the solubility of sodium chloride at room room temperature is 35.8g of nacl per 135.8 g of solution and the density of this solution is 1.20g/ml. calculate the molarity of a saturated solution of nacl

The pH after the addition of 0.100 mol NaOH to 1.00 L of a buffer that is 0.700 M HF and 0.553 M NaF. The pH  is 7.031.

To calculate the pH after the addition of 0.100 mol NaOH to 1.00 L of a buffer that is 0.700 M HF and 0.553 M NaF, we can use the Henderson-Hasselbalch equation.

The Henderson-Hasselbalch equation is: pH = pKa + log ([A-]/[HA])

Where [A-] is the concentration of the anion (in this case, NaF) and [HA] is the concentration of the acid (in this case, HF).

pKa for HF is 7.1 x 10-4

Before we add the 0.100 mol NaOH, the pH of the buffer is:

pH = 7.1 x 10-4 + log ([0.553 M NaF]/[0.700 M HF])

= 7.1 x 10-4 + log(0.787)

= 7.1 x 10-4 + -0.103

= 6.997

Now, let's calculate the concentration of NaOH after we add 0.100 mol of it to the buffer. We know that 1 mole of NaOH will produce 1 mole of OH- ions, so the concentration of OH- ions is 0.100 M.

Since the buffer already contains HF and NaF, the total concentration of anions is 0.653 M.

We can now calculate the new pH using the Henderson-Hasselbalch equation:

pH = 7.1 x 10-4 + log([0.653 M anions]/[0.700 M HF])

= 7.1 x 10-4 + log(0.933)

= 7.1 x 10-4 + -0.069

= 7.031

Therefore, the pH of the buffer after the addition of 0.100 mol NaOH is 7.031.

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The equilibrium potential of a copper wire immersed in 0.0007 M CuSO₄ is 1.191 V.

To calculate the equilibrium potential of a copper wire immersed in 0.0007 M CuSO₄ solution, we need to use the Nernst equation. The Nernst equation is:

E = E° - (RT/nF) * ln(Q)

Where E is the equilibrium potential (in volts), E° is the standard electrode potential, R is the gas constant (8.314 J K-1 mol-1), T is the temperature (in Kelvin), n is the number of electrons transferred (2 in this case), F is Faraday’s constant (96485 C mol-1), and Q is the reaction quotient.

In this case, E° = 0.34 V, T = 298 K, n = 2, F = 96485 C mol-1, and Q = 0.0007 M CuSO4. Therefore, the equilibrium potential of the copper wire is:

E = 0.34 V - (8.314 J K-1 mol-1 * 298 K / (2 * 96485 C mol-1)) * ln(0.0007 M CuSO4)
E = 0.34 V - (-0.851 V)
E = 1.191 V

Therefore, the equilibrium potential of the copper wire immersed in 0.0007 M CuSO₄ solution at 25°C is 1.191 V.

Complete question:

Calculate the equilibrium potential of a copper wire immersed in 0.0007 M CuSO4 solution. The standard electrode potential for the reaction Cu2+ + 2e- = Cu0 at 25°C is 0.34 V (NHE).

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There are 567.8 mmHg in 75.7 kpa.

Pressure is the amount of force that is applied over a given area divided by the size of this area.

The units of pressure are as follows:

In SI units, pressure is measured in pascals where;

1 kPa = 760/101.325 = 7.5 mmHg

Hence, 75.7kpa is equal to 567.8 mmHg

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The total number of chain bonds in an average molecule is 4.

To calculate the repeat unit molecular weight, you need to add the atomic weights of all the atoms in the molecule and multiply the sum by the number of repeat units in the molecule.

How to calculate the repeat unit molecular weight?

Step 1: Determine the molecular formula of the polymer unit.

Step 2: Find the atomic weights of all the atoms present in the repeat unit.

Step 3: Add up the atomic weights of all the atoms in the molecule.

Step 4: Multiply the sum by the number of repeat units in the molecule.

