awastecontains 100 mg/l ethylene glycol(c2h6o2) and50 mg/l nh3–n. determine the theo-retical carbonaceous and the theoretical nitrogenousoxygen demand of the waste

For standard 1, the volume of stock solution required is 2.00 mL, while for standard 2, it is 4.00 mL.

In order to calculate the volume of 0.0315 m Bromocresol green (HBCG) standard stock solution required to make 10.00 ml of the three standards, we need to use the formula:

M1V1 = M2V2

Where M1 is the concentration of the stock solution,

V1 is the volume of the stock solution required,

M2 is the concentration of the final solution, and

V2 is the final volume of the solution.

To calculate the volume of 0.0315 m Bromocresol green (HBCG) stock solution required to make 10.00 ml of 0.00630 m Bromocresol green (HBCG) standard 1, we can plug in the values into the formula as:

M1V1 = M2V2V1 = (M2V2)/M1= (0.00630 mol/L x 0.01000 L)/0.0315 mol/L= 0.00200 L = 2.00 mL

Therefore, the volume of 0.0315 m Bromocresol green (HBCG) stock solution required to make 10.00 ml of 0.00630 m Bromocresol green (HBCG) standard 1 is 2.00 mL.

To calculate the volume of 0.0315 m Bromocresol green (HBCG) stock solution required to make 10.00 ml of 0.0126 m Bromocresol green (HBCG) standard 2, we can use the same formula as above:

M1V1 = M2V2V1 = (M2V2)/M1= (0.0126 mol/L x 0.01000 L)/0.0315 mol/L= 0.00400 L = 4.00 mL

Therefore, the volume of 0.0315 m Bromocresol green (HBCG) stock solution required to make 10.00 ml of 0.0126 m Bromocresol green (HBCG) standard 2 is 4.00 mL.

In conclusion, we can use the formula M1V1 = M2V2 to calculate the volume of 0.0315 m Bromocresol green (HBCG) stock solution required to make different standards.

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Answer:

1.204x10²⁴ molecules of water

Explanation:

The ethane dioic acid is the same oxalic acid. The molecular formula of oxalic acid is:

C2H2O4

Usually, the oxalic acid is dihydrated, that means contains 2 moles of water per mole of acid.

Using Avogadro's number, 1 mole of ethane dioic acid contains 6.022x10²³ molecules of ethane dioic acid. As we have 2 molecules of water per molecule of ethane dioic acid, the molecules of water are.

6.022x10²³ molecules ethane dioic acid * (2molecules Water / molecule Ethane Dioic acid) =

C: carbon is getting oxidised in the reaction. Oxidation and reduction both takes place in a redox reaction.

In k2Cr2O7 , the K is in + 1 oxidation state , Cr is in + 6 oxidation state, and O is in -2 oxidation state

On the right hand side Cr is in +3 oxidation state that means Cr is getting reduced in the reaction.

In C3H8O2 , H and O are ih their general oxidation states +1 and -2 respectively. Here alcoholic group is getting oxidised to carboxylic group. So C is getting oxidised.

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Answer:

C3h8o2

Explanation: oxidation and reduction simultaneously occurs in a redox reaction. Atoms in a molecule during oxidation and reduction changes the oxidation state, in case of oxidation, it increases while in the case of reduction it decreases.

In the given reaction K,Cr, and O have oxidation number +1,+6 and -2 in K2Cr2O7. But on the right side, the oxidation state of Cr changes to +3 and gets reduced in the reaction. Similarly, C3H8O2 undergoes oxidation to C3H8O4,where the alcoholic group get oxidized to carboxylic acid.

A. The mole of water formed is 0.137 mole

B. The moles of butane burned is 0.027 mole

C. The mass of butane burned is 1.566 g

D. The moles of oxygen used up is 0.178 mole

E. The mass (in grams) of oxygen used up is 5.696 g

We can obtain the mole of water formed as shown below:

Mole = mass / molar mass

Mole of water = 2.46 / 18

Mole of water = 0.137 mole

The mole of butane burned can be obtained as follow:

2C₄H₁₀ + 13O₂ -> 8CO₂ + 10H₂O

From the balanced equation above,

10 moles of H₂O were obtained from 2 moles of butane, C₄H₁₀

Therefore,

0.137 mole of H₂O will obtain from = (0.137 × 2) / 10 = 0.027 mole of butane, C₄H₁₀

Thus, the mole of butane, C₄H₁₀ is 0.027 mole

The mass of butane burned can be obtained as illustarted below:

Mole = mass / molar mass

Cross multiply

Mass = Mole × molar mass

Mass of buatne = 0.027 × 58

Mass of butane = 1.566 g

The mole of oxygen used up can be obtained as follow:

2C₄H₁₀ + 13O₂ -> 8CO₂ + 10H₂O

From the balanced equation above,

10 moles of H₂O were obtained from 13 moles of oxygen, O₂

Therefore,

0.137 mole of H₂O will obtain from = (0.137 × 13) / 10 = 0.178 mole of oxygen, O₂

Thus, the mole of oxygen, O₂ is 0.178 mole

The mass of oxygen used up can be obtained as illustarted below:

Mole = mass / molar mass

Cross multiply

Mass = Mole × molar mass

Mass of oxygen = 0.178 × 32

Mass of oxygen = 5.696 g

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Answer:

Stronger

And its science

Explanation:

Answer:

stronger for top.

weaker for bottom.

passed all sciences in freshman year :D

In order to calculate the volume of a sample of hydrogen at standard temperature and pressure (STP), we must first consider the ideal gas law equation.

An ideal gas is a hypothetical gas composed of molecules that follow a set of postulates which greatly simplifies the equations of state of gases that describe the behavior of a real gas. It is assumed that the molecules are point-like, have no volume, and have no interactions except for perfectly elastic collisions. The ideal gas postulates are as follows: (1) the molecules are small compared to the average distance between them; (2) the molecules do not interact except for perfectly elastic collisions; (3) the molecules move in random, straight line trajectories; and (4) the pressure exerted by the gas is independent of the amount of gas present. This allows us to calculate properties such as the pressure, volume, temperature, and molar mass of a hypothetical ideal gas.

This equation is PV = nRT, where P is the pressure, V is the volume, n is the number of moles of gas, R is the ideal gas constant, and T is the temperature. Since the temperature and pressure are changing, we must use the combined gas law equation. The equation is PV1/T1 = PV2/T2, where P1 and T1 are the initial pressure and temperature, and P2 and T2 are the final pressure and temperature.

We can plug in the values given in the question. P1 = 621 mm Hg, V1 = 798 mL, T1 = -105 ºC, and P2 = 760 mm Hg and T2 = 0 ºC. When the equation is solved, we find V2 = 836.1 mL. This is the volume of the sample of hydrogen at standard temperature and pressure.

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B) I. Point I will experience the largest electric field because it is closest to the charged wires. The electric field magnitude increases with increasing proximity to the conductors.

The surface known as the equipotential surface is the location of all points at the same potential. A charge can be moved effortlessly from one place on the equipotential surface to another. To put it another way, an equipotential surface is one that has the same electric potential at each point throughout.

Using web technologies, desktop GUI apps can be created using the open-source framework Electron. The Chromium rendering engine and the Node. js runtime are used in its development and maintenance, processing and rendering HTML and CSS.

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An increase in pressure would not significantly affect the [H2] in the given reactions.

In the reactions provided, the concentration of hydrogen gas ([H2]) is not directly affected by changes in pressure. This is because [H2] is not a reactant or product whose concentration is influenced by changes in pressure, according to the balanced chemical equations.

In the first reaction, the combustion of hydrogen gas (2H2(g) + O2(g) → 2H2O(g)), increasing the pressure would not alter the concentration of hydrogen gas. The stoichiometric coefficients of hydrogen gas remain unchanged.

