Line currents of 15<(30)a are supplied to a balanced 3-phase load from a 7200<(0)v source. what is the total 3 phase power from the source?

Answer:

640 grams

Explanation:

look up Solubility table in wikipedia for potassium nitrate (KNO3)

32 grams of potassium nitrate (KNO3) water solubility at 20 degrees celsius (room temperature) can be dissolved in 100 milliliters (0.1 L) of water.

2 liters = 2000 milliliters

32 grams / 100 milliliters = x grams / 2000 milliliters

cross-multiply

100 * x = 32 * 2000

x = (32 * 2000) / 100

x ≈ 640 grams

chatgpt

Answer:

Carbon dioxide and water vapor

Explanation:

Answer:

Its believe its water vapor and carbon dioxide

Kill him or PS5? Which one?

When a hydrogen atom's electron is excited to the n=4 level and then loses energy, it can emit multiple wavelengths of light. To determine the number of different wavelengths, we need to consider the possible transitions between energy levels.

In the case of the n=4 level, the electron can transition to the n=3, n=2, or n=1 levels. Each of these transitions corresponds to a different energy difference and, therefore, a different wavelength of light.

The formula to calculate the wavelength of light emitted is given by the Rydberg formula:
\(\frac{1}{λ} = R * (\frac{1}{n_f^2} -\frac{1}{n_i^2})\)
where λ is the wavelength, R is the Rydberg constant, and \(n_f\) and \(n_i\) are the final and initial energy levels, respectively.

To find the number of different wavelengths, we can substitute the values of \(n_f =\) 3, 2, and 1 into the formula. By calculating the wavelength for each transition, we can determine how many different wavelengths of light can be emitted as the excited atom loses energy. The Rydberg formula applies specifically to hydrogen atoms. The actual values of the wavelengths will depend on the specific energy levels and the corresponding Rydberg constant.

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The value of g would be 4 solving the equation 4.7g + 7 = 2.7g + 15 out to find the value of g .

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Explanation:

A.

AgNO₃ + KCl → AgCl + KNO₃ (Already Balanced)

B.

H₂O + SO₃ → H₂SO₄ (Already balanced)

C.

2KI + Cl₂ → 2KCl + I₂

D.

2NaHCO₃ → Na₂CO₃ + H₂O + CO₂

E.

Zn + 2HCl → ZnCl₂ + H₂

F.

BaCl₂ + Na₂SO₄ → BaSO₄ + 2NaCl

G.

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

H.

2Al + 3CuCl₂ → 2AlCl₃ + 3Cu

The enthalpy change for the given reaction is 9.16 kJ/mol, which is positive, indicating that the reaction is endothermic.

To find the enthalpy change (ΔH0) for the given reaction, we need to use the standard enthalpies of formation (ΔHf0) for each species involved in the reaction. We can use the following equation:

ΔH0 = ΣnΔHf0(products) - ΣmΔHf0(reactants)

where n and m are the stoichiometric coefficients of the products and reactants, respectively.

The standard enthalpies of formation for O2 (g), N2 (g), and N2O4 (g) are 0 kJ/mol, 0 kJ/mol, and 9.16 kJ/mol, respectively. Therefore, the enthalpy change for the reaction can be calculated as follows:

ΔH0 = [1 x ΔHf0(N2O4)] - [1 x ΔHf0(O2) + 2 x ΔHf0(N2)]

ΔH0 = [1 x 9.16 kJ/mol] - [1 x 0 kJ/mol + 2 x 0 kJ/mol]

ΔH0 = 9.16 kJ/mol

Therefore, the enthalpy change for the given reaction is 9.16 kJ/mol, which is positive, indicating that the reaction is endothermic (absorbs heat).

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Compounds can have different empirical formulas if the ratio of their elements is not the same. To identify which compounds do not have the same empirical formula, we need to compare the ratios of elements in each compound.

The empirical formula of a compound represents the simplest ratio of elements present in the compound. For compounds to have the same empirical formula, the ratios of their elements must be equal.

To determine which compounds do not have the same empirical formula, we need to compare the ratios of elements in each compound. This can be done by analyzing the chemical formulas of the compounds.
For example, let's consider two compounds, Compound A with the formula C2H4O and Compound B with the formula CH2O. To find their empirical formulas, we simplify the ratios of elements. In Compound A, the ratio of carbon to hydrogen to oxygen is 2:4:1, which simplifies to C1H2O0.5. In Compound B, the ratio is 1:2:1, which simplifies to CH2O.

