If you change the solution volume but keep the solute amount the same what happens to the Molarity?

A Circular buffer, also known as a circular queue, cyclic buffer, or ring buffer, is a data structure used in computer programming to manage a fixed-sized buffer. Unlike a linear buffer, the circular buffer has a circular or cyclic nature, allowing it to efficiently store and retrieve data in sequential order.

The buffer is divided into two regions, a read and write region, and the data is read from and written to the buffer in a circular fashion. When the buffer is full, the oldest data is overwritten by the newest data, making it a useful tool for streaming and real-time applications.

Circular buffers are widely used in embedded systems, multimedia, networking, and other applications where a continuous flow of data is needed. The circular buffer can also be used to implement algorithms like the producer-consumer problem, where data is produced at a certain rate and consumed at another rate.

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The mass fraction of nitrogen in the ideal gas mixture is 21.6%. The mass fraction of nitrogen in the ideal gas mixture can be calculated using the mole fraction of nitrogen and the molar masses of nitrogen and the other two gases.

First, we need to find the mole fraction of the other two gases by subtracting the mole fraction of nitrogen from 1.

Mole fraction of other two gases = 1 - 0.26 = 0.74

Now, we can use the concept of apparent molar mass to calculate the molar masses of the other two gases.

Apparent molar mass of mixture = (0.26 x molar mass of nitrogen) + (0.74 x molar mass of other two gases)

36 kg/mol = (0.26 x 28 kg/mol) + (0.74 x molar mass of other two gases)

Solving for the molar mass of the other two gases, we get:

The molar mass of the other two gases = 48 kg/mol

Finally, we can use the molar masses and mole fractions to calculate the mass fraction of nitrogen:

Mass fraction of nitrogen = (0.26 x 28 kg/mol) / [(0.26 x 28 kg/mol) + (0.74 x 48 kg/mol)]

Mass fraction of nitrogen = 0.216 or 21.6%

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The correct option is b. Venus. Of the four terrestrial planets (Mercury, Venus, Earth, and Mars), Venus has the most dense atmosphere.

Of the four terrestrial planets, Venus has the most dense atmosphere. The atmosphere of Venus is about 90 times denser than the atmosphere of Earth.

The atmosphere of Venus is composed mainly of carbon dioxide with small amounts of nitrogen and other gases. The thick atmosphere of Venus produces a strong greenhouse effect, which traps heat and causes the planet's surface temperature to be very hot - around 460°C (860°F), making it the hottest planet in our solar system.

In contrast, Mars has a very thin atmosphere, which is about 1% the density of Earth's atmosphere. Mars' atmosphere is primarily composed of carbon dioxide with small amounts of nitrogen and argon. Due to the thin atmosphere, Mars does not have a strong greenhouse effect and its surface temperature is much colder than Earth.

Jupiter, on the other hand, is not a terrestrial planet but a gas giant. It has a very thick atmosphere, which is mainly composed of hydrogen and helium, with small amounts of methane, ammonia, and water vapor.

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The correct balanced equation for the reaction of sulfurous acid with lithium hydroxide to form water and lithium sulfite is: B) H2SO3(aq) + 2 LiOH(aq) → 2 H2O(l) + Li2SO3(aq).

Lithium hydroxide is an inorganic compound with the formula LiOH. It can exist as anhydrous or hydrated, and both forms are white hygroscopic solids. They are soluble in water and slightly soluble in ethanol. Both are available commercially. While classified as a strong base, lithium hydroxide is the weakest known alkali metal hydroxide.

Sulfurous acid (H2SO3) reacts with lithium hydroxide (LiOH) in a 1:2 ratio to produce water (H2O) and lithium sulfite (Li2SO3). The balanced equation ensures that the number of atoms for each element is the same on both sides of the equation.

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The formula for density is: density = mass/volume In this case, the mass of the object is given as 149.8 g, and the volume can be determined by measuring the volume of water displaced when the object is placed in a graduated cylinder, which is 12.1 mL. So, the density can be calculated as density = 149.8 g / 12.1 mL density = 12.4 g/mL Therefore, the correct answer is e. 12.4 g/mL.

To determine the density of an object with a mass of 149.8 g that displaces 12.1 mL of water when placed in a graduated cylinder, follow these steps:

1. Obtain the mass of the object (149.8 g).
2. Measure the volume of water displaced by the object using a graduated cylinder (12.1 mL).
3. Use the formula for density: density = mass/volume.

Now, let's calculate the density:
density = 149.8 g / 12.1 mL = 12.38 g/mL

Based on the given options, the closest answer is:
e. 12.4 g/mL

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Polonium-212 is considered to be much more radioactive than Uranium-238.

Polonium-212 is more radioactive than Uranium-238. This is because the shorter the half-life of a radioactive isotope, the more radioactive it is.

Polonium-212 has a half-life of only 0.0000003 seconds, which means that it decays extremely quickly, releasing a large amount of radiation in a short period of time.

In contrast, Uranium-238 has a much longer half-life of 4.5 billion years, which means that it decays much more slowly, releasing a smaller amount of radiