if a substance x has a solubility of 1.9×10−4 mg l−1, and a molar mass of 132 g mol−1, what is the molar solubility of the substance?

The volume of the 11 M hydrochloric acid solution needed to make 2.50 L with a molarity of 4.5 M is 1.02 L

The following data were obtained from the question:

Dilution equation is written as follow:

M₁V₁ = M₂V₂

Inputting the given parameters, the volume of the stock solution of the hydrochloric acid needed can be obtained as follow:

11 × V₁ = 4.5 × 2.5

Divide both sides by 4.67

V₁ = (4.5 × 2.5) / 11

V₁ = 1.02 L

Thus, we can conclude that the volume of the hydrochloric acid soluton needed is 1.02 L

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Complete question:

A solution of hydrochloric acid has a molarity of 11 M. What volume of this hydrochloric acid solution is needed to make 2.50L with a molarity of 4.5?

The percentage yield of Cu3(PO4)2 is approximately 98.56%.  To determine the percentage yield of Cu3(PO4)2, we need to first identify the limiting reactant.

The limiting reactant is the reactant that is completely consumed in the reaction and determines the maximum amount of product that can be formed.

Let's calculate the moles of each reactant:

Moles of CuCl2 = 1.1780g / 170.48 g/mol = 0.006907 mol

Moles of Na3PO4 = 2.2773g / 380.12 g/mol = 0.005998 mol

From the balanced equation, we can see that the stoichiometric ratio between CuCl2 and Cu3(PO4)2 is 3:1. Therefore, 0.006907 mol of CuCl2 would theoretically produce 0.006907/3 = 0.002302 mol of Cu3(PO4)2.

Similarly, the stoichiometric ratio between Na3PO4 and Cu3(PO4)2 is 1:1. So, 0.005998 mol of Na3PO4 would theoretically produce 0.005998 mol of Cu3(PO4)2.

Since 0.002302 mol is less than 0.005998 mol, CuCl2 is the limiting reactant.

Next, let's calculate the theoretical yield of Cu3(PO4)2 using the limiting reactant:

The molar mass of Cu3(PO4)2 is 434.60 g/mol.

Theoretical yield of Cu3(PO4)2 = 0.002302 mol × 434.60 g/mol = 1.000 g

Given that the actual yield of Cu3(PO4)2 is 0.9856 g, we can now calculate the percentage yield:

Percentage yield = (actual yield / theoretical yield) × 100

Percentage yield = (0.9856 g / 1.000 g) × 100

Percentage yield = 98.56%

Therefore, the percentage yield of Cu3(PO4)2 is approximately 98.56%.

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Compounds that are ionic or polar in nature tend to dissolve in water, while compounds that are nonpolar or have strong intermolecular forces are less likely to dissolve in water.

When determining if a compound will dissolve in water, it is important to consider the nature of the compound's chemical bonds and intermolecular forces. Ionic compounds, such as sodium chloride (NaCl), readily dissolve in water due to the attraction between the charged ions and the polar water molecules. The positive ends of water molecules (hydrogen atoms) are attracted to the negative ions, while the negative ends of water molecules (oxygen atoms) are attracted to the positive ions.

Similarly, polar compounds, such as ethanol (C2H5OH), also dissolve in water because they have a positive and a negative region, allowing them to interact with water molecules through hydrogen bonding and dipole-dipole interactions.

On the other hand, nonpolar compounds, such as hydrocarbons like hexane (C6H14), have little to no polarity and do not readily interact with polar water molecules. Therefore, they are insoluble or have very low solubility in water.

In summary, compounds that are ionic or polar generally dissolve in water due to the favorable interactions between their chemical properties and the polar nature of water molecules. Nonpolar compounds, on the other hand, tend to be insoluble in water because they lack the necessary polarity to interact with water molecules effectively.

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The [OH-] concentration of the solution is approximately 3.3 × 10^-10 M (rounded to two significant figures).

To find the [OH-] of the solution, we can use the ion product constant of water (Kw), which is defined as Kw = [H3O+][OH-] and has a value of 1.0 × 10^-14 at 25 degrees Celsius.

