what is the penalty for exceeding nitrogen spec in waste water used for sckools and cemetaries in new mecico

Answer:

true (in some labs you aren't allowed to have any drinks so i dont know if that applies to your lab or not )

Explanation:

The value of ΔG at 298 K for a reaction mixture containing 1.9 atm N2, 1.6 atm H2, and 0.65 atm, the answer is (a) -3.86 × 10^3.

NH3 can be calculated using the equation:

ΔG = ΔG° + RT ln(Q)

where ΔG is the standard Gibbs free energy change, ΔG° is the standard Gibbs free energy change at standard conditions, R is the gas constant, T is the temperature in Kelvin, and Q is the reaction quotient.

In this case, we are given ΔG° as -33.3 kJ/mol. To calculate Q, we need to use the partial pressures of the gases in the reaction mixture. The reaction stoichiometry tells us that the ratio of the partial pressures of N2, H2, and NH3 is 1:3:2. Therefore, we can write:

Q = (P(NH3))^2 / (P(N2) * P(H2)^3)

Plugging in the given values of P(N2) = 1.9 atm, P(H2) = 1.6 atm, and P(NH3) = 0.65 atm, we can calculate Q. Then, using the value of R = 8.314 J/(mol·K) and the temperature T = 298 K, we can substitute these values into the equation and solve for ΔG.

The calculated value of ΔG at 298 K for the given reaction mixture is approximately -3.86 × 10^3 J/mol. This value is equivalent to -3.86 kJ/mol. Therefore, the answer is (a) -3.86 × 10^3.

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At the end of the 3-hour time period, the concentrations of Mn2+ and Au3+ in the two compartments will depend on various factors such as the initial concentrations, the rate of diffusion, and the reaction kinetics. Without knowing the specific conditions of the experiment, it is difficult to provide an exact answer.

However, it is important to note that both Mn2+ and Au3+ ions have different chemical properties and may react differently with the substances present in the compartments. It is also possible that the concentrations of the ions may reach equilibrium over time. To accurately determine the concentrations of Mn2+ and Au3+ in the compartments, further experiments and measurements would be required. Additionally, factors such as temperature, pressure, and the presence of other ions may also affect the final concentrations.

Therefore, it is important to carefully design experiments and consider all relevant factors when studying the behavior of chemical species in different compartments.

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Assuming the volume change upon dissolving the sodium acetate is negligible, the pH of the solution is 3.41.

To calculate the pH of the solution, we need to use the acid dissociation constant (Ka) of acetic acid, which is given as 1.75 x 10^-5.

First, we need to determine the concentration of acetic acid in the solution. We know that the volume of acetic acid is 67.5 mL, or 0.0675 L, and its concentration is 0.15 M. So, the amount of acetic acid in the solution is:

⇒ \(0.0675*0.15=0.010125\) mol

Next, we need to calculate the moles of sodium acetate that was dissolved in the solution. The molar mass of sodium acetate is 82.03 g/mol, so the number of moles of sodium acetate is:

⇒ \(\frac{1.60}{82.03} =0.0195\) mol

Since sodium acetate is a salt, it will dissociate in water to form sodium ions and acetate ions. The acetate ions will react with the acetic acid to form more water and acetic acid will dissociate to form hydrogen ions and acetate ions. Therefore, the concentration of acetate ions in the solution is:

⇒ \(\frac{0.0195}{0.0675} = 0.289M\)

Now we can use the equilibrium expression for acetic acid to calculate the concentration of hydrogen ions:

\($\mathrm{K_a} = \frac{[\mathrm{H^+}][\mathrm{CH_3COO^-}]}{[\mathrm{CH_3COOH}]}$\)

Assuming that the concentration of acetic acid has decreased negligibly due to the dissociation, we can substitute the concentrations of acetate and acetic acid into the expression and solve for [H+]:

⇒ \(1.75*10^{-5}=\frac{{(H^+}*0.289)}{0.010125}\)

⇒ [H+] = 0.000387 M

Finally, we can use the definition of pH to calculate the pH of the solution:

⇒ \($\mathrm{pH} = -\log[\mathrm{H^+}]$\)

⇒ \($\mathrm{pH} = -\log[\mathrm{0.000387}]$\)

⇒ \($\mathrm{pH}=3.41\)

Therefore, the pH of the solution is 3.41.

