A mixture of 100 g of k2cr207 and 200 g of water is stirred at 60 °c until no more of the salt dissolves. the resulting solution is poured off, leaving the undissolved solid behind. the solution is now cooled to 20°c. what mass of k2cr207 crystallizes from the solution during the cooling?

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.

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.

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Answer:

Kinetic energy of a system .

Explanation:

Hello,

The temperature is a measure of the average kinetic energy of the atoms or molecules in the system as long as it accounts for the movement of the particles composing the system, thus, the higher the temperature, the higher the kinetic energy as the atoms or molecules could move faster and vigorously.

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Paramagnetic can be detected by the unpaired electrons in the electron configuration. In this case, out of Sc³⁺, Br⁻, Mg²⁺, and Se, only Se (Selenium) is paramagnetic because it has unpaired electrons.

The substances in question are Sc³⁺ (Scandium ion), Br⁻ (Bromide ion), Mg²⁺ (Magnesium ion) and Se (Selenium). To determine if these species are paramagnetic, we need to look at their electronic configurations.

Sc³⁺ (Scandium ion) has an atomic number of 21. When it loses 3 electrons to form the cation Sc³⁺, it has the same electron configuration as Argon, with all shells filled, therefore it is diamagnetic, not paramagnetic.

Br⁻ (Bromide ion) is the ion formed when Bromine (atomic number 35) gains one electron. This results in a complete electron shell, and so bromide ion is diamagnetic, not paramagnetic.

The  Mg²⁺ (Magnesium ion) is formed when Magnesium (atomic number 12) loses 2 electrons. This results in a complete electron shell, making this species diamagnetic as well.

Lastly, Se (Selenium) with an atomic number of 34 has 4 unpaired electrons in its ground state and therefore is paramagnetic.

So, out of Sc³⁺, Br⁻, Mg²⁺, and Se, only Se (Selenium) is paramagnetic.

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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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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 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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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

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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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.

Final answer:

Vinegar contains acetic acid, an organic compound classified as a carboxylic acid, due to the presence of a carboxyl group with both a carbonyl and a hydroxyl group attached to the same carbon atom.

Explanation:

Vinegar is an example of an organic compound called a carboxylic acid. Its functional group is known as the carboxyl group, which features a carbonyl group (a carbon atom double-bonded to an oxygen atom) and a hydroxyl group (an oxygen atom single-bonded to a hydrogen atom) attached to the same carbon. This unique structure with the general formula RCOOH is characteristic of carboxylic acids such as acetic acid (CH₃COOH), which gives vinegar its sour taste and pungent smell.

The presence of the carboxylic acid functional group distinguishes it from other compounds like amines, which contain nitrogen; aldehydes, which have the carbonyl group bonded to at least one hydrogen atom; and esters, which have the carbonyl group bonded to an oxygen atom that is in turn connected to another carbon group.

Acetic acid is a well-known carboxylic acid found in vinegar and has been used since ancient times for various purposes, from a condiment and a preservative to even an antibiotic and a detergent.

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.