Calculate the mass percent by
volume of 281.1 g of glucose
(C6H12O6, MM = 180.2 g/mol) in
325 mL of solution.

The solubility of the salts is affected by the temperature changes. 1. NaCl is least affected by temperature. 2. supersaturated. 3. 60 grams KBr. 4. Ethanol has both polar and non-polar groups. 5. Mixing and shaking.

A KBr solution with 90 gm solute in 100 grams of water at 50 degrees is classified as supersaturated. 60 grams of KBr are needed to make a saturated solution in 100 gm of water at 30 degrees.

Ethanol is a general solvent due to the presence of both the polar and the non-Polar groups. As a result, it is easier to dissolve both polar molecules and non-Polar molecules. The dissolving rate can be increased by mixing or shaking the solution. Also, the sugar dissolves faster in hot than cold tea.

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Theoretically, if 20 grams of hydrogen reacts then 112.5 grams of ammonia is produced.

The balanced chemical equation can be given as:

N₂+3H₂→ 2NH₃

From stoichiometry, 2 mol of NH₃is produced from 3 mol of H₂

5 mol of NH₃ will be produced from = 3/2×5 = 7.5 mol of H₂

∴mass of H₂=7.5×2= 15gm of H₂.

Excess reagents are those reactants in a chemical reaction that are not exhausted at the end of the reaction. A completely exhausted or reacted reagent is called a limiting reagent because its amount limits the number of products formed. In this reaction, the excess reagent is Nitrogen as 35 grams of nitrogen and 15 grams of hydrogen react to produce 34 grams of ammonia.

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The molarity of the calcium ion (Ca²⁺) in the solution is 0.411 M and the molarity of the hydroxide ions (OH⁻) in the solution is 0.822 M.

The molarity of ions in a solution can be determined by considering the dissociation of the compound into its constituent ions. In the case of Ca(OH)₂, it dissociates into one calcium ion (Ca²⁺) and two hydroxide ions (OH⁻) per formula unit.

Since the solution is 0.411 M Ca(OH)₂, the molarity of the calcium ion (Ca²⁺) would also be 0.411 M because there is one calcium ion for every formula unit of Ca(OH)₂. The molarity of the hydroxide ions (OH⁻) would be twice that of the Ca²⁺ ion because there are two hydroxide ions per formula unit of Ca(OH)₂.

The molarity of the hydroxide ions = 2 × 0.411 M = 0.822 M.

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To calculate the pH of the solution obtained by mixing HCl and NaOH, we need to consider the neutralization reaction between the two compounds. The reaction between HCl (hydrochloric acid) and NaOH (sodium hydroxide) produces water (H₂O) and forms a salt (NaCl).

Given:

Volume of HCl solution (V₁) = 40 cm³

Concentration of HCl solution (C₁) = 0.2 M

Volume of NaOH solution (V₂) = 30 cm³

Concentration of NaOH solution (C₂) = 0.1 M

1. Determine the moles of HCl and NaOH used:

Moles of HCl = Concentration (C₁) × Volume (V₁)

Moles of HCl = 0.2 M × 0.04 L (converting cm³ to L)

Moles of HCl = 0.008 mol

Moles of NaOH = Concentration (C₂) × Volume (V₂)

Moles of NaOH = 0.1 M × 0.03 L (converting cm³ to L)

Moles of NaOH = 0.003 mol

2. Determine the limiting reagent:

The stoichiometry of the reaction between HCl and NaOH is 1:1, meaning that they react in a 1:1 ratio. Whichever reactant is present in a smaller amount will be the limiting reagent.

In this case, NaOH is present in a smaller amount (0.003 mol), which means it will be fully consumed during the reaction.

3. Determine the excess reagent and its remaining moles:

Since NaOH is the limiting reagent, we need to find the remaining moles of HCl.

