Write the net cell equation for this electrochemical cell. phases are optional. do not include the concentrations. sn(s)||sn2+(aq, 0.0155 m)∥∥ag+(aq, 3.50 m)||ag(s) net cell equation: sn +2ag^{+}->sn^{2+} +2ag sn+2ag+⟶sn2++2ag special δσω λμπ reset ( ) [ ] xyxyyyx⟶↽−−⇀ • (s) (l) (aq) (g) calculate e∘cell, δg∘rxn, δgrxn, and ecell at 25.0 ∘c, using standard potentials as needed

Final answer:

The beta decay of thallium-210 results in the conversion of the thallium nucleus into a lead-210 nucleus, with the emission of a beta particle and an antineutrino.

Explanation:

The beta decay of thallium-210 involves the transformation of a neutron into a proton, with the emission of an electron (beta particle) and an antineutrino. The balanced nuclear equation for this process can be written as follows:

^{210}_{81}Tl →
^{210}_{82}Pb +
^-1_{0}e (or β^-) + ar{ν}_e

Here, thallium-210 (Tl-210) undergoes beta decay to become lead-210 (Pb-210), with the emission of a beta particle (β^-) and an antineutrino (ν_e).

Answer is 
      Ni²⁺(aq) + 6NH₃(aq) ⇌ [Ni(NH₃)₆]²⁺

The equilibrium constant expression for the reaction 2HI (g) ⇌ H₂ (g) + I₂ (g) and given concentrations of H₂ and I₂ , we calculated the concentration of HI at equilibrium to be approximately 3.24 × 10⁻³ M. We used the equilibrium constant Keq = 1.67 × 10⁻² to solve for [HI].

To find the concentration of HI at equilibrium for the reaction 2HI (g) ⇌ H₂. (g) + I₂ (g) given Keq, [H₂] and [I₂], we can use the equilibrium constant expression:

Given data:

Plug these values into the equilibrium expression and solve for [HI]:

First, calculate the numerator:

Now plug this back into the equation:

Solving for [HI]₂ :

Taking the square root to find [HI]:

Correct question is: At 350°C , keq = 1.67 × 10⁻² for the reversible reaction 2HI (g) ⇌ H₂ (g) + I₂ (g). what is the concentration of hi at equilibrium if [ H₂ ] is 2.44 × 10⁻³ m and [ I₂ ] is 7.18 × 10⁻⁵ m?

To find the number of gram atoms in 7.5g of gold, divide the mass by the molar mass of gold, which gives approximately 0.03807 moles of gram atoms.

The student was asked to calculate the number of gram atoms in 7.5g of gold where the atomic weight of gold is given as 197.2 amu (atomic mass unit). We use the concept of molar mass to answer this question. The molar mass of gold (Au) is about 197.0g/mol, and we know that one mole of any substance contains Avogadro's number of atoms, which is 6.022 x 1023 atoms/mol.

To find the number of moles (or gram atoms) in 7.5g of gold, we use the formula:

Number of moles = Mass (g) / Molar mass (g/mol)

Substituting the values, we get:

Number of moles = 7.5g / 197.0g/mol ≈ 0.03807 mol

Therefore, there are about 0.03807-gram atoms in 7.5g of gold.

The overall reaction for the galvanic cell has been [tex]\rm Cu^2^+\;+\;Ni\;\rightarrow\;Ni^2^+\;+\;Cu[/tex].

The galvanic cell has been given as the electrochemical cell that converts chemical energy of the reaction into electrical energy.

The galvanic cell has anode as the oxidizing electrode, whee the loss of electrons takes place, and cathode as the reducing electrode where the gain of electrons takes place.

The cathodic reaction in the cell has been:

[tex]\rm Cu^2^+\;\rightarrow\;Cu\;(s)[/tex]

The anodic reaction in the cell has been:

[tex]\rm Ni\;(s)\;\rightarrow\;Ni^2^+[/tex]

The overall reaction for the galvanic cell has been:

[tex]\rm Cu^2^+\;+\;Ni\;\rightarrow\;Ni^2^+\;+\;Cu[/tex]

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Answer:

True, they are closely related.