Given that [H3O+] = 3.0 × 10^-5 M, we can rearrange the equation to solve for [OH-]:

Kw = [H3O+][OH-]

1.0 × 10^-14 = (3.0 × 10^-5)([OH-])

Divide both sides of the equation by 3.0 × 10^-5:

[OH-] = (1.0 × 10^-14) / (3.0 × 10^-5)

[OH-] ≈ 3.3 × 10^-10 M

Therefore, the [OH-] concentration of the solution is approximately 3.3 × 10^-10 M (rounded to two significant figures).

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The missing reaction partner needed to carry out the reaction is alcohol, and the correct choice is ethyl alcohol (ethanol).

The given compounds in options (a), (b), (c), and (d) are esters, which have the general structure RCO2R', where R and R' are alkyl groups. Esters can undergo hydrolysis reactions in the presence of water and an acid or a base to form carboxylic acids and alcohols. In this case, the reaction requires an alcohol to carry out the hydrolysis.

Option (a) ethyl formate, (b) diethyl carbonate, and (c) diethyl oxalate are all esters, but they do not contain alcohol groups. Therefore, they cannot act as the missing reaction partner. Option (d) ethyl acetate is an ester that contains an alcohol group (CH3CO2Et), specifically an ethyl alcohol group (EtOH). Ethyl acetate can undergo hydrolysis with water and an acid or a base to form acetic acid and ethanol.

Thus, the correct choice for the missing reaction partner, in this case, is ethyl acetate (CH3CO2Et).

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The product obtained from the reaction of (S)-2-butanol with tosyl chloride and pyridine, followed by exposure to bromide, is (R)-2-bromobutane.

When (S)-2-butanol is treated with tosyl chloride (TsCl) and pyridine, followed by exposure to bromide (Br⁻), the product obtained is (R)-2-bromobutane.

The reaction proceeds as follows:

(S)-2-butanol + TsCl + pyridine → (S)-2-tosyloxybutane + HCl

(S)-2-tosyloxybutane + Br⁻ → (S)-2-bromobutane + TsO⁻

In the first step, (S)-2-butanol reacts with tosyl chloride and pyridine to form (S)-2-tosyloxybutane. This reaction involves the substitution of the hydroxyl group of the alcohol with the tosyl group (tosyloxy group).

In the second step, (S)-2-tosyloxybutane reacts with bromide (Br⁻) to undergo a nucleophilic substitution reaction, where the tosyl group is replaced by a bromide ion. This results in the formation of (R)-2-bromobutane.

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The boiling point of a substance is determined by the strength of intermolecular forces between its molecules. The higher the intermolecular forces, the higher the boiling point. To compare the boiling points of the given pairs of substances, we need to consider their intermolecular forces.

(a) O2 or N2:

Both oxygen (O2) and nitrogen (N2) are diatomic molecules and belong to the same group in the periodic table. They have similar molecular weights and London dispersion forces are the dominant intermolecular forces between them. However, oxygen (O2) has a higher boiling point than nitrogen (N2) due to its higher electron density and greater polarizability, which increases the strength of London dispersion forces. Therefore, O2 has a higher boiling point than N2.

(b) SO2 or CO2:

Sulfur dioxide (SO2) and carbon dioxide (CO2) have different molecular structures. SO2 is a bent molecule with a polar S-O bond, while CO2 is linear with polar C=O bonds. Both substances have London dispersion forces, but SO2 also exhibits dipole-dipole interactions due to its polar bonds. As a result, SO2 has stronger intermolecular forces and a higher boiling point compared to CO2.

(c) HF or HI:

Hydrogen fluoride (HF) and hydrogen iodide (HI) are both hydrogen halides. HF has strong hydrogen bonding due to the high electronegativity difference between hydrogen and fluorine, resulting in strong dipole-dipole interactions. On the other hand, HI also exhibits dipole-dipole interactions, but its boiling point is lower than that of HF due to the larger size of iodine, which leads to weaker intermolecular forces.

Therefore, HF has a higher boiling point than HI.

