The normal boiling point of water is 100.0 °c and its molar enthalpy of vaporization is 40.67 kj/mol. what is the change in entropy in the system in j/k when 51.1 grams of steam at 1 atm condenses to a liquid at the normal boiling point?

Answer: 1 M

Explanation:

Molarity : It is defined as the number of moles of solute present per liter of the solution.

Formula used :

[tex]Molarity=\frac{n\times 1000}{V_s}[/tex]

where,

n= moles of solute

[tex]Moles=\frac{\text{Given mass}}{\text{Molar mass}}=\frac{85.0g}{170g/mol}=0.5moles[/tex]  

[tex]V_s[/tex] = Volume of solution in ml

Now put all the given values in the formula of molarity, we get

[tex]Molarity=\frac{0.5moles\times 1000}{500ml}=1mole/L[/tex]

Therefore, the molarity of solution will be 1 M.

Complete the following table.

Acid Molarity Moles of H⁺ released per liter

HCl 1

H2SO4 1

H3PO4 1

H2SO4 0.5

H3PO4 3

HNO3 2

[Answer]

1

2

3

1

9

2

The answer is:

the mass of the excess reactant  leftover after the reaction has reached completion is 90.135 grams

The explanation:

According to the reaction equation:

 2C₂H₆(g) + 7O₂(g) → 4CO₂(g) + 6H₂O(g)

when m is the mass of C₂H₆ m(C₂H₆) = 152 g

So we need to get the number of moles of C₂H₆

n(C₂H₆) = mass C₂H₆ / molar mass of C₂H₆(M)

and when the molar mass of C₂H₆ = 30 g/mol

so, by substitution:

n(C₂H₆) = m(C₂H₆) / M(C₂H₆).

n(C₂H₆) = 152 g / 30 g/mol.

n(C₂H₆) = 5.067 mol.

Then

when the mass of O₂ m(O₂) = 231 g

so we need to get the number of moles of O₂

when nO₂ = mass O₂/ molar mass of O₂

when molar mass of O₂ = 32 g /mol

So, by substitution:

n(O₂) = 231 g / 32 g/mol.

n(O₂) = 7.218 mol

So O₂ is the limiting reactant

according to the chemical reaction we can get the molar ratio between the O₂and C₂H₆:

n O₂ : n C₂H₆    →  7.218 : n C₂H₆

      7 : 2                   7      :   2

∴ n(C₂H₆) = 2 * 7.218 mol / 7

∴ n(C₂H₆) = 2.0625 mol

The number of moles remaining n(C₂H₆) = 5.067 mol - 2.0625 mol

∴ n (C₂H₆) = 3.0045 mol

So the mass remains = moles remains * molar mass of C₂H₆

∴ m (C₂H₆) = 3.0045 mol * 30 g/mol = 90.135 g

Answer:

Potassium dicarbonatedifluoroplatinate (II)

Explanation:

Hello,

Based on the IUPAC rules, the given compound is a complex called potassium dicarbonatodifluoroplatinate (II) as long as the carbonate ion is present twice as a monodentate ligand therefore it is preceded by a "di" prefix. In addition, two fluorines are also present with the "di" prefix as well as a platinum which complete the anionic section including +2 as the platinum oxidation state.

Best regards.

Answer:

The answer is smooth and Involuntary.

Explanation:

Smooth muscles are involuntary which means they automatically do a function without the brain telling it to.

For light 3 to function, switch D does not have to be on. This diagram shows a parallel circuit that provides more than one way for the current to return to the power source.

Answer:

E°cell = 0.94 V

Ecell = 1.00 V

ΔG = -1.9 × 10⁵ J

ΔG° = -1.8 × 10⁵ J

Explanation:

Let's consider this electrochemical cell:

Sn(s)|Sn²⁺(aq,0.0155M)||Ag⁺(aq, 3.50M)|Ag(s)

The corresponding half-reactions are:

Oxidation (anode): Sn(s) → Sn²⁺(aq) + 2 e⁻              E°red = -0.14 V

Reduction (cathode): 2 Ag⁺(aq) + 2 e⁻ → 2 Ag(s)    E°red = 0.80 V

The standard cell potential (E°cell) is the difference between the standard reduction potential of the cathode and the standard reduction potential of the anode.