Here's an example:

Polymer unit: CH2CHCl

Atomic weights: C = 12.01 g/mol, H = 1.008 g/mol, Cl = 35.45 g/mol

Molecular weight = (12.01 x 2) + 1.008 + 35.45

= 60.49 g/mol

Repeat unit = (CH2CHCl)n

Repeat unit molecular weight = 60.49 x n, where n is the number of repeat units in the molecule.

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Compounds that can undergo an addition and elimination reaction mechanism are OH and OCH3 since they are the ones that have a nucleophilic site or a leaving group.

The compounds that can undergo an addition and elimination reaction mechanism are listed below: OH - It is a hydroxyl group and is a nucleophile, which means it has an electron pair available for donation. OCH3 - Methoxy group, also known as OCH3, is a leaving group.

Addition reactions occur when two or more reactants combine to form a single product. They typically involve unsaturated compounds like alkenes or alkynes, which have double or triple carbon-carbon bonds. Elimination reactions, on the other hand, involve the removal of elements from a reactant to create a more unsaturated product, typically forming a double bond.

OH: This group represents an alcohol functional group. Alcohols can undergo elimination reactions, such as dehydration, to form alkenes.
OCH: This seems to be an incomplete functional group, as it is missing a carbon or hydrogen. If it's meant to represent an ether functional group (OCH3 or OCH2R, where R is an alkyl group), ethers generally do not undergo addition or elimination reactions.

In conclusion, without further information about compounds A, B, C, D, and E, we can only determine that a compound containing an OH functional group (an alcohol) can undergo elimination reactions, while the given OCH functional group does not undergo addition or elimination reactions.

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The star is moving by a velocity of 3 *10^{5}.

The formula for the Doppler shift is given by

f2/f1 = (c-v)/c,

where c is the speed of light, v is the velocity of the moving object, and f1 and f2 are the emitted and received frequencies of light, respectively.

The Doppler effect occurs when the light source and the observer are moving relative to one another, giving the impression that the light's frequency has changed.

The Doppler effect alters the frequency of light from a moving source, shifting it either to the red or blue. This resembles (but does not necessarily mimic) the behavior of other types of waves, such as sound waves.

The star is moving away from the observer because the wavelength of the spectral line has shifted to a longer wavelength.

doppler shift

Thus, the velocity is given by the formula

:v/c = (Δλ/λ)

where  is the rest wavelength and  is the change in wavelength.

v/c = (Δλ/λ)v/c = (1000.1 - 1000.0)/1000.0v/c = 0.0001/1000.

0v/c = 1e-7v = (1e-7) × c = 300 × 1e-7 = 3e-5

The star is moving away from the observer at a velocity of[tex]3 *10^{5}[/tex]m/s.

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0.81 moles of O2 would be produced if 1.62 moles of H2O2 were to undergo decomposition

The balanced chemical equation for the decomposition of hydrogen peroxide (H2O2) is:

2 H2O2 → 2 H2O + O2

This means that for every 2 moles of hydrogen peroxide, 1 mole of oxygen gas is produced. So to calculate the number of moles of O2 produced when 1.62 moles of H2O2 decompose, we need to use a proportion:

2 mol H2O2 : 1 mol O2 = 1.62 mol H2O2 : x mol O2

where "x" is the number of moles of O2 produced.

To solve for "x", we can cross-multiply and simplify:

2 mol H2O2 * x mol O2 = 1 mol O2 * 1.62 mol H2O2

2x = 1.62

x = 0.81

Therefore, 0.81 moles of O2 would be produced if 1.62 moles of H2O2 were to undergo decomposition.

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The concentration of dpip is 0.02 mmol/L.

To calculate the concentration of dpip, we can use the Beer-Lambert Law, which states that the absorbance (A) of a solution is directly proportional to the concentration (c) of the absorbing species and the path length (l) of the sample cell:

A = εcl

where;

In this case, we are given the absorbance value (A) and the molar extinction coefficient (ε), so we can rearrange the equation to solve for the concentration (c):

c = A / (εl)

Substituting the given values, we get:

c = 0.426 / (21.3/(mm cm) x 1 cm)

c = 0.426 / 21.3

c = 0.02 mmol/L

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The molecular equation for the gas evolution reaction between aqueous hydrobromic acid (HBr) and aqueous lithium sulfite (Li2SO3) is as follows:  2 HBr (aq) + [tex]Li_{2} So_{3}[/tex] (aq) → 2 LiBr (aq) + [tex]H_{2} So_{3}[/tex] (aq)

In this reaction, hydrobromic acid (HBr) reacts with lithium sulfite ([tex]Li_{2} So_{3}[/tex]) to form lithium bromide (LiBr) and sulfurous acid ([tex]H_{2} So_{3}[/tex]). The sulfurous acid is unstable and decomposes into water( [tex]H_{2o[/tex]) and sulfur dioxide gas ([tex]So_{2}[/tex]):

[tex]H_{2} So_{3}[/tex] (aq) → [tex]H_{2} 0[/tex]l) + [tex]So_{2}[/tex] (g)

The overall reaction is:

2 HBr (aq) + [tex]Li_{2} So_{3}[/tex] (aq) → 2 LiBr (aq) + [tex]H_{2} o[/tex] (l) + [tex]So_{2}[/tex] (g)

In this gas evolution reaction, the mixing of the two aqueous solutions results in the formation of a new compound, lithium bromide, which remains dissolved in the solution. The other product, sulfurous acid, decomposes into water and sulfur dioxide gas, which is released as bubbles in the solution. This release of gas is the characteristic feature of gas evolution reactions.

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The concentration of sodium chloride (NaCl) in the solution is 840,000 parts per billion (ppb).

To calculate this, divide the mass of sodium chloride (8424 mg) by the mass of water (0.1711 kg), then multiply the result by 1 billion (10^9).

To calculate the concentration of a solution, you must first determine the mass of the solute (NaCl in this case). The mass of the solute is given in the question as 8424 mg.

The mass of the solvent (water) is given as 0.1711 kg.

To calculate the concentration of the solution, divide the mass of the solute by the mass of the solvent, and then multiply the result by 1 billion (10^9).

In this example, 8424 mg divided by 0.1711 kg is equal to 49,336,297, which multiplied by 1 billion is equal to 49,336,297,000,000, or 840,000 parts per billion (ppb).

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As you rise from low light intensity to higher light intensity, the rate of photosynthesis will increase because there is more light available to drive the reactions of photosynthesis.

Under standard conditions (298 K and 1 atm), neither statement is true.

Diamond and graphite are both forms of carbon and are in a state of equilibrium under standard conditions. This means that neither diamond nor graphite will spontaneously convert to the other form.

Therefore, the correct answer is option (c): none of the above.

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The thermodynamic equilibrium constant In a chemical equilibrium, K is the appropriate quotient of species activities. Under normal temperatures and pressures, an activity cannot be very many orders of magnitude more than 1.

The definition of thermodynamic properties is "system characteristics that can specify the state of the system." Certain constants, like R, are not attributes since they do not describe the state of a system.

Thermodynamics states that the conversion of diamond to graphite occurs spontaneously and is favourable. Yet, this reaction moves extremely slowly because kinetics, not thermodynamics, regulates it. As a result, diamond is thermodynamically unstable but kinetically stable.

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The forward reaction is favored when the graph shows that the reactant concentration is higher than the product concentration.

To determine which reaction is favored, examine the graph and look at the concentrations of reactants and products at equilibrium. If the reactant concentration is higher, the forward reaction is favored. Conversely, if the product concentration is higher, the reverse reaction is favored.

A graph can help you visualize the reactants and products of a reaction at equilibrium. The y-axis of the graph typically indicates the concentration of the reactants or products, and the x-axis of the graph indicates the reaction rate.

At equilibrium, the reaction rate is 0, meaning that the reactants and products are neither increasing nor decreasing in concentration. By looking at the concentrations of the reactants and products at equilibrium on the graph, you can determine which reaction is favored.

If the reactant concentration is higher than the product concentration, then the forward reaction is favored. This means that the forward reaction occurs more quickly than the reverse reaction when the concentrations of the reactants and products are equal.

Conversely, if the product concentration is higher than the reactant concentration, then the reverse reaction is favored.

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The ratio of [HCO₃⁻] to [HCO₂H] in an acetate buffer is 5.01.

The ratio of [HCO₃⁻] to [HCO₂H] (formic acid) in an acetate buffer at pH 4.50 is determined by the Henderson-Hasselbalch equation:

pH = pKa + log ([HCO₃⁻]/[HCO₂H]).
[HCO₃⁻]/[HCO₂H] = 10^(pH-pKa)
= 10^(4.50 - 3.80)
= 5.01


To further understand the buffering capacity of an acetate buffer, we must first understand the role of formic acid and bicarbonate in an acetate buffer.

Formic acid is an organic acid and bicarbonate is a salt of carbonic acid. Both of these species can form and break down as needed to maintain the pH of the buffer.

As the pH of the buffer is increased, the formic acid will break down, forming more bicarbonate.

On the other hand, as the pH of the buffer is decreased, more formic acid will form, resulting in fewer bicarbonate ions.

The buffering capacity of an acetate buffer is dependent on the relative concentrations of formic acid and bicarbonate ions, and these concentrations can vary depending on the pH of the buffer.

In summary, the ratio of [HCO₃⁻] to [HCO₂H] is found to be 5.01 in an acetate buffer at pH 4.50.

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The concentration of sulfuric acid that remains after neutralization is 0.056 M.

To find out what concentration of sulfuric acid remains after neutralization, you will need to use the balanced equation for the reaction:

H2SO4 + 2KOH → K2SO4 + 2H2O

First, you will need to determine the moles of each reactant in the solution.

Moles can be determined using the formula:

moles = concentration x volume

In this case:

moles of H2SO4 = 0.850 L x 0.400 M = 0.34 mol

moles of KOH = 0.800 L x 0.250 M = 0.2 mol

Since the reaction is a 1:2 ratio, you will need to determine which reactant is limiting the reaction.

To do this, compare the mole ratios of the reactants:

0.34 mol H2SO4 : 0.2 mol KOH = 1.7 : 1

Since the ratio of H2SO4 to KOH is greater than 1:2, KOH is the limiting reactant. Therefore, all of the KOH is used up in the reaction, leaving some H2SO4 unreacted.

To find the amount of H2SO4 remaining, you will need to use the mole ratio of H2SO4 to KOH.

Since 2 moles of KOH react with 1 mole of H2SO4, you can use the mole ratio:

0.2 mol KOH x (1 mol H2SO4 / 2 mol KOH) = 0.1 mol H2SO4 remaining

Finally, you can determine the concentration of the H2SO4 remaining:

concentration = moles / volume

concentration = 0.1 mol / (0.850 L + 0.800 L)

concentration = 0.056 M

Therefore, the concentration of sulfuric acid that remains after neutralization is 0.056 M.

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The Ultimate BOD ( Biochemical Oxygen Demand) of the waste is 81.3 mg/L when the 5-day BOD is 20 mg/L and the BOD rate coefficient is 0.15/day.

The Ultimate Biochemical Oxygen Demand (BOD) is defined as he quantity of oxygen required to stabilize or,

eliminate biodegradable organic matter in wastewater by the action of aerobic microorganisms under controlled laboratory conditions at a specified temperature over a period of time.

The 5-day BOD is measured by calculating the amount of oxygen consumed by microorganisms over a period of five days.

The Ultimate BOD of a waste can be determined by knowing the 5-day BOD and BOD rate coefficient. The following formula is used to determine the Ultimate BOD:

Ultimate BOD = 5-day BOD × [(e^(k×t))-1] / e^(k×t)Where k is the BOD rate coefficient and t is the time required to reach the Ultimate BOD.

The Ultimate BOD of the waste as follows: 5-day BOD = 20 mg/L k = 0.15/day t = ? Ultimate BOD = 5-day BOD × [(e^(k×t))-1] / e^(k×t) Ultimate BOD = 20 × [(e^(0.15×t))-1] / e^(0.15×t)

The Ultimate BOD is reached after 20 days. Ultimate BOD = 20 × [(e^(0.15×20))-1] / e^(0.15×20) Ultimate BOD = 81.3 mg/L

Therefore, the Ultimate BOD of the waste is 81.3 mg/L when the 5-day BOD is 20 mg/L and the BOD rate coefficient is 0.15/day. The coefficient is the numerical multiplier of a variable or quantity that follows a term or a factor.

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The isotope that yields four neutrons and the artificial isotope dubnium-260 when bombarded with nitrogen-15 is curium-244.

Curium-244 is a transuranic element of the actinide series. When bombarded with nitrogen-15, a nucleus of curium-244 splits into two smaller nuclei, releasing four neutrons in the process.

This process is called nuclear fission. The nucleus of nitrogen-15 is then combined with the two smaller nuclei to form dubnium-260, which is an artificially produced isotope.

Nuclear fission of curium-244 is a common process used in nuclear power plants. In nuclear power plants, uranium-235 is bombarded with neutrons, causing a chain reaction that produces energy and more neutrons.

The neutrons then bombard other uranium-235 nuclei, continuing the process. By bombarding curium-244 with nitrogen-15, a similar chain reaction is created that produces dubnium-260.

The production of dubnium-260 through nuclear fission of curium-244 can be used for various scientific and industrial purposes.

It can be used in the production of nuclear weapons, nuclear fuel, medical isotopes, and in other research activities.

In addition, it can be used as a catalyst for chemical reactions, to produce high energy radiation for sterilization, and for other industrial processes.

In conclusion, curium-244 yields four neutrons and the artificial isotope dubnium-260 when bombarded with nitrogen-15.

This process, known as nuclear fission, can be used in a variety of scientific and industrial applications.

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Answer:

The reaction will reach equilibrium faster.

Explanation:

just took this lesson

At the equivalence point of a titration between 24.6 mL of 0.389 M ethylamine, C2H5NH2, and 0.325 M hydroiodic acid, the pH is 0.

At the equivalence point of a titration between 24.6 mL of 0.389 M ethylamine, C2H5NH2, and 0.325 M hydroiodic acid, the pH is 0. The equation for the reaction is:

C2H5NH2 + HI → C2H5NH3+ + I-

The number of moles of hydroiodic acid, HI, needed to reach the equivalence point is equal to the number of moles of ethylamine, C2H5NH2. To calculate this, use the following equation:

Moles of HI = Moles of C2H5NH2

Volume of C2H5NH2 x Molarity of C2H5NH2 = Volume of HI x Molarity of HI

24.6 mL x 0.389 M = Volume of HI x 0.325 M

Volume of HI = 24.6 mL x 0.389 M / 0.325 M

Volume of HI = 30.53 mL

At the equivalence point, the pH of the solution is 0.

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Answer:

pressure on a balloon holding 433 ml of an ideal gas is increased from 688 torr to 1.00 atm. what is the newpressure on a balloon holding 433 ml of an ideal gas is increased from 688 torr to 1.00 atm. what is the new volume of the balloon (in ml) at constant temperature

When determining the energy effect of a chemical reaction, the system is/are reactants and products and the surroundings are everything outside the system.

When determining the energy effect of a chemical reaction, the system and surroundings are involved. The system refers to the reactants and products involved in the chemical reaction, whereas the surroundings refer to everything else outside the system, including the temperature, pressure, and any other factors that can affect the reaction.

The energy effect of a chemical reaction can be determined by calculating the difference between the energy of the products and the energy of the reactants. This difference is known as the energy change or the enthalpy change of the reaction.

If the energy change is positive, it means that the reaction is endothermic, and energy is absorbed from the surroundings. This results in a decrease in the temperature of the surroundings.

On the other hand, if the energy change is negative, it means that the reaction is exothermic, and energy is released into the surroundings. This results in an increase in the temperature of the surroundings.

It is important to note that the energy effect of a chemical reaction can also be affected by external factors such as pressure, temperature, and the presence of a catalyst.

In conclusion, the system is the reactants, and the products and surroundings are factors like temperature and pressure, i.e., everything outside the system.

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For precipitation to occur, the value of Qsp (the ion product constant) should be greater than the solubility product constant (Ksp).

Precipitation is the conversion of a dissolved substance into a solid, which then settles out of a solution. Precipitation occurs when a liquid solution is cooled or heated, causing it to become super-saturated with one or more solutes. A solution's super-saturation means that it contains more of a solute than it can contain at equilibrium.

A tiny seed crystal of the solute is added to the solution to kick off the precipitation. The seed crystal provides a template for the rest of the solute to nucleate and form a solid. For precipitation to occur, the value of Qsp (the ion product constant) should be greater than the solubility product constant (Ksp). When Qsp is greater than Ksp, the solution is supersaturated and precipitates are formed. If Qsp is less than Ksp, the solution is unsaturated and no precipitation occurs.

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Acid rain effects are detrimental to the environment. The most common method to neutralize acid rain is Lime Neutralization. When the pH level decreases, Acid rain becomes too acidic, and it can have an adverse effect on the environment

The acid rain causes the water to become too acidic and the pH level decreases. It is very harmful to plants and wildlife. It damages buildings and monuments and also affects the water bodies. It affects the aquatic life, and the creatures living in it. It is essential to prevent this from happening.

Several methods can be used to neutralize acid rain. They are as follows:

Lime neutralization: It is one of the most common methods to neutralize acid rain. It is a process in which lime is added to the water body to neutralize the acid content.

Buffering: The buffering capacity is used to treat the water. Buffering capacity is the ability of the water to neutralize acid. The water with a higher buffering capacity will neutralize more acid than the water with less buffering capacity.

Gas scrubbing: It is a process in which the smokestacks from factories and other industries are fitted with scrubbers. These scrubbers help in trapping the pollutants that are released into the atmosphere.

The pH level change plays a significant role in acid rain. When the pH level decreases, it becomes too acidic, and it can have an adverse effect on the environment. It can cause the aquatic life to die, and it can damage the buildings and monuments. It is crucial to control the pH level to prevent such damage. The pH level of 7 is considered neutral. The pH level lower than 7 is acidic, and higher than 7 is alkaline. Hence, it is essential to control the pH level to prevent damage from acid rain.

Thus, Acid rain effects are detrimental to the environment. And it is very important to prevent and control the pH level to prevent damage from acid rain.

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

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

M1V1 = M2V2

Where M1 is the concentration of the stock solution,

V1 is the volume of the stock solution required,

M2 is the concentration of the final solution, and

V2 is the final volume of the solution.

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

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

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

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

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

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

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

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This herd of cattle released 8.44 metric tons of methane (CH4) into the atmosphere. Methane is composed of one atom of carbon and four atoms of hydrogen, so this 8.44 metric tons of methane contained (8,440 kg) x (12.01/16.05) g/kg = 6,309 kg (6.31 metric tons).

To answer the given question, we need to know the molecular formula of methane, which is CH4. The atomic mass of carbon is 12.01 g/mol and the atomic mass of hydrogen is 1.01 g/mol. Therefore, the molecular mass of methane is:

Molecular mass of CH4 = (1 x 12.01) + (4 x 1.01) = 16.05 g/mol
Now, we need to convert the amount of methane released into metric tons.
1 metric ton = 1,000 kg
8.44 metric tons = 8.44 x 1,000 = 8,440 kg

To convert the mass of methane into mass of carbon, we need to use the ratio of the molecular masses of carbon and methane.

1 mol of CH4 contains 1 mol of carbon
1 mol of CH4 has a mass of 16.05 g
1 mol of carbon has a mass of 12.01 g

Therefore,
16.05 g of CH4 contains 12.01 g of carbon
1 kg of CH4 contains (12.01/16.05) g of carbon

To convert the mass of methane into mass of carbon, we need to multiply it by the ratio of the molecular masses of carbon and methane.
Mass of carbon = (8,440 kg) x (12.01/16.05) g/kg
= 6,309 kg

Therefore, the herd of cattle released 6,309 kg (or 6.31 metric tons) of carbon into the atmosphere through the release of 8.44 metric tons of methane.

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Potassium hydrogen phthalate (KHP) is used to standardize the NaOH solution instead of measuring out a known mass of solid NaOH to be dissolved in water because KHP has several benefits that make it a better option.

Potassium hydrogen phthalate (KHP) is a crystalline powder that has a chemical formula of KHC8H4O4. It is a primary standard for acid-base titrations, meaning that its molar mass and purity are known to a high degree of accuracy.

KHP is stable when exposed to air and is easy to obtain.4. KHP is inexpensive in comparison to other primary standards, such as sodium carbonate, which is costly and difficult to prepare.KHP reacts with NaOH in a 1:1 molar ratio, so one mole of KHP reacts with one mole of NaOH. Because the amount of KHP used in the titration is known, the concentration of the NaOH solution can be determined mathematically.

A secondary standard, such as NaOH, can be standardized using KHP, which is a primary standard. As a result, it is preferred to use KHP to standardize NaOH rather than measuring out a known mass of solid NaOH to be dissolved in water.

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The mass of methanol in a 55.0-g aqueous solution of methanol, CH4O, is 5.53 g and the mass of water is 27.91 g. when the mole fraction of methanol is 0.100.

The mass of each component in a 55.0-g aqueous solution of methanol, CH4O, can be found by using the mole fraction of methanol (0.100).

First, calculate the total number of moles of the solution:
55.0 g x (1 mol/32.04 g) = 1.72 moles

Then, calculate the number of moles of methanol:
1.72 moles x (0.100 mole fraction) = 0.172 moles

Finally, calculate the mass of each component:
Methanol mass: 0.172 moles x (32.04 g/mol) = 5.53 g
Water mass: 1.72 moles - 0.172 moles = 1.55 moles x (18.02 g/mol) = 27.91 g

Therefore, the mass of methanol in a 55.0-g aqueous solution of methanol, CH4O, is 5.53 g and the mass of water is 27.91 g.

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Yes, it is true that the electronic configuration of O2- is 1s2 2s2 2p6.

Arrangement of electrons in orbitals around atomic nucleus is called electronic configuration and describes how electrons are distributed in its atomic orbitals.

When oxygen atom gains two electrons to form an O2- ion, the two electrons occupy the lowest energy level available, which is the 2s orbital. Therefore, the electronic configuration of O2- is the same as that of neon (1s2 2s2 2p6), which has a full outermost shell of electrons. This noble gas configuration makes the O2- ion stable and less likely to react with other elements.

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The energy of a photon with a wavelength of 410 nm is equal to 3.03 eV.

To calculate this, you can use the formula E = hc/λ, where h is Planck's constant, c is the speed of light, and λ is the wavelength. Plugging in the values, you get E = (6.626x10⁻³⁴J·s)(3.0x10⁸m/s)/(410x10⁻⁹m) = 4.839 × 10-19 J = 3.03 eV.

An atomic transition produces a photon with a wavelength of 410 nm. The energy of this photon is 3.03 eV.

The following formula can be used to calculate the energy of a photon.

Energy = Planck's constant x (speed of light/wavelength).

Here, Planck's constant is (h) = 6.626 × 10⁻³⁴ J s. The speed of light is (c) = 3 × 10⁸m/s (in a vacuum). The wavelength of the photon is (λ) = 410 nm.

So, let's first convert the wavelength to meters (1 nm =10⁻⁹ m).

So, 410 nm = 410 × 10⁻⁹ m = 4.10 × [tex]10^{-7}[/tex]m. Now, we can calculate the energy of the photon using the formula.

Energy = h x (c/λ)

Energy = 6.626 × 10⁻³⁴ J s x (3 × 10⁸ m/s / 4.10 × [tex]10^{-7}[/tex] m)

Energy = 4.839 × [tex]10^{-19}[/tex] J (joules)

One electron volt is equal to 1.6 × [tex]10^{-19}[/tex]J.

So, we can convert the energy from joules to electron volts.

Energy (in eV) = Energy (in J) / (1.6 × [tex]10^{-19}[/tex]J/eV)

Energy (in eV) = 4.839 × [tex]10^{-19}[/tex]J / (1.6 × [tex]10^{-19}[/tex]J/eV)

Energy (in eV) = 3.03 eV

Therefore, the energy of the photon is 3.03 eV.

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