what volume of 0.500 m hydrochloric acid solution needs to be added to excess sodium carbonate in order to cause the evolution of 14.5 l of carbon dioxide gas at stp?

The volume of ammonium sulfate required for the reaction can be calculated as follows: V = n/CV = 0.00705/0.429V = 16.42 mL Therefore, 16.42 mL of ammonium sulfate solution is required to react completely with 32.9 mL of 1.03 M sodium hydroxide solution.

The balanced equation for the reaction between ammonium sulfate and sodium hydroxide is;NH4+​(aq) + OH−(aq) → NH3​(g) + H2​O(l) Volume of sodium hydroxide solution used = 32.9 m LC = 1.03 M Molarity of sodium hydroxide solution, M = 1.03 M Volume of sodium hydroxide solution, V = 32.9 mL By using the above information, we can determine the number of moles of sodium hydroxide used in the reaction; n = cVn = 0.429 x Vn = 0.429 x 32.9/1000n = 0.0141 moles Number of moles of ammonium sulfate used in the reaction is 1/2 times the moles of sodium hydroxide used. n = 0.0141/2n = 0.00705 moles.

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Answer:

ADP is similar to a drained battery, while ATP is like to a charged battery. With the addition of water to the substrate, ATP can be hydrolyzed into ADP, releasing energy.

Explanation:

The volume occupied by 1.78 mol of nitrogen gas at a pressure of 1.51 atm and a temperature of 322 K is 30.57 liters.

To calculate the volume occupied by 1.78 mol of nitrogen gas at a pressure of 1.51 atm and a temperature of 322 K, we can use the Ideal Gas Law equation: PV = nRT.

P represents the pressure, V represents the volume, n represents the number of moles of gas, R is the ideal gas constant, and T represents the temperature in Kelvin.

First, let's convert the pressure from atm to Pascals (Pa) by using the conversion factor: 1 atm = 101325 Pa.

So, the pressure of 1.51 atm is equal to 1.51 × 101325 = 152,928.75 Pa.

Next, let's convert the temperature from Kelvin to Celsius by subtracting 273.15. Thus, 322 K = 48.85 °C.

Now, we need to convert the temperature from Celsius to Kelvin by adding 273.15. Therefore, 48.85 °C = 322 K.

Now, we can substitute the values into the Ideal Gas Law equation: PV = nRT.

V × 152,928.75 = 1.78 × 8.314 × 322.

Simplifying the equation, we have V × 152,928.75 = 4679.67.

To solve for V, divide both sides of the equation by 152,928.75.

V = 4679.67 / 152,928.75.

Calculating the value, V = 0.03057 m³.

Finally, let's convert the volume from cubic meters (m³) to liters (l) by using the conversion factor: 1 m³ = 1000 l.

Thus, 0.03057 m³ = 0.03057 × 1000 = 30.57 l.

Therefore, the volume occupied by 1.78 mol of nitrogen gas at a pressure of 1.51 atm and a temperature of 322 K is approximately 30.57 liters (l).

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1. Short time scale change on the ecosystem:

One example of a short time scale change on the ecosystem is the oil spill that occurred in the Gulf of Mexico in 2010. This disaster resulted in the contamination of the water, which ultimately led to the death of marine life and affected the food chain. The spill had a significant impact on the fishing industry in the region and took years to recover from.

2. The law of unintended consequences:

The law of unintended consequences states that actions always have consequences that are not anticipated. One example is the use of pesticides in farming. Although the use of pesticides reduces the risk of crop damage and increases yield, it also has unintended consequences. The pesticides can harm other organisms that come into contact with the crops, including bees and other beneficial insects that play a critical role in pollination.

3. Disposal sanitary method:

The disposal of waste is a significant problem in today's world, and sanitary landfill is a popular method of disposal. It involves the burial of waste in a landfill that is lined with materials that prevent leaching of harmful chemicals into the soil. This method of disposal is effective but has disadvantages. It produces greenhouse gases and requires large amounts of land.

4. Causes of Acid Rain:

The causes of acid rain include emissions from industrial activity, power generation, and transportation. These emissions release sulfur dioxide and nitrogen oxides, which react with the atmosphere to form acid rain. Acid rain has significant consequences for aquatic life and forests, as well as buildings and infrastructure.

5. Greenhouse gases:

Greenhouse gases are gases that trap heat in the atmosphere, causing the earth's temperature to rise. Examples include carbon dioxide, methane, and water vapor. Greenhouse gases are primarily produced by human activity, including transportation, industrial processes, and deforestation. The increase in greenhouse gas concentrations has resulted in climate change.

6. Effect of Ozone problem on Human:

The ozone layer is essential because it absorbs harmful UV rays from the sun. The depletion of the ozone layer due to human activity has resulted in an increase in skin cancer and other health problems. Exposure to UV radiation can also cause damage to the eyes and the immune system.

7. Genetic Mutation causes:

Genetic mutations can occur naturally or due to human activity, including exposure to radiation, chemicals, and toxins. Genetic mutations can cause health problems, including cancer and developmental disorders. Mutations can also have positive effects, such as improving immunity to certain diseases.

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The recommended maximum PO2 for recreational enriched air nitrox diving is 1.4 ATA with a contingency of 1.6 ATA.

The partial pressure of oxygen or PO2 is a measure of the amount of oxygen in the breathing gas mixture. It is used in diving as a safety parameter to ensure that divers don't breathe gas mixtures that can cause oxygen toxicity. Enriched air nitrox is a gas mixture that has a higher concentration of oxygen than normal air.

The recommended maximum PO2 for recreational enriched air nitrox diving is 1.4 ATA. This means that the partial pressure of oxygen in the gas mixture should not exceed 1.4 atmospheres absolute. This is a conservative limit that is designed to reduce the risk of oxygen toxicity. However, there is a contingency of 1.6 ATA that allows for a higher PO2 in case of emergency situations.

This contingency is included to ensure that divers have access to a higher concentration of oxygen if they need it to decompress after a deep dive or if they experience other problems while diving. However, it should only be used in emergency situations as breathing gas with a higher PO2 can be dangerous.

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It takes about 15.02 seconds for the drug concentration to reach 0.24 g/cm3.

Drug concentration after 10 seconds:

C(t)= 0.2(1 + 3e^-0.3t)

Given t = 10 seconds

C(10) = 0.2 (1 + 3e^-0.3*10)≈ 0.75 g/cm32.

Rate of change between 30 and 45 seconds (g/cm3/sec)

Rate of change of C with respect to t is given by d

C/dt = -0.18e^-0.3t

When t = 30,dC/dt = -0.18e^-0.3*30 = -0.0104 g/cm3/sec

When t = 45,dC/dt = -0.18e^-0.3*45 = -0.0015 g/cm3/sec3.

Time taken for drug concentration to reach 0.24 g/cm3

To find the time it takes for the drug concentration to reach 0.24 g/cm3,

we solve the equation C(t) = 0.24 g/cm3 for t.

C(t) = 0.2 (1 + 3e^-0.3t)0.24

      = 0.2 (1 + 3e^-0.3t)1.2

      = 1 + 3e^-0.3t0.2

      = 3e^-0.3tln(0.2/3)

      = -0.3tln(0.2) - ln(3)

     = -0.3tln(3) + ln(0.2)

     = 0.3tApproximately, t

    ≈ 15.02 seconds,

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1. The balanced chemical equation for the reaction between hydrochloric acid and magnesium carbonate is:

[tex]MgCO^{3}[/tex] + 2HCl → MgCl2 + CO2 + H2O

The reaction involves a 1:2 mole ratio between [tex]MgCO^{3}[/tex] and HCl.

Thus, the number of moles of [tex]MgCO^{3}[/tex] can be calculated by dividing the mass of the compound by its molar mass.

Mass of [tex]MgCO^{3}[/tex] = 7.27 g

Molar mass of [tex]MgCO^{3}[/tex] = 24.31 + 12.01 + (3 x 16.00) = 84.31 g/mol

Number of moles of [tex]MgCO^{3}[/tex] = Mass/Molar mass = 7.27/84.31 = 0.086 moles

The volume of 0.409 M HCl required to react with this amount of [tex]MgCO^{3}[/tex] can be calculated using the equation below:

Volume of HCl = (Number of moles of MgCO3) x (Volume of HCl required per mole of MgCO3)

Volume of HCl = 0.086 x (2 x 1000) = 172 mL (since 2 moles of HCl react with one mole of [tex]MgCO^{3}[/tex])

Therefore, the volume of 0.409 M HCl required to react completely with 7.27 g of [tex]MgCO^{3}[/tex] is 172 mL.

2. The molar mass of manganese(II) nitrate can be calculated as follows:

Manganese(II) nitrate = Mn(NO3)2

Molar mass of Mn = 54.94 g/mol

Molar mass of N = 14.01 g/mol

Molar mass of O = 16.00 g/mol

Molar mass of Mn(NO3)2 = (54.94) + 2(14.01 + 3(16.00)) = 178.96 g/mol

To calculate the molarity of manganese(II) nitrate, we use the following formula:

Molarity = Number of moles of solute/Volume of solution in liters

Number of moles of solute can be calculated using the mass and molar mass of the solute as shown below:

Number of moles of solute = Mass of solute/Molar mass of solute

Mass of manganese(II) nitrate = 13.7 g

Volume of solution = 500. mL = 0.5 LV = 0.5 L

The molarity of manganese(II) nitrate can be calculated as follows:

Molarity of Mn(NO3)2 = (13.7/178.96) / 0.5 = 0.152 M

The molarity of Mn2+ is the same as that of Mn(NO3)2 since the ion is not complexed with any other ligand in the solution.

Molarity of Mn2+ = 0.152 M

The molarity of NO3- can be calculated as follows:

Each mole of Mn(NO3)2 dissociates to form two moles of NO3-.

Molarity of NO3- = 2 x Molarity of Mn(NO3)2 = 2 x 0.152 = 0.304 M

3. The molecular formula of copper(II) iodide is CuI2.

The number of moles of copper(II) iodide required can be calculated using the molarity and volume of the solution.

Molarity = Number of moles of solute/Volume of solution in liters

Number of moles of solute = Molarity x Volume of solution in liters

Volume of solution = 250 mL = 0.250 L

The number of moles of copper(II) iodide required = 0.155 x 0.250 = 0.0388 moles

The mass of copper(II) iodide required can be calculated using the following formula:

Mass of solute = Number of moles of solute x Molar mass of solute

Molar mass of CuI2 = (63.55 + 2(126.90)) = 317.35 g/mol

Mass of copper(II) iodide required = 0.0388 x 317.35 = 12.32 g

Thus, the mass of copper(II) iodide that must be added to a 250-mL volumetric flask in order to prepare 250 mL of a 0.155 M aqueous solution of the salt is 12.32 g.

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The specific heat (c) of the substance, obtained by absorbing 2805 J of heat and experiencing a temperature change from 21.5°C to 149.0°C, is approximately 1.18 J/g°C.

To calculate the specific heat (c) of a substance, we can use the formula:

Heat absorbed (Q) = mass (m) × specific heat (c) × temperature change (ΔT)

First, we need to determine the temperature change of the substance:

ΔT = final temperature - initial temperature

ΔT = 149.0°C - 21.5°C = 127.5°C

Next, we substitute the given values into the formula:

2805 J = 28.50 g × c × 127.5°C

To isolate the specific heat (c), we divide both sides of the equation by (28.50 g × 127.5°C):

c = 2805 J / (28.50 g × 127.5°C)

c ≈ 1.18 J/g°C

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The heat in kJ required to raise 1,290 g of water from 27°C to 74°C is 236.69 kJ. Here's how it can be calculated:

First, we need to determine the heat energy required to raise 1 g of water by 1°C.

Given that the specific heat capacity of water is 4.184 J/g°C, we multiply this value by the mass of water (1,290 g) to obtain the heat energy required for a 1°C increase:

4.184 J/g°C × 1,290 g = 5,390.16 J

Next, we utilize the formula Q = mcΔT, where Q represents the heat energy, m is the mass of water, c is the specific heat capacity of water, and ΔT is the change in temperature. Substituting the given values, we find:

Q = (1,290 g) × (4.184 J/g°C) × (74°C - 27°C)

Q = 236,689.76 J

To convert this value to kJ, we divide it by 1,000:

Q = 236,689.76 J ÷ 1,000 = 236.69 kJ

The heat in kJ required to raise 1,290 g of water from 27°C to 74°C is 236.69 kJ.

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The diagram that best depicts the result of combining the contents of the two flasks into one 4 L container is the one where the gas fills the entire container without any empty space.

In order to determine the diagram that accurately represents the result of combining the contents of two sealed gas containers into one 4 L container at the same temperature, we need to consider the principles of gas behavior.          The principles of gas behavior, also known as the gas laws, describe the relationships between the physical properties of gases, such as pressure, volume, and temperature.                                                                                                                                  When two gas containers are combined, the total volume of the gas will be the sum of the individual volumes.                                        In a sealed container, the gas particles are confined, and there is no exchange with the surroundings.                                                As a result, the pressure of the gas remains constant because there is no change in the force exerted by the gas on the container walls.                                                                                                                                                                        Additionally, since the gas is at the same temperature, there are no temperature-induced variations in pressure according to the ideal gas law.                                                                                                                                                                               In this case, both containers have a volume of 2 L, resulting in a total volume of 4 L.                                                                                  Since the containers are sealed and the gas is at the same temperature, the pressure of the gas remains constant. Consequently, the combined gas will uniformly occupy the entire 4 L container, with no empty space.                                              Hence, the most suitable diagram should depict the gas filling the entirety of the container, without any vacant areas.

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The amount of salt dissolved in each liter of seawater is 36.7 g/L.

Archimedes' Principle states that the buoyant force on an object immersed in a fluid is equivalent to the weight of the displaced fluid and is aimed upward.

This principle is named after the ancient Greek scientist Archimedes, who discovered that the volume of an object submerged in water could be determined using this principle. This principle is used to evaluate the relative density of objects immersed in a fluid in the modern era.

Sailors on a trans-Pacific solo voyage observe one day that if they place 694 ml of fresh water into a 25.0 g plastic cup, the cup floats in the seawater around their boat with the fresh water inside the cup at the same level as the seawater outside the cup.

We must calculate the amount of salt dissolved in each liter of seawater.To solve the problem, we can use the following steps: We'll start by calculating the mass of water displaced by the cup using Archimedes' principle.Buoyant force = Weight of displaced water, Fb = W Water displaced = mWater * g Buoyant force = mCup * g, where mCup is the mass of the cupWe may express the density of seawater, ρSw, in terms of the salt dissolved in it using the following formula:ρSw = ρfw + Δρ, where Δρ is the increase in density due to salt.[tex]Δρ = ρSw - ρfw[/tex].

The volume of water displaced by the cup is equal to the volume of fresh water it contains. Thus: [tex]ρCup * Vfw = (mCup + mWater) / ρSw[/tex], where Vfw is the volume of fresh water, mWater is the mass of the water, and ρCup is the density of the cup.

Rearranging the formula gives:[tex]ρSw = (mCup + mWater) / (ρCup * Vfw) + ρfw[/tex]. Substituting the given values into the formula yields: [tex]ρSw = (25.0 g + 694.0 g) / (ρCup * 694.0 mL) + 0.999 g/mLρSw = (719.0 g) / (ρCup * 0.6940 L) + 0.999 g/mLρSw = (719.0 g) / (ρCup * 694.0 mL) + 0.999 g/mLρSw = (719.0 g) / (ρCup * 6.940 × 10-4 L) + 0.999 g/mLρSw = (719.0 g) / (ρCup * 0.0006940 L) + 0.999 g/mLρSw = 1.0358 g/mL.[/tex].

The mass of salt in each liter of seawater, mSalt, can be calculated using the formula:m [tex]Salt = Δρ / ρSw * 1000 g/LmSalt = (1.0358 - 0.9990) / 1.0358 * 1000 g/LmSalt = 36.7 g/L[/tex]. Therefore, the amount of salt dissolved in each liter of seawater is 36.7 g/L.

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The glucose concentration of the unknown sample is estimated to be 255 mg/dL based on the absorbance values of 0.50 (standard) and 0.85 (unknown) using the Beer-Lambert Law.

To calculate the glucose concentration of the unknown sample, we can use the Beer-Lambert Law and set up a proportion based on the absorbance values.

According to the Beer-Lambert Law, the absorbance (A) is directly proportional to the concentration (C) of a substance, multiplied by the path length (b) and the molar absorptivity (ε) of the substance.

Mathematically, it can be expressed as:

A = ε * C * b

Given that the standard absorbance (A1) is 0.50, the unknown absorbance (A2) is 0.85, and the glucose concentration of the standard (C1) is 150 mg/dL, we can set up the following proportion:

A1 / A2 = C1 / C2

Plugging in the values, we have:

0.50 / 0.85 = 150 mg/dL / C2

Simplifying the proportion, we can solve for C2 (glucose concentration of the unknown sample):

C2 = (0.85 * 150 mg/dL) / 0.50

C2 = 255 mg/dL

Therefore, the estimated glucose concentration of the unknown sample is 255 mg/dL.

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3. Number of phosphorus atoms, will be  3.07 × [tex]10^{16}[/tex] 9. Mass of two moles would be 39.098 u. 10. Atomic number of carbon is 6. The atomic mass of carbon is 12, which is the sum of the number of protons and neutrons.

The number of phosphorus atoms contained in 1.58 × [tex]10^{-6}[/tex] g of phosphorus is as follows:From the periodic table, the atomic mass of phosphorus is 30.974 u. Hence, the number of moles in 1.58 ×[tex]10{-6}[/tex] g of phosphorus is:Number of moles = Mass of sample/Molar mass= 1.58 × 10{6} g/ 30.974 u/mol= 5.1 × [tex]10^{-8}[/tex]mol

The number of phosphorus atoms in the sample is obtained by multiplying the number of moles by Avogadro's number: Number of atoms = Number of moles × Avogadro's number= 5.1 × [tex]10^{-8}[/tex] mol × 6.022 × 10^{23} atoms/mol≈ 3.07 × 10^{16} atoms9. To determine the mass of 2 moles of potassium, use the following formula:Mass = Number of moles × Molar massFrom the periodic table, the atomic mass of potassium is 39.098 u.

Hence, the molar mass of potassium is: Molar mass of potassium = 39.098 g/molUsing the formula above, the mass of 2 moles of potassium atoms is given by:Mass = Number of moles × Molar mass= 2 mol × 39.098 g/mol= 78.196 g

Atomic number is the number of protons present in the nucleus of an atom while atomic mass is the sum of the number of protons and neutrons present in the nucleus of an atom. Let us consider an example using carbon.

Carbon has 6 protons and 6 neutrons in its nucleus, hence the atomic number of carbon is 6. The atomic mass of carbon is 12, which is the sum of the number of protons and neutrons. The formula for calculating the atomic mass is:Atomic mass = Number of protons + Number of neutrons.

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The subsequent mistakes made during titration can have an impact on the estimated percent acidity. The impact can be influenced by factors such as

If the buret's tip isn't entirely filled, it can lead to an inaccurate volume measurement of the titrant added to the solution. This can result in an incorrect calculation of the acid's concentration and subsequently affect the estimated percent acidity.

If the flask used in the titration leaks a small amount of the acid sample, it can lead to a loss of the analyte. This loss can cause a decrease in the amount of acid reacted with the base, resulting in an underestimation of the acid's concentration and the estimated percent acidity.

3. Utilizing a lower molarity of the base solution in the computation compared to the actual molarity can result in an incorrect stoichiometric ratio between the acid and base. This will lead to an inaccurate determination of the acid's concentration and subsequently affect the estimated percent acidity.

Overall, these mistakes can introduce errors and inaccuracies in the titration process, affecting the estimation of percent acidity. It is crucial to minimize these mistakes and ensure proper technique and equipment usage during titration to obtain reliable and accurate results.

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The [H⁺] concentration is 10⁻¹⁴ M, the [OH⁻] concentration is 10⁻³⁷ M, and the pH of the solution is 3.37 at 25°C.

To determine the [H⁺], [OH⁻], and pH of the solution, we need to use the relationship between pH and pOH. The pH and pOH are related by the equation:

pH + pOH = 14

Given that the pOH is 10.63, we can subtract it from 14 to find the pH:

pH = 14 - 10.63 = 3.37

The pH represents the negative logarithm (base 10) of the [H⁺] concentration. Therefore, we can calculate the [H⁺] concentration using the formula:

[H⁺] = 10(-pH)

[H⁺] = 10(-3.37) = 4.83 × 10(-4) M

Similarly, we can find the [OH⁻] concentration using the equation:

[OH⁻] = 10(-pOH)

[OH⁻] = 10(-10.63) = 3.37

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Nucleophilic displacement reactions involve the substitution of one nucleophile by another. This process occurs in three steps: initiation, nucleophilic attack, and elimination.

Nucleophilic displacement reactions are a fundamental class of reactions in organic chemistry. These reactions involve the substitution of a nucleophile (an atom or group with an unshared pair of electrons) for a leaving group (a group that can depart with its pair of electrons). The three steps involved in these reactions are initiation, nucleophilic attack, and elimination.

In the initiation step, a nucleophile reacts with a suitable reagent or catalyst, which provides the necessary conditions for the reaction to occur. This initiation step prepares the nucleophile for the subsequent attack on the substrate.

The second step is the nucleophilic attack. Here, the activated nucleophile, now possessing a partial negative charge, attacks the electrophilic carbon of the substrate, displacing the leaving group. This attack results in the formation of a new bond between the nucleophile and the substrate, while the leaving group is expelled.

Finally, in the elimination step, the leaving group is eliminated, typically accompanied by the formation of a new bond or rearrangement of existing bonds. This step completes the nucleophilic displacement reaction and yields the final product.

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The energy required to heat 1.60 kg of mercury from −9.2∘C to 11.1∘C is 4.51 J.

The energy required to heat a substance can be calculated using the following formula;

Q = mc∆T

Where;

According to this question, 1.60 kg of mercury is heated from −9.2∘C to 11.1∘C. The amount of energy required can be calculated as follows:

Q = 1.6 × 0.139 × {284.1 - 263.8}

Q = 4.51 J

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To answer this question, we need more information.

To calculate the percent by mass of KBr in a saturated solution at 10 °C, we need to know the solubility of KBr at that temperature. The solubility of KBr in water varies with temperature. Without this information, we cannot provide an accurate calculation.

The solubility of KBr at 10 °C can be determined from experimental data or reference sources. Once the solubility is known, the percent by mass of KBr can be calculated using the formula:

Percent by mass of KBr = (mass of KBr / mass of solution) × 100

Where the mass of KBr is the mass of KBr dissolved in the given amount of water to form a saturated solution. The mass of the solution is the total mass of the solution, which includes the mass of both KBr and water.

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Based on the average properties found in the appendix, the modulus of elasticity of steel (Esteel) is generally greater than that of plastics (Eplastics).

According to the average properties found in the appendix, the modulus of elasticity (E) of steel is generally greater than that of plastics.

The modulus of elasticity, also known as Young's modulus, measures the stiffness or rigidity of a material. It quantifies how much a material deforms under an applied load.

Steel is known for its high strength and stiffness, and it typically has a higher modulus of elasticity compared to plastics.

On the other hand, plastics have a wide range of modulus of elasticity values depending on their composition and structure.

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The correct option is E) None of the above, as none of the provided answer choices matches the calculated density. To convert the density of substance A from g/cm³ to lbm/ft³, we need to use the appropriate conversion factors.

1 g/cm³ is equal to 62.43 lbm/ft³.

Therefore, the density of substance A in lbm/ft³ is:

Density in lbm/ft³ = Density in g/cm³ × Conversion factor

Density in lbm/ft³ = 1.34 g/cm³ × 62.43 lbm/ft³

Density in lbm/ft³ ≈ 83.6102 lbm/ft³

Rounded to two decimal places, the density of substance A is approximately 83.61 lbm/ft³.

Therefore, the correct option is E) None of the above, as none of the provided answer choices matches the calculated density.

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The amount of 0.000875 M KOH required would be 1.18 ml to titrate the 40.00 mL sample of acid rain with an sulphuric acid concentration of 1.290 × 10⁻⁴ M.

To calculate the volume of 0.000875 M KOH required to titrate a 40.00 mL sample of acid rain with an sulphuric acid  concentration of 1.290 × 10⁻⁴ M, we need to use the concept of stoichiometry and the balanced chemical equation for the reaction between sulfuric acid (H2SO4) and potassium hydroxide (KOH).

From the balanced equation, we can see that the stoichiometric ratio between sulphuric acid  and KOH is 1:2. This means that 1 mole of sulphuric acid  reacts with 2 moles of KOH.

First, let's calculate the number of moles of sulphuric acid  in the 40.00 mL sample of acid rain: moles sulphuric acid  = concentration of sulphuric acid × volume of acid rain sample = (1.290 × 10⁻⁴ M) × (40.00 mL / 1000 mL/ L) = 5.16 × 10⁻⁶ moles

Since the stoichiometric ratio between sulphuric acid and KOH is 1:2, we need twice as many moles of KOH to completely neutralize the sulphuric acid. Therefore, the number of moles of KOH required is:

Moles KOH = 2 × moles sulphuric acid = 2 × 5.16 × 10⁻⁶ moles = 1.032 × 10⁻⁵ moles Now, let's calculate the volume of 0.000875 M KOH required to contain 1.032 × 10⁻⁵ moles of KOH:

Volume KOH = moles KOH / concentration of KOH = (1.032 × 10⁻⁵ moles) / (0.000875 M) = 1.18

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Given data: N2 = 6.65 × 102 gH2 = 1.04 × 102 g , The balanced chemical equation for the reaction of nitrogen and hydrogen to form ammonia is: N2(g) + 3H2(g) → 2NH3(g)

From the balanced equation, it is evident that the ratio of N2 and NH3 is 1:2 while the ratio of H2 and NH3 is 3:2.Therefore, the number of moles of N2 = 6.65 × 102 g / 28 g/mol = 23.75 molThe number of moles of H2 = 1.04 × 102 g / 2 g/mol = 52 mol.From the balanced equation, we can see that 1 mole of N2 reacts with 3 moles of H2 to give 2 moles of NH3, thus limiting the reaction to the availability of N2. Therefore, the maximum number of moles of NH3 that can be produced from N2 = 23.75 mol x (2/1) = 47.5 molTherefore, the maximum mass of ammonia that can be produced = 47.5 mol x 17 g/mol (molecular weight of NH3) = 807.5 g.Thus, the maximum mass of ammonia that can be produced from the given quantity of N2 and H2 is 807.5 g. The reaction is limited by N2, so the H2 is in excess. Amount of H2 reacted with 1 mole of N2 = (3/1) x 2 g/mol = 6 g/molThe amount of H2 reacted with 23.75 moles of N2 = 23.75 moles x 6 g/mol = 142.5 g. Therefore, the mass of H2 left unreacted = 104 - 142.5 = - 38.5 g.

Since the mass of H2 left unreacted is negative, there is no H2 left unreacted but some H2 has been consumed more than its available quantity.

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HNO3 transfers a hydrogen ion to H2O.

When the H2O molecule is brought close to the HNO3 molecule, a transfer of a hydrogen ion (H+) occurs from HNO3 to H2O.

This transfer leads to the formation of H3O+ (hydronium ion) and NO3- (nitrate ion).

HNO3, also known as nitric acid, is a strong acid composed of hydrogen (H), nitrogen (N), and oxygen (O) atoms. H2O, or water, consists of two hydrogen (H) atoms and one oxygen (O) atom.

In the presence of an acid like HNO3, water can act as a base and accept a proton (H+) from the acid.

As HNO3 donates a proton to H2O, the hydrogen ion (H+) bonds with one of the lone pairs of electrons on the oxygen atom in H2O, forming the hydronium ion (H3O+).

This process is represented by the equation: HNO3 + H2O -> H3O+ + NO3-.

The resulting hydronium ion, H3O+, is a positively charged ion with a central hydrogen atom bonded to three hydrogen atoms and a lone pair of electrons.

The nitrate ion, NO3-, carries a negative charge and consists of one nitrogen atom bonded to three oxygen atoms.

In summary, when the H2O molecule is brought near the HNO3 molecule, a transfer of a hydrogen ion occurs, resulting in the formation of hydronium ion (H3O+) and nitrate ion (NO3-).

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A 0.221 g sample of antacid is found to neutralize 23.8 ml of 0.1m hcl. If one tablet has a mass of 750 mg, it can neutralize about 0.0214 L of stomach acid.

Mass is the measure of the amount of matter in an object. It is a scalar quantity usually measured in kilograms or grams.

The number of moles of HCl neutralized by the antacid can be calculated using the following equation:

moles of HCl = M x V

where M is the molarity of the HCl solution and V is the volume of the HCl solution in liters.

Converting the volume of the HCl solution from milliliters to liters:

V = 23.8 mL = 0.0238 L

Substituting the given values:

moles of HCl = 0.1 M x 0.0238 L = 0.00238 moles

The number of moles of antacid that reacted with the HCl can be calculated using the following equation:

moles of antacid = moles of HCl

Substituting the given mass of antacid:

moles of antacid = 0.221 g / 103.3 g/mol = 0.00214 moles

Since the number of moles of antacid that reacted with the HCl is equal to the number of moles of HCl, we can use the following equation to calculate the volume of stomach acid that could be neutralized by one tablet of antacid:

V = moles of HCl / M

Substituting the given values:

V = 0.00214 moles / 0.1 M

= 0.0214 L

Converting the volume from liters to milliliters:

V = 21.4 mL

Therefore, one tablet of antacid having mass 750mg could neutralize 21.4 mL of stomach acid.

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When the piece of glass is submerged in water, it will weigh approximately 47.14 kilograms.

The specific gravity of a substance is the ratio of its density to the density of a reference substance. In this case, the specific gravity of the glass is 2.55, which means it is 2.55 times denser than the reference substance, which is usually water.

To determine the weight of the glass when submerged in water, we need to consider the concept of buoyancy. Buoyancy is the upward force exerted on an object submerged in a fluid, which opposes the force of gravity. When an object is immersed in a fluid, it displaces an amount of fluid equal to its own volume.

Since the glass has a specific gravity greater than 1, it will sink in water. However, the buoyant force will act on the glass, reducing the net force of gravity. The buoyant force is equal to the weight of the water displaced by the submerged glass.

To find the weight of the glass when submerged in water, we need to calculate the weight of the water displaced by the glass. The weight of the water displaced is equal to the volume of the glass multiplied by the density of water (which is approximately 1000 kg/m³).

We can calculate the volume of the glass by dividing its weight by its density, which is equal to the specific gravity multiplied by the density of water. Then, we can calculate the weight of the water displaced by the glass by multiplying the volume by the density of water.

Finally, to find the weight of the glass when submerged, we subtract the weight of the water displaced from the original weight of the glass.

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The empirical formula of the compound is C3H5O.

To determine this, we need to find the simplest whole number ratio of atoms in the compound.

Assuming a 100 g sample, we have:

- 48.64 g C

- 8.16 g H

- 43.2 g O

Next, we need to convert these masses to moles:

- C: 48.64 g / 12.01 g/mol = 4.05 mol

- H: 8.16 g / 1.01 g/mol = 8.07 mol

- O: 43.2 g / 16.00 g/mol = 2.70 mol

Now we need to divide each of these values by the smallest number of moles (which is 2.70) to get the simplest whole number ratio:

- C: 4.05 mol / 2.70 mol = 1.50 ≈ 3

- H: 8.07 mol / 2.70 mol = 2.99 ≈ 5

- O: 2.70 mol / 2.70 mol = 1

Therefore, the empirical formula is C3H5O.

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The compound  [tex]PCL_{5}[/tex] is named phosphorus pentachloride according to the rules for naming binary molecular compounds

When it comes to naming binary molecular compounds, there are a few general rules to follow. The first element in the formula is named first and it is followed by the second element named as if it is a monatomic anion.

For the second element in the compound, the suffix “-ide” is added to the root of the element name. If there are multiple atoms of the first or second element in the formula, the prefixes mono-, di-, tri-, tetra-, penta-, hexa-, hepta-, octa-, nona-, and deca- are added to the element name to indicate the number of atoms.

Therefore, the name for the compound {PCl5} is phosphorus pentachloride.  [tex]PCL_{5}[/tex] is a colorless, solid or a yellowish-green liquid that fumes in the air because it reacts with moisture to give HCl gas. It is a highly reactive compound with phosphorus in the +5 oxidation state and has a trigonal bipyramidal shape, with three equatorial P–Cl bonds with a bond length of 204 pm and two axial P–Cl bonds with a bond length of 207 pm.

It is important for various applications like as an intermediate for the production of phosphoric acid and other phosphorus compounds, as a chlorinating agent, and as a catalyst in organic synthesis.In summary, {  [tex]PCL_{5}[/tex]} is named phosphorus pentachloride according to the rules for naming binary molecular compounds.

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A 250.0 mL 1.25 M copper (II) sulphate pentahydrate solution requires 78.35 grammes. To make 250.0 mL of 0.25 M solution, add 0.050 L of the 1.25 M stock solution. Dissolving 75.831 g of copper (II) sulphate pentahydrate in 250.0 mL of distilled water yields a 1.210 M solution.

To calculate the mass of copper (II) sulfate pentahydrate needed, we can use the formula:

Concentration (in moles/L) × Volume (in L) × Molar mass (in g/mol) = Mass (in grammes).

Mass (in grammes) = 1.25 mol/L × 0.250 L × 249.68 g/mol.

Results calculation: Mass (g) = 78.35

To make a 1.25 M solution in 250.0 mL, you would need 78.35 grammes of copper (II) sulphate pentahydrate.

The 1.25 M stock solution of copper (II) sulphate needed to create 250.0 mL of a 0.25 M solution can be calculated using the dilution equation M1V1 = M2V2. To make 250.0 mL of 0.25 M solution, add 0.050 L (50 mL) of the 1.25 M stock solution.

We must convert 75.831 g of copper (II) sulphate pentahydrate to moles and divide by 250.0 mL of distilled water to compute the solution's molarity. Calculate copper (II) sulphate pentahydrate moles:

75.831 g/249.68 g/mol = moles.

We calculate solution molarity:

Molarity = Moles / Volume = (75.831 g/249.68 g/mol) / 0.250 L

Calculating result: 1.210 M.

Thus, 75.831 g of copper (II) sulphate pentahydrate dissolved in 250.0 mL of distilled water yields a 1.210 M solution.

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Using the equations in the pre-lab, the steps in the procedure, and observations made during lab, develop a model for the experiment, Therefore :

1. Adding HCl to ammonium oxalate forms NH₄Cl and H₂C₂O₄, creating a cloudy solution.

2. HCl reacts with calcium carbonate to produce CaCl₂ and CO₂, resulting in a cloudy solution with CO₂ bubbles.

3. Urea decomposition in water yields NH₃ and CO₂ gases, with NH₃ bubbling out and CO₂ dissolving, causing a warm reaction.

1. Molecular-level picture of the contents of the Ammonium oxalate solution (NH₄​)₂​C₂​O₄​ after HCl is added

The molecular-level picture of the contents of the ammonium oxalate solution (NH₄​)₂C₂​O₄​ after HCl is added would show the following:

2. Molecular-level picture of the contents of the unknown solution after HCl is added

The molecular-level picture of the contents of the unknown solution after HCl is added would show the following:

3. Molecular-level picture to describe what happens as the urea is decomposed

The molecular-level picture to describe what happens as the urea is decomposed would show the following:

The urea would react with water molecules to form ammonia and carbon dioxide gases. The ammonia gas would bubble out of the solution, and the carbon dioxide gas would dissolve in the solution.

Here are some additional details about the physical and chemical changes that occur in each of the reactions:

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The Haber-Bosch process is a crucial industrial process. The process is employed in the manufacture of ammonia, which is an important nitrogen-based compound.

Nitrogen is abundant in the air, comprising around 80% of the earth's atmosphere. The problem is that atmospheric nitrogen is very inert and does not readily react with other elements or molecules, making it very difficult to produce nitrogen-based compounds such as ammonia. The Haber-Bosch process involves the reaction of hydrogen and nitrogen gas to produce ammonia through a multi-step process. The first step in the process is the reaction of nitrogen and hydrogen to produce ammonia.

This reaction is exothermic and releases energy, which is used to drive the reaction forward. The second step is the removal of the ammonia from the reaction mixture. This is done by cooling the reaction mixture to a temperature where ammonia condenses into a liquid, which is then removed from the reaction mixture. The third step is the separation of the unreacted nitrogen and hydrogen gases from the ammonia product. This is done by passing the reaction mixture through a series of scrubbers that remove the unreacted gases from the ammonia product.

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