E°cell = E°red, cat - E°red, an = 0.80 V - (-0.14 V) = 0.94 V

We can find the cell potential using the Nernst equation.

Ecell = E°cell - (0.05916/n) . log Q

Ecell = 0.94 V - (0.05916/2) . log ([Sn²⁺]/[Ag⁺]²)

Ecell = 1.00 V

We can find ΔG and ΔG° using the following expressions.

ΔG = -n.F.Ecell = (-2mol).(96468J/mol.V).(1.00V) = -1.9 × 10⁵ J

ΔG° = -n.F.E°cell = (-2mol).(96468J/mol.V).(0.94V) = -1.8 × 10⁵ J

The net cell equation for the electrochemical cell is Sn(s) + 2Ag+ -> Sn₂+ + 2Ag. To find the standard cell potential, standard free energy change, free energy change, and cell potential at 25.0 °C, we use standard reduction potentials and the Nernst equation. Standard free energy change is calculated with ΔG°rxn = -nFE°cell, while the cell potential under non-standard conditions is found using the Nernst equation.

The net cell equation for the given electrochemical cell is Sn(s) + 2Ag+ (aq) → Sn₂+ (aq) + 2Ag(s). To calculate the standard cell potential (E°cell), standard free energy change (ΔG°rxn), free energy change (ΔGrxn), and the cell potential (Ecell) at 25.0 °C, we can use the standard reduction potentials and the Nernst equation. The standard cell potential is calculated by subtracting the standard reduction potential of the anode from that of the cathode.

The standard free energy change can be calculated from the standard cell potential using the formula ΔG°rxn = -nFE°cell, where n is the number of moles of electrons transferred in the reaction, and F is the Faraday constant. The cell potential under non-standard conditions (Ecell) can be determined using the Nernst equation, which incorporates the concentration of the ionic species involved in the half-reactions.

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The nuclide produced when uranium-238 decays by alpha emission is thorium-234. This process involves the emission of an alpha particle, which causes uranium-238 to lose 2 protons and 2 neutrons, resulting in thorium-234.

The nuclide produced when uranium-238 decays by alpha emission is thorium-234. This decay process involves the release of an alpha particle from uranium-238 nucleus. An alpha particle is equivalent to a helium nucleus - it contains 2 protons and 2 neutrons. Therefore, when uranium-238 (which has 92 protons and 146 neutrons) emits an alpha particle, it loses 2 protons and 2 neutrons, transforming into a new element with 90 protons and 144 neutrons, which is thorium-234 (234 90Th).

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The freezing point is the temperature at which the fluid freezes to a solid form. The freezing point of the solution is -0.435 degrees celsius.

The freezing point is the product of the molality, van 't Hoff factor, and the cryoscopic constant. It is given as,

[tex]\rm \Delta T_{F} = K_{F} \times b\times i[/tex]

Given,

Mass of water = 0.1 kg

Mass of sucrose = 8.0 gm

Moles of sucrose are calculated as:

[tex]\begin{aligned}\rm moles &= \dfrac{8.0}{342.3}\\\\&= 0.0233 \;\rm mol \end{aligned}[/tex]

The molality of sucrose is calculated as:

[tex]\begin{aligned}\rm b &= \dfrac{\text{moles of sucrose}}{\text{mass of water}}\\\\&= \dfrac{0.0233 \;\rm mol}{0.1}\\\\&= 0.233 \;\rm m\end{aligned}[/tex]

The freezing point depression is calculated as:

[tex]\begin{aligned}\rm \Delta T &= 0.233 \;\rm m \times 1.86\; ^{\circ} \;\rm C/m\\\\&= 0.435 ^{\circ}\;\rm C\end{aligned}[/tex]

Therefore, the freezing point of a solution is -0.435 degrees celsius.

