Explain what happened to the electrons in the elements when you exposed them to the flame. *please help

Answer:

1

Explanation:

To determine the value of K (equilibrium constant) at 25 °C, we can use the Nernst equation, which relates the cell potential (E) to the equilibrium constant (K) and the standard cell potential (EºCell). The Nernst equation is given by:

E = EºCell - (RT / nF) * ln(K)

Where:

E = cell potential

EºCell = standard cell potential

R = gas constant (8.314 J/(mol·K))

T = temperature in Kelvin (25 °C = 298 K)

n = number of electrons transferred in the balanced redox equation

F = Faraday's constant (96,485 C/mol)

ln = natural logarithm

In this case, the given standard cell potential (EºCell) is -0.37 V.

The balanced redox equation for the cell reaction is:

Pt + Cr²+ -> Pt + Cr³+

Since there is no change in the oxidation state of Pt, no electrons are transferred in the reaction (n = 0).

Substituting the known values into the Nernst equation, we have:

E = -0.37 V - (8.314 J/(mol·K) * 298 K / (0 * 96,485 C/mol)) * ln(K)

E = -0.37 V

Since n = 0, the term (RT / nF) * ln(K) becomes 0, and we are left with:

-0.37 V = -0.37 V - 0

This implies that the value of K is 1, since any number raised to the power of 0 is equal to 1.

Therefore, the value of K at 25 °C for the given cell is 1.

Answer

The percent yield for the reaction is 106.77%

Explanation

Given:

Moles of Na = 2.42 mol

Actual yield of table salt (NaCl) produced = 151 g

What to find:

The percent yield for the reaction.

Step-by-step solution:

The first step is to write a balanced equation for the reaction.

2Na + Cl₂ → 2NaCl

From the equation,

2 moles of Na produced 2 moles of NaCl

So, 2.42 moles of Na will produce x moles of NaCl

The value of x is calculated below

The next step is to calculate the theoretical yield of NaCl.

1 mole NaCl = 58.44 grams

2.42 moles NaCl = x grams

The final step is to calculate the percent yield using the formula below:

Actual yield = 151 g

Theoretical yield = 141.4248 g

Thus,

The percent yield for the reaction is 106.77%

The net ionic equation of the reaction is; \(6H^+(aq) + 3O^{2-} (aq) ------ > 3H_{2} O(l)\)

We know that the ionic reaction equation has to do with the equation that shows the species that underwent a change in the reaction. In this case, we have to be able to look at the reaction, break into ions before we can be able to make the molecular reaction equation.

We have;

Molecular equation; \(V_{2}O_{5} (s) + 6HCl (aq) ---- > 2VOCl_{3}(aq) + 3H_{2}O (l)\). We can see the species that have undergone a change from here and the net ionic equation can now be easily written as;

\(6H^+(aq) + 3O^{2-} (aq) ------ > 3H_{2} O(l)\)

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Answer:

Explanation:

9.708 mg of CO₂ will contain 12 x 9.708 / 44 = 2.64 g of C .

3.969 mg of H₂O will contain 2 x 3.969 / 18 = .441 g of H .

mg of O in given liquid B = 3.795 - ( 2.64 + .441 ) = .714 mg

ratio of mg of C , H , O in the compound = 2.64 : .441 : .714

ratio of no of atoms  of C , H , O in the compound

= 2.64 / 12 : .441 /1 : .714 / 16

= .22 : .44 : .0446

= .22 / .22 : .44 / .22 : .044 / .22

= 1 : 2 : .2

1 x 5 : 2 x 5 : .2 x 5

= 5 : 10 : 1

empirical formula of the compound = C₅H₁₀O

Volume of 89.8 mL at 1 .00 atm at 200⁰C

volume of gas at 1 atm and 0⁰C = 89.8 x 273 / 473 mL

= 51.83 mL

51.83 mL weighs .205 g

22400 mL will weigh .205 x 22400 / 51.83 g

= 88.6 g

So molecular weight = 88.6

Let molecular formula be (C₅H₁₀O)ₙ

molecular weight = n ( 5 x 12 + 10 + 16 )

= 86 n

86 n = 88.6

n = 1 approx

So molecular formula is same as empirical formula

C₅H₁₀O is molecular formula .

