pure solvent

P', vapor pressure of solution

Xm, mole fraction of solute

i, Van't Hoff factor.

Let's determine the Xm.

Moles of solute = mass / molar mass → 34.2 g / 58.45 g/mol = 0.585 moles

Moles of solvent = 375 g / 18 g/mol = 20.83 moles

Mole fraction = 0.585 mol / 0.585 mol + 20.83 mol → 0.0273

Let's replace the data in the formula:

118.1 Torr - P' = 118.1 Torr . 0.0273 . 1.9

118.1  Torr - 6.12 Torr = P'

Vapor pressure of solution = 111.98 Torr

Answer : The number of mmol of bromobenzene is, 20.6 mmol

Explanation :

The balanced chemical reaction will be:

[tex]C_6H_5Br+HNO_3\rightarrow C_6H_4NO_2Br[/tex]

First we have to calculate the moles of p-bromonitrobenzene.

Moles of p-bromonitrobenzene =[tex]\frac{\text{ given mass of p-bromonitrobenzene}}{\text{ molar mass of p-bromonitrobenzene}}[/tex]

Molar mass of p-bromonitrobenzene = 202 g/mol

Moles of p-bromonitrobenzene = [tex]\frac{4.17g}{23g/mole}=0.0206moles[/tex]

Now we have to calculate the mmol of bromobenzene.

From the balanced chemical reaction we conclude that,

As, 1 mole of p-bromonitrobenzene produced from 1 mole of bromobenzene

So, 0.0206 mole of p-bromonitrobenzene produced from 0.0206 mole of bromobenzene

Moles of bromobenzene = 0.0206 moles = 20.6 mmol

Conversion used : 1 moles = 1000 mmol

Thus, the number of mmol of bromobenzene is, 20.6 mmol

To find out how many millimoles of bromobenzene were used to synthesize 4.17 grams of p-bromonitrobenzene, divide the mass of p-bromonitrobenzene by its molecular weight (202 g/mol) and convert the result to millimoles, yielding 20.6 mmoles.

If you synthesized 4.17 grams of p-bromonitrobenzene, the number of millimoles (mmoles) of bromobenzene you started with can be calculated using the molecular weight of p-bromonitrobenzene. The molecular weight of p-bromonitrobenzene (C6H4BrNO2) is approximately 202 g/mol. Assuming 100% conversion, the moles of p-bromonitrobenzene synthesized would be the mass of the final compound divided by its molecular weight.

Number of moles =  rac{4.17 g}{202 g/mol} = 0.0206 moles

Since 1 mole equals 1000 mmoles, the amount in mmoles would be:

Number of mmoles = 0.0206 moles  imes 1000 = 20.6 mmoles

Therefore, you would have started with 20.6 mmoles of bromobenzene to synthesize 4.17 grams of p-bromonitrobenzene assuming 100% conversion in the reaction.

Answer: The molarity of saline solution is 0.909 M

Explanation:

We are given:

5.21 w/w % NaCl

This means that 5.21 grams of NaCl is present in 100 grams of solution

To calculate mass of a substance, we use the equation:

[tex]\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}[/tex]

Density of solution = 1.02 g/mL

Mass of solution = 100 g

Putting values in above equation, we get:

[tex]1.02g/mL=\frac{100g}{\text{Volume of solution}}\\\\\text{Volume of solution}=\frac{100g}{1.02g/mL}=98.04mL[/tex]

To calculate the moalrity of solution, we use the equation:

[tex]\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}[/tex]

Given mass of NaCl = 5.21 g

Molar mass of NaCl = 58.44 g/mol

Volume of solution = 98.04 mL

Putting values in above equation, we get:

[tex]\text{Molarity of solution}=\frac{5.21\times 1000}{58.44g/mol\times 98.04}\\\\\text{Molarity of solution}=0.909M[/tex]

Hence, the molarity of saline solution is 0.909 M

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:

The density of solution is 1.283 g/mL.