In summary:

(a) O2 has a higher boiling point than N2.

(b) SO2 has a higher boiling point than CO2.

(c) HF has a higher boiling point than HI.

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The final concentrations of Ni2+ and Zn2+ are [Ni2+] = 0.0204 M and [Zn2+] = 10.19 M, respectively. To find the final concentrations of Ni2+ and Zn2+ when the cell potential is 0.45V, we need to use the Nernst equation:

Ecell = E°cell - (RT/nF)ln(Q)

Where Ecell is the cell potential, E°cell is the standard cell potential, R is the gas constant, T is the temperature in Kelvin, n is the number of electrons transferred in the cell reaction, F is Faraday's constant, and Q is the reaction quotient.

For this specific cell, the standard cell potential E°cell is 1.10V. We also know that n = 2 since two electrons are transferred in the cell reaction. Plugging in the given values and solving for ln(Q), we get:

ln(Q) = (2 * 0.45V - 1.10V) / ((8.314 J/K*mol) * (298 K) / (2 * 96485 C/mol))

Simplifying this equation gives us:

ln(Q) = -1.846

Solving for Q gives us:

Q = e^(-1.846) = 0.157

We can then use the equation for the reaction quotient to find the final concentrations of Ni2+ and Zn2+:

Q = [Ni2+]/[Zn2+]

0.157 = [Ni2+]/0.130

Thus, [Ni2+] = 0.0204 M.

Finally, we can use the equation for the reaction quotient again to solve for [Zn2+]:

0.157 = 1.60/[Zn2+]

Thus, [Zn2+] = 10.19 M.

Therefore, the final concentrations of Ni2+ and Zn2+ are [Ni2+] = 0.0204 M and [Zn2+] = 10.19 M, respectively.

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The complete table for the number of particles are;

Formula   Molar Mass   # of particles      # of moles      Mass (grams)

                   (g/mol)

A. H2O2      34.0              6.02×10²³               1                    34.02

B. H2O2      34.0              1.204×10²⁴              2                    68.04

C. CO2         44.0             4.515×10²³             0.750                33.0

D.Na2O        62.0             9.03×10²³              1.5                    93.0

E. CO2          44.0            3.01×10²³               0.500                22.0

For us too calculate the number of particles, we use the following formulas

The number of particles is given by the formula n=N/Na.

Number of particles = Moles × Avogadro's number

Avogadro's number is 6.02×10²³.

For example 0.750 × 6.0210²³ = 4.51510²³

                    1.5 × 6.0210²³ = 9.0310²³

To calculate the mass, we use the following formula:

Mass = Moles × Molar mass

Molar mass is the mass of one mole of a substance. It should be written in grams per mole (g/mol).

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The number of d electrons in the metal ion of [Fe(CN)6]3- is 5. The metal ion is Fe3+, which has lost 3 electrons from its 4s and 3d orbitals.

To determine the number of d electrons in the metal ion of the complex [Fe(CN)6]3-, we first identify the metal ion and its oxidation state. In this case, the metal ion is Fe (iron). The complex has a charge of -3, and each CN- ligand contributes -1 charge, giving a total of -6 from the six ligands. To balance this, the oxidation state of the Fe ion must be +3 (Fe3+).

The electronic configuration of Fe is [Ar] 3d6 4s2. When it loses 3 electrons to become Fe3+, it loses the 2 electrons from the 4s orbital and 1 electron from the 3d orbital. Thus, Fe3+ has the configuration [Ar] 3d5. There are 5 d electrons in the metal ion of the complex [Fe(CN)6]3-.

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The bullet will be moving at a speed of 810 meters per second.


To calculate the speed of the bullet, we first need to convert the weight of the bullet to kilograms (150 grains = 0.00972 kg). Next, we need to calculate the energy released by the combustion of nitroglycerin, which is 5.56 kJ/g. Therefore, the total energy released is 13.9 kJ (2.5 g x 5.56 kJ/g).
Now, we can calculate the kinetic energy transferred to the bullet, which is 60% of the total energy released. This is equal to 8.34 kJ (0.6 x 13.9 kJ).
Finally, we can use the kinetic energy formula (1/2mv^2) to calculate the velocity of the bullet, where m is the mass of the bullet and v is the velocity. Solving for v, we get 810 meters per second.

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Chromium (Cr) is the most suitable metal coating among the options listed to prevent the corrosion of iron.

To prevent the corrosion of iron, a metal coating should act as a sacrificial anode or provide a protective barrier. Out of the options provided, the metal that would prevent the corrosion of iron when coated onto it is chromium (Cr).

Chromium can form a thin, passive oxide layer (chromium oxide, Cr₂O₃) on the surface of iron, which acts as a protective barrier against further corrosion. This oxide layer prevents the direct contact of iron with the surrounding environment, thus inhibiting the corrosion process.

Magnesium (Mg) and copper (Cu), on the other hand, would not be effective in preventing the corrosion of iron. Magnesium can also act as a sacrificial anode but would quickly corrode itself in the presence of moisture or electrolytes. Copper, while resistant to corrosion itself, does not provide effective protection to iron against corrosion.

Therefore, chromium (Cr) is the most suitable metal coating among the options listed to prevent the corrosion of iron.

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The IUPAC name for the given compound is n-ethylcyclohexanamine. In organic chemistry, the IUPAC name is a systematic method of naming a compound based on its molecular structure.                                                                              

The name consists of several parts, each representing a specific feature of the compound. In this case, "n-ethyl" indicates that there is an ethyl group attached to the nitrogen atom of the amine group. "Cyclohexane" represents the six-membered carbon ring, and "amine" indicates the presence of a nitrogen atom. Altogether, the compound is named as n-ethylcyclohexanamine.
In this compound, the functional group is an amine (-NH2) attached to a cyclohexane ring. The prefix "N" indicates that the ethyl group (C2H5) is bonded to the nitrogen atom of the amine group. This IUPAC name follows the systematic naming rules and provides a clear and accurate description of the compound's structure, making it easy for scientists to identify and work with the compound in various applications.

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The molarity of the solution, when 5.85 g of NaCl (molecular weight = 58.5) is dissolved in water to make a 0.5-liter solution, will be 0.4 M. So the correct option is option B.

To calculate the molarity, we need to determine the number of moles of NaCl present in the solution and divide it by the volume of the solution in liters.

First, we calculate the number of moles of NaCl:

moles = mass / molar mass

moles = 5.85 g / 58.5 g/mol

moles = 0.1 mol

Next, we calculate the molarity using the formula:

molarity = moles / volume (in liters)

molarity = 0.1 mol / 0.5 L

molarity = 0.2 M

Therefore, the molarity of the solution is 0.2 M, which corresponds to option (a). However, there seems to be a discrepancy between the calculated molarity and the given options. It appears that the correct answer may not be present among the given choices.

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2 Fe2O3 + 3 C → 4 Fe + 3 CO2

2 SO2 + O2 → 2 SO3

SO3 + H2O → H2SO4

2 C2H6 + 7 O2 → 4 CO2 + 6 H2O

6 CO2 + 6 H2O → C6H12O6 + 6 O2 (photosynthesis to produce glucose sugar)

C6H12O6 → 2 C2H5OH + 2 CO2 (fermentation of glucose sugar to ethyl alcohol)

In the first equation, to balance the number of iron (Fe) atoms on both sides, a coefficient of 4 is placed in front of Fe. To balance the number of carbon (C) atoms, a coefficient of 3 is placed in front of C. This results in the formation of 4 Fe and 3 CO2.

In the second equation, a coefficient of 2 is placed in front of SO2 to balance the number of sulfur (S) and oxygen (O) atoms. For the reaction to be balanced, O2 is added to the right side to form 2 SO3.

In the third equation, 2 H2O is added to the left side to balance the number of oxygen (O) atoms and create H2SO4.

The fourth equation is balanced by placing a coefficient of 4 in front of CO2 and 6 in front of H2O to equalize the carbon (C) and hydrogen (H) atoms on both sides.

