KNO3(s) --->ž K+(aq) + NO3-(aq)This reaction was carried out in a Styrofoam insulated calorimeter and the following data were recorded:Mass of solid KNO3 dissolved10.1 gMass of aqueous solution (c = 4.18 J/gºC)100. gT initial30.0ºCT final21.6ºCMolar mass of KNO3101 g/molIf the mass of KNO3 solid dissolved were doubled while all other experimental conditions were kept the same, what change would occur in delta T, J per reaction, J/g of KNO3, and kJ/mol KNO3?Larger delta T, larger J/reaction, larger J/g, no change kJ/molLarger delta T, no change J/reaction, no change J/g, no change kJ/molLarger delta T, larger J/reaction, no change J/g, no change kJ/molLarger delta T, larger J/reaction, larger J/g, larger kJ/mol

Answer:

26%

Explanation:

Chlorination is a reaction that substitutes an atom of hydrogen for an atom of chlorine in a hydrocarbon. Pentane has 5 carbons and 12 hydrogens, and the chlorination can happen at the carbons 1,2 or 3. If it happens at carbon 4, the structure will be the same as carbon 2, and if it happens at carbon 5, will be the same as carbon 1.

Then, the percentage of 2-chloropentane and 3-chloropentane is 100 - 22 = 78%. As stated above, 4 hydrogens can be substituted to form 2-chloropentane (two at carbon 2, and two at carbon 4), and only two for 3-chloropentane. So, the percentage of secondary hydrogens to form the structure wanted is:

2/6 = 1/3

Thus, the percentage of it is:

(1/3)*78% = 26%

Answer:

B

Explanation:

If the sample is reweighed after it is removed from the furnace without being cooled to room temperature, its percentage water content will be too low as there would be almost no water if crystalization at such high temperature. However, when it is cooled to room temperature, the actual percentage of water contained in the sample can be accurately determined by weighing.

Final answer:

The procedural mistake that would result in determining a percentage of water that is too low is heating the sample in a closed container. Mistake I prevents the water from fully evaporating, which would cause an undercalculation of the water content.

Explanation:

When determining the percentage of water in a hydrated salt through heating, two procedural mistakes could lead to an inaccurate calculation of a lower percentage of water than the actual value. The correct answer to the given options is A. I only.

Heating the sample in a closed container (Mistake I) could prevent all the water from escaping, meaning that some water may remain inside, leading to an underestimation of the percentage of water in the salt.

In contrast, re-weighing the sample before it has cooled to room temperature (Mistake II) would not cause an underestimation but rather an overestimation, as the sample would weigh more due to the warmth. Therefore, Mistake II would not result in a lower percentage but rather potentially a higher percentage if the heat impacted the scale's reading.

Answer:

the gas it runs on it

Explanation:

Answer:

Using valence bond theory, HgCl2 is sp hybridized while PCl3 is sp3 hybridized.

Explanation:

sp hybridized molecules are largely covalent and cannot conduct electricity hence the nonconducting nature of HgCl2. PCl3 contains a lone pair and three unpaired electrons in sp3 hybridized orbitals.

Answer:

a. [tex]4 C_3H_5N_3O_9 (l)\rightarrow 6N_2 (g) + O_2 (g) + 10 H_2O (g) + 12 CO_2 (g)[/tex]

b. 146.0 g

Explanation:

Question 1 (a). Just as the problem states, liquid nitroglycerin decomposes into nitrogen gas [tex]N_2[/tex], oxygen gas [tex]O_2[/tex], water vapor [tex]H_2O[/tex] and carbon dioxide [tex]CO_2[/tex]. Let's write the decomposition of nitroglycerin into these 4 components:

[tex]C_3H_5N_3O_9 (l)\rightarrow N_2 (g) + O_2 (g) + H_2O (g) + CO_2 (g)[/tex]

Now we need to balance the equation. Firstly, notice we have 3 carbon atoms on the left and 1 on the right, so let's multiply carbon dioxide by 3:

[tex]C_3H_5N_3O_9 (l)\rightarrow N_2 (g) + O_2 (g) + H_2O (g) + 3 CO_2 (g)[/tex]

Now, we have 3 nitrogen atoms on the left and 2 on the right, so let's multiply nitrogen on the right by [tex]\frac{3}{2}[/tex]:

[tex]C_3H_5N_3O_9 (l)\rightarrow \frac{3}{2}N_2 (g) + O_2 (g) + H_2O (g) + 3 CO_2 (g)[/tex]

We have 5 hydrogen atoms on the left, 2 on the right, so let's multiply the right-hand side by [tex]\frac{5}{2}[/tex]:

[tex]C_3H_5N_3O_9 (l)\rightarrow \frac{3}{2}N_2 (g) + O_2 (g) + \frac{5}{2} H_2O (g) + 3 CO_2 (g)[/tex]

Finally, count the oxygen atoms. We have a total of 9 on the left. On the right we have (excluding oxygen molecule):