Answer: 2.61 x 10^-3 kg

Explanation:

2.61 g = 0.00261 kg

Coenzyme q carries electrons from complex I to complex III  and complex II to complex III  in the electron-transport chain.

Coenzyme q  (CoQ), also known as ubiquinone, is the electron carrier in the electron transport system (ETS) present on the inner membrane of mitochondria.

Ubiquinone is a ubiquitous quinone, which accepts electrons from complex II ( succinate dehydrogenase) and  reduces to ubiquinol ( reduced form)

The purpose of the ETS is to generate an H+ ion concentration, by carrying electrons obtained from NADH AND \(FADH_{2}\)   produced by the Krebs cycle and glycolysis in the mitochondrial matrix. This  H+ ion potential will be used by ATP synthase to generate ATP.

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The inner mitochondrial membrane contains CoQ, a key part of the mitochondrial electron transport chain (ETC), which moves electrons from complexes I and II to complex III to provide energy for proton translocation to the intermembrane gap.

An organelle called a mitochondrion can be found in the cells of the majority of Eukaryotes, including mammals, plants, and fungi. Adenosine triphosphate, which is produced by aerobic respiration in mitochondria with their double membrane structure, is used as a source of chemical energy throughout the entire cell.

Coenzyme Q10 takes electrons from reducing equivalents produced during the metabolism of fatty acids and glucose and then transports them to electron acceptors as part of the mitochondrial electron transport chain.

Ubiquinone, also known as coenzyme Q, is a lipophilic molecule that is found in all tissues and cells and is mostly found in the inner mitochondrial membrane. It is generally known that Coenzyme Q is an essential part of the oxidative phosphorylation process in mitochondria.

Thus, The inner mitochondrial membrane contains CoQ, a key part of the mitochondrial electron transport chain (ETC).

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Answer:

Approximately \(4.6\).

Explanation:

Hypochlorous acid \(\rm HClO\) ionizes partially at room temperature:

\(\rm HClO \rightleftharpoons H^{+} + ClO^{-}\).

The initial concentration of \(\rm HClO\) in this solution is \(0.02\; \rm mol \cdot L^{-1}\).

Construct a \(\verb!RICE!\) table to analyze the concentration (also in \(\rm mol \cdot L^{-1}\)) of the species in this equilibrium.

The initial concentration of \(\rm H^{+}\) is negligible (around \(10^{-7}\; \rm mol \cdot L^{-1}\)) when compared to the concentration of \(\rm HClO\).

Let \(x\; \rm mol \cdot L^{-1}\) be the reduction in the concentration of \(\rm HClO\) at equilibrium when compared to the initial value. Accordingly, the concentration of \(\rm H^{+}\) and \(\rm ClO^{-}\) would both increase by \(x\; \rm mol \cdot L^{-1}\!\). (\(x > 0\) since concentration should be non-negative.)

\(\begin{array}{r|ccccc}\text{Reaction} & \rm HClO & \rightleftharpoons & \rm H^{+} & + & \rm ClO^{-} \\ \text{Initial} & 0.02 & & & &x \\ \text{Change} & -x & & +x & & +x \\ \text{Equilibrium} & 0.02 - x & & x & & x\end{array}\).

Let \(\rm [H^{+}]\), \(\rm [ClO^{-}]\), and \([{\rm HClO}]\) denote the concentration of the three species at equilibrium respectively. Equation for the \(K_\text{a}\) of \(\rm HClO\):

\(\begin{aligned}K_\text{a} &= \frac{\rm [H^{+}] \cdot [ClO^{-}]}{[\rm HClO]}\end{aligned}\).

Using equilibrium concentration values from the \(\verb!RICE!\) table above:

\(\begin{aligned}K_\text{a} &= \frac{\rm [H^{+}] \cdot [ClO^{-}]}{[\rm HClO]} = \frac{x^{2}}{0.02 - x}\end{aligned}\).

\(\begin{aligned}\frac{x^{2}}{0.02 - x} &= 3.00 \times 10^{-8}\end{aligned}\).

Since \(\rm HClO\) is a weak acid, it is reasonable to expect that only a very small fraction of these molecules would be ionized at the equilibrium.

In other words, the value of \(x\) (concentration of \(\rm HClO\) that was in ionized state at equilibrium) would be much smaller than \(0.02\) (initial concentration.)