For the fifth equation, 6 CO2 and 6 H2O are formed on the left side to represent the process of photosynthesis, producing C6H12O6 and 6 O2.

Lastly, the sixth equation represents fermentation, where a coefficient of 2 is placed in front of C2H5OH and CO2 to balance the number of carbon (C) and oxygen (O) atoms.

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While cirrocumulus clouds may have some minor impact on the Earth's energy balance by reflecting sunlight, their role in the greenhouse effect is negligible compared to the greenhouse gases themselves.

Cirrocumulus clouds do not directly play a significant role in the greenhouse effect. The greenhouse effect refers to the process by which certain gases in the Earth's atmosphere trap heat from the sun, leading to an increase in the planet's temperature. The primary greenhouse gases responsible for this effect are carbon dioxide, methane, and water vapor.Cirrocumulus clouds, also known as high-level clouds, are composed of ice crystals and form at altitudes above 20,000 feet. They are thin, white, and often appear as small, rounded puffs or ripples in the sky. While these clouds can reflect some sunlight back into space, their overall impact on the greenhouse effect is minimal.In contrast, low-level clouds such as stratus or cumulus clouds can have a more significant influence on the greenhouse effect. These clouds have a higher potential to reflect incoming solar radiation and cool the Earth's surface, thus partially counteracting the warming effect caused by greenhouse gases.

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The estimated half-life of Po-218 is approximately 3.03 minutes.

To estimate the half-life of polonium-218 (Po-218), we can use the concept of half-life, which is the time it takes for half of the radioactive substance to decay.

Given that 62.0% of the sample remains after 2.14 minutes, it means that 38.0% of the sample has decayed. Since half of the sample decays in one half-life, we can set up the following equation:

0.380 = (1/2)^n

Where n is the number of half-lives. We can solve for n by taking the logarithm of both sides:

log(0.380) = n * log(1/2)

n ≈ log(0.380) / log(1/2)

n ≈ 1.415

Since n represents the number of half-lives, we can estimate the half-life of Po-218 by multiplying the time interval by the number of half-lives:

half-life ≈ 2.14 minutes * 1.415 ≈ 3.03 minutes

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To calculate the solubility product constant (Ksp) for Mn(OH)2, we need to first determine the concentration of Mn^2+ and OH^- ions in the saturated solution of Mn(OH)2.

The balanced chemical equation for the dissociation of Mn(OH)2 is:

Mn(OH)2(s) ⇌ Mn^2+(aq) + 2OH^-(aq)

From the equation, we can see that one mole of Mn(OH)2 produces one mole of Mn^2+ and two moles of OH^-.

Given the solubility of Mn(OH)2 in pure water as 7.18 × 10^(-1) g/L, we can convert this into moles per liter (M) by using the molar mass of Mn(OH)2.

Molar mass of Mn(OH)2:

M(Mn) = 54.94 g/mol

M(O) = 16.00 g/mol

M(H) = 1.01 g/mol

Molar mass of Mn(OH)2 = M(Mn) + 2 * (M(O) + M(H))

= 54.94 + 2 * (16.00 + 1.01)

= 54.94 + 2 * 17.01

= 54.94 + 34.02

= 88.96 g/mol

Now, we can calculate the concentration of Mn^2+ ions in the saturated solution:

Concentration of Mn^2+ = solubility of Mn(OH)2 / molar mass of Mn(OH)2

= (7.18 × 10^(-1) g/L) / (88.96 g/mol)

= 8.07 × 10^(-3) mol/L

Since the concentration of Mn^2+ ions is equal to the concentration of OH^- ions (according to the stoichiometry of the equation), we can say:

[OH^-] = 8.07 × 10^(-3) mol/L

Finally, we can calculate the Ksp for Mn(OH)2 by multiplying the concentrations of Mn^2+ and OH^- ions:

Ksp = [Mn^2+][OH^-]

= (8.07 × 10^(-3) mol/L)(8.07 × 10^(-3) mol/L)

= 6.51 × 10^(-5) mol^2/L^2

Therefore, the Ksp for Mn(OH)2 is 6.51 × 10^(-5) mol^2/L^2.