[tex]\frac{5}{2} + 6 = 8.5[/tex]

This leaves [tex]9 - 8.5 = 0.5 = \frac{1}{2}[/tex] of oxygen. Since oxygen is diatomic, we need to take one fourth of it to get one half in total:

[tex]C_3H_5N_3O_9 (l)\rightarrow \frac{3}{2}N_2 (g) + \frac{1}{4} O_2 (g) + \frac{5}{2} H_2O (g) + 3 CO_2 (g)[/tex]

To make it look neater without fractional coefficients, multiply both sides by 4:

[tex]4 C_3H_5N_3O_9 (l)\rightarrow 6N_2 (g) + O_2 (g) + 10 H_2O (g) + 12 CO_2 (g)[/tex]

Question 2 (b). Now we can make use of the balanced chemical equation and apply it for the context of this separate problem. We're given the following variables:

[tex]V_{CO_2} = 41.0 L[/tex]

[tex]T = -14.0^oC + 273.15 K = 259.15 K[/tex]

[tex]p = 1 atm[/tex]

Firstly, we may find moles of carbon dioxide produced using the ideal gas law [tex]pV = nRT[/tex].

Rearranging for moles, that is, dividing both sides by RT (here R is the ideal gas law constant):

[tex]n_{CO_2} = \frac{pV_{CO_2}}{RT} = \frac{1 atm\cdot 41.0 L}{0.08206 \frac{L atm}{mol K}\cdot 259.15 K} = 1.928 mol[/tex]

According to the stoichiometry of the balanced chemical equation:

[tex]4 C_3H_5N_3O_9 (l)\rightarrow 6N_2 (g) + O_2 (g) + 10 H_2O (g) + 12 CO_2 (g)[/tex]

4 moles of nitroglycerin (ng) produce 12 moles of carbon dioxide. From here we can find moles o nitroglycerin knowing that:

[tex]\frac{n_{ng}}{4} = \frac{n_{CO_2}}{12} \therefore n_{ng} = \frac{4}{12}n_{CO_2} = \frac{1}{3}\cdot 1.928 mol = 0.6427 mol[/tex]

Multiplying the number of moles of nitroglycerin by its molar mass will yield the mass of nitroglycerin decomposed:

[tex]m_{ng} = n_{ng}\cdot M_{ng} = 0.6427 mol\cdot 227.09 g/mol = 146.0 g[/tex]

Answer:

C₂H₃Cl

Explanation:

We can calculate the compound's molar mass using the data given by the problem and Graham's law:

Rate₁/Rate₂ = [tex]\sqrt{\frac{M_{2}}{M_{1}} }[/tex]

In this case the subscript 1 refers to the compound and 2 refers to Ar.

Keeping in mind that Rate = volume/time, and that the volume is the same for both compounds, we can rewrite the equation as:

Time₂/Time₁ = [tex]\sqrt{\frac{M_{2}}{M_{1}} }[/tex]

6.18/7.73 =  [tex]\sqrt{\frac{39.95}{M_{1}} }[/tex]

M₁ = 62.5 g/mol

Now we determine the molecular formula by using the elemental % analysis:

Assuming we have 1 mol of the compound:

C ⇒ 62.5 g * 38.4/100 = 24 g C = 2 mol C

H ⇒ 62.5 g * 4.8/100 = 3 g H = 3 mol H

C ⇒ 62.5 g * 56.8/100 = 35.5 g Cl = 1 mol Cl

Thus the molecular formula is C₂H₃Cl

Answer:

2.1 × 10⁻¹⁹ M/s

Explanation:

Let's consider the degradation of CF₃CH₂F by OH radicals.

CF₃CH₂F + OH → CF₃CHF + H₂O

Considering the order of reaction for each reactant is 1 and the rate constant is 1.6 × 10⁸ M⁻¹s⁻¹, the rate law is:

r = 1.6 × 10⁸ M⁻¹s⁻¹.[CF₃CH₂F].[OH]

where,

r is the rate of the reaction

If the tropospheric concentrations of OH and CF₃CH₂F are 8.1 × 10⁵ and 6.3 × 10⁸ molecules/cm³, respectively, what is the rate of reaction at this temperature in M/s?

The Avogadro's number is 6.02 × 10²³ molecules/mole.

The molar concentration of OH is:

[tex]\frac{8.1 \times 10^{5}molecules}{cm^{3}}.\frac{1mol}{6.02 \times 10^{23}molecules  }.\frac{1000cm^{3} }{1L}  =1.3 \times 10^{-15} M[/tex]

The molar concentration of CF₃CH₂F is:

[tex]\frac{6.3 \times 10^{8}molecules}{cm^{3}}.\frac{1mol}{6.02 \times 10^{23}molecules  }.\frac{1000cm^{3} }{1L}  =1.0 \times 10^{-12} M[/tex]

r = 1.6 × 10⁸ M⁻¹s⁻¹ × 1.0 × 10⁻¹² M × 1.3 × 10⁻¹⁵ M = 2.1 × 10⁻¹⁹ M/s

Answer:

The correct option are:

1.  (b) current

2. (c) potential

3.  (b) coulombs (C)

4.  (a) amperes (A)

5.  (d) volts (V)

Explanation:

Electric charge is a property of a particle, like electron and proton. The subatomic particles, electrons and protons are negatively and positively charged particles, respectively.

