Calculate the maximum overflow rate for a horizontal sedimentation basin designed to remove a 0.1 mm particle (ps = 2.65 g/cm3) at 10 °C (μ=1.307 x 10^-3 kg m-1 s-1). 6.88 x 10-3 m s-1.

Answer:

I'm guessing the alimentary canal

Explanation:

1. 1NH4NO3 = 1N2O+ 2H2O

2. 2k + 2H2O = 1H2 + 2KOH

According to Boyle's law which is a law for constant temperature the the total equilibrium pressure of the system if the volume of the reaction vessel is reduced to 1.0 l is 2.4 atmospheres.

Boyle's law is an experimental gas law which describes how the pressure of the gas decreases as the volume increases. The statement's formula is: "Certain a given mass of just an ideal gas, the absolute pressure it exerts is proportional to its size. amount of gas remains unchanged.

P∝1/V or PV=K. For a given amount of gas, the equation asserts that perhaps the product of volume and pressure is constant. The equation is valid as long as the temperature is maintained constant.

If both gases are taken into consideration, the equation allows for the determination of the variable pressure and size of any one gas. That is,

P₁V₁=P₂V₂

Substitution in above equation gives P₂=1.2×2/1=2.4 atmospheres.

Hence, the total equilibrium pressure of the system if the volume of the reaction vessel is reduced to 1.0 l is 2.4 atmospheres.

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According to Boyle's law which is a law for constant temperature the the total equilibrium pressure of the system if the volume of the reaction vessel is reduced to 1.0 l is 2.4 atmospheres.

Boyle's law is an experimental gas law which describes how the pressure of the gas decreases as the volume increases. The statement's formula is: "Certain a given mass of just an ideal gas, the absolute pressure it exerts is proportional to its size. amount of gas remains unchanged.

P ∝ 1/V

or, PV = K

For a given amount of gas, the equation asserts that perhaps the product of volume and pressure is constant. The equation is valid as long as the temperature is maintained constant.

If both gases are taken into consideration, the equation allows for the determination of the variable pressure and size of any one gas. That is,

P₁V₁ = P₂V₂

initial pressure (P₁) = 1.2 atm

initial volume (V₁) = 2.0 L

final volume (V₂) = 1.O L

final pressure (P₂) = ?

Substitution the values in above equation we get -

P₂ = (1.2×2)/ 1

P₂ =2.4 atmospheres.

Hence, the total equilibrium pressure of the system if the volume of the reaction vessel is reduced to 1.0 l is 2.4 atmospheres.

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Considering the Dalton's partial pressure, the partial pressure of Xe is 1.65 atm.

The pressure exerted by a particular gas in a mixture is known as its partial pressure.

So, Dalton's law states that the total pressure of a gas mixture is equal to the sum of the pressures that each gas would exert if it were alone:

\(P_{T}\)= P₁ + P₂ + ...+ Pₙ

where n is the amount of gases present in the mixture.

This relationship is due to the assumption that there are no attractive forces between the gases.

In this case, you know:

The total pressure of the mixture is:

\(P_{T}\)= partial pressure of He + partial pressure of Ar + partial pressure of Xe

Substituting the corresponding values you get:

2.30 atm= 0.450 atm + 0.200 atm + partial pressure of Xe

Solving:

2.30 atm= 0.650 atm + partial pressure of Xe

2.30 atm - 0.650 atm= partial pressure of Xe

1.65 atm= partial pressure of Xe

Finally, this means that the partial pressure of Xe is 1.65 atm.

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The combustion reaction of propanol, represented as 2C₄H₁₀(1) + 13O₂(g) → 8CO₂(g) + 10H₂O(1), involves the transfer of heat from the surroundings to the system. This exothermic process releases energy, as indicated by the negative enthalpy value of -442 kJ.

The combustion of propanol results in the formation of carbon dioxide (CO₂) and water (H₂O), while consuming oxygen (O₂) in the process.

This reaction is commonly used to provide heat in various applications, such as heating systems and fuel combustion.

By understanding the thermodynamics of this reaction, it becomes evident that it releases a substantial amount of energy in the form of heat, making it a valuable source of warmth and power.

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The pressure that would need to be applied for ammonia to boil at 25.00°C is approximately 1.9 *10^{-6} atm.

The Clausius-Clapeyron equation is given as ln(P2/P1) = (ΔHvap/R) × (1/T1 - 1/T2), where P1 and P2 are the initial and final pressures, ΔHvap is the enthalpy of vaporization, R is the ideal gas constant, T1 is the initial temperature, and T2 is the final temperature.

Given:

T1 = -33.34°C (converted to Kelvin: 239.81 K)

T2 = 25.00°C (converted to Kelvin: 298.15 K)

ΔHvap = 23.35 kJ/mol (converted to J/mol: 23,350 J/mol)

To solve for the pressure (P2), we rearrange the equation as follows:

ln(\frac{P2}{P1}) = (\frac{ΔHvap}{R}) * (\frac{1}{T1} -\frac{ 1}{T2})

Substituting the values, we have:

ln(\frac{P2}{1 atm }) = (\frac{23,350 J/mol }{ 8.314 J/(mol·K)}) * (\frac{1}{239.81 K }- \frac{1}{298.15 K})

After solving the equation, we find that ln(\frac{P2}{1 atm }) ≈ -12.526.

Taking the antilog of both sides, we have:

\frac{P2}{1 atm }≈ e^(-12.526) = 1.9 *10^{-6} atm

Therefore, the pressure that would need to be applied for ammonia to boil at 25.00°C is approximately 1.9 *10^{-6} atm.

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Answer:

4 NH3 (g) + 5 O2 (g) → 4 NO (g) + 6 H2O (l)

This is an oxidation-reduction (redox) reaction:

4 N-III - 20 e- → 4 NII

(oxidation)

10 O0 + 20 e- → 10 O-II

(reduction)

NH3 is a reducing agent, O2 is an oxidizing agent.

Explanation:

good luck

The pH of a solution containing 0.195 M HC2H3O2 and 0.115 M KC2H3O2 is 4.41, while the pH of a solution containing 0.245 M CH3NH2 and 0.135 M CH3NH3Br is 10.34.

A) The pKa of HC2H3O2 is 4.76. Using the Henderson-Hasselbalch equation, pH = pKa + log([A-]/[HA]), where [A-] is the concentration of the acetate ion (C2H3O2-) and [HA] is the concentration of acetic acid (HC2H3O2).

pH = 4.76 + log(0.115/0.195) = 4.76 - 0.351 = 4.41

Therefore, the pH of the solution is approximately 4.41.

B) The pKa of CH3NH3+ is 10.70. Using the Henderson-Hasselbalch equation, pH = pKa + log([A-]/[HA]), where [A-] is the concentration of the methylamine ion (CH3NH2) and [HA] is the concentration of methylammonium ion (CH3NH3+).

pH = 10.70 + log(0.135/0.245) = 10.70 - 0.362 = 10.34

Therefore, the pH of the solution is approximately 10.34.

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Answer:

bob up and down

Explanation:

i think it is the answer i am not sure

Answer:d

Explanation:

The black lines in the Sun's spectrum are caused by gases on, or above, the Sun's surface that absorb some of the emitted light.

Answer:

I think that in a pressure cooker, the liquid reaches a higher temp. because of the fact that in a pressure cooker, the pressure of the air is just circulating within, and in an unsealed pot the air/heat can ease out causing it to keep the temp. stable.

Explanation:

1. The coefficient (green) is 45

2. The exponent (yellow) is 9

To convert from giga grams (Gg) to grams (g), the following coversion scale can be use:

1 Gg = 10⁹ g

With the above convesion scale, we can convert 45 Gg to g as follow

1 Gg = 10⁹ g

Therefore,

45 Gg = 45×10⁹ g

Thus, 45 Gg is equivalent to 45×10⁹ g. Hence, we can conclude as follow:

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When placed at the same temperature, the entropy of 10 mole of Ar(g) at 10.0 atm and the 10 mole of Ar(g) at 0.5 atm. The system has a higher entropy is 10 mole of Ar(g) at 0.5 atm.

The entropy is the measure the randomness of the of the system.  the measure of the system's thermal energy per unit the temperature that is not available for doing the useful work.

The Entropy will increases with temperature at the constant pressure. The  pressure increases leads to the higher degree of the order in the molecular arrangement. so, the entropy decreases with the increasing pressure.

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To determine the pressure of the ammonia gas at a new volume and temperature, we can use the combined gas law, which states that the ratio of the initial pressure, volume, and temperature is equal to the ratio of the final pressure, volume, and temperature.

Using the combined gas law equation: (P1 * V1) / T1 = (P2 * V2) / T2

Given:

P1 = 585 torr (initial pressure)

V1 = 20.0 ml (initial volume)

T1 = 20.0 °C + 273.15 = 293.15 K (initial temperature)

V2 = 50.0 ml (final volume)

T2 = 50.0 °C + 273.15 = 323.15 K (final temperature)

We need to solve for P2 (final pressure).

Rearranging the equation, we have:

P2 = (P1 * V1 * T2) / (V2 * T1)

Substituting the given values into the equation:

P2 = (585 torr * 20.0 ml * 323.15 K) / (50.0 ml * 293.15 K)

Calculating this expression gives us the final pressure (P2) of the ammonia gas at the new volume and temperature.

In summary, using the combined gas law equation, we can determine the pressure of the ammonia gas at a new volume and temperature. By substituting the given values into the equation and performing the calculation, we can find the final pressure of the gas.

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2 C4H10 + 13 O2 → 8 CO2 + 10 H2O.

Answer:

B. Ca

Explanation:

Let's look at the electron configurations of all four elements (I am going to write it in noble gas configuration to make it simpler):

Mg electron configuration: [Ne]3s2

Ca electron configuration: [Ar]4s2

Ar electron configuration: [Ar]

K electron configuration: [Ar] 4s1

We notice that Ca has two electrons in the 4s sublevel, which satisfies what the question is asking for.

The answer is thus B. Ca.

We can use the radioactive decay formula to determine the amount of cadmium-109 remaining after a certain amount of time:

N = N0 (1/2)^(t/T)

where:

N is the amount of cadmium-109 remaining after time t

N0 is the initial amount of cadmium-109

t is the elapsed time

T is the half-life of cadmium-109

We know that the half-life of cadmium-109 is T = 450 days. Therefore, after 450 days, the amount of cadmium-109 remaining will be half of the initial amount. After another 450 days (900 days total), the amount of cadmium-109 remaining will be half of the amount remaining after 450 days, or one quarter of the initial amount. After another 450 days (1350 days total), the amount of cadmium-109 remaining will be half of the amount remaining after 900 days, or one eighth of the initial amount.

So, using the formula above, we can calculate the amount of cadmium-109 remaining after 1350 days:

N = N0 (1/2)^(t/T)

N = 200 g (1/2)^(1350 days / 450 days)

N = 200 g (1/2)^3

N = 200 g (1/8)

N = 25 g

Therefore, approximately 25 grams of cadmium-109 would remain after 1,350 days.

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Answer:

Ions are formed when atoms lose or gain electrons in order to fulfill the octet rule and have full outer valence electron shells.

Explanation:

When they lose electrons, they become positively charged and are named cations. When they gain electrons, they are negatively charged and are named anions.

Answer:

here is your answer.

Explanation:

atom.

thanks for asking:)

The best option to be a recrystallization solvent for sodium benzoate is water. Recrystallization is a purification technique that involves dissolving an impure sample in a solvent and then allowing the solute to slowly crystallize out of the solution under controlled conditions. The pure crystals can be separated from the remaining liquid through filtration.

The most effective solvent for recrystallization is one in which the compound is only slightly soluble at low temperatures but very soluble at high temperatures. By dissolving the compound in a hot solvent and then allowing the solvent to cool, the compound will slowly crystallize out of the solution while any impurities remain dissolved. These impurities are then removed by filtration.

Benzene is no longer used as a solvent because of its toxicity. Ethanol is not the best solvent for recrystallizing sodium benzoate, as it does not have a large enough temperature range to allow for sufficient crystallization and purity. Hexane is non-polar, whereas sodium benzoate is polar, making it ineffective as a solvent for recrystallization. Therefore, the best option to be a recrystallization solvent for sodium benzoate is water.

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The concentration of the solution is 6.56 M.

The concentration of a solution is typically expressed in units of moles per liter (M) or grams per liter (g/L). To determine the concentration of this solution, we need to first calculate the number of moles of NH₃ present in the solution:

Molar mass of NH₃ = 14.01 g/mol + 1.01 g/mol

                                = 17.02 g/mol

Number of moles of NH₃ = 55.82 g / 17.02 g/mol

                                           = 3.28 mol

We need to convert the volume of the solution from milliliters to liters:

500 mL = 0.5 L

Finally, we can calculate the concentration of the solution in units of M:

Concentration = Number of moles / Volume in liters

Concentration = 3.28 mol / 0.5 L = 6.56 M

As a result, the solution concentration is 6.56 M.

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The heat of the brass is approximately -2702.75 J.

To calculate the heat (q) of the brass, we can use the equation:

q = m × c × ΔT

where q is the heat, m is the mass of the brass, c is the specific heat capacity of brass, and ΔT is the change in temperature.

First, we need to determine the specific heat capacity of brass. The specific heat capacity of brass varies depending on the composition of the brass, but we can use an approximate value of 0.38 J/g°C.

Given:

Mass of brass (m) = 25.0 g

Change in temperature (ΔT) = 42.0 °C - 325 °C = -283 °C

Now we can calculate the heat of the brass:

q = (25.0 g) × (0.38 J/g°C) × (-283 °C)

q = -2702.75 J

The negative sign indicates that heat is being transferred from the brass to the surroundings (water in this case). Therefore, the heat of the brass is approximately -2702.75 J.