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A metric prefix is a unit prefix used to denote a multiple or submultiple of a basic unit of measure. Today's metric prefixes are all decadic. Each prefix has its own distinctive sign, which is added before any unit symbol.

Since humans have been measuring things for thousands of years, the units we use to represent those measurements have changed. Numerous units are now available to describe physical amounts.

Longitude, for instance, can be expressed in terms of the foot, metre, fathom, chain, parsec, league, and so on.

In the SI, multiples and subdivisions of any unit can be designated by combining the prefixes deka, hecto, and kilo, which represent 10, 100, and 1000 respectively, and deci, centi, and milli, which represent correspondingly, with the name of the unit.

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The molar mass of Na₂SO₄, or sodium sulfate, is 142.04 g/mol that can be calculated by adding the atomic masses of each element in the compound.

Na₂SO₄ contains two sodium atoms, one sulfur atom, and four oxygen atoms. The atomic mass of sodium is 22.99 g/mol, sulfur is 32.06 g/mol, and oxygen is 16.00 g/mol. Therefore, the molar mass of Na₂SO₄ can be calculated as follows:

(2 x 22.99 g/mol) + (1 x 32.06 g/mol) + (4 x 16.00 g/mol) = 142.04 g/mol

So the molar mass of Na₂SO₄ is 142.04 g/mol.

The molar mass is an important concept in chemistry because it allows us to convert between the mass of a substance and the number of moles of that substance. One mole of a substance contains Avogadro's number of particles, which is approximately 6.02 x 10²³. Therefore, the molar mass of a substance in grams is equal to the mass of one mole of that substance.

In the case of Na₂SO₄ the molar mass of 142.04 g/mol means that one mole of Na₂SO₄ weighs 142.04 grams. This can be useful in chemical reactions where the reactants and products are measured in moles. For example, if we wanted to react one mole of Na₂SO₄ with another substance, we would need 142.04 grams of Na₂SO₄.

Therefore, the molar mass of Na₂SO₄ is 142.04 g/mol, which is calculated by adding the atomic masses of the elements in the compound. The concept of molar mass is important in chemistry because it allows us to convert between the mass of a substance and the number of moles of that substance.

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Answer:

It's (A)

Because: No, heat can only travel through solids by conduction.

Answer:

Heat ALWAYS flows from warmer objects to cooler objects.

Explanation:

Heat always transfers from warmer objects to cooler objects. For example, if you put a piece of ice in your drink, heat from your drink transfers into the ice, thus, cooling your drink and melting the ice.

Answer:

orange juice because it a type of acid.

The following has the potential to change the equilibrium in a reaction involving gases: decreasing the volume of the system to half of its original value, increasing the volume of the system to twice its original value, doubling the amount of one of the components in the system, and doubling the amount of all of the components in the system.

In chemical reactions involving gases, the concentrations of the reactants and products determine the position of equilibrium. The position of equilibrium, in turn, is affected by any changes made to the concentrations of these reactants and products, such as by changing the volume of the system or altering the amount of one or more of the components in the system.

A change in the volume of a gas system can alter the position of equilibrium in a reaction. For example, decreasing the volume of the system to half of its original value can increase the concentration of the gases and shift the position of equilibrium to the side with fewer moles of gas. Conversely, increasing the volume of the system to twice its original value can decrease the concentration of the gases and shift the position of equilibrium to the side with more moles of gas.

Doubling the amount of one of the components in the system can also change the position of equilibrium in a reaction involving gases. Increasing the amount of one of the reactants will result in the formation of more products, which will shift the position of equilibrium to the side with fewer moles of gas. Similarly, doubling the amount of all of the components in the system will result in the formation of more products, which will shift the position of equilibrium to the side with fewer moles of gas.

In summary, the position of equilibrium in a reaction involving gases can be altered by changes in volume or by changes in the amounts of the components in the system.