Similarly, in the second reaction (4HCl(g) + Fe(s) → 2H2(g) + FeCl3(s)), altering the pressure would not affect the concentration of hydrogen gas. The stoichiometric coefficients for hydrogen gas again remain constant.

Lastly, in the third reaction (H2(g) + Cl2(g) → 2HCl(g)), increasing the pressure would not directly modify the concentration of hydrogen gas. The balanced equation already accounts for the appropriate stoichiometric coefficients.

It's important to note that while an increase in pressure may impact other aspects of these reactions (such as the equilibrium position or reaction rates), the concentration of hydrogen gas ([H2]) would remain unaffected.

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Sodium (Na) and chlorine (Cl) are the largest dissolved components in ocean water due to the abundance of sodium and chloride ions in the Earth's crust and the continuous input of these elements into the oceans through various processes. Sodium is one of the most common elements in the Earth's crust, and chlorine is widely distributed in rocks, minerals, and salts.

Over millions of years, weathering of rocks, volcanic activity, and erosion release these elements into rivers and ultimately into the oceans. The combination of sodium and chlorine ions results in the formation of sodium chloride, which is commonly known as table salt and contributes to the salinity of seawater.

The most abundant dissolved gas in ocean water is carbon dioxide (CO2). Carbon dioxide dissolves in the surface waters of the ocean through gas exchange with the atmosphere. It plays a crucial role in regulating the pH of seawater and is an essential component of the carbon cycle. Carbon dioxide is involved in various biological and chemical processes in the ocean, including photosynthesis by marine plants and the formation of calcium carbonate shells by marine organisms. Additionally, the increase in atmospheric carbon dioxide due to human activities has led to ocean acidification, which is a significant concern for marine ecosystems.

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Due to the fact that they combine to form the ionic compound sodium chloride (NaCl), also known as salt, sodium (Na) and chlorine (Cl) are the two most abundant dissolved elements in ocean water.

Thus, Salts are among the many dissolved compounds that water from rivers and streams transports into the ocean.

In particular, sodium and chloride ions have accumulated in the ocean throughout time, leading to the high concentration of these elements in seawater. Magnesium, calcium, potassium, and sulphate ions are among the other dissolved substances in ocean water.

Oxygen  is the dissolved gas that is most prevalent in ocean water.

Thus, Due to the fact that they combine to form the ionic compound sodium chloride (NaCl), also known as salt, sodium (Na) and chlorine (Cl) are the two most abundant dissolved elements in ocean water.

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The cation that is isoelectronic with neon is the one that has the same number of electrons as neon. Since neon has 10 electrons, we need to find a cation that also has 10 electrons. An ion is isoelectronic with neon if it has the same number of electrons as neon, even though it has a different number of protons.

The electron configuration of neon is 1s2 2s2 2p6. Among the given options, the one that is isoelectronic with neon is "all of the above" - sodium ion (has lost one electron from its outer shell, so its electron configuration is 1s2 2s2 2p6), magnesium ion (has lost two electrons from its outer shell, so its electron configuration is also 1s2 2s2 2p6), and aluminum ion (has lost three electrons from its outer shell, so its electron configuration is 1s2 2s2 2p6). Therefore, all of these ions have the same number of electrons as neon and are isoelectronic with it.

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The standard enthalpy change for reaction (3) is 117.1 kJ.

The standard enthalpy change for reaction (3) can be calculated by using the enthalpy changes of reactions (1) and (2) and applying Hess's Law.

To do this, we need to manipulate the given equations so that the desired reaction (3) can be obtained.