Explanation:

Because they both sometimes are related to the same thing.

Answer : The theoretical yield of [tex]K_2CO_3[/tex] = 27.089 g

The percent yield of [tex]K_2CO_3[/tex]  is, 80.47 %

Explanation :  Given,

Mass of [tex]KO_2[/tex] = 27.9 g

Volume of [tex]CO_2[/tex] = 29.0 L  (At STP)

Molar mass of [tex]KO_2[/tex] = 71.10 g/mole

Molar mass of [tex]CO_2[/tex] = 44 g/mole

Molar mass of [tex]K_2CO_3[/tex] = 138.21 g/mole

First we have to calculate the moles of [tex]CO_2[/tex] and [tex]KO_2[/tex].

At STP,

As, 22.4 L volume of [tex]CO_2[/tex] present in 1 mole of [tex]CO_2[/tex]

So, 29.0 L volume of [tex]CO_2[/tex] present in [tex]\frac{29.0}{22.4}=1.29[/tex] mole of [tex]CO_2[/tex]

[tex]\text{Moles of }KO_2=\frac{\text{Mass of }KO_2}{\text{Molar mass of }KO_2}=\frac{27.9g}{71.10g/mole}=0.392mole[/tex]

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

[tex]4KO_2+2CO_2\rightarrow 2K_2CO_3+3O_2[/tex]

From the balanced reaction we conclude that

As, 4 moles of [tex]KO_2[/tex] react with 2 mole of [tex]CO_2[/tex]

So, 0.392 moles of [tex]KO_2[/tex] react with [tex]\frac{2}{4}\times 0.392=0.196[/tex] moles of [tex]CO_2[/tex]

From this we conclude that, [tex]CO_2[/tex] is an excess reagent because the given moles are greater than the required moles and [tex]KO_2[/tex] is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of [tex]K_2CO_3[/tex].

As, 4 moles of [tex]KO_2[/tex] react to give 2 moles of [tex]K_2CO_3[/tex]

So, 0.392 moles of [tex]KO_2[/tex] react to give [tex]\frac{2}{4}\times 0.392=0.196[/tex] moles of [tex]K_2CO_3[/tex]

Now we have to calculate the mass of [tex]K_2CO_3[/tex].

[tex]\text{Mass of }K_2CO_3=\text{Moles of }K_2CO_3\times \text{Molar mass of }K_2CO_3[/tex]

[tex]\text{Mass of }K_2CO_3=(0.196mole)\times (138.21g/mole)=27.089g[/tex]

The theoretical yield of [tex]K_2CO_3[/tex] = 27.089 g

The actual yield of [tex]K_2CO_3[/tex] = 21.8 g

Now we have to calculate the percent yield of [tex]K_2CO_3[/tex]

[tex]\%\text{ yield of }K_2CO_3=\frac{\text{Actual yield of }K_2CO_3}{\text{Theoretical yield of }K_2CO_3}\times 100=\frac{21.8g}{27.089g}\times 100=80.47\%[/tex]

Therefore, the percent yield of [tex]K_2CO_3[/tex]  is, 80.47 %

The theoretical yield of K₂CO₃ is 54.17 g and the percent yield is 40.24%.

Percent yield is the percent ratio of actual yield to the theoretical yield. It is calculated to be the experimental yield divided by theoretical yield multiplied by 100%. If the actual and theoretical yield ​are the same, the percent yield is 100%

In chemistry, yield is a measure of the quantity of moles of a product formed in relation to the reactant consumed, obtained in a chemical reaction, usually expressed as a percentage.

Given,

Mass of K₂O = 27.9g

Molar Mass of K₂O = 71.1 g/mol

Mass of CO₂ = 29.01g

The reaction can be written as -

K₂O + CO₂ = K₂CO₃

Moles of K₂O = 27.9 / 71.1 = 0.392 moles

Moles of CO₂ = 29 / 22.4 = 1.29 moles

Since moles of K₂O is lesser, it is the limiting reagent.