It's important to note that the negative sign does not change the magnitude of the heat, it simply indicates the direction of heat flow. In this case, the brass is losing heat to the water, resulting in a decrease in its temperature and a corresponding increase in the temperature of the water.

So, the heat of the brass is -2702.75 J.

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Answer: -2688.5. I urge you to see what I have to say below, however, as it is very easy to understand how to do this once you get the simple equation down.

Explanation:

An easy way to get these answers down yourself without having to reach out for help is to do this equation:

Mass of the metal * Specific heat of the metal * (Final Temperature - Initial Temperature). Doing that got me the answer.

In this case, 25*0.380*(42-325) will get you the answer, 25 being the mass, 0.380 being the specific heat of the brass, and 42-325 being the final temperature of 42 degrees celsius minus the initial temperature of 325 degrees celsius. Hope this helped so you can do your best in the future.

Answer:1:water molecule 2:E coli bacteria 3:skin cells ,4:grain of salt 5:human

Explanation:

There u go buddy

For the first unknown concentration with an absorbance of 1.72, the concentration will be, c = 1.72/(ɛ × b). For the second unknown concentration with an absorbance of 0.75, the concentration will be: c = 0.75/(ɛ × b).

Beer lambert's law states that the concentration of a solution is directly proportional to the absorbance of a solution. Mathematically, Beer's Law: A = εlc

where, A is absorbance, ε is the molar absorptivity, l is the path length, and c is the concentration.

We can rewrite the equation as, c = A / εl

where, c is the concentration, A is the absorbance, ε is the molar absorptivity, and l is the path length.

We have two absorbance values, which we will use to determine the concentration of the unknowns. Let's substitute the given values into the equation to determine the concentration of the first unknown.

where, c₁ = A₁ / εlc₁ = 1.72 / εl (1)

Now, let's substitute the second absorbance value to determine the concentration of the second unknown.

c₂ = A₂ / εlc₂ = 0.75 / εl(2)

The concentrations of the unknowns are c₁ and c₂, which we have expressed in terms of the concentration of the solution. The total concentration of the solution is not provided. Thus, we cannot determine the concentration of the unknown solutions.

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Two possible sources of systematic (determinate) error in the experiment are; Incorrect calibration of volumetric glasswareIncorrect mass of NaCl

To calculate the concentration of NaCl in the resulting solution in mg/L NaCl, we can use the formula; Concentration (mg/L) = (Mass of solute ÷ Volume of solution in L) × 1000 g / 1 mg NaCl is present in the stock solution of 25 mL. So, the mass of NaCl in the solution would be;0.0842 g ÷ 25 mL = 0.00337 g/mL. Now, in the resulting solution, a 1.00 mL aliquot of the stock solution is transferred to a 50.00 mL volumetric flask and diluted to volume. Therefore, the volume of the resulting solution is 50.00 mL. We will substitute these values in the formula, Concentration (mg/L) = (0.00337 g/mL ÷ 50 mL) × 1000 g / 1 mg concentration (mg/L) = 67.4 mg/L. Therefore, the concentration of NaCl in the resulting solution in mg/L NaCl is 67.4 mg/L.7.  Concentration = 67.4 mg/LTolerance = 4.28 mg/LTotal concentration = 67.4 + 4.28 mg/L = 71.68 mg/LWe round off this value to one decimal place; Total concentration = 71.7 mg/LTherefore, the concentration of NaCl in the resulting solution using propagation of error through the calculation is 67.4+0.4 mg/L.8. Two possible sources of random (indeterminate) error in the experiment are; Errors in temperature measurement. Errors in measurement of water volume. Two possible sources of systematic (determinate) error in the experiment are; Incorrect calibration of volumetric glasswareIncorrect mass of NaCl.

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Answer:

550 K

Explanation:

Given the following data;

Original Volume, V1 = 20mL

Original Temperature, T1 = 200K

New Temperature, V2 = 55mL

To find new temperature T2, we would use Charles' law.

Charles states that when the pressure of an ideal gas is kept constant, the volume of the gas is directly proportional to the absolute temperature of the gas.

Mathematically, Charles is given by;

\( VT = K\)

\( \frac{V1}{T1} = \frac{V2}{T2}\)

Making T2 the subject of formula, we have;

\(T_{2} = \frac{V2}{V1} * T_{1}\\\)

Substituting into the equation;

\(T_{2} = \frac{55}{20} * 200\\T_{2} = \frac{11000}{20}\\T_{2} = 550K\)

Therefore, the new temperature is 550K.