The electric charge of a particle can be measured in coulomb, denoted by C.

Current or electric current is defined as the net flow of the electric charges through a given region, per unit time. Electric current can be measured in ampere, denoted by A.

Potential or Electric potential is described as the work done to displace the electric charges from one point to another. Potential is the driving force for the movement of charges and can be calculated in volt, denoted by V.

The flow of charge per unit time is current, measured in amperes (A), while the driving force for electron flow is electric potential, measured in volts (V), and charge itself is quantified in coulombs (C).

Understanding Electrical Quantities

The amount of charge that passes per unit time is called current, and is measured in amperes (A). The driving force that causes electrons to flow, known as the electric potential, is measured in volts (V). Charge itself is measured in coulombs (C).

Answering the questions provided:

The amount of charge that passes per unit time is called current.

The driving force for the electrons is measured by potential.

Charge is measured in coulombs (C).

Current is measured in amperes (A).

Potential is measured in volts (V).

Explanation:

Octahedral complexes will be favoured over tetrahedral ones because:

It is more favourable to form six bonds rather than four

The crystal field stabilisation energy is usually greater for octahedral than tetrahedral complexes.

The transition metals Co, Rh, Lr are in group 9 of d block and they have 3d, 4d, and 5d orbitals respectively

Metals like Co, Rh, and Lr form octahedral complexes due to their electron configurations and the stability achieved with six ligands. This coordination results in a geometry that minimizes electron repulsions and maximizes stability.

Transition metals like Co (Cobalt), Rh (Rhodium), and Lr (Lawrencium) typically form octahedral complexes due to their electron configurations and positions in the periodic table. These metals have available d-orbitals that can accommodate six ligands, resulting in a coordination number of six which is most stable in an octahedral geometry.

For instance, Co³⁺ has an electron configuration that favors the formation of octahedral complexes due to the stabilization of its d-electrons in a ligand field that splits the d-orbitals into two energy levels, with four orbitals at lower energy and two at higher energy. Similarly, Rh³⁺ and Lr³+ configurations also favor six-coordinate octahedral structures, providing maximum separation of electron pairs and minimizing electron-electron repulsions.

Examples:

[Co(NH₃)₆]³⁺: An example of an octahedral complex where the cobalt ion is bonded to six ammonia ligands.

[RhCl₆]³⁺: An octahedral complex where the rhodium ion is coordinated by six chloride ligands.

Answer: The cell potential of the cell is +0.118 V

Explanation:

The half reactions for the cell is:

Oxidation half reaction (anode):  [tex]Ag(s)+Cl^-(aq.)\rightarrow AgCl(s)+e^-[/tex]

Reduction half reaction (cathode):  [tex]AgCl(s)+e^-\rightarrow Ag(s)+Cl^-(aq.)[/tex]

In this case, the cathode and anode both are same. So, [tex]E^o_{cell}[/tex] will be equal to zero.

To calculate cell potential of the cell, we use the equation given by Nernst, which is:

[tex]E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Cl^{-}]_{diluted}}{[Cl^{-}]_{concentrated}}[/tex]

where,

n = number of electrons in oxidation-reduction reaction = 1

[tex]E_{cell}[/tex] = ?

[tex][Cl^{-}]_{diluted}[/tex] = 0.0222 M

[tex][Cl^{-}]_{concentrated}[/tex] = 2.22 M

Putting values in above equation, we get:

[tex]E_{cell}=0-\frac{0.0592}{1}\log \frac{0.0222M}{2.22M}[/tex]

[tex]E_{cell}=0.118V[/tex]

Hence, the cell potential of the cell is +0.118 V

The concentration cell involving Al3+ has a standard cell potential (E°) of -1.66 V, and the cell potential (E) is approximately -1.54 V. The correct option is D: E° = -1.66 V and E = -1.54 V.

To determine the standard cell potential (E°) and the cell potential (E) for the given concentration cell, we can use the Nernst equation and the standard reduction potential (E°red) for the half-reaction involved.

The shorthand notation for the concentration cell is given as:

[tex]\[ \text{Al(s) | Al}^{3+} (1.0 \times 10^{-5} \, \text{M}) \, || \, \text{Al}^{3+} (0.100 \, \text{M}) \, | \, \text{Al(s)} \][/tex]

Firstly, let's identify the two half-reactions occurring in the cell. The half-reaction for the left side (anode) is [tex]\(\text{Al} \rightarrow \text{Al}^{3+} + 3e^-\)[/tex], and for the right side (cathode), it is [tex]\(\text{Al}^{3+} + 3e^- \rightarrow \text{Al}\)[/tex].