Explanation:

Molarity of the KOH before dilution = [tex]M_1[/tex]

Volume of the solution before dilution = [tex]V_1=30.2 mL[/tex]

Molarity of the KOH after dilution = [tex]M_2=0.173 M[/tex]

Volume of the solution after dilution = [tex]V_2=1.20 L=1200 mL[/tex]

[tex]M_1V_1=M_2V_2[/tex]

[tex]M_1=\frac{M_2V_2}{V_1}=\frac{0.173 M\times 1200 mL}{30.2 mL}[/tex]

[tex]M_1=6.8742 M[/tex]

[tex]Molarity=\frac{moles}{Volume (L)}[/tex]

[tex]V_1=30.2 mL=0.0302 L [/tex] (1 mL = 0.001 L)

[tex]M_1=\frac{n}{V_1}[/tex]

[tex]n=M_1\times V_1=6.8742 M\times 0.0302 L=0.2076 mol[/tex]

Mass of 0.2076 moles of KOH:

0.2076 mol × 56 g/mol = 11.6256 g

Mass of KOH is solution = 11.6265 g

Mass of the solution = M

Mass percentage of solution = 30.0% of KOH

[tex]30.0\%=\frac{11.6265 g}{M}\times 100[/tex]

M = 38.755 g

Density of the solution , d= [tex]\frac{M}{V_1}[/tex]

[tex]d=\frac{38.755 g}{30.2 mL}=1.283 g/mL[/tex]

The density of solution is 1.283 g/mL.

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:

As explained

Explanation:

My reaction will be positive and I will feel elated that a new elements has been discovered as this will be a modification of the earlier claim by berzelius and the likes that there are over 106 known elements ; Metal, Non Metals and Metalloids.

However, fitting the new discovered element in between Tin and antimony may be an issue, as the physical properties of the new elements has to be known, its melting point, metals or non metals as this is paramount to be able to classify the elements. if the elements fits in between tin and antimony, it is an indication that the elements is mot likely a metal or non-mental and may belong to group IVB or VB. Another issue is, the discoverer will have to create his own periodic table to be able to stand in with the discovery of Dimitriy Mendelev.

Final answer:

Skepticism would be the initial reaction to a claim of a new element discovered between tin and antimony, as it contradicts the known ordered structure of the periodic table, where all elements are arranged based on their atomic numbers.

Explanation:

If a claim was made that a new element had been discovered that fits between tin (Sn) and antimony (Sb) in the periodic table, my reaction would be skeptical. Based on the well-established structure of the periodic table, it is unlikely that such a discovery would go unnoticed as the elements are arranged in sequential order of atomic number, and any gaps within this order have been filled by either known elements or placeholders for elements that have yet to be observed in nature or synthesized in a lab. Additionally, Mendeleev's predictions of eka-aluminum and eka-silicon were fulfilled by gallium and germanium, respectively, because their properties matched the predictions so well that they were accepted as the elements Mendeleev had predicted to exist.

Scientifically, it would be extraordinary to find an element that has been missed in such a well-studied part of the periodic table. Chemistry professionals would expect substantial evidence, including reproducible experimental results and peer-reviewed research, to support such an extraordinary claim. The element's discovery would also need to be in accordance with the properties of other elements around it, considering trends across periods and groups within the periodic table.

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:

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...

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:

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Explanation:

Final answer:

To calculate the mass of NaCl needed to lower the water's freezing point to -14.4°C, we use the freezing point depression formula with the known freezing point depression constant for water. After finding the molality, we can calculate the required mass of NaCl using its molar mass.

Explanation:

The student has asked about calculating the mass of salt (NaCl) needed to add to a specific volume of water to achieve a desired freezing point depression for making ice cream. This involves a concept called freezing point depression, which is a colligative property and is part of the solution chemistry topics in high school.

To calculate the mass of NaCl required to lower the freezing point of water to -14.4°C, we can use the freezing point depression equation ΔT = Kf × m × i, where ΔT is the freezing point depression, Kf is the freezing point depression constant of the solvent (water in this case, which is 1.86°C/m), m is the molality of the solution, and i is the van 't Hoff factor, which is 2 for NaCl due to its complete dissociation into Na+ and Cl- ions.

To find the molality (m), we rearrange the equation to m = ΔT / (Kf × i). With ΔT being 14.4°C (since water normally freezes at 0°C and we want -14.4°C), Kf is 1.86°C/m, and i is 2, we get m = 14.4°C / (1.86°C/m × 2) = 3.87 mol/kg. The mass of NaCl needed can then be calculated by converting molality to moles and then to grams using the molar mass of NaCl (58.44 g/mol).

Final answer:

Calculate the molality using the freezing point depression formula, convert to moles, and then to mass, resulting in 330 grams of NaCl needed to lower the freezing point of 1.46L of water to -14.4°C.