Moles of HCl remaining = Moles of HCl initially - Moles of NaOH reacted

Moles of HCl remaining = 0.008 mol - 0.003 mol

Moles of HCl remaining = 0.005 mol

4. Calculate the concentration of HCl in the resulting solution:

Volume of resulting solution = Volume of HCl solution + Volume of NaOH solution

Volume of resulting solution = 0.04 L + 0.03 L

Volume of resulting solution = 0.07 L

Concentration of HCl in the resulting solution = Moles of HCl remaining / Volume of resulting solution

Concentration of HCl in the resulting solution = 0.005 mol / 0.07 L

Concentration of HCl in the resulting solution ≈ 0.071 M

5. Calculate the pH of the resulting solution:

pH = -log[H⁺]

pH = -log(0.071)

Using logarithm properties, we can determine the pH value:

pH ≈ -log(0.071)

pH ≈ -(-1.147)

pH ≈ 1.147

Therefore, the pH of the solution obtained by mixing 40 cm³ of 0.2 M HCl and 30 cm³ of 0.1 M NaOH is approximately 1.147.

The complete reaction, according to the law of conservation of mass is:

XX + YY → 2XY

The Law of Conservation is a fundamental principle in chemistry and physics. It states that in a closed system, mass cannot be created or destroyed during a chemical reaction or a physical change. The total mass of the substances involved before the reaction or change must equal the total mass of the substances after the reaction or change.

This principle is based on the understanding that atoms are not created or destroyed, but they can combine or separate to form different substances.

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To determine the molal concentration of a solution, we need to know the amount of solute (ethanol) in moles and the mass of the solvent (usually water) in kilograms.

Given that the solution is 30% ethanol, it means that there are 30 grams of ethanol in 100 grams of the solution. Let's assume we have 100 grams of the solution.

To find the amount of ethanol in moles, we need to convert grams to moles using the molar mass of ethanol (C2H5OH).

The molar mass of C2H5OH:

2 * atomic mass of carbon (C) = 2 * 12.01 g/mol = 24.02 g/mol

6 * atomic mass of hydrogen (H) = 6 * 1.01 g/mol = 6.06 g/mol

1 * atomic mass of oxygen (O) = 1 * 16.00 g/mol = 16.00 g/mol

1 * atomic mass of hydrogen (H) = 1 * 1.01 g/mol = 1.01 g/mol

Total molar mass of C2H5OH = 24.02 + 6.06 + 16.00 + 1.01 = 47.09 g/mol

Now, let's calculate the amount of ethanol in moles:

30 grams ethanol * (1 mol / 47.09 g) = 0.637 moles ethanol

Next, we need to determine the mass of the solvent (water) in kilograms. Let's assume we have 100 grams of the solution, so the mass of water would be 100 - 30 = 70 grams.

Converting the mass of water to kilograms:

70 grams * (1 kg / 1000 grams) = 0.07 kg

Finally, we can calculate the molal concentration (m) using the formula:

molal concentration (m) = moles of solute/mass of solvent in kilograms

m = 0.637 moles / 0.07 kg ≈ 9.10 mol/kg

Therefore, the molal concentration of the 30% ethanol solution (C2H5OH) is approximately 9.10 mol/kg.

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The Statement 3 (the products have more potential energy than the reactants in an exothermic reaction) is partially correct because the products do have lower potential energy than the reactants in an exothermic reaction.

The correct statement regarding endothermic and exothermic reactions is:

Energy is absorbed in an endothermic reaction.

In an endothermic reaction, energy is taken in from the surroundings, usually in the form of heat. The reactants have a lower energy level than the products, so energy must be absorbed to reach the higher energy state of the products. This energy absorption causes a decrease in the temperature of the surroundings, making the reaction feel cold.

On the other hand, in an exothermic reaction, energy is released. The reactants have a higher energy level than the products, so energy is released during the reaction, usually in the form of heat. This energy release causes an increase in the temperature of the surroundings, making the reaction feel warm or hot.

Therefore, statement 2 (energy is released in an endothermic reaction) and statement 4 (the products have more potential energy than the reactant in an endothermic reaction) are incorrect.

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The mass of CaCl₂ required  to make a 1L solution of 3M CaCl₂ is equal to 332.94 g, hence option C is correct.

To find the mass of CaCl₂ required to make a 3M solution, it considers the molar mass of CaCl2 and the desired concentration.

The molar mass of CaCl₂ can be observed as follows:

Molar mass (CaCl₂) = (molar mass of Ca) + 2 × (molar mass of Cl)

= (40.08 g/mol) + 2 × (35.45 g/mol)

= 40.08 g/mol + 2 × 35.45 g/mol

= 40.08 g/mol + 70.90 g/mol

= 110.98 g/mol

Now, by using the formula for molarity to find the mass of CaCl₂ required:

Molarity (M) = (moles of solute) / (volume of solution in liters)

Arrange the formula to solve for moles of solute:

(moles of solute) = (Molarity) × (volume of solution in liters)

It is required to make a 1L solution of 3M CaCl₂:

(moles of CaCl2) = (3 mol/L) × (1 L)

= 3 mol

Finally, find the mass of CaCl₂ using the moles and molar mass:

(mass of CaCl2) = (moles of CaCl₂ × (molar mass of CaCl₂)

= 3 mol × 110.98 g/mol

= 332.94 g

Thus, the mass of CaCl2 required to make a 1L solution of 3M CaCl₂ is  332.94 g.

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An atom has 17 protons and 17 electrons. The atom's charge is neutral. The positive charge of the 17 protons in this atom is balanced by the negative charge of the 17 electrons.

The ratio of an atom's protons, which have a positive charge, to its electrons, which have a negative charge, determines the charge of the atom. The quantity of protons in an electrically neutral atom is equal to the quantity of electrons.

The positive charge of the 17 protons in this atom is balanced by the negative charge of the 17 electrons, since there are 17 protons and 17 electrons in it. Consequently, the atom is electrically neutral or has a net charge of zero.

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1.To calculate the concentration of ammonium ions in water, we need to determine the number of moles of ammonium ions and then convert it to milligrams per liter (mg/L).

Given:

Volume of water = 30 ml

Volume of ammonia released = 200 ml

First, we need to convert the volume of ammonia released to the volume of water. Since the ammonia was released from the reaction with sodium hydroxide, the volume of ammonia released is equivalent to the volume of water used. Therefore, the volume of water used is 200 ml.

Next, we'll calculate the number of moles of ammonium ions:

Molar volume of water = 18.015 g/mol

Volume of water used = 200 ml = 0.2 L

The molar ratio between sodium hydroxide and ammonium ions is 1:1. Therefore, the number of moles of ammonium ions is equal to the number of moles of sodium hydroxide used.

Now, let's calculate the number of moles of sodium hydroxide used:

Molar mass of sodium hydroxide (NaOH) = 22.99 g/mol + 16.00 g/mol + 1.01 g/mol = 39.99 g/mol

The concentration of sodium hydroxide in water is not provided. If you have the concentration of sodium hydroxide, we can use it to determine the number of moles of sodium hydroxide used. Without that information, we cannot calculate the number of moles of ammonium ions and, subsequently, the concentration of ammonium ions in water.

2. To determine the mass of barium in 1 ml of drinking water, we'll use the information given:

Volume of drinking water = 20 ml

Mass of barium carbonate precipitate formed = 20 mg

We need to calculate the mass of barium in the precipitate and then convert it to milligrams per milliliter (mg/ml).

The molar mass of barium carbonate (BaCO₃) is:

Molar mass of barium (Ba) = 137.33 g/mol

Molar mass of carbonate (CO₃) = 12.01 g/mol + (3 × 16.00 g/mol) = 60.01 g/mol

Molar mass of barium carbonate (BaCO₃) = 137.33 g/mol + 60.01 g/mol = 197.34 g/mol

The molar ratio between barium carbonate and barium is 1:1. Therefore, the number of moles of barium in the precipitate is equal to the number of moles of barium carbonate formed.

Now, let's calculate the number of moles of barium carbonate:

Mass of barium carbonate precipitate formed = 20 mg = 0.020 g

Number of moles of barium carbonate = Mass of barium carbonate / Molar mass of barium carbonate

= 0.020 g / 197.34 g/mol

Finally, we'll calculate the mass of barium in 1 ml of drinking water:

Volume of drinking water = 20 ml

Mass of barium in 1 ml of drinking water = (Number of moles of barium carbonate / Volume of drinking water) × Molar mass of barium

= (0.020 g / 197.34 g/mol) / 20 ml × 137.33 g/mol

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Different gases effuse at different rates due to the relationship between their molecular masses, average velocities, and kinetic energy.

Lighter gases have higher average velocities and effuse more rapidly, while heavier gases have lower average velocities and effuse at slower rates. Graham's law of effusion provides a quantitative explanation for this phenomenon.

Different gases effuse at different rates due to variations in their molecular masses and average velocities. Effusion is the process by which gas molecules escape through a small opening or porous barrier into a vacuum or a region of lower pressure.

According to Graham's law of effusion, the rate of effusion of a gas is inversely proportional to the square root of its molar mass. Mathematically, it can be expressed as:

Rate A / Rate B = √(Molar mass B / Molar mass A)

This means that lighter gas molecules, with lower molar masses, effuse faster compared to heavier gas molecules. The reason behind this can be understood by considering the kinetic theory of gases.

Gas molecules are in constant random motion, colliding with each other and the walls of the container. The average velocity of gas molecules is directly related to their kinetic energy, which depends on their mass and temperature. Lighter gas molecules have higher average velocities due to their lower mass and therefore higher kinetic energy.

During effusion, gas molecules near the opening of the container collide with the walls more frequently and possess higher velocities. Lighter gas molecules have a higher chance of having a velocity that exceeds the escape velocity threshold, allowing them to effuse more easily.

On the other hand, heavier gas molecules have lower average velocities and collide less frequently with the walls. They require more energy or higher velocities to overcome intermolecular forces and effuse through the opening.

In summary, different gases effuse at different rates due to the relationship between their molecular masses, average velocities, and kinetic energy. Lighter gases have higher average velocities and effuse more rapidly, while heavier gases have lower average velocities and effuse at slower rates. Graham's law of effusion provides a quantitative explanation for this phenomenon.

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If it travells at 330m/s, and it has to travel 5100m just

5100 ÷ 330

because time = distance ÷ speed

= 15.45 s

Answer:

15.4545455 seconds

Explanation:

(5100 m) / (330 (m / s))

5100/330 = 15.4545455 seconds

The amount of time it will take for the reaction to consume 6.02 x 10²³ molecules of reactant is 3.67 × 10² seconds.

The amount of time it will take for a molecule to react can be calculated by dividing the number of molecules in the substance by the rate of time as follows;

Time taken = no of molecules ÷ no of molecules/seconds

According to this question, if a chemical reaction consumes reactants at a steady rate of 1.64 x 10²¹ molecules per second, the amount of time it will take for the reaction to consume 6.02 x 10²³ molecules of reactant is as follows!

Time = 6.02 x 10²³ molecules ÷ 1.64 x 10²¹ molecules per second

Time = 3.67 × 10² seconds

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Answer:

An atom contains three basic particles namely protons, neutrons and electrons. The nucleus of the atom contains protons and neutrons where protons are positively charged and neutrons are neutral. The electrons are located at the outermost regions called the electron shell.

The sample containing 7.1 × 10^19 molecules of H2O corresponds to approximately 1.18 × 10^(-4) moles of H2O.

To determine the number of moles of H2O in a sample containing 7.1 × 10^19 molecules, we need to use Avogadro's number, which states that 1 mole of any substance contains 6.022 × 10^23 molecules.

Given that there are 7.1 × 10^19 molecules of H2O in the sample, we can calculate the number of moles using the following formula:

Moles = Number of molecules / Avogadro's number

Moles = 7.1 × 10^19 / 6.022 × 10^23

Moles ≈ 1.18 × 10^(-4) moles

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The ideal gas law and stoichiometry must be used to calculate the volume of carbon dioxide gas produced by the breakdown of 4.09 g of calcium carbonate at STP (Standard Temperature and Pressure).