The standard cell potential (E°) can be calculated using the standard reduction potentials [tex](\(E°_{\text{red}}\))[/tex] of the half-reactions:

[tex]\[ E° = E°_{\text{cathode}} - E°_{\text{anode}} \][/tex]

Given[tex]\(E°_{\text{red}} = -1.66 \, \text{V}\) for \(\text{Al}^{3+}/\text{Al}\)[/tex], we can substitute this into the equation to find E°. However, we need to consider the fact that the half-reaction for the anode is reversed in the standard reduction potential table. Therefore, \(E°_{\text{anode}} = -(-1.66 \, \text{V}) = 1.66 \, \text{V}\).

[tex]\[ E° = 0 - 1.66 \, \text{V} = -1.66 \, \text{V} \][/tex]

Now, to find the cell potential (E), we use the Nernst equation:

[tex]\[ E = E° - \frac{0.0592}{n} \log\left(\frac{[\text{Al}^{3+}]_{\text{cathode}}}{[\text{Al}^{3+}]_{\text{anode}}}\right) \][/tex]

Here, n is the number of electrons transferred (which is 3 for both half-reactions), and[tex]\([X]_{\text{cathode}}\[/tex] ) and [tex]\([X]_{\text{anode}}\)[/tex]are the concentrations of X at the cathode and anode, respectively.

For the given cell, plugging in the values:

[tex]\[ E = -1.66 - \frac{0.0592}{3} \log\left(\frac{0.100}{1.0 \times 10^{-5}}\right) \][/tex]

Calculating this gives approximately -1.54 V.

Therefore, the correct answer is option D: [tex]\(E° = -1.66 \, \text{V}\) and \(E = -1.54 \, \text{V}\)[/tex].

Complete question :- Given that E°red = -1.66 V for Al3+/ Al at 25°C, find E° and E for the concentration cell expressed using shorthand notation below.

Al(s) | Al3+(1.0 × 10-5 M) || Al3+(0.100 M) | Al(s)

A) E° = 0.00 V and E = +0.24 V

B) E° = 0.00 V and E = +0.12 V

C) E° = -1.66 V and E = -1.42 V

D) E° = -1.66 V and E = -1.54 V

the energy difference is negative, it means that the band at 263.5 nm has a lower energy than the absorption maximum at 559.1 nm.

To calculate the energy difference between two absorption bands, we can use the equation:

[tex]\[ \Delta E = \frac{hc}{\lambda} \][/tex]

Where:

- [tex]\( \Delta E \)[/tex] is the energy difference in joules (J)

- [tex]\( h \)[/tex] is Planck's constant [tex](\(6.62607015 \times 10^{-34} \, \text{J} \cdot \text{s}\))[/tex]

- c is the speed of light in a vacuum [tex](\(3.00 \times 10^8 \, \text{m/s}\))[/tex]

- [tex]\( \lambda \)[/tex] is the wavelength in meters (m)

First, we need to convert the wavelengths from nanometers (nm) to meters (m):

[tex]\[ \lambda_1 = 559.1 \, \text{nm} \times \frac{1 \, \text{m}}{10^9 \, \text{nm}} = 5.591 \times 10^{-7} \, \text{m} \][/tex]

[tex]\[ \lambda_2 = 263.5 \, \text{nm} \times \frac{1 \, \text{m}}{10^9 \, \text{nm}} = 2.635 \times 10^{-7} \, \text{m} \][/tex]

Now, we can calculate the energy difference (\( \Delta E \)):

[tex]\[ \Delta E = \frac{(6.62607015 \times 10^{-34} \, \text{J} \cdot \text{s})(3.00 \times 10^8 \, \text{m/s})}{5.591 \times 10^{-7} \, \text{m}} - \frac{(6.62607015 \times 10^{-34} \, \text{J} \cdot \text{s})(3.00 \times 10^8 \, \text{m/s})}{2.635 \times 10^{-7} \, \text{m}} \][/tex]

[tex]\[ \Delta E = (3.00 \times 10^8 \, \text{m/s})\left(\frac{6.62607015 \times 10^{-34} \, \text{J} \cdot \text{s}}{5.591 \times 10^{-7} \, \text{m}} - \frac{6.62607015 \times 10^{-34} \, \text{J} \cdot \text{s}}{2.635 \times 10^{-7} \, \text{m}}\right) \][/tex]

[tex]\[ \Delta E = (3.00 \times 10^8 \, \text{m/s})(2.3723 \times 10^{-19} \, \text{J} - 2.5161 \times 10^{-19} \, \text{J}) \][/tex]

[tex]\[ \Delta E \approx (3.00 \times 10^8 \, \text{m/s})(-0.1438 \times 10^{-19} \, \text{J}) \][/tex]

[tex]\[ \Delta E \approx -4.314 \times 10^{-12} \, \text{J} \][/tex]

[tex]\[ \Delta E \approx -4314 \, \text{pJ} \][/tex]

Since the energy difference is negative, it means that the band at 263.5 nm has a lower energy than the absorption maximum at 559.1 nm.