By comparing the simplified ratios, we can see that the empirical formulas of Compound A and Compound B are different. Therefore, these compounds do not have the same empirical formula.

In conclusion, to identify compounds that do not have the same empirical formula, we need to compare the ratios of elements in each compound. If the ratios are not equal, the compounds will have different empirical formulas.

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Answer:

catalpa

Explanation:

i took the test

The Ca2+ concentration in a groundwater that is in equilibrium with calcite and has a PCO2 of 10-2.5 (atm) is 1.6 × 10-3 M.

The equilibrium expression for the dissolution of calcite (CaCO3) in water, assuming that CO2 is the only acidic gas present, is : CaCO3(s) + H2O + CO2(g) ↔ Ca2+(aq) + 2HCO3-(aq)

The equation for the relationship between the PCO2 of a gas and the concentration of dissolved CO2 in a solution in equilibrium with the gas is as follows : PCO2 = K(H2CO3) × [H2CO3] where,

K(H2CO3) is the Henry's law constant for CO2

H2CO3 is the concentration of dissolved CO2 in equilibrium with the gas.

We can estimate the concentration of H2CO3 as follows :

H2CO3 = α(CO2) × PCO2 where α(CO2) is the solubility coefficient of CO2 in water, which is a function of temperature, pressure, and salinity.

To solve the problem, we need to know the values of the following constants :

K(H2CO3) at 25 °C is 1.20 × 10-3 atm/(mol/L).

α(CO2) at 25 °C is 3.37 × 10-2 mol/L/atm.

Substitute the values into the equation :

PCO2 = K(H2CO3) × [H2CO3]10-2.5 atm = (1.20 × 10-3 atm/(mol/L)) × [H2CO3]

H2CO3 = (10-2.5 atm) / (1.20 × 10-3 atm/(mol/L)) = 8.33 × 10-3 M

Substitute the value of H2CO3 into the equilibrium expression and solve for the concentration of Ca2+ :

Ksp = [Ca2+][HCO3-]2 = 4.86 × 10-9

Ksp = [Ca2+][HCO3-]2[Ca2+] = Ksp / [HCO3-]2[Ca2+] = (4.86 × 10-9) / (8.33 × 10-3)2 = 1.6 × 10-3 M

Therefore, the Ca2+ concentration in a groundwater that is in equilibrium with calcite and has a PCO2 of 10-2.5 (atm) is 1.6 × 10-3 M.

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Answer:

One Barr body

Explanation:

Answer:

Al

Explanation:

Al = Aluminum

Among the given options, NH3 (ammonia) can dissolve in water.

NH3 is a polar molecule, meaning it has a partial positive charge on the hydrogen atoms and a partial negative charge on the nitrogen atom. Water (H2O) is also a polar molecule, with the oxygen atom being partially negative and the hydrogen atoms partially positive.

BH3 (borane) is a nonpolar molecule. It does not possess a significant charge separation and does not readily form hydrogen bonds with water molecules. Therefore, BH3 is not expected to dissolve in water to a significant extent.

Therefore, NH3 (ammonia) can dissolve in water, while BH3 (borane) does not readily dissolve in water.

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2-methylpropan-2-ol is a secondary alcohol, the correct answer is option (d)

The term "2-methylpropan-2-ol" refers to a specific type of organic compound that belongs to the alcohol family. The compound is also known by its IUPAC name, which is "2-methylpropan-2-ol."

The "2-methyl" part of the name indicates that the compound contains a methyl group (-CH3) attached to the second carbon atom of the parent chain, which is propane. The "-ol" suffix indicates that the compound belongs to the alcohol family, meaning it contains a hydroxyl group (-OH) attached to one of the carbon atoms in the molecule.

The term "secondary alcohol" refers to the specific type of alcohol that has a hydroxyl group (-OH) attached to a secondary carbon atom, which is a carbon atom that is bonded to two other carbon atoms. In the case of 2-methylpropan-2-ol, the hydroxyl group is attached to the second carbon atom of the parent chain, which is a secondary carbon atom, and thus, it is considered a secondary alcohol.