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Answer:

they are seed producing so can populate more areas

Answer:

7th of March and have been trying the other half way round to work out how the histor y worked

Explanation:

cool samsung just a

The change in the cell voltage when the ion concentrations in the anode half-cell are increased by a factor of 10 is decreased by 0.030 V.

The electrochemical cell has both species the Fe²⁺ and Ag⁺ at 1 M concentration . the cell potential is given the standardbred cell potential E°.

at anode if the concentration increased by 10 then:

ΔE = -(R T / nF ) ln Q

now the reactions are :

at anode : Fe  --->   Fe²⁺  + 2e

at cathode : 2 ( Ag⁺ + e --->  Ag)

overall : Fe + 2Ag⁺  ---->   Fe²⁺  +  2Ag

Q =  [Fe²⁺ ] / [Ag⁺]²

now, n = 2 , the change in cell potential is given as:

ΔE = ( -RT / nF ) ln [Fe²⁺ ] / [Ag⁺]²

[Fe²⁺ ]  = 10 M

ΔE = -(8.314 × 298 ) / 2 × 96485 ln [10] / [1]

ΔE = - 0.030 V

Thus, the cell potential at anode half cell will decreased by 0.030 V.

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Answer: The empirical formula of the compound becomes \(TiCl_5\)

Explanation:

The empirical formula is the chemical formula of the simplest ratio of the number of atoms of each element present in a compound.

Given values:

Mass of Ti = 0.72 g

Mass of Cl = 2.85 g

The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:

\(\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}\) ......(1)

To formulate the empirical formula, we need to follow some steps:

Molar mass of Ti = 47.9 g/mol

Molar mass of Cl = 35.6 g/mol

Putting values in equation 1, we get:

\(\text{Moles of Ti}=\frac{0.72g}{47.9g/mol}=0.015 mol\)

\(\text{Moles of Cl}=\frac{2.85g}{35.6g/mol}=0.080 mol\)

Calculating the mole fraction of each element by dividing the calculated moles by the least calculated number of moles that is 0.015 moles

\(\text{Mole fraction of Ti}=\frac{0.015}{0.015}=1\)

\(\text{Mole fraction of Cl}=\frac{0.080}{0.015}=5.33\approx 5\)

The ratio of Ti : Cl = 1 : 5

Hence, the empirical formula of the compound becomes \(Ti_1Cl_5=TiCl_5\)

Answer:

3,1,1 would work

Explanation:

The diagrammatic expression for the solutions in the two arms of the U-tube separated by a membrane that is permeable to water and glucose but not to sucrose is shown in the image attached below.

When the system illustrated above reaches equilibrium, the sugar concentrations on both sides of the U-tube will be 1.5 M sucrose, 1.5 M glucose

From the image attached, if the system approaches equilibrium;

The movement of the solute(i.e. one molecule of sucrose) have the potential to proceed and move to the other arm of the U-shaped tube but with that, the total overall concentration cannot change in the whole U-tube structure due to the fact that the volume of the solvent(water) is constant.

Similarly, both arms of the U-tube shape have the same concentration as well as the same net Kinetic energy of solutes giving them the isotonic condition property.

Now, at equilibrium, the concentration of the sugar on both sides is:

= \((\dfrac{1M +2M}{2})\)

\(\mathsf{=\dfrac{3M}{2}}\)

= 1.5 M sucrose, 1.5 M glucose

Therefore, we can conclude that the concentration of the sugar on both sides at equilibrium is 1.5 M sucrose, 1.5 M glucose.

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Based on the scenario provided, it is possible that the student did not add enough 6M HNO₃ to the test tube containing 15 M NH₃ ,the unknown metal cation, and chloride ions.

The lack of a precipitate after adding ~20 drops of acid could be due to the incomplete neutralization of NH₃ or insufficient interaction between HNO₃ and the metal cation to form a precipitate.

The student may need to add more HNO₃ until the precipitate appears, ensuring proper neutralization and formation of the expected product.

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Answer:

10.1 M

Explanation:

Applying,

pH = -log(H⁺).................... Equation 1

But,

[H⁺][OH⁻] = 1×10⁻¹⁴................ Equation 2

Where [H⁺] = Hydrogen ion concentration, [OH⁻] = Hydroxyl ion concentration.