Learn more about a freezing point here:

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Molten barium chloride is separetes:
BaCl₂(l) → Ba(l) + Cl₂(g), 
but first ionic bonds in this salt are separeted because of heat: 
BaCl₂(l) → Ba²⁺(l) + 2Cl⁻(l).

Reaction of reduction at cathode(-): Ba²⁺(l) + 2e⁻ → Ba(l).

Reaction of oxidation at anode(+): 2Cl⁻(l) → Cl₂(g) + 2e⁻.

The anode is positive and the cathode is negative.

The electrolysis of molten barium chloride involves reduction of barium ions and oxidation of chloride ions, requiring two electrons to move through the circuit for one unit of the reaction.

During the electrolysis of molten barium chloride (BaCl2), barium ions (Ba2+) are reduced at the cathode, and chloride ions (Cl-) are oxidized at the anode. The half-reactions, showing the movement of electrons through the circuit, are as follows:

The overall cell reaction is obtained by combining the half-reactions and balancing the electrons:

Ba2+(l) + 2Cl-(l) → Ba(l) + Cl2(g)

For each mole of barium ions reduced, two electrons are involved in the transfer. Similarly, for each mole of chlorine gas produced, two electrons are given up by chloride ions. As such, there would be two electrons that moved through the circuit for one unit of the reaction to occur.

Not filtering a saturated solution before titrating will likely cause the calculated Ksp to be a) too high

When a student does not filter their saturated solution before titrating, the calculated Ksp (solubility product constant) will likely be affected.

Specifically, the Ksp will probably be too high.

This is because the undissolved solid present in the unfiltered solution will falsely contribute to the ionic concentration being measured during the titration, leading to an overestimation of the dissolved ions and thus a higher calculated Ksp.

As a result, to ensure accurate determination of the Ksp, it is essential to filter the solution to remove any undissolved solute before conducting the titration.

Complete question is - A student does not filter his/her saturated solution before titrating. will the calculated ksp probably be (a) too high, (b) too low, or (c) unaffected? why?

If you cut the height of an object in half, its gravitational potential energy (GPE) will also halve.

If you cut the height of an object in half, its gravitational potential energy (GPE) will also halve.

GPE is determined by the height of the object and its mass. When you cut the height in half, the GPE is reduced by half because the potential energy is directly proportional to the height.

For example, if a book has a certain GPE at a certain height, cutting the height in half will result in the book having half the GPE it had before.

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Answer:

The correct answer is 1. Moseley started his investigations by reviewing the work of those who contributed to the previous versions of the table.

Explanation:

Henry Gwyn Jeffreys Moseley was a British physicist. His most important scientific contribution was the proof of the correctness of the concept of atomic numbers in chemistry.

In 1913, using X-ray spectroscopy, he found a systematic relationship between the wavelength and the atomic number. Previously, it was assumed that the atomic number was an arbitrary number based on the order of the atomic masses, but had to be changed to bring an element into the right place in the periodic table. Moseley's discovery showed that atomic numbers had an experimentally measurable basis. In addition, he showed that there were gaps in numbers 43, 61 and 75 (known today as the radioactive elements technetium and promethium, as well as the stable but rare rhenium). His work was further proof of the then controversial atomic theory.

The colors of the compounds in the beakers could potentially be: [CoF6]3– (green), [Co(NH3)6]3+ (yellow-orange), and [Co(CN)6]3– (red), although colors can vary based on different conditions.

The question asks us to identify the compounds present in three beakers based on the color of the solution. While this is generally impossible to answer with absolute certainty without more information or additional tests, we can make a fair guess based on some known color-characteristics of these compounds.

[CoF6]3– is generally green due to the color of the Cobalt(III) ion.

[Co(NH3)6]3+ is usually yellow to orange in color.

FInally, [Co(CN)6]3– commonly has a strong red color.

Note that in reality, actual colors can vary depending on concentration, temperature, and other factors.

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