Explanation:

To find the mass of salt (NaCl) to be added to water in an ice-cream maker to achieve a freezing point depression to -14.4° C, we will use the freezing point depression formula ΔTf = i * Kf * m, where ΔTf is the freezing point depression, i is the van't Hoff factor (number of particles the solute breaks into), Kf is the cryoscopic constant of the solvent (water), and m is the molality of the solution.

For NaCl, which dissociates into Na+ and Cl-, the van't Hoff factor (i) is 2. Since 1L of water is approximately 1000g and the density of water is 1.00 g/mL, we have 1.46 kg of water. The freezing point depression (freezing point change) ΔTf is 14.4° C because pure water freezes at 0° C and we're freezing at -14.4° C. Kf for water is given as 1.86°C/m.

Solving the equation for m (molality), we have m = ΔTf / (i * Kf) = 14.4° C / (2 * 1.86°C/m) = 3.87 m (molality). To convert molality to moles of salt, we multiply by the mass of the solvent in kg: 3.87 m * 1.46 kg = 5.65 moles of NaCl. The molar mass of NaCl is approximately 58.44 g/mol, so the mass of NaCl needed is 5.65 moles * 58.44 g/mol = 330 grams.

Answer:

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Explanation:

Final answer:

Henry Moseley discovered that the periodic table should be arranged by atomic number, not atomic mass, by observing the frequencies of x-ray emissions. This corrected previous discrepancies and properly sequenced elements like argon and potassium.

Explanation:

Henry Moseley's study of atomic x-ray spectra led to a significant change in how elements are ordered in the periodic table. Previously, the elements were arranged by atomic mass, but Moseley discovered that the atomic number, which is the number of protons in an atom's nucleus, was a more accurate way to organize them. This realization occurred when he noticed that the frequencies of x-rays emitted by elements, when bombarded with electrons, formed a consistent pattern in relation to their atomic number instead of their mass.

An example sequence that was confirmed by Moseley's finding is the correct placement of argon before potassium in the periodic table. Despite argon having a higher atomic mass, its atomic number is 18, while potassium's is 19, following Moseley's rule of increasing atomic numbers. This corrected discrepancies in Mendeleev's periodic table, such as the sequence of cobalt (atomic number 27) being placed before nickel (atomic number 28), despite their atomic masses suggesting the opposite order.

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

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:

Specific gravity of the sample = 8.947

Explanation:

Specific gravity of a substance is defined as the density of that substance divided by the density of water.

Density of water = 1000g/l

Density of substance = mass/volume

= 85/9.5 x 10^-3

= 8947.37 g/l

SG = 8947.37/1000

= 8.947

Final answer:

To calculate the specific gravity of a sample, divide its density by the density of water. With a given mass of 85 grams and a volume of 9.5 mL, the sample's density is 8.9474 g/mL. Consequently, the specific gravity is 8.9474, as it's a dimensionless ratio.

Explanation:

The specific gravity of a sample is the ratio of the density of that sample to the density of a reference material, usually water. To determine the specific gravity of the sample given, you need to first calculate its density using the mass and volume. The mass of the sample is given as 85 grams and the volume is given as 9.5 mL.

First, calculate the density of the sample using the formula density = mass/volume. Density = 85 g / 9.5 mL = 8.9474 g/mL. Since the density of water at 4 degrees Celsius (which is typically used as the reference density) is 1 g/mL, the specific gravity of the sample is the ratio of the sample's density to that of water. Therefore, specific gravity = sample density / water density = 8.9474 g/mL / 1 g/mL = 8.9474. This means that the specific gravity of the sample is 8.9474, which is a dimensionless number.

Answer:

m = 341.22 g

Explanation:

∴ V = 242 mL

∴ m = (δ)×(V)

⇒ m = (1.41 g/mL)×(242 mL)

⇒ m = 341.22 g

The mass of 242 mL of sulfuric acid with a density of 1.41 g/mL is 341.22 g.

To calculate the mass of 242 mL of sulfuric acid, we can use the density of the acid, which is 1.41 g/mL. The formula for calculating mass is:

Mass = Volume x Density

Substituting the given values into the formula:

Mass = 242 mL x 1.41 g/mL

Calculating the multiplication:

Mass = 341.22 g

Therefore, the mass of 242 mL of sulfuric acid is 341.22 g.

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