From the question,

Given: [OH⁻] = 1.25×10⁻⁴ M

Substitute into equation 2

[H⁺][1.25×10⁻⁴] = 1×10⁻¹⁴

[H⁺] = 1×10⁻¹⁴/1.25×10⁻⁴

[H⁺] = 0.8×10⁻¹⁰ M

[H⁺] = 8×10⁻¹¹ M

Also, Substitute the value of [H⁺] into equation 1

pH = -log[8×10⁻¹¹]

pH = 10.1 M

Answer:

1.72 M

Explanation:

Molarity is the molar concentration of a solution. It can be calculated using the formula a follows:

Molarity = number of moles (n? ÷ volume (V)

According to the information provided in this question, the solution has 58.7 grams of MgCl2 in 359 ml of solution.

Using mole = mass/molar mass

Molar mass of MgCl2 = 24 + 35.5(2)

= 24 + 71

= 95g/mol

mole = 58.7g ÷ 95g/mol

mole = 0.618mol

Volume of solution = 359ml = 359/1000 = 0.359L

Molarity = 0.618mol ÷ 0.359L

Molarity = 1.72 M

Answer:

A) 2SO2 + O2 = 2SO3

B) Al2O3 + O2 = Al2O3

C) Ca + O2 = CaO2

D) Na (s) + Cl2 (g) = NaCl (s)

E) K + O2 = K2O2

brainliest please

Gold is found in various ionic compounds, but one of the most well-known and commercially significant compounds is gold chloride, also known as auric chloride or gold(III) chloride.

The chemical formula for gold chloride is AuCl₃. Gold chloride is an ionic compound composed of gold cations (Au³⁺) and chloride anions (Cl-). It is a yellow-orange solid that is highly soluble in water. Gold chloride can be formed by reacting the gold metal with chlorine gas or by dissolving the gold metal in aqua regia, which is a mixture of concentrated nitric acid and hydrochloric acid.

Gold chloride has several uses and applications. It is commonly used in the field of nanotechnology for the synthesis of gold nanoparticles. These nanoparticles have unique optical, electronic, and catalytic properties, making them valuable in various fields such as medicine, electronics, and materials science.

In addition to gold chloride, gold can also form other ionic compounds with different anions, such as gold bromide (AuBr), gold iodide (AuI), gold sulfide (Au2S), and gold cyanide (AuCN). These compounds have their own unique properties and applications.

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Answer:

mass number = protons + neutrons.

Explanation:

Together, the number of protons and the number of neutrons determine an element's mass number: mass number = protons + neutrons. If you want to calculate how many neutrons an atom has, you can simply subtract the number of protons, or atomic number, from the mass number.

To determine the percent yield, we need to first calculate the theoretical yield of the reaction using stoichiometry, and then divide the actual yield by the theoretical yield and multiply by 100%. The percent yield of the reaction is approximately 84.9%.

Percent yield is a measure of the efficiency of a chemical reaction, calculated by dividing the actual yield of a reaction by the theoretical yield and multiplying by 100%. It represents the percentage of the theoretical amount of product that was actually obtained in a reaction.

The balanced chemical equation is:

2Al + Cr₂O₃ → Al₂O₃ + 2Cr

The molar mass of Cr₂O₃ is 152 g/mol, the molar mass of Al is 27 g/mol, and the molar mass of Cr is 52 g/mol.

We need to determine which reactant is limiting, so we can calculate the theoretical yield based on the amount of limiting reactant. We can do this by calculating the number of moles of each reactant using their molar masses and dividing by their stoichiometric coefficients in the balanced equation:

moles of Cr₂O₃= 44.1 g / 152 g/mol = 0.29 mol

moles of Al = 35.0 g / 27 g/mol = 1.30 mol

From the balanced equation, we see that 1 mole of Cr2O3 reacts with 2 moles of Cr. Therefore, the theoretical yield of Cr is:

moles of Cr produced = 0.29 mol Cr₂O₃x (2 mol Cr / 1 mol Cr₂O₃) = 0.58 mol Cr

mass of Cr produced = 0.58 mol Cr x 52 g/mol = 30.16 g Cr

The percent yield is:

% yield = (actual yield / theoretical yield) x 100%

% yield = (25.6 g Cr / 30.16 g Cr) x 100% = 84.9%

Therefore, the percent yield of the reaction is approximately 84.9%.