The energy difference between the absorption maximum at 559.1 nm and the band at 263.5 nm is 240 kJ/mol.

To calculate the energy difference between the absorption maximum at 559.1 nm and a band at 263.5 nm in kilojoules per mole, follow these steps:

Step 1. Convert Wavelengths to Energy:

Use the equation [tex]\( E = \frac{hc}{\lambda} \)[/tex], where:

(E) is the photon's energy.

Planck's constant [tex](\(6.626 \times 10^{-34}\) J.s)[/tex] is represented as (h).

The speed of light is represented as (c) = [tex](3.00 \times 10^8\)[/tex] m/s.

The wavelength in meters is [tex](\lambda \))[/tex].

Step 2. Calculate Energy for 559.1 nm:

[tex]\[ \lambda_1 = 559.1 \text{ nm} = 559.1 \times 10^{-9} \text{ m} \][/tex]

[tex]\[ E_1 = \frac{6.626 \times 10^{-34} \times 3.00 \times 10^8}{559.1 \times 10^{-9}} = 3.556 \times 10^{-19} \text{ J} \][/tex]

Step 3. Calculate Energy for 263.5 nm:

[tex]\[ \lambda_2 = 263.5 \text{ nm} = 263.5 \times 10^{-9} \text{ m} \][/tex]

[tex]\[ E_2 = \frac{6.626 \times 10^{-34} \times 3.00 \times 10^8}{263.5 \times 10^{-9}} = 7.552 \times 10^{-19} \text{ J} \][/tex]

Step 4. Find the Energy Difference:

[tex]\[ \Delta E = E_2 - E_1 = 7.552 \times 10^{-19} \text{ J} - 3.556 \times 10^{-19} \text{ J} = 3.996 \times 10^{-19} \text{ J} \][/tex]

Step 5. Convert to Kilojoules per Mole:

[tex]\[ 1 \text{ photon} = 3.996 \times 10^{-19} \text{ J} \][/tex]

[tex]\[ 1 \text{ mole of photons} = 3.996 \times 10^{-19} \text{ J} \times 6.022 \times 10^{23} \text{ photons/mol} \][/tex]

[tex]\[ \Delta E_{\text{mol}} = 2.406 \times 10^{5} \text{ J/mol} = 240.6 \text{ kJ/mol} \][/tex]

So, the energy difference between the absorption maximum at 559.1 nm and the band at 263.5 nm is 240.6 kJ/mol.

Answer:

0

Explanation:

The relation between Kp and Kc is given below:

[tex]K_p= K_c\times (RT)^{\Delta n}[/tex]

Where,  

Kp is the pressure equilibrium constant

Kc is the molar equilibrium constant

R is gas constant

T is the temperature in Kelvins

Δn = (No. of moles of gaseous products)-(No. of moles of gaseous reactants)

For the first equilibrium reaction:

[tex]SO_3_{(g)}+NO_{(g)}\rightleftharpoons SO_2_{(g)}+NO_2_{(g)}[/tex]

Δn = (2)-(2) = 0

Thus, Kp is:

[tex]K_p= Kc\times \times (RT)^{0}[/tex]

[tex]K_p= Kc[/tex]

Final answer:

The Δn represents the difference in moles of gaseous products and reactants for a reaction when relating Kc to Kp. For the reaction SO3(g) + NO(g) ↔ SO2(g) + NO2(g), Δn would be 0. Kp is related to Kc by the equation Kp = Kc(RT)Δn.

Explanation:

The question relates to the concept of the reaction quotient (Δn) when relating the equilibrium constants Kc (equilibrium constant in terms of concentration) to Kp (equilibrium constant in terms of partial pressure) for a given chemical reaction involving gases. The value of Δn is the difference in the sum of the moles of gaseous products and the sum of the moles of gaseous reactants in a balanced chemical equation. In the example given:

[tex]SO_{3}[/tex] (g) + NO(g) ↔ [tex]SO_{2}[/tex] (g) + [tex]NO_{2}[/tex] (g), Δn would be (1 + 1) - (1 + 1) = 0.

1[tex]N_{2}[/tex](g) + [tex]2H_{2} O[/tex](g) ↔ 2NO(g) + [tex]2H_{2}[/tex] (g), Δn would be (2 + 2) - (1 + 2) = 1.

To relate Kc and Kp, we use the equation Kp = Kc(RT)Δn, where R is the gas constant and T is the temperature in Kelvins. In cases where Δn is zero, as in the first equation, Kp will be equal to Kc because (RT)0 equals 1.

The correct steps in the activation of PKA after the adenylyl cyclase reaction are: cAMP binding to and activating the regulatory subunits of PKA, the catalytic subunits of PKA phosphorylating target proteins, and PKA being inactivated when cAMP is hydrolyzed by phosphodiesterase.