From the reaction, 1 mole of K₂O gives 1 mole of K₂CO₃

so, 0.392 moles of K₂CO₃ is produced.

Theoretical yield of K₂CO₃ = 0.392 × 138.21 = 54.17 g

Actual yield = 21.8 g

Percent yield = (21.8 / 54.17) × 100

= 40.24%

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its defiantly a solid, because the particles are packed in close together and it has a definite shape and volume. water is lose and not packed in like this. and with gas, the particles would juts fly everywhere and plasma is electricity and a wild card. so its definitely a solid, trust me

Nuclear reaction: ¹⁰⁶Ru → ¹⁰⁶Rh + e⁻(electron) + ve(electron antineutrino).

Beta decay is radioactive decay in which a beta ray and a neutrino are emitted from an atomic nucleus.There are two types of beta decay: beta minus and beta plus.

Ionizing radiation is most damaging to soft tissue among the materials listed (soft tissue, paper, wood, lead) due to its ability to break chemical bonds and cause cell malfunctions. However, the impact largely depends on the type of radiation and the characteristics of the material.

The effects of ionizing radiation on materials largely depend on the type of radiation and the characteristics of the material. Ionizing radiation is harmful as it can ionize molecules or break chemical bonds, causing malfunctions in cell processes. This can lead to somatic or genetic damage, particularly in rapidly reproducing cells.

In terms of the materials listed, soft tissue is the most susceptible to damage from ionizing radiation. Materials like paper, wood, and lead have different degrees of resistance to ionizing radiation. Paper and wood can block alpha and beta particles, low-energy forms of ionizing radiation, while metal or lead can stop gamma radiation, a high-energy form of ionizing radiation, more effectively.

However, it's crucial to note that ionizing radiation has its greatest effect on cells that rapidly reproduce. Therefore, in living organisms, areas with rapidly dividing cells, such as the skin or the lining of the stomach or intestines, are especially susceptible to damage from ionizing radiation.

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Final answer:

After 3.7 days, approximately 3.54mg of the original 227.0mg Sodium-24 sample will remain when considering its half-life of 14.8 hours. This is calculated by determining the number of half-life periods that occur in 3.7 days and applying the power of this number to half the mass to represent the consistent halving of the sample

Explanation:

The question is about the concept of half-life and how it applies to a sample of Sodium-24. Sodium-24 has a half-life of 14.8 hours, which means every 14.8 hours, half of the Sodium-24 atoms decay. Therefore, to find out how much of a 227.0 mg sample remains after 3.7 days, we need to first convert the days into hours (3.7 days * 24 hours/day = 88.8 hours) and then divide this by the half-life of Sodium-24 (88.8 hours / 14.8 hours/half-life = ~6 half-lives).

Given that half of the material decays each half-life period, we calculate: 227.0 mg * (1/2)^6 = ~3.54 mg. So, approximately 3.54 mg of the original Sodium-24 sample would remain after 3.7 days.



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The concentration of the HBr solution is approximately 0.3125 M.

To find the concentration of the HBr solution, we can use the equation:

M1V1 = M2V2

where M1 is the concentration of the KOH solution (given as 0.25 M), V1 is the volume of the KOH solution (given as 15.0 mL), M2 is the concentration of the HBr solution (what we're trying to find), and V2 is the volume of the HBr solution (given as 12.0 mL).

Plugging in the values:

(0.25 M)(15.0 mL) = M2(12.0 mL)

Simplifying:

3.75 = 12M2

Dividing both sides by 12:

M2 = 3.75/12 ≈ 0.3125 M

Therefore, the concentration of the HBr solution is approximately 0.3125 M.

Methanol (CH₃OH) can act as both a hydrogen bond donor and acceptor due to its hydrogen atom bonded to electronegative oxygen and the oxygen's two lone pairs of electrons. This enables methanol to form a network of hydrogen bonds with water, affecting its physical properties.