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Answer: wait im gonna search some info.

Explanation:I come back with an asnwer for you

Answer: B disposed of according to your instructor's

directions. :)

Answer:

B. disposed of according to your instructor's  directions.

Explanation:

After completing an experiment all chemical waste should be disposed of according to your instructor's directions.

Chemical wastes should be carefully handled in the laboratory so as not to carelessly release them into the environment. If not properly treated and they are discharged into the environment, they can lead to serious hazards and even pollution. Contamination is another very serious problem attributed to disposing chemical waste carelessly.

Instructor's guides should be followed to dispose chemical wastes.

Answer:

Power

Explanation:

#platofam

Answer:

Final pressure = 362.7 Pa

Explanation:

Given that,

Initial volume, V₁ = 930 ml

Initial pressure P₁ = 156 Pa

Final volume, V₂ = 400 mL

We need to find the final pressure. We know that the relation between volume and pressure is inverse i.e.

\(V\propto \dfrac{1}{P}\\\\\dfrac{V_1}{V_2}=\dfrac{P_2}{P_1}\\\\P_2=\dfrac{V_1P_1}{V_2}\\\\P_2=\dfrac{930\times 156}{400}\\\\P_2=362.7\ Pa\)

So, the final pressure is equal to 362.7 Pa.

The pH of the buffer is 4.60.

To calculate the pH of a buffer, we can use the Henderson-Hasselbalch equation:

\(pH = pKa + log([A-]/[HA])\)

where pKa is the dissociation constant of the weak acid, \([A-]\) is the concentration of the conjugate base, and \([HA]\) is the concentration of the weak acid.

In this case, the weak acid is acetic acid\((HC2H3O2)\), the conjugate base is acetate \((C2H3O2-)\), and the dissociation constant (Ka) is \(1.8 × 10^-5\).

First, we need to calculate the ratio of \([A-]/[HA]\):

\([A-]/[HA] = (0.162 M)/(0.225 M) = 0.72\)

Next, we can substitute the values into the Henderson-Hasselbalch equation:

\(pH = pKa + log([A-]/[HA])\\pH = -log(1.8 × 10^-5) + log(0.72)\)

pH = 4.74 + (-0.14)

pH = 4.60

Therefore, the pH of the buffer is 4.60.

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Answer:

Explanation:

Decomposition reaction, also known as analysis or dissociation, is a type of chemical reaction in which a compound breaks down into simpler substances or elements. In this reaction, a single reactant undergoes a chemical change and produces two or more products.

The decomposition reaction can be represented by the general equation:

AB → A + B

Where AB is the reactant, and A and B are the products. The reactant AB is usually a compound, and it breaks down into its constituent elements or simpler compounds.

There are different types of decomposition reactions, including:

Thermal decomposition: It occurs when a compound is heated, resulting in its decomposition into simpler substances. For example, the thermal decomposition of calcium carbonate (CaCO3) produces calcium oxide (CaO) and carbon dioxide (CO2):

CaCO3 → CaO + CO2

Electrolytic decomposition: It takes place when an electric current is passed through an electrolyte, causing it to break down into its component ions. For instance, the electrolysis of water (H2O) leads to the decomposition into hydrogen gas (H2) and oxygen gas (O2):

2H2O → 2H2 + O2

Photochemical decomposition: It occurs when a compound undergoes decomposition due to exposure to light energy. Chlorine gas (Cl2) can decompose into chlorine atoms (Cl) under the influence of light:

Cl2 → 2Cl

These are just a few examples of decomposition reactions. They are important in various chemical processes and are used in industries, laboratory experiments, and natural phenomena. By understanding and controlling decomposition reactions, scientists can gain insights into the behavior of different compounds and develop practical applications in fields such as chemistry, materials science, and environmental science.

Answer:

Explanation:

reaction in which a compound breaks down into simpler substances or elements