The correct steps in the activation of PKA, after the adenylyl cyclase reaction, are:

Overall, the activation of PKA by cAMP allows for the regulation of various cellular processes, including metabolism, gene expression, and cell signaling.

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The correct steps in the activation of PKA after the adenylyl cyclase reaction are: adenylyl cyclase converts ATP to cAMP, cAMP binds to and activates Protein Kinase A (PKA), and activated PKA phosphorylates serine and threonine residues of target proteins, activating them.

The correct steps in the activation of PKA after the adenylyl cyclase reaction are:

These steps occur in the cAMP second messenger system and are responsible for the effects of cAMP within the cell.

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Answer:

t = 55 min ⇒ m = 1497.692 g

Explanation:

polyaramide fiber:

∴ L = 1 m

∴ d = 710 μm

∴ m = 0.059 g

∴ t = 0.13 s

mass flow:

∴ mf = 0.059 g / 0.13 s = (0.454 g/s)×(60s/min) = 27.23 g/min

If t = 55 min:

⇒ m = mf×t = (27.23 g/min)×(55 min) = 1497.692 g

Answer:

[tex]m_{fiber}=1497.7g[/tex]

Explanation:

Hello,

In this case, as both the diameter and the length of the fiber is the same for the second case, one applies a simple rule of three to compute the mass of fiber that can be spun in 55 min as shown below:

[tex]m_{fiber}=\frac{55min*0.059g}{0.13s*\frac{1min}{60s} }\\m_{fiber}=1497.7g[/tex]

Best regards.

Answer:

6,04x10⁻³M

Explanation:

For the reaction:

Fe(s) + CuSO₄(aq) → Cu(s) + FeSO₄(aq)

The precipitate of Cu(s) weights 96,0 mg. In moles:

Moles of Cu(s):

0,096g×(1mol/63,546g) = 1,51x10⁻³ moles of Cu(s). If you see the balanced equation 1 mole of CuSO₄ produce 1 mole of Cu(s). That means moles of CuSO₄ are the same of Cu(s), 1,51x10⁻³ moles of CuSO₄

As volume of the solution is 250 mL, 0,250L, the molar concentration of the original solution is:

1,51x10⁻³ moles of CuSO₄ / 0,250L = 6,04x10⁻³M

I hope it helps!

Answer:

1. 1M for zinc and copper solutions. No concentration was needed for the salt bridge.

2. See explanation.

Explanation:

Hello,

1. In this case, and considering the experiment you performed, the concentration of both copper solution and zinc solution were measured as 1 M, while the concentration of salt bridge was not taken into account.

2. However, as the reaction proceeded, the transfer of anions allowed the total charges in solution to remain neutral until the end, thus, the net movement of anions produced a concentration gradient between the solutions. It means that, after a while the net concentration of both anions and cations in the zinc oxidation section became greater than in the copper reduction section, making the concentration gradient opposite to the movement of anions. In such a way, as anions moved, cations moved to the right as well.

Best regards.

The concentrations of the zinc, copper, and salt bridge solutions are dependent on the amount of solute dissolved in the solvent. Standard state conditions, used as a reference point, include specific constant values for pressure, temperature, and concentration. Factors such as temperature and ion product of water can significantly affect a solution's properties and how they align to these standard conditions.

The question asked about the concentrations of different solutions in standard state conditions. Based on the information provided, it's difficult to conclusively determine the concentration of each solution as the specific concentrations are not detailed. However, you can generally understand that a solution's concentration describes the amount of solute dissolved in a solvent. For example, for Zn(OH)₂(s) in a solution buffered at a pH of 11.45, this suggests a certain quantity of Zn (zinc) ions in the solution.

The standard state condition refers to a specific set of conditions including pressure (1 bar), temperature (298.15 K), and concentration (1M). For example, Zn(s) + 2HCl(aq) ZnCl₂ (aq) + H₂(g) details a reaction involving zinc and hydrochloric acid producing zinc chloride and hydrogen gas. This standardized condition forms the basis for the determination of properties such as enthalpy, entropy, and Gibbs free energy. Therefore, whether the solution is consistent with the standard state condition would be based on if the pressure, temperature, and concentration are adhering to these standard parameters.

Many factors such as temperature, ion product of water, solutes in the solutions can affect a solution's properties and its alignment to standard state conditions. Thus, it's important that these factors are closely monitored and regulated in a reaction setup.