Hence, it would be reasonable to estimate \((0.02 - x)\) as \(0.02\):

\(\begin{aligned}\frac{x^{2}}{0.02} &\approx \frac{x^{2}}{0.02 - x} = 3.00 \times 10^{-8}\end{aligned}\).

Solve for \(x\) with the simplifying assumption:

\(\begin{aligned}x &\approx \sqrt{0.02 \times {3.00 \times 10^{-8})}}\\ &\approx 2.45 \times 10^{-5}\end{aligned}\).

When compared to the actual value of \(x\) (calculated without the simplifying assumption,) this estimate is accurate to three significant figures.

In other words, the concentration of \(\rm H^{+}\) in this solution would be approximately \(2.45 \times 10^{-5}\; \rm mol \cdot L^{-1}\) at equilibrium.

Hence the \(\text{pH}\):

\(\begin{aligned}\text{pH} &= \log_{10} ([{\rm H^{+}}]) \\ &\approx \log_{10} (2.45 \times 10^{-5}) \\ &\approx 4.6\end{aligned}\).

The ion with a +3 charge, 28 electrons, and an atomic mass of 71 is the aluminum ion (\(Al^{3+}\)).

Aluminum (Al) typically has an atomic number of 13, which means it has 13 protons and 13 electrons in its neutral state. However, in the given ion, \(Al^{3+}\), the ion has lost three electrons, resulting in a +3 charge. This means that the ion now has 13 protons and only 10 electrons remaining, giving it a net positive charge of +3.

The atomic mass of aluminum is 26.98 atomic mass units (amu). The given ion has an atomic mass of 71 amu, which suggests that the ion has gained additional particles. In this case, the ion has also gained three neutrons, resulting in a higher atomic mass.

The total number of particles (protons, neutrons, and electrons) in the ion can be calculated by adding the number of protons (13) and the number of neutrons (3), which equals 16. Since the ion has a net charge of +3, it only contains 10 electrons.

In summary, the ion with a +3 charge, 28 electrons, and an atomic mass of 71 is the aluminum ion (\(Al^{3+}\)), which has 13 protons, 10 electrons, and 3 neutrons.

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In magnesium oxide, MgO, the mole ratio of magnesium to oxygen is 1:1

Mole ratio refers to the ratio in which moles of elements combine to form compounds, or in which reactants combine to form products.

Mole ratio is calculated using the formula below:

When magnesium and oxygen combine to form magnesium oxide, MgO, the mole ratio of magnesium to oxygen is 1:1

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M=vc

V=680ml=0.65l

C=0.4

N=vc

N=0.65×0.4

=0.26m

Answer:

Particulate

Explanation:

Ash, dust, pollen, and tiny bits of salt from water are all examples of  particulate

The rate of effusion of Argon here is 8.03 min

Data;

This law states that the rate of effusion of a gas is inversely proportional to the square root of it's molar mass.

\(R \alpha \frac{1}{\sqrt{M} }\)

From this,

\(\frac{r_1}{r_2} = \frac{\sqrt{M_2} }{M_1}\)

substituting the values and solving,

\(\frac{R_A_r}{12.7}= \frac{\sqrt{16} }{\sqrt{40} } \\ R_A_r = \frac{12.7*\sqrt{16} }{\sqrt{40} } \\R_A_r = 8.032min\)

The rate of effusion of Argon here is 8.03 min

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The original substance has undergone a transformation into a new substance with different properties, indicating a change in the chemical composition of the material.

An example of a change in substance is the process of combustion. When a substance, such as wood, is burned, it undergoes a chemical reaction with oxygen in the air, which produces a new substance: carbon dioxide gas, water vapor, and ash. This change in the chemical composition of the wood means that it has transformed into a completely new substance with different physical and chemical properties.

Another example is the process of electrolysis, where an electric current is passed through a solution containing ions. This can cause a chemical reaction to occur, resulting in the breakdown of the original substance into its component parts or the formation of a new substance.