The compound that can act as both a hydrogen bond donor and a hydrogen bond acceptor is methanol (CH₃OH). It contains a hydrogen atom attached to oxygen (making it a hydrogen bond donor) and two lone pairs of electrons on the oxygen (making it a hydrogen bond acceptor). A substance, like methanol, that can both donate a hydrogen atom and accept a hydrogen bond due to these features can participate in hydrogen bonding with water.

To assess whether a compound can act as a hydrogen bond donor or acceptor, one should look for a hydrogen atom bonded to a highly electronegative atom like oxygen, nitrogen, or fluorine; this structure facilitates hydrogen bond donation. For a compound to act as an acceptor, one should identify the presence of lone pairs of electrons on a highly electronegative atom, which can attract the hydrogen atom from another molecule.

Methanol's ability to act in both capacities allows it to form a network of hydrogen bonds with water molecules, thus affecting properties such as the boiling point and solubility. When drawing the hydrogen-bonded structure, we would show lines or dotted lines between the hydrogen of one methanol molecule and the oxygen of another methanol molecule or of a water molecule to represent the hydrogen bonds.

Sulfuric acid loses a proton and acts as a Brønsted-Lowry acid, while ammonia gains a proton and becomes a Brønsted-Lowry base in their respective reactions.

In the reaction between sulfuric acid (H2SO4) and ammonia (NH3), when H2SO4 becomes HSO4-, it loses a proton. This change makes it a Brønsted-Lowry acid because a Brønsted-Lowry acid donates a proton in an acid-base reaction. On the other hand, when NH3 becomes NH4+, it gains a proton, which makes it a Brønsted-Lowry base, since a Brønsted-Lowry base accepts a proton.

The minimum energy of a photon that can produce one hydrogen atom from water is [tex]\boxed{7.755 \times {{10}^{ - 19}}{\text{ J}}}[/tex]

The frequency of the photon is [tex]\boxed{1.17 \times {{10}^{15}}{\text{ }}{{\text{s}}^{ - 1}}}[/tex].

The wavelength of the photon is [tex]\boxed{2.563 \times {{10}^{ - 7}}{\text{ m}}}[/tex].

Further Explanation:

Frequency[tex]\left( \nu  \right)[/tex] is defined as number of times n event occurs in unit time. It is generally applied to waves including light, sound, and radio waves. It is denoted by [tex]{\nu }}[/tex] and its SI unit is Hertz (Hz).

Wavelength is the characteristic property of a wave. It is defined as the distance between two successive crests or troughs. A crest is that point where there is maximum displacement of the medium whereas trough is a point that has minimum displacement of the medium. It is represented by [tex]\lambda[/tex] and its SI unit is meter (m).

First, the energy required to produce one hydrogen atom in joule can be calculated as follows:

[tex]E\left({\text{J}}\right)=\frac{{E\left( {{\text{kJ}}}\right)\times\left({\frac{{{{10}^3}{\text{ J}}}}{{{\text{kJ}}}}}\right)}}{{{{\text{N}}_{\text{A}}}}}[/tex]           …… (1)

Here, [tex]{{\text{N}}_{\text{A}}}[/tex] is an Avogadro's number and has a value [tex]6.022 \times {10^{23}}{\text{ mo}}{{\text{l}}^{ - 1}}[/tex].

Substitute [tex]467{\text{ kJ/mol}}[/tex] for [tex]E\left( {{\text{kJ}}} \right)[/tex] and [tex]6.022 \times {10^{23}}{\text{ mo}}{{\text{l}}^{ - 1}}[/tex] for [tex]{{\text{N}}_{\text{A}}}[/tex] in equation (1).

[tex]\begin{aligned}E\left( {\text{J}} \right)&=\frac{{\left({467{\text{ kJ/mol}}} \right) \times \left( {\frac{{{{10}^3}{\text{ J}}}}{{{\text{kJ}}}}}\right)}}{{6.022\times {{10}^{23}}{\text{ mo}}{{\text{l}}^{-1}}}}\\&=7.755\times{10^{-19}}{\text{ J}}\\\end{aligned}[/tex]

Thus the energy of a photon that can free one atom of hydrogen is [tex]7.755 \times {10^{ - 19}}{\text{ J}}[/tex].