First, we reverse reaction (1) to get the formation of C2H4(g) from C2H6(g):

C2H4(g)C2H6(g) ΔH° = -52.3 kJ

Next, we multiply reaction (2) by 2 and reverse it to obtain 2 moles of C2H6(g) reacting to form 3 moles of H2(g):

2C2H6(g)2C(s) + 3H2(g) ΔH° = 169.4 kJ

Now, we add the two modified equations together:

C2H4(g)C2H6(g) ΔH° = -52.3 kJ
2C2H6(g)2C(s) + 3H2(g) ΔH° = 169.4 kJ

When adding these equations, the C2H6(g) on the left side cancels out with the C2H6(g) on the right side, leaving us with the desired reaction (3):

C2H4(g) + H2(g)C2H6(g) ΔH° = -52.3 kJ + 169.4 kJ = 117.1 kJ

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Answer:

D.

Hope it helps!!!!!!!!!

a. Under equilibrium conditions, the ratio of Pyruvate to PEP is \(3.78 * 10^6.\)

b. Under standard conditions for equilibrium, the ratio of ATP to ADP is 441.

c. This equation shows that PEP can drive the synthesis of ATP from ADP and Pi. REACTION 1 is in the forward direction, while REACTION 2 is in the reverse direction.

d. The overall net coupled reaction is exergonic or spontaneous because the ΔG' is negative (-92.4 kJ/mol). This means that the reaction releases energy and can proceed spontaneously without the addition of energy.

e. At equilibrium, the concentrations of PEP, ADP, Pi, H+, Pyruvate, and ATP will be equal.

a. The equilibrium constant (Keq) for REACTION 1 can be calculated using the equation:

ΔG° = -RT ln(Keq)

where R is the gas constant (8.314 J mol^-1 K^-1), T is the temperature in Kelvin (assumed to be 298 K), and ΔG° is the standard free energy change. Rearranging the equation to solve for Keq gives:

Keq = e^(-ΔG°/RT)

Substituting the given values, we get:

Keq =\(e^{(-(-61.9 kJ mol^{-1})/(8.314 J mol^{-1} K^{-1} * 298 K))}\) = \(2.10 * 10^8\)

The equilibrium constant expression for the hydrolysis of PEP can be written as:

Keq = [Pyruvate][Pi][\(H^+\)] / [PEP][\(H_2O\)]

At equilibrium, the ratio of products to reactants is equal to Keq. Therefore, the ratio of Pyruvate to PEP is:

[Pyruvate] / [PEP] = Keq / ([Pi][\(H^+\)]/[\(H_2O\)])

Substituting the given values, we get:

[Pyruvate] / [PEP] = \((2.10 * 10^8) / ((1 M)(10^{-7} M) / (55.5 M))\)

[Pyruvate] / [PEP] = \(3.78 * 10^6\)

b. The equilibrium constant (Keq) for REACTION 2 can be calculated in the same way as in part (a):

Keq = e^(-ΔG°/RT) = \(e^{(-(-30.5 kJ mol^{-1})/(8.314 J mol^{-1} K^{-1} * 298 K))} = 1.26 * 10^5\)

The equilibrium constant expression for the hydrolysis of ATP can be written as:

Keq = [ADP][Pi][\(H^+\)] / [ATP][\(H_2O\)]

At equilibrium, the ratio of products to reactants is equal to Keq. Therefore, the ratio of ATP to ADP is:

[ATP] / [ADP] = [\(H_2O\)] / ([Pi][\(H^+\)]/([ATP]Keq))

Substituting the given values, we get:

[ATP] / [ADP] = \((55.5 M) / ((1 M)(10^{-7} M)/(1 M)(1.26 * 10^5))\)

[ATP] / [ADP] = 441

c. The net coupled chemical equation for the synthesis of ATP from ADP and inorganic phosphate using the hydrolysis of PEP into Pyruvate can be written as:

PEP + ADP + Pi --> Pyruvate + ATP

This equation shows that PEP can drive the synthesis of ATP from ADP and Pi. REACTION 1 is in the forward direction, while REACTION 2 is in the reverse direction.

d. To calculate the ΔG' for the overall net coupled reaction, we need to sum up the ΔG' of the individual reactions.