Use the molar mass of calcium carbonate (CaCO3) to determine how many moles it contains. CaCO3 has a molar mass of 100.09 g/mol.

CaCO3 mass divided by its molar mass equals the number of moles of CaCO3: 4.09 g/100.09 g/mol.

The number of moles of carbon dioxide (CO2) generated may be calculated using the stoichiometric ratio from the balancing equation. By using the equation:

A unit of CaCO3 and CO2 is produced.

CO2 moles equal the same number of moles of CaCO3.

Use the ideal gas law to translate the volume of carbon dioxide into moles.

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Answer:  505 grams K3PO4 x (3 x 222 grams HCI)/ (3 x K3PO4) = 555.5 grams KCl

Explanation:

The freezing point of a 0.66 m solution of  [tex]C_4H_{10[/tex]  in benzene is approximately 2.1208 °C.

To calculate the freezing point of a solution we can use the below formula

ΔT = K * m

where ΔTthe change in freezing point, K is the cryoscopic constant, and m is the molality of the solution.

Given:

Freezing point of benzene = 5.50 °C

Cryoscopic constant of benzene  = 5.12 °C/m

Molality of the solution= 0.66 m

Substituting the values into the formula:

ΔT = 5.12 °C/m * 0.66 m

Calculating the value:

ΔT = 3.3792 °C

We have to subtract the calculated change in freezing point from the freezing point of pure benzene to find the freezing point of the solution

The freezing point of solution = Freezing point (benzene) - ΔT

Freezing point of solution = 5.50 °C - 3.3792 °C

Calculating the value:

Freezing point of solution = 2.1208 °C

Therefore, the freezing point of a 0.66 m solution of [tex]C_4H_{10[/tex] in benzene is approximately 2.1208 °C.

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The chromium ions with a +3 oxidation state are reduced to chromium atoms with an oxidation state of 0.The reduction of Cr^3+ to Cr in this chemical equation is an example of a reduction reaction.

The chemical equation Cr^3+ + 3e^- = Cr represents the reduction of chromium ions (Cr^3+) to elemental chromium (Cr). In this reaction, the chromium ions gain three electrons to form neutral chromium atoms. Reduction reactions involve the gain of electrons and a decrease in the oxidation state of an element.

During the reduction process, the chromium ions are undergoing a change in their electronic configuration, gaining three electrons to achieve a stable configuration. This reduction reaction typically occurs in the presence of a reducing agent that donates electrons, allowing the chromium ions to be reduced. By gaining three electrons, the chromium ions are reduced to their elemental form, which has a neutral charge and an oxidation state of 0.

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For the following questions:

2. The periodic table was first arranged by increasing atomic mass. False

The periodic table was first arranged by increasing atomic number. This was done by Dmitri Mendeleev in 1869.

3. The short configuration of Hf is [Xe] 6s2 4f14 5s1. True

The short configuration of Hf is [Xe] 6s2 4f14 5s1. This is because Hf has 72 electrons, and the electron configuration of Xe is [Kr] 5s2 4d10 5p6. So, the electron configuration of Hf can be written as [Xe] 6s2 4f14 5s1.

4. 150 LX 4.0 moles would equate to a molarity of 0.0266 mol/L. False

Molarity is defined as the moles of solute per liter of solution. So, to calculate the molarity of a solution, we need to divide the moles of solute by the volume of solution in liters. In this case, we have 150 L of solution and 4.0 moles of solute. So, the molarity of the solution is 4.0 moles / 150 L = 0.0266 mol/L.

5. A block 1.35 m x 2.467 m = 3.3 m². False

The area of a rectangle is calculated by multiplying the length by the width. So, the area of a block that is 1.35 m long and 2.467 m wide is 1.35 m x 2.467 m = 3.319 m².

6. The density of an unknown solid weighs 3.00 g in 5.0 mL = 0.60 g/mL. True

Density is defined as mass per unit volume. So, to calculate the density of a substance, we need to divide the mass of the substance by the volume of the substance. In this case, we have a solid that weighs 3.00 g and has a volume of 5.0 mL. So, the density of the solid is 3.00 g / 5.0 mL = 0.60 g/mL.

7. 2 moles of helium would occupy 50 L of a balloon filled with it at STP. True

At STP, one mole of any gas occupies 22.4 L. So, two moles of helium would occupy 2 x 22.4 L = 44.8 L.

8. The empirical formula of a compound is half of the molecular? False

The empirical formula of a compound is the simplest whole-number ratio of the atoms in the compound. The molecular formula of a compound is the actual number of atoms in the compound. So, the empirical formula of a compound is not necessarily half of the molecular formula.

9. Multiple compounds can have the same empirical formulas? True

Multiple compounds can have the same empirical formulas. For example, the empirical formula of methane, ethane, and propane are all CH₃. However, the molecular formulas of methane, ethane, and propane are CH₄, C₂H₆, and C₃H₈, respectively.

10. Stoichiometric calculations can only be achieved by converting to moles? False

Stoichiometric calculations can be achieved by converting to moles, but they can also be achieved by using other units, such as grams or liters.

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To calculate the pH of a solution of NaOH (sodium hydroxide), we need to consider that NaOH is a strong base that dissociates completely in water, producing hydroxide ions (OH⁻).

Given:

Concentration of NaOH = 0.005 M

Since NaOH dissociates into one hydroxide ion (OH⁻) per molecule, we can determine the concentration of hydroxide ions in the solution, which will allow us to calculate the pOH. Then, we can convert the pOH to pH using the relationship: pH + pOH = 14.

1. Calculate the concentration of hydroxide ions (OH⁻):

The concentration of OH⁻ ions will be the same as the concentration of NaOH since NaOH dissociates completely.

Concentration of OH⁻ = 0.005 M

2. Calculate the pOH:

pOH = -log[OH⁻]

pOH = -log(0.005)

Using logarithm properties, we can determine the pOH value:

pOH = -log(0.005)

pOH = -(-2.301)

pOH = 2.301

3. Calculate the pH:

pH = 14 - pOH

pH = 14 - 2.301

pH ≈ 11.699

Therefore, the pH of a 0.005 M NaOH solution is approximately 11.699.

The pH of a 0.005 M concentration of NaOH ( sodium hydroxide ) solution is approximately 11.70.

The pH of a solution is defined as the logarithm of the reciprocal of the hydrogen ion concentration  [H+] of the given solution.

From the formula;

pH = -log[ H⁺ ]

pOH = -log[ OH⁻ ]

pH + pOH = 14

Given that; the concentration of solution (molarity) ( OH⁻ ) is 0.005 M.

First, we determine the pOH.

pOH = -log[ OH⁻ ]

Plug in ( OH⁻ ) = 0.005

pOH = -log[ 0.005 ]

pOH = 2.30

Now, plug pOH = 2.30 into the above formula and solve for the pH:

pH + pOH = 14

pH + 2.30 = 14

Subtract 2.30 from both sides:

pH + 2.30 - 2.30 = 14 - 2.30

pH = 14 - 2.30

pH = 11.7

Therefore, the pH of the solution is 11.7.

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7.54 *[tex]10^8[/tex] meters is  75.4 nanometers.

To convert 7.54 *  [tex]10^8[/tex] meters to nanometers, you can multiply the value by [tex]10^9[/tex]

as,  [tex]10^9[/tex]nanometers = 1  meter.

7.54 * [tex]10^8[/tex] m * [tex]10^9[/tex] =  7.54 x [tex]10^1[/tex] nm

Therefore, 7.54 *[tex]10^8[/tex] meters is equal to 75.4 nanometers.

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To convert 7.54 x 10^-8 meters to nanometers, you multiply 7.54 x 10^-8 by 1 x 10^9 to get 75.4 nanometers.

To convert meters to nanometers, you need to know that 1 meter is equivalent to 1 x 109 nanometers. Therefore, if you were to convert 7.54 x 10-8 m to nanometers, you would multiply 7.54 x 10-8 by 1 x 109.