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Answer:

The molar concentration of hydrogen peroxide is 0.01164 M

Explanation:

Step 1: Data given

Volume of gas yielded = 75.3 mL = 0.0753 L

Temperature = 25.0 °C

Atmosphere = 0.976 atm

(Water = 0.032 atm at 25°C)

The original volume of H2O2 is 500 mL

Molar mass of O2 =32 g/mol

Step 2: The balanced equation

2H2O2(aq) → 2 H2O(l) + O2(g)

Step 3: Calculate pressure of O2

P = P(02) + P(water)

P(O2) = P - P(water)

P(O2) = 0.976 atm - 0.032 atm

P(O2) = 0.944 atm

Step 4: Calculate moles O2

p*V=n*R*T

⇒ with p = the pressure of O2 gas = 0.944 atm

⇒ with V= the volume of O2 = 0.0753 L

⇒ with n = the number of moles of O2

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

⇒ with T = the temperature = 25°C = 298 Kelvin

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

n = (0.944 * 0.0753)/(0.08206*298K)

n = 0.00291 moles O2

Step 5: Calculate moles of H2O2

For 1 mole of O2 produced, we need 2 moles of H2O2

For 0.00291 moles O2 we need 2*0.00291 = 0.00582 moles H2O2

Step 6: Calculate molar concentration of H2O2

Molar concentration = moles / volume

Molar concentration = 0.00582 moles / 0.500L

Molar concentration = 0.01164 M

The molar concentration of hydrogen peroxide is 0.01164 M

Answer:

Answer is in the explanation.

Explanation:

Thin layer chromatography is a chromatographic technique used to separate the components of a mixture using a thin stationary phase supported by an inert backing and a mobile phase. The separation principle is in the different affinities between the components of the mixture and the stationary or mobile phase.

The affinity in mobile phase could be improved changing the polarity of this phase. In this case, you could change proportion of hexane/ethyl acetate to change polarity of mobile phase and the affinity of the different compounds to mobile or stationary phase.

I hope it helps!

Thin-layer chromatography with its adjustable hexane/ethyl acetate ratio, serves as a precise and versatile tool for optimizing separation conditions and exploring diverse interactions in chromatography.

Thin-layer chromatography (TLC) is a chromatographic method that segregates mixture components via a slender stationary phase on an inert support and a mobile phase. The separation hin-ges on distinct affinities between mixture constituents and the stationary or mobile phase. Modifying the polarity of the mobile phase can enhance its affinity.

By adjusting the hexane/ethyl acetate ratio, the mobile phase's polarity transforms, influencing compounds' interactions with the stationary and mobile phases. This dynamic shift in affinity leads to differential migration rates, facilitating component separation.

TLC stands as a versatile tool in analytical chemistry, where subtle adjustments in solvent composition yield nuanced variations in separation patterns.

Fine-tuning the hexane/ethyl acetate proportions allows for targeted optimization of separation conditions, offering a precise means to explore and exploit the diverse interactions governing the chromatographic process.

To learn more about Thin-layer chromatography

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Answer:

See below

Explanation:

Lets choose a baseball ball which has to have a minimum of  diameter of 1.43 inches ( 1.43 in x 2.54 cm/in = 3.63 cm )

ratio electron/nucleus =  10−10 m / 10−15 m = 100000

distance "e" = 3.63 cm x 100000 = 363000 cm ( 3630 m or 3.630 km )

Imagine to bat a baseball 3.6 km away !

Answer:

D

Explanation:

There are principally three states of matter. These are the solid, liquid and gaseous states. The gaseous state has the highest degree of disorderliness as gas particles can move randomly while the solid state has the highest level of compactness.

Hence, we need to be adequately fed with information as regards the phase change to know if entropy has decreased or increased.

A. is wrong

Evaporation is a change of state to the gaseous state meaning there is an increased entropy.

B. is wrong

Sublimation is a change of state which means a solid substance like iodine or naphthalene changes its state directly to the gaseous state. There is an increased entropy here too.

C is wrong

Melting of ice means going from ice block to liquid water. This is synonymous to going from the solid state to the liquid state which is an increased entropy

D is correct

Condensation involves going from the gaseous state to the liquid state. This means going from a less ordered state to a more ordered state. This is accompanied by an entropy decrease.

E is wrong

While there are some processes that increase entropy, we also have some process that decrease entropy.

Final answer:

The process that represents a decrease in entropy is the condensation of steam on a kitchen window. This is because it involves the transition from a disordered gas state to a more ordered liquid state.

Explanation:

The question asks which of the given scenarios represents a decrease in entropy. Entropy, in simple terms, is a measure of the disorder or randomness of a system. A decrease in entropy means the system is becoming more ordered.

The evaporation of perfume - This process increases entropy as the perfume molecules spread out.The sublimation of carbon dioxide - Sublimation, the process of transitioning directly from a solid to a gas phase, results in an increase in entropy due to the increased dispersion of molecules.The melting of ice - This process increases entropy because the water molecules move from a highly ordered solid state to a less ordered liquid state.The condensation of steam on a kitchen window - This process decreases entropy. When steam (gas) condenses into liquid water on a window, it transitions from a highly disordered state to a more ordered state. Therefore, this is the correct answer.It is not possible to have a decrease in entropy - This statement is not accurate as there are processes, such as condensation, that can result in a decrease in entropy.

Thus, the correct answer is d) The condensation of steam on a kitchen window.

Answer:

potassium, nitrogen and phosphorous cycle

Explanation:

A fertilizer is a substance which is applied on the plants by farmers to increase the supply of nutrients for the plants. Fertilizers have known to be toxic in many ways such as they alter the potassium, nitrogen and phosphorus cycles. Nitrogen, potassium and phosphorus are present in abundant amounts in the fertilizers. Draining of these fertilizers into rivers and ponds is toxic for the aquatic life. Hence, the use of fertilizers disrupts the natural cycles and is toxic for many aquatic plants and animals.

Answer: moles of Ne = 0.149 moles

moles of F₂ = 0.221 moles

Explanation:

The process occurs at costant volume.

Neon is a monoatomic gas, Cv =3/2R, Flourine is a diatomic gas, Cv = 5/2R

n = PV/RT ; where n is total = number of moles, P is total pressure = 3.32atm, V is volume = 2.5L, R is molar gas constant = 0.08206 L-atm/mol/K = 8.314 J/mol/K, T is temperature = 0.0°C = 273.15K

n(total) = (3.32 atm)(2.5 L)/(0.08206 L-atm/mol/K)(273.15) = 0.3703 mol

For one mole heated at constant volume,

Change in entropy, ∆S = ∫dq/T = ∫(Cv/T)dT

From T1 to T2, Cvln(T2/T2) = Cvln(288.15/273.15) = 0.05346•Cv

So, for 0.3703 moles,

∆S = (0.3703 mol)(0.05346)Cv = 0.345 J/K

⇒ Cv = 17.43 J/mol/K for the Ne/F₂ mixture.

For pure Ne, Cv = (3/2)R = 1.5 • 8.314 J/mol/K = 12.471 J/mol/K

For pure F₂, Cv = (5/2)R = 2.5 • 8.314 J/mol/K = 20.785 J/mol/K

If Y is the mole fraction of Ne, Y can be determined by setting the observed entropy change (∆S) to the weighted average of the entropy changes expected for the two different gases in the mixture:

17.43 J/mol/K = Y * 12.471 J/mol/K + (1 – Y) * 20.785 J/mol/K

⇒ 20.785 J/mol/K – 8.314 J/mol/K * Y = 17.43J/mol/K

Y = 0.403 ; 1 – Y = 0.597

moles of Ne = (0.403)(0.3703 mol) = 0.149 moles

moles of F₂ = (0.597)(0.3703 mol) = 0.221 moles

Answer:

A and B only

Explanation:

Both Scintillation counter and Geiger counter are used  for instantaneous measures of radioactivity.

Both are used to detect and quantify the amount of radiation. GM counter can detect all kind of radiation that alpha, beta and gamma radiation. Whereas  Scintillation counter can detect only ionization radiation.

Answer:

Non of these are true

Explanation:

To form a solution, solute-solvent interaction must exceed solute-solute and solvent-solvent interaction. Hence a new attraction leading to solvation of the solid in the solvent is set up.

Answer:

D) 5.36 hours

Explanation:

According to mole concept:

1 mole of an atom contains [tex]6.022\times 10^{23}[/tex] number of particles.

We know that:

Charge on 1 electron = [tex]1.6\times 10^{-19}C[/tex]

Also, copper will produce 2 electrons. So, out of 5 moles of copper, 10 moles of electrons will be produced.

So,

Charge on 10 mole of electrons = [tex]10\times 1.6\times 10^{-19}\times 6.022\times 10^{23}=9.6352\times 10^5C[/tex]

To calculate the time required, we use the equation:

[tex]I=\frac{q}{t}[/tex]

where,

I = current passed = 50.0 A

q = total charge = [tex]9.6352\times 10^5C[/tex]

t = time required = ?

Putting values in above equation, we get:

[tex]50.0A=\frac{9.6352\times 10^5C}{t}\\\\t=\frac{9.6352\times 10^5C}{50.0A}=19270.4s[/tex]

Converting this into hours, we use the conversion factor:

1 hr = 3600 seconds

So, [tex]19270.4s\times \frac{1hr}{3600s}=5.36hr[/tex]

Hence, the amount of time needed is 5.36 hrs.

Answer:

b

Explanation:

bc

Answer:

Hi, no structure in the question was given to the answer the question.

Explanation:

Certain molecules or groups, when attached to an organic compound can make it become more acidic or less acidic. These groups are classified into 2

Electron withdrawing groups makes the carboxylic more acidic. Also, the closer the electron withdrawing groups to the proton in the acid also makes the acid more acidic. This electron withdrawing group allows the electron density around the oxygen atom to be decreased.

Electron donating group makes the carboxylic acid less acidic. The more alkyl groups in a carboxylic acid, the less acidic the acid becomes. This allows the electron density around the oxygen atom in to be increased.

Answer:

All the 8.82*10^-5 moles of lead present is removed.

Explanation:

The total amount of lead II ion present is less than the amount of sulphide ion present and they react in 1:1 ratio so all the lead II ions are removed from the solution. See attached image.

Answer:

Phosphorus can act as an acceptor while galium can act as a donor

Explanation:

In electronics, pentavalent atoms are used as acceptors while trivalent atoms are used donors. Hence the answers provided above.