It is important to note that the classification of alcohols as primary, secondary, or tertiary depends on the number of carbon atoms bonded to the carbon atom bearing the hydroxyl group. Primary alcohols have one carbon atom bonded to the carbon bearing the hydroxyl group, while tertiary alcohols have three carbon atoms bonded to the carbon bearing the hydroxyl group. Therefore the correct option is d.

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The ΔG value for the hypothetical reaction \(2A(s) + B_2(g)\) → \(2AB(g)\) can be calculated by summing the individual ΔG values of the given reactions.

To find the overall ΔG for the reaction, we need to combine the two given reactions and cancel out the common intermediate species, AB(g). We can achieve this by the second reaction and multiplying the first reaction by 2.

The first reaction, \(A(s) + B_2(g)\) → \(AB_2(g)\), has a ΔG of -215.6 kJ. By reversing the second reaction, \(2AB(g) + B_2(g)\)→ \(2AB_2(g)\), the ΔG becomes +672.7 kJ.

Now, we can add the two reactions together:

\(2A(s) + B_2(g)\)→ 2AB(g) (ΔG = -215.6 kJ)

\(2AB(g) + B_2(g)\) → \(2AB_2(g)\) (ΔG = +672.7 kJ)

By summing the ΔG values, we have -215.6 kJ + 672.7 kJ = 457.1 kJ.

Thus, the ΔG for the hypothetical reaction \(2A(s) + B_2(g)\)→ 2AB(g) is 457.1 kJ.

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The concentration of Ni(H2O)62+ ions at equilibrium in the solution formed is 0.00153 M.

The expression for the dissociation constant to determine the concentration of Ni(H2O)62+ ions at equilibrium in the solution formed in Part 1 is: Kd = [Ni(H2O)62+] / ([Ni(H2O)62+] + [Ni(NH3)62+])

In this equation, [Ni(H2O)62+] represents the concentration of hydrated nickel(II) ions, and [Ni(NH3)62+] represents the concentration of complexed nickel(II) ions with ammonia ligands.

To calculate the concentration of Ni(H2O)62+ ions at equilibrium in the solution, we need to consider the stoichiometry of the reaction and the values given.

From the stoichiometry of the equation, we know that 1 mole of Ni(NO3)2 yields 1 mole of Ni(H2O)62+ ions. Given that there are 0.00153 mol of Ni(NO3)2, the concentration of Ni(H2O)62+ ions is also 0.00153 M.

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Answer:

I think it would be 0 since cations and anions must always combine in such a way so that their charges cancel.

If you started with 2.864 g of sodium bicarbonate and heated it to constant mass, 0.59 gram would be the mass that residue after the reaction.

Stoichiometry helps us use the balanced chemical equation to measure quantitative relationships and it is to calculate the amounts of products and reactants that are given in a reaction.

The balanced chemical equation is the equation in which the number of atoms on the reactant side is equal to the number of atoms on the product side in an equation.

Now we have to write the balanced equation

2NaHCO₃  → Na₂CO₃ + CO₂ + H₂O

According to Stoichiometry

= \(2.864\ g\ NaHCO_{3} \times \frac{1\ \text{mole}\ Na_2CO_3}{105.9\ g/mol} \times \frac{1\ \text{mole}\ CO_2}{2\ \text{mole}\ NaHCO_3} \times \frac{44\ g \ CO_2}{1\ \text{mole}\ CO_2}\)

= 0.59 g

Thus from the above conclusion we can say that If you started with 2.864 g of sodium bicarbonate and heated it to constant mass, 0.59 gram would be the mass that residue after the reaction.

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Disclaimer: The question was given incomplete on the portal. Here is the complete question.

Question: If you started with 2.486 g of sodium bicarbonate and heated it to constant mass, what would the mass of the residue be after the reaction? Show your work.

NaHCO₃  → Na₂CO₃ + CO₂ + H₂O  

Answer:

Answers Below

Explanation:

Answer:

Sodium Chloride and Water

The Fahrenheit temperatures corresponding to Celsius 31°C, 21°C, -9°C, and -15°C are approximately 88°F, 70°F, 16° F, 5°F.

To find the equation relating Fahrenheit temperature (F) to Celsius temperature (C), we can use the given freezing and boiling points of water on both scales.

We know that at the freezing point of water:

Fahrenheit temperature = 32°F

Celsius temperature = 0°C

And at the boiling point of water:

Fahrenheit temperature = 212°F

Celsius temperature = 100°C

We can use these two points to find the equation of the line relating F and C.