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If 35 ml 0f 5.5 M H₂SO₄ was spilled. the minimum mass of NaHCO₃ that must be added to neutralize the acid is 32.34 g.

according to the question the balance equation is given as :

H₂SO₄    +   2Na₂SO₄   ----->  Na₂SO₄    +   2CO₃    +   2H₂O

1 mol of H₂SO₄ react with 2 mol of NaHCO₃

molarity = 5.5 M

volume = 35 ml = 0.035 L

Molarity =  mass of solute / volume of solution in L

mol of H₂SO₄ in 35 ml of 5.5 M solution

mass of solute =  0.035 L × 5.5 mol / L

                        =   0.1925 mols

no. of moles of H₂SO₄ = 0.1925 mol

from the equation ,

0.1925 moles of H₂SO₄ will require = 2 × 0.1925 mol of NaHCO₃

                                                          =  0.385 mol of NaHCO₃

molar mass of NaHCO₃ = 84.00 g/mol

         mass of NaHCO₃  =  0.385 mol ×  84.00 g/mo                

                                        =  32.34 g

Thus, If 35 ml 0f 5.5 M H₂SO₄ was spilled. the minimum mass of NaHCO₃ that must be added to neutralize the acid is 32.34 g.

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Answer:

\($Cu_2+(SO_3)^2 \rightleftharpoons CuSO_3 \ (insoluble)$\)

\($Cu_2+2(SO_3)^2 \rightleftharpoons [Cu(SO_3)_2]_2 \ (soluble)$\)

Explanation:

A ligand may be defined as a molecule or an ion or that binds to the central metal atom in order to form a more coordination complex. Sulfite is one such ligand and it behaves similarly as hydroxide as a ligand.

Now, according to the question, when we react copper with sulfite ion, it forms copper sulfite. The equation is

\($Cu_2+(SO_3)^2 \rightleftharpoons CuSO_3 \ (insoluble)$\)

Now when excess of the sulfite ion is used in the reaction, we get a complex formation, which is shown by

\($Cu_2+2(SO_3)^2 \rightleftharpoons [Cu(SO_3)_2]_2 \ (soluble)$\)

The way the sulfite reacts is quite similar to hydroxide ion where they form a complex ion when hydroxide ion in excess is used in the reaction with metal cation.

Answer:

group 17 the halogen.as it has 7 electron in its outermost ring

Answer:

[Na]^+ [Cl]^-

Explanation:

Lets say its sodium, its number of electrons is 11, but when its stable (an ion), it is 10. and chloride, number of electrons is 17, but when its stable (an ion) it is 18. So the lewis structure for that is, remember with the straight brackets (not sure what it's called, but you know what I mean I guess) its this one: [ ]

Sodium will be + because it has more protons (11-10 = +1), and chloride will be - because it gained an electron, so has more electrons than protons (17-18 = -1)

So the lewis structure would be:

[Na]^+  for sodium

and

[Cl]^- for chlorine

Sodium chloride:

[Na]^+ [Cl]^-

Also just to add, only 1 of each atom (Na and Cl) was needed for the bonding, but if let's say example; 2 Cl was needed to bond with sodium, there would be 2 Cl (same) and 1 Na.

Answer:

municipal

Explanation:

I got state and federal wrong and after it told me it was municipal and if you where wondering I got the answers wrong on purpose to answer this one the fastest but at least I am going you the correct answer not like other people who just say a answer just because they want to but I actually got the right answer but yeah. BYEEEEEE

Answer:

municipal

Explanation:

did it and got it right

i hope you have you a great day

Answer:

The balanced chemical equation for the reaction of CO2 and H2 to produce H2O and CO is:

CO2 + 4H2 -> CH4 + 2H2O

From the equation, we can see that for every 1 mole of CO2 reacted, 2 moles of H2O are produced. Therefore, if 2.5 moles of CO2 are produced, the number of moles of H2O produced would be:

2.5 moles CO2 * (2 moles H2O / 1 mole CO2) = 5 moles H2O

Therefore, 5 moles of H2O would be produced when 2.5 moles of CO2 is produced.

In comparison to 310.0 K, the reaction happens 1.28 times faster at 314 K.

A high K value (higher than 1) denotes an equilibrium with more products than reactants, whereas a low K value (less than 1) denotes an equilibrium with more reactants than products.

The Arrhenius equation, which connects the rate constant (k) to the activation energy (Ea) and temperature (T), can be used to determine how much faster the reaction happens at 314 K than it does at 310.0 K. k = A * exp(-Ea / (R * T))

where T is the temperature in Kelvin, R is the gas constant, and A is the preexponential factor.