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The correct answer is: mol [tex]l^-2[/tex]

The solubility product (Ksp) is a constant that describes the equilibrium of a sparingly soluble or insoluble salt in water, and it is expressed as the product of the concentrations of the ions raised to their stoichiometric coefficients.

The units of solubility product depend on the specific reaction being described.

Generally, for a salt AB that dissociates into A+ and B-, the units of Ksp are (mol/L)^2, since Ksp = [A+][B-].

Therefore, the correct answer is: mol [tex]l^-2[/tex]

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Answer:

d

Explanation:

Based on the analysis, none of the given changes (a, b, c, d) would cause the cell potential to increase (become more positive). The cell potential is determined by the standard cell potential (E°cell) and the concentrations of the species involved in the half-reactions.

To determine which change to the galvanic cell would cause an increase in the cell potential (become more positive), we need to examine the half-reactions and the Nernst equation.

The given galvanic cell can be represented as:

Zn(s) | Zn2+(aq) || Pb2+(aq) | Pb(s)

The reduction half-reaction occurring at the cathode (positive electrode) is:

Pb2+(aq) + 2e- → Pb(s) (Reduction)

The oxidation half-reaction occurring at the anode (negative electrode) is:

Zn(s) → Zn2+(aq) + 2e- (Oxidation)

The cell potential (Ecell) can be determined using the Nernst equation:

Ecell = E°cell - (RT/nF) * ln(Q)

Where:

Ecell is the cell potential

E°cell is the standard cell potential

R is the ideal gas constant (8.314 J/(mol·K))

T is the temperature in Kelvin

n is the number of electrons transferred in the balanced half-reaction

F is the Faraday constant (96,485 C/mol)

Q is the reaction quotient

Since we want to determine which change will increase the cell potential, let's analyze each option:

a) Increase the Zn2+: If the concentration of Zn2+ is increased, it will affect the reaction quotient (Q) by increasing the concentration of Zn2+ in the anode half-cell. According to the Nernst equation, an increase in Q will result in a decrease in the cell potential (more negative), so this change would not increase the cell potential.

b) Increase the Pb2+: Similarly, increasing the concentration of Pb2+ will affect the reaction quotient (Q) by increasing the concentration of Pb2+ in the cathode half-cell. According to the Nernst equation, an increase in Q will result in a decrease in the cell potential (more negative), so this change would not increase the cell potential.

c) Increase the mass of Zn: The mass of Zn does not directly affect the cell potential. The concentration of Zn2+(aq) would remain the same, as the concentration is determined by the concentration of Zn2+(aq) and not the mass of Zn. Therefore, increasing the mass of Zn would not increase the cell potential.

d) Decrease the mass of Zn: Similarly, the mass of Zn does not directly affect the cell potential. The concentration of Zn2+(aq) would remain the same, so decreasing the mass of Zn would not increase the cell potential.

Based on the analysis, none of the given changes (a, b, c, d) would cause the cell potential to increase (become more positive). The cell potential is determined by the standard cell potential (E°cell) and the concentrations of the species involved in the half-reactions.

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The matchings for each chromatography are:

1. Gas Chromatography: Vapor travels over a column of tiny beads.

2. Liquid Chromatography: Solvent travels over a column of tiny beads.

3. Thin-layer Chromatography: A sheet of cellulose is placed in a liquid.

4. Paper Chromatography: Solid particles are spread over a flat glass.

Chromatography is a versatile separation technique used to separate and analyze mixtures of substances into their individual components. It is widely used in various fields, including chemistry, biochemistry, forensics, and environmental science.

Chromatography works on the principle of differential migration of components in a mixture due to their interactions with a stationary phase and a mobile phase.

Gas Chromatography: Vapor travels over a column of tiny beads.

Liquid Chromatography: Solvent travels over a column of tiny beads.

Thin-layer Chromatography: A sheet of cellulose is placed in a liquid, which travels up the sheet.