Here's how you'd do it: 7.54 x 10-8 m * 1 x 109 nm/m = 75.4 nm. So, 7.54 x 10-8 meters is equivalent to 75.4 nanometers.

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When the temperature is lowered to -76 °C and the pressure is 0.561 atm, the volume of the balloon will be approximately 179 L.

To solve this problem, we can use the combined gas law equation, which relates the initial and final conditions of a gas sample:

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

where P1, V1, and T1 are the initial pressure, volume, and temperature, respectively, and P2, V2, and T2 are the final pressure, volume, and temperature, respectively.

Substituting the given values:

(P1 * 266 L) / (38 + 273.15 K) = (0.561 atm * V2) / (-76 + 273.15 K)

Simplifying the equation:

(0.995 atm * 266 L) / (311.15 K) = (0.561 atm * V2) / (197.15 K)

Solving for V2:

V2 = [(0.995 atm * 266 L) / (311.15 K)] * (197.15 K / 0.561 atm)

V2 ≈ 179 L

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[tex]\Large \textsf{$\boxed{\boxed{\rm (NH_4)_2CO_3}}$}[/tex]

When working with percentage compositions, we can say, "let the mass of the compound be 100 grams."

[tex]\large \textsf{$\therefore$ There is 29.15 g of nitrogen, 8.41 g of hydrogen, 12.50 g of carbon, }\\ \large \textsf{\ \ \ and 49.9 g of oxygen in 100 g of compound.}[/tex]

The empirical formula of a compound is its formula in which the constituent elements are in the simplest mole ratio.

To find the number of moles of each element (denoted by symbol [tex]\textsf{$n$}[/tex]), we can divide the mass of each element (in grams, denoted by symbol [tex]\large \textsf{$m$}[/tex]), by the molar mass of each element (in g/mol, denoted by symbol [tex]\textsf{$M$}[/tex]), which can be found on an international standard IUPAC Periodic Table.

[tex]\Large \textsf{$\therefore \rm number\ of\ moles=\frac{mass\ present}{molar\ mass}$}[/tex]

[tex]\Large \textsf{$\implies \boxed{n= \frac{m}{M}}$}[/tex]

Now we can apply this to the above masses of each element:

[tex]\large \textsf{$n(\rm N) = \frac{29.15}{14.01}$}\\\\\large \textsf{$\phantom{n(\rm N)}=2.0807\ \rm mol$}\\\large \textsf{$n(\rm H) = \frac{8.41}{1.008}$}\\\\\large \textsf{$\phantom{n(\rm H)}=8.3433\ \rm mol$}\\\\\large \textsf{$n(\rm C) = \frac{12.50}{12.01}$}\\\\\large \textsf{$\phantom{n(\rm C)}=1.0408\ \rm mol$}\\\\\large \textsf{$n(\rm O) = \frac{49.9}{16.00}$}\\\\\large \textsf{$\phantom{n(\rm O)}=3.1188\ \rm mol$}\\[/tex]

[tex]\large \text{$\therefore $ the ratio of N : H : C : O}\\\\ \large \text{$\Rightarrow$2.0807 : 8.3433 : 1.0408 : 3.1188}[/tex]

Simplifying this ratio by dividing all parts by 2.0807:

[tex]\large \text{$\therefore$ 1 : 4.0098 : 0.5002 : 1.4989}\\\\\large \text{$\implies$ 1 : 4 : 0.5 : 1.5}[/tex]

Since the mole ratio is displayed in integers, multiply this result by 2:

[tex]\large \text{$\therefore$ 2 : 8 : 1 : 3 is the final mole ratio.}\\\\\\ \large \text{$\boxed{\boxed{\implies \rm N_2H_8CO_3$ or $\rm (NH_4)_2CO_3}}$}[/tex]

Note: the compound found, is a common ionic compound known as ammonium carbonate.

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The calculate the age of the rock sample values, the age of the rock sample can be determined.

we can use the concept of radioactive decay and the ratio of 87Sr to 87Rb. Since 87Sr is a stable product of the beta decay of 87Rb, the increase in the ratio of 87Sr to 87Rb over time reflects the decay of 87Rb.

The ratio of 87Sr to 87Rb in ancient samples is given as 0.0205. This means that for every 0.0205 moles of 87Rb, there is one mole of 87Sr.

Since the half-life of 87Rb is 4.75×10^10 years, after each half-life, half of the 87Rb would have decayed into 87Sr. Therefore, the ratio of 87Sr to 87Rb increases by a factor of 2.

To determine the age of the rock sample, we can calculate the number of half-lives that have occurred based on the change in the ratio. The ratio of 0.0205 corresponds to 1 half-life, 0.041 corresponds to 2 half-lives, 0.082 corresponds to 3 half-lives, and so on.

By taking the logarithm of the ratio change and dividing it by the logarithm of 2 (since the ratio doubles with each half-life), we can find the number of half-lives.

Using this information, the age of the rock sample can be calculated as follows:

Age (in years) = number of half-lives × half-life of 87Rb

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Answer:

Phosphorus(P) and Oxygen(O)=Covalent bond

Chlorine(Cl) and Sodium(Na) = Ionic bond

Silver (Ag) and Silver (Ag)= Metallic bond

The conjugate acid-base pairs in the reaction between ammonia and hydrofluoric acid in aqueous solution are NH3/NH4+ and HF/F-.

In the reaction between ammonia (NH3) and hydrofluoric acid (HF) in aqueous solution, the following conjugate acid-base pairs can be identified:

NH3 (ammonia) and NH4+ (ammonium ion):

Ammonia (NH3) acts as a base by accepting a proton (H+) from hydrofluoric acid (HF) to form the ammonium ion (NH4+). In this reaction, ammonia acts as a Lewis base by donating an electron pair to the proton, resulting in the formation of the ammonium ion as the conjugate acid.

HF (hydrofluoric acid) and F- (fluoride ion):

Hydrofluoric acid (HF) acts as an acid by donating a proton (H+) to ammonia (NH3) to form the fluoride ion (F-).

In this reaction, hydrofluoric acid acts as a Lewis acid by accepting an electron pair from ammonia, resulting in the formation of the fluoride ion as the conjugate base.

To summarize, in the reaction NH3 (aq) + HF (aq) = NH4+ (aq) + F- (aq), the conjugate acid-base pairs are NH3/NH4+ and HF/F-. Ammonia (NH3) is the base that forms its conjugate acid, the ammonium ion (NH4+), while hydrofluoric acid (HF) is the acid that forms its conjugate base, the fluoride ion (F-).

It is important to note that in an aqueous solution, ammonia is present as NH3 molecules, and hydrofluoric acid dissociates into H+ and F- ions. The resulting ammonium ion (NH4+) and fluoride ion (F-) remain in the solution.

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Answer:

The concentration of the nitric acid (HNO3) solution is 72 M.

Explanation:

To determine the concentration of the nitric acid solution, we can use the concept of stoichiometry and the equation of the neutralization reaction between nitric acid (HNO3) and sodium hydroxide (NaOH):

HNO3 + NaOH → NaNO3 + H2O

The balanced equation shows that the molar ratio between HNO3 and NaOH is 1:1. This means that 1 mole of HNO3 reacts with 1 mole of NaOH.

Given:

Volume of HNO3 solution = 10.0 ml

Volume of NaOH solution = 3.6 ml

Molarity of NaOH solution = 0.2 M

To find the concentration of the HNO3 solution, we need to calculate the number of moles of NaOH used in the neutralization reaction:

moles of NaOH = volume of NaOH solution * molarity of NaOH solution

= 3.6 ml * 0.2 M

= 0.72 mmol (millimoles)

Since the molar ratio between HNO3 and NaOH is 1:1, the number of moles of HNO3 in the solution is also 0.72 mmol.

Now, we can calculate the concentration of the HNO3 solution using the formula:

concentration (in M) = moles of solute / volume of solution (in L)

concentration = 0.72 mmol / 0.010 L

= 72 mmol/L

= 72 M

Therefore, the concentration of the nitric acid (HNO3) solution is 72 M.