First, we find the slope of the line:

Slope = (Change in Fahrenheit temperature) / (Change in Celsius temperature)

= (212°F - 32°F) / (100°C - 0°C)

= 180°F / 100°C

Next, we find the y-intercept of the line:

Using the freezing point of water (0°C, 32°F):

32°F = Slope * 0°C + y-intercept

32°F = 0.18 * 0°C + y-intercept

y-intercept = 32°F

Therefore, the equation relating Fahrenheit temperature (F) to Celsius temperature (C) is:

F = 1.8C + 32

Now, we can use this equation to find the Fahrenheit temperatures corresponding to the given Celsius temperatures:

For 31°C:

F = 1.8 * 31 + 32 = 87.8 ≈ 88°F

For 21°C:

F = 1.8 * 21 + 32 = 69.8 ≈ 70°F

For -9°C:

F = 1.8 * -9 + 32 = 15.8 ≈ 16°F

For -15°C:

F = 1.8 * -15 + 32 = 5 ≈ 5°F

Therefore, the Fahrenheit temperatures corresponding to 31°C, 21°C, -9°C, and -15°C are approximately 88°F, 70°F, 16° F, 5°F.

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Answer:

-2247 kJ.

Explanation:

If you want to calculate the free energy change of a reaction at 25 ∘C, you need to follow these simple steps:

Congratulations! You have successfully calculated the free energy change of a reaction at 25 ∘C using some basic chemistry concepts and formulas. Now you can impress your friends and family with your newfound knowledge and skills!

The percent of the mass in a sample of zinc sulfate that comes from zinc would be 40.50%. Option 5.

The percent composition of a component in a substance is the ratio of the mass of the component and the mass of the substance itself.

This can be mathematically expressed as;

Pecent composition= mass of component/mass of substance x 100%

In this case, the substance is zinc sulfate, a compound that contains zinc, sulfur, and oxygen in a ratio 1:1:4.

The molar weight of zinc is 65.38 while the molar mass of the entire zinc sulfate is 161.442.

Thus, the percent of the mass of any sample of zinc sulfate that comes from zinc would be:

Percent zinc = 65.38/161.442 x 100

                    = 40.50%

In other words, the percent of zinc in any zinc sulfate sample would be 40.50%

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Answer:

physical

Explanation: water can change from solid to liquid to a gas

Answer:

The atomic theory has changed over time as new technologies have become available. Scientific knowledge builds on past research and experimentation. 400 B.C. Democritus, a philosopher in ancient Greece, began the search for a description of matter.

Explanation:

Chlorhexidine and alexidine are: used in mouthrinses and are effective against plaque formation.

Chlorhexidine is a disinfectant and antiseptic. It is used for preventing dental plaque, treating yeast infections of the mouth and cleaning wounds.

Chlorhexidine interacts with components of toothpaste and forms salts of low solubility and antibacterial activity.

Alexidine is an antimicrobial agent, used in mouth washes as antiplaque cleaner. It is better for use than chlorhexidine because it is relatively non-toxic

Chlorhexidine and alexidine were synthesized in 1950s.

They inhibit a dental plaque formation by 70%.

Dental plaque is a biofilm of microorganisms that grows in the mouth.

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Answer:

See the answer below

Explanation:

P can displace Q from its salt solution means that P is more reactive than Q.

Q cannot displace R from its salt solutions means that R is more reactive than Q.

a) There is no information as to which of P and R has the capacity to displace one another from their salt solutions. Hence, the most reactive metal can be either of P and R.

b) When P displaces Cu from CuSO4 solution, the equation can be represented below (assuming that P is monoatomic):

\(P + CuSO_4 --> P_2SO_4 + Cu\)

Hence, the product formed is \(P_2SO_4\)

Emf generated is 4.124 v by an pt-cu electrode pair in a 1 molar solution of ions at standard temperature

The average kinetic energy of all the atoms or molecules in a substance is what we refer to as the substance's temperature.  Temperature decreases as a result of atoms and molecules' reduced motion as a result of removing energy (cooling). Fahrenheit, Celsius, and Kelvin are the three temperature scales currently in use.

E° cell = E° cathode- E° anode

= 1.2-(-2.924)

=1.2+2.924

=4.124 v

E° cell = 4.124 v

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