ΔG'net = ΔG'1 + ΔG'2

ΔG'1 = -61.9 kJ/mol

ΔG'2 = -30.5 kJ/mol

ΔG'net = -61.9 kJ/mol + (-30.5 kJ/mol)

ΔG'net = -92.4 kJ/mol

e. The overall net coupled reaction can be written as follows:

PEP + ADP + Pi + \(H^+\) → Pyruvate + ATP

The ratio of products and reactants for the overall net coupled reaction can be calculated using the equilibrium constant (Keq). The equilibrium constant is defined as the ratio of the concentrations of products to the concentrations of reactants at equilibrium.

Keq = [Pyruvate][ATP]/[PEP][ADP][Pi][\(H^+\)]

At equilibrium, Keq = 10^(ΔG'net/(-RT))

where R is the gas constant (8.314 J/molK), T is the temperature (in Kelvin), and ΔG'net is the standard free energy change of the reaction.

Assuming standard conditions of 25°C (298 K), we get:

Keq = \(10^{(-92400/(8.314*298))\)

Keq = \(2.1 * 10^{27\)

The ratio of products to reactants is given by the coefficients in the balanced equation:

PEP : ADP : Pi : H+ : Pyruvate : ATP = 1 : 1 : 1 : 1 : 1 : 1

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Answer:

Can u somehow make it bigger please?

Explanation:

Fe(ii) and Fe(iii) ions differ by E) 1 electron.

Ferric is the term used for an iron atom that has lost three electrons from the valence shell, on the other hand, ferrous is the term used for an iron atom that has lost two electrons from the valence shell.

As one shares two electrons while the other shares three electrons therefore they differ by one electron.

Fe(iii) is more stable due to its half-filled subshell while Fe(ii) is unstable in nature. There is a partially-filled d orbital in Fe(ii) resulting in its instability as half-filled and completely filled orbitals are more stable as compared to the partially filled orbitals.

Although a part of your question is missing, you might be referring to this question:

Iron forms compounds as both a ferrous Fe(II) and a ferric Fe(III) ion. These two ions differ by:

A) 1 proton

B) 1 neutron

C) 2 protons

D) 2 electrons

E) 1 electron

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Answer:

 a supply of electric charges which are free to flow, some form of push to move the charges through the circuit and a pathway to carry the charges.

Explanation:

There's your answer have a good day

Red = Green (complementary color). Blue = Orange (complementary color). Orange/Yellow = Blue (colors opposite on the colour wheel). Red = Cyan/Blue-Green (opposite hues on the colour wheel) are colours that a solution can absorb.

It relies on the wavelength of light that a substance or solution absorbs. The brightness and vibrancy of the colour will increase with increasing light intensity.

The complementary colour, or the colour across from the absorbed colour on the colour wheel, is what we see when a sample absorbs light of a specific colour. If a sample absorbs red light, for instance.

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The Bohr model of the atom is a simple model developed by Niels Bohr in 1913 to explain the structure of the hydrogen atom. It proposed that electrons in atoms exist in certain discrete energy levels or "shells" and can jump between these levels by absorbing or emitting a photon of light.

The Bohr model can be used to calculate the radius and energy of the He1 ion in the n=5, l=5 state. The radius is given by the formula:

r = \(0.529 * 10^-^1^0 m * (n^2)\)

where n is the principal quantum number. Plugging in n=5, we get:

r = \(0.529 * 10^-^1^0 m * (5^2) = 2.646 * 10^-^9 m\)

The energy of the He1 ion in the n=5, l=5 state is given by the formula:

E = -2.18 * \(10^-^1^8\) J * (\(\frac{1}{n^{2} }\))

Plugging in n=5, we get:

E = -\(2.18 * 10^-^1^8 J * (1/5^2) = -1.31 * 10^-^1^7 J\)

To remove an electron from 1 mol of He1 in this state, the energy required is:

E = n * E = 6.022 * \(10^2^3\) * -1.31 * \(10^-^1^7\) J = -7.8 * \(10^6\) J

The frequency of the light emitted can be calculated using the formula:

nu = E / h

where h is Planck's constant and nu is the frequency of the light.

nu = 2.4 * \(10^-^1^8\) J / 6.626 * \(10^-^3^4\) J.s = 3.6 * \(10^1^5\) Hz

The wavelength of the light emitted can be calculated using the formula:

lambda = c / nu

where c is the speed of light.

lambda = 3*\(10^8\) m/s / 3.6 * \(10^1^5\) Hz = 8.3 * \(10^-^8\) m or 830 nm

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The percent yield for this reaction is 50.8%.

First, we need to determine the moles of p-phenetidine used:

Moles of p-phenetidine = mass / molar mass

Moles of p-phenetidine = 6.48 g / 137.18 g/mol

Moles of p-phenetidine = 0.0472 mol

Since p-phenetidine is the limiting reagent, the moles of acetophenetidin produced should be equal to the moles of p-phenetidine used.

The theoretical yield of acetophenetidin can be calculated using the stoichiometry of the reaction:

Theoretical yield = Moles of p-phenetidine × (molar mass of acetophenetidin / molar mass of p-phenetidine)

Theoretical yield = 0.0472 mol × (179.22 g/mol / 137.18 g/mol)

Theoretical yield = 0.0616 mol

Now we can calculate the percent yield:

Percent yield = (actual yield / theoretical yield) × 100

Percent yield = (5.72 g / (0.0616 mol × 179.22 g/mol)) × 100

Percent yield = (5.72 g / 11.2512 g) × 100

Percent yield = 50.8%

Therefore, the percent yield for this reaction is 50.8%.

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NOTES TO TAKE DOWN:

Apparent brightness is a human measurement, and it would change for each star if the measurement were taken from another location. The more precise counterpart of apparent brightness is called absolute brightness (or absolute magnitude) and is the measure of the luminosity of a star, but on a common scale.

ANSWER:

Absolute brightness is the actual amount of light produced by the star, whereas apparent brightness changes with distance from the observer.

The complete balanced equation for the neutralization reaction between HCl and NaHCO₃ is:

HCl + NaHCO₃ → NaCl + H₂O + CO₂

The reaction between hydrochloric acid (HCl) and sodium bicarbonate (NaHCO₃) is known as a neutralization reaction. In this reaction, HCl and NaHCO₃ combine to produce NaCl, water, and carbon dioxide.

The reaction can be represented by the following equation: HCl + NaHCO₃ → NaCl + H₂O + CO₂

This reaction already the balanced chemical equation for the reaction since the number of each element in the reactant side is equal to the number of each element in the product side.

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Solvolysis reaction is a type of substitution reaction in which a nucleophile (solvent) replaces a leaving group from the carbon atom of an organic molecule.

Here are the expected products for each of the given solvolysis reactions:

1. Br EtOH heat: The reaction will lead to the formation of ethyl bromide and HBr.

The reaction follows the following mechanism: 2. Cl MeOH heat: The reaction will lead to the formation of methyl chloride and HCl.

The reaction follows the following mechanism:3. Br Cl MeOH heat: The reaction will lead to the formation of methyl chloride, ethyl chloride, and HCl.

The reaction follows the following mechanism:4. 2 MeOH heat (d):The reaction will lead to the formation of methyl alcohol and ethyl alcohol. The reaction follows the following mechanism:

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Answer:

They have the same mass but different electrical charges.

Explanation:

Protons have a positive charge while neutrons have a neutral charge. Although they have different charges, they have the same amount of mass.

The protons and neutrons have the same mass but different electrical charges. THerefore, option (D) is correct.

A proton is described as a stable subatomic particle, a chemical symbol with +1e elementary charge. The mass of the proton is approximately equal to the mass of the neutron but 1836 times the mass of an electron.