The expression of frequency and energy is as follows:

[tex]E=hv[/tex]              …… (2)

Here, [tex]v[/tex] is a frequency of photon and h is a Plank’s constant and has a value [tex]\left({6.626\times{{10}^{-34}}{\text{ Js}}}\right)[/tex].

Rearrange equation (2) to calculate the frequency of the photon as follows:

[tex]v=\frac{E}{h}[/tex]               …… (3)

Substitute [tex]6.626\times{10^{-34}}{\text{ J}}\cdot{\text{s}}[/tex] for h and [tex]7.755 \times {10^{-19}}{\text{ J}}[/tex] for E in equation (3).

[tex]\begin{aligned}v&=\frac{{7.755\times{{10}^{-19}}{\text{ J}}}}{{6.626\times{{10}^{-34}}{\text{ Js}}}}\\&=1.17\times{10^{15}}{\text{}}{{\text{s}}^{-1}}\\\end{aligned}[/tex]

Thus the frequency of photon is [tex]1.17\times{10^{15}}{\text{}}{{\text{s}}^{-1}}[/tex].

The expression to calculate the wavelength from energy of the photon is as follows:

[tex]E = \frac{{h{\text{c}}}}{{\lambda }}}[/tex]                  …… (4)

Here [tex]{\lambda }}[/tex] is a wavelength of a photon and c is a speed of light.

Rearrange equation (4) to calculate wavelength of the photon as follows:

[tex]{\lambda }}=\frac{{h{\text{c}}}}{E}[/tex]                                 …… (5)

Substitute [tex]6.626 \times {10^{ - 34}}{\text{ J}} \cdot {\text{s}}[/tex] for h, [tex]3.0 \times {10^8}{\text{ m/s}}[/tex] for c and [tex]7.755 \times {10^{ - 19}}{\text{ J}}[/tex] for E in equation (5).

[tex]\begin{aligned}{\lambda}}=\frac{{\left({6.626\times {{10}^{-34}}{\text{ J}}\cdot {\text{s}}} \right)\left( {3.0 \times {{10}^8}{\text{ m/s}}}\right)}}{{\left({7.755\times {{10}^{ - 19}}{\text{ J}}} \right)}}\\=2.563\times {10^{-7}}{\text{m}}\\\end{aligned}[/tex]

Hence wavelength of the photon is equal to [tex]2.563 \times {10^{ - 7}}{\text{ m}}[/tex].

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Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Structure of atom

Keywords: Hydrogen atom, cheapest source of h2, 467 kj, 1 mol of h atom, frequency wavelength.

Explanation:

its the third one down :)

The first carbon in CH3CN is sp3 hybridized, due to its four single bonds and tetrahedral geometry, while the second carbon is sp hybridized, having a triple bond and a single bond resulting in a linear geometry.

To determine the hybridization of the carbon atoms in CH3CN, we must examine the bonding around each carbon. Starting with the CH3 group, the first carbon is bonded to three hydrogen atoms and has a single bond with the second carbon. This suggests an sp3 hybridization, which is typical for carbons with four single bonds and leads to a tetrahedral geometry. Proceeding to the carbon in the CN group, it has a triple bond with nitrogen and a single bond to the first carbon. This configuration means it is sp hybridized, as there are two electron groups around it resulting in a linear geometry.

Therefore, from left to right, the hybridization of the carbons in CH3CN are sp3 and sp.

The electron configuration for Ni²⁺ is [Ar] 3d⁸.

Nickel (Ni) has an atomic number of 28, which means that it has 28 protons in its nucleus. It also has 28 electrons, which are arranged in shells around the nucleus.

The electron configuration for a neutral nickel atom is:

[Ar] 3d⁸ 4s²

The argon (Ar) core is filled with 18 electrons, and the remaining 10 electrons are arranged in the 3d and 4s orbitals.

When nickel loses two electrons to form Ni²⁺, the two electrons that are lost are the two 4s electrons. This leaves the 8 3d electrons behind, which gives Ni²⁺ the electron configuration shown above.

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