Paper Chromatography: Solid particles are spread over a flat glass or plastic surface and a solvent is allowed to travel up through the solid particles.

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Carbon dioxide cannot be liquefied above the critical temperature, even when high pressure is applied. The critical temperature is the highest temperature at which a substance can be liquefied by increasing the pressure. For carbon dioxide, the critical temperature is approximately 31.1°C (87.98°F). Above this temperature, carbon dioxide remains in the gaseous state regardless of the pressure applied.

Carbon dioxide or carbonic acid is a chemical compound consisting of two oxygen atoms covalently bonded to a carbon atom. It is a gas at standard temperature and pressure conditions and is present in the Earth's atmosphere.

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The Lewis structure for BCl3 is:

B Cl

| |

Cl-B-Cl

In BCl3, boron (B) is the central atom, and it forms three single bonds with chlorine (Cl) atoms. Each chlorine atom contributes one valence electron, while boron contributes three valence electrons. The octet rule does not apply to boron, as it is an exception and can have an incomplete octet.

In the Lewis structure, the boron atom is surrounded by three chlorine atoms, with each chlorine atom bonded to the boron atom through a single bond. The boron atom has only six electrons around it, which is less than the octet rule. However, boron is stable with only six valence electrons due to its unique electronic configuration.

It's important to note that the Lewis structure does not provide information about the molecular shape or bond angles. It simply shows the arrangement of atoms and their bonding.

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The energy required to vaporize 99 g of water is approximately 4019.33 kJ. To calculate the energy required to vaporize a given mass of water, we can use the following formula:

Energy = (mass of water) x (heat of vaporization)

We are given the mass of water as 99 g, and the molar mass of water as 18.02 g/mol. We can use this information to calculate the number of moles of water:

moles of water = (mass of water) / (molar mass of water)

= 99 g / 18.02 g/mol

= 5.491 mol

We are also given the heat of vaporization of water as 40.67 kJ/mol.

Now we can use the above formula to calculate the energy required to vaporize 99 g of water:

energy = (mass of water) x (heat of vaporization)

= 99 g x (40.67 kJ/mol)

= 4019.33 kJ

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The remaining mass of the isotope after 30 years will be 12.5 g.

To calculate the remaining mass of an isotope after a certain time, we can use the formula:

Remaining mass = Initial mass × (1/2)^(time elapsed / half-life)

In this case, we have a 100 g sample of an isotope with a half-life of 10 years, and we want to find the remaining mass after 30 years.

Using the formula, we can substitute the values:

Remaining mass = 100 g × (1/2)^(30 years / 10 years)

Simplifying the equation, we have:

Remaining mass = 100 g × (1/2)^3

Calculating the value inside the parentheses, we get:

Remaining mass = 100 g × (1/8)

Therefore, the remaining mass of the isotope after 30 years will be:

Remaining mass = 12.5 g

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The molarity of the diluted solution is approximately 0.22 M.

M1V1 = M2V2

(12 M)(22 ml) = M2(1200.0 ml)

Simplifying the equation:

M2 = (12 M)(22 ml) / (1200.0 ml)

M2 ≈ 0.22 M

Molarity is a unit of concentration used in chemistry to express the amount of a solute dissolved in a solvent. It is defined as the number of moles of solute per liter of solution. The formula to calculate molarity is: Molarity (M) = moles of solute/liters of solution

Molarity is commonly denoted by the symbol "M" and is expressed in moles per liter (mol/L or M). It is a fundamental concept in chemistry and is used in various applications, such as determining reaction rates, preparing solutions of known concentration, and performing quantitative analysis. Molarity provides a standardized way to quantify the concentration of a substance in a solution. It allows scientists to accurately measure and compare the amount of a solute present in different solutions.

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The molal concentration of the aqueous solution containing a non-electrolyte is 2.15 m.

To find the molal concentration of the solution, we need to use the freezing point depression equation, which is ΔTf = Kf x molal concentration. Here, ΔTf is the difference between the freezing point of the pure solvent (water) and the freezing point of the solution, which is 4.0°C (since the freezing point of the solution is -4.0°C), and Kf is the freezing point depression constant of water, which is 1.86 °C/m.