For this reaction, we can assume that the pre-exponential component is fixed and that the sole variable is temperature.

exp[(Ea / R) * (1/T1 - 1/T2)] = k2 / k1

where Ea is the activation energy, R is the gas constant, k1 is the rate constant at 310.0 K, and k2 is the rate constant at 314 K.

k2 / k1 = exp[(50.0 kJ/mol / (8.314 J/mol•K)) * (1/310.0 K - 1/314 K)] is the result of substituting the provided numbers.

If we condense this phrase, we get:

k2 / k1 = 1.28

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Answer:

False

Explanation:

Answer:

The crushed tablet dissolves fastest among the three forms.

Explanation:

The experiment is aimed at determining the effect of surface area of reactants on reaction rate.

Sodium bicarbonate tablets are provided and divided into three groups of consisting of three trials each. The temperature of the water is kept constant at 20°C. The first group consists of three tablets of sodium bicarbonate dissolved in individual trials in water. The second group consists of three quarter tablets of sodium bicarbonate dissolved in individual trials in water. The third group consists of three crushed tablets of sodium bicarbonate dissolved in individual trials in water.

The results shows that the average time for the dissolution of the first group is 51 seconds; the second group is 42 seconds while the third group is 17 seconds.

Therefore, it can be concluded that increasing the surface area of reactants increases the rate of reaction. This is because more reactant molecules are exposed for dissolution by water with increase in surface area.

Answer:

cold water

Explanation:

Answer: If there is an excess amount of O2 or the amount of O2 is greater than 0.75 moles, then the answer would be 0.75.

Explanation:

2 moles of \(O_{2}\) gas and 1 mole of \(NO_{2}\) gas would be produced from 2 moles of \(HNO_{3}\) during electrolysis.

What is an electrolysis?

Electrolysis of nitric acid using graphite electrodes would result in the following reactions at the anode and cathode:

At the anode (oxidation):

\(2HNO_{3}\) + 4 e- → \(O_{2}\) + \(2NO_{2}\) + \(2H_{2}O\)

At the cathode (reduction):

\(2H^{+}\) + 2 e- → \(H_{2}\)(g)

Overall reaction:

\(2HNO_{3}\) (aq) +\(2H^{+}\)(aq) + 4 e- →  \(O_{2}\)(g) + \(2NO_{2}\)(g) + \(2H_{2}O\)(l)

This means that for every 2 moles of nitric acid, 1 mole of \(O_{2}\) gas is produced. The products of the electrolysis are  \(O_{2}\) gas, \(NO_{2}\) gas, and \(H_{2}O\).

The concentration of the nitric acid (4 mol/L) indicates that there are 4 moles of \(HNO_{3}\) in 1 liter of solution. To calculate the number of moles of  \(HNO_{3}\) in a given volume of solution, we can use the following formula:

moles of solute = concentration × volume (in liters)

For example, if we have 500 mL (0.5 L) of the 4 mol/L nitric acid solution, the number of moles of \(HNO_{3}\) present would be:

moles of \(HNO_{3}\) = 4 mol/L × 0.5 L = 2 moles

Therefore, 2 moles of \(O_{2}\) gas and 1 mole of \(NO_{2}\) gas would be produced from 2 moles of \(HNO_{3}\) during electrolysis.

It's worth noting that the oxidation of nitric acid to form nitrogen dioxide is an exothermic reaction that can produce heat, so the electrolysis may need to be performed under controlled conditions to prevent overheating. Additionally, nitrogen dioxide is a toxic gas that should be handled with care in a well-ventilated area.

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Complete question is: An aqueous solution of 4mol/L nitric acid is electrolysed in an electronic cell using graphite electrodes produced 2 moles of \(O_{2}\) gas and 1 mole of \(NO_{2}\) gas from 2 moles of \(HNO_{3}\) during electrolysis.

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Remember that ions contain charges, those charges could be positive or negative. So, the process of becoming atoms and molecules a charged would be ionizing radiation, because the concept of ionizing radiation is radiation with enough energy to remove tightly bound electrons from the orbit of an atom, causing that atom to become charged or ionized.

We have another type of radiation that can cause this too and it is particulate radiation because particulate radiation consists of ionizing radiation which contains alpha, beta, neutrons, and positrons particles and those particles have charges.

Explanation:

It is known that the chemical formula of aluminum sulfite is given by:

The sulfite ion = SO3-2, so it has 3 x 1 SO3-2 = 3 sulfite ions

Answer: 3