Protons and neutrons are jointly together termed "nucleons" each with masses of approximately one atomic mass unit (a.m.u.). There will be one or more than one proton present in the nucleus of an atom. There is an attractive electrostatic central force that holds the atomic electrons.

The number of protons in the atom is referred to as the atomic number and is expressed by the symbol Z. Since each element in the periodic table has a unique number of protons, determines the chemical characteristics of the element.

As the neutrons present in the nucleus of an atom are neutral and do not carry any charge while the protons are positively charged.

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Answer:

The Chemical Formula for Cadmium Fluoride is CdF2. What is the formula for francium permanganate? The compound francium permanganate has the chemical formula FrMnO4.

Explanation:

There are two types of chemical compound one is covalent compound and other is ionic compound, covalent compound formed by sharing of electron and ionic compound formed by complete transfer of electron. Therefore, FrF is the formula of francium & fluoride.

Chemical Compound is a combination of molecule, Molecule forms by combination of element and element forms by combination of atoms in fixed proportion.

An ionic compound is a metal and nonmetal combined compound.  Ionic compound are very hard. They have high melting and boiling point because of strong ion bond.

The chemical compound formed using  francium & fluoride is FrF. Flourine and Francium is very reactive element.

Therefore, FrF  is the formula of francium & fluoride.

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Answer:

Atomic theory is. A

Explanation:

Subject to change if new information is discovered. B. A solution to the problem of differing isotopes. C. Unchanged able D. A descriptive table that lists all of the element

I hope this answered your question.

The sample that the student has is a metal because it has luster.

Metals are those elements that occur towards the left hand side of the periodic table. They have the following properties;

From the few properties listed above, the sample that the student has is a metal because it has luster.

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Answer:

\(m_{undissolved}=27g\)

Explanation:

Hello,

In this case, we first define the solubility as the maximum amount of a solute that is completely dissolved in an specific amount of solvent and it is temperature-dependent. Thus for potassium iodide, its solubility at 30°C is 153 g per 100 cm3 of water, therefore, with the given amount, the undissolved amount results:

\(m_{undissolved}=180g-153g=27g\)

Best regards.

Answer: So the answer would be um I dont

know the answer right now but message me later on and I will help

Explanation: The equation that relates heat  (q) to specific heat (c_p) , mass (m), and temperature change  (Delta{T}) is shown below.

q=c_p times m times Delta{T}  The heat that is either absorbed or released is measured in joules. The mass is measured in grams. The change in temperature is given by Delta{T}= T_f - T_i , where  T_f is the final temperature and  T_i is the initial temperature. Step 1: List the known quantities and plan the problem .

Known

heat = q = 134 J

mass =  m = 15.0 g

Delta{text{T}} = 62.7^circ text{C} - 24.0^circ text{C} = 38.7^circ text{C}

Unknown

c_p text{of cadmium}= ? text{J}/ text{g}^circ text{C}

The specific heat equation can be rearranged to solve for the specific heat.

Step 2: Solve .

c_p=frac{q}{m times Delta{T}}=frac{134 text{ J}}{15.0 text{ g} times 38.7^circ text{C}}=0.231 text{ J/g}^circ text{C}

Step 3: Think about your result .

The specific heat of cadmium, a metal, is fairly close to the specific heats of other metals. The result has three significant figures.

Since most specific heats are known, they can be used to determine the final temperature attained by a substance when it is either heated or cooled. Suppose that a 60.0 g sample of water at 23.52°C was cooled by the removal of 813 J of heat. The change in temperature can be calculated using the specific heat equation.

Delta{T}=frac{q}{c_p times m}=frac{813 text{ J}}{4.18 text{ J/g}^circ text{C} times 60.0 text{ g}}=3.24^circ text{C}

Since the water was being cooled, the temperature decreases. The final temperature is:

T_f=23.52^circ text{C} - 3.24^circ text{C}=20.28^circ text{C}