Substituting these values in the equation, we get:
4.0°C = 1.86 °C/m x molal concentration

Solving for molal concentration, we get:
molal concentration = 4.0°C / 1.86 °C/m = 2.15 m

The concept of molality is used to express the concentration of a solution in terms of the number of moles of solute per kilogram of solvent. It is a useful measure because it is temperature-independent and therefore more accurate than molarity in many cases. The molality of a solution is related to its freezing point depression, which is the lowering of the freezing point of the solvent caused by the presence of the solute. This phenomenon occurs because the solute molecules disrupt the crystal lattice of the solvent, making it more difficult for the solvent to freeze. The extent of the freezing point depression depends on the concentration of the solute in the solution and the nature of the solute.

Aqueous solutions are those in which water is the solvent. In such solutions, the freezing point depression is influenced not only by the concentration of the solute but also by whether the solute is an electrolyte or a non-electrolyte. Electrolytes are substances that dissociate into ions in solution and therefore increase the number of particles in the solution, whereas non-electrolytes do not dissociate and therefore do not increase the number of particles. As a result, a solution containing an electrolyte will have a greater freezing point depression than a solution containing a non-electrolyte at the same molality.

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The radius of the unknown atom in the primitive cubic unit cell is approximately 1.425 Å

To determine the radius of the unknown atom in a primitive cubic unit cell, we can use the relationship between the length of the unit cell edge (a) and the radius (r) of the atoms in the cell.

For a primitive cubic unit cell, the atom is located at the corners, and the unit cell edge length (a) is equal to two times the radius (2r) of the atom.

Therefore, we can write the equation as:

a = 2r

Given that the length of the unit cell edge (a) is 2.85 Å, we can solve for the radius (r) by rearranging the equation:

r = a / 2

r = 2.85 Å / 2

r ≈ 1.425 Å

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So, the weekly exposure of each dye from this source in grams (g) can be calculated by multiplying the mass of dye in 0.65 liters of drink per day by 7 days.

The mass of dye per volume of cherry-flavored drink can be calculated as follows:

First, we need to convert the volume from liters to milliliters. 1 liter = 1000 milliliters. So, 0.65 liters = 0.65 x 1000 = 650 milliliters.

Next, we need to calculate the mass of dye per liter of drink. This can be done by dividing the total mass of dye by the total volume of drink.

The total mass of dye can be calculated by multiplying the mass of dye per gram by the total number of grams of dye.

The total number of grams of dye can be calculated by multiplying the total volume of drink by the mass of dye per liter of drink.

So, the mass of dye per liter of drink is:

Mass of dye per liter = Mass of dye per gram x Total number of grams of dye

Next, we need to calculate the mass of dye per milliliter of drink. This can be done by dividing the mass of dye per liter of drink by the total volume of drink in milliliters.

The total volume of drink in milliliters can be calculated by multiplying the total volume of drink in liters by 1000.

So, the mass of dye per milliliter of drink is:

Mass of dye per milliliter = Mass of dye per liter of drink / Total volume of drink in milliliters

Next, we need to calculate the mass of dye per 0.65 liters of drink per day. This can be done by multiplying the mass of dye per liter of drink by the total volume of drink in milliliters per day.

The total volume of drink in milliliters per day can be calculated by multiplying the total volume of drink in liters per day by 1000 milliliters per liter.

So, the mass of dye per 0.65 liters of drink per day is:

Mass of dye per 0.65 liters per day = Mass of dye per liter of drink x Total volume of drink in milliliters per day

The mass of dye in 0.65 liters of drink per day can be calculated by multiplying the mass of dye per milliliter of drink by the total volume of drink in milliliters per day.

So, the mass of dye in 0.65 liters of drink per day is:

Mass of dye in 0.65 liters per day = Mass of dye per milliliter of drink x 0.65 liters per day

Finally, we need to calculate the weekly exposure of each dye from this source in grams (g) by multiplying the mass of dye in 0.65 liters of drink per day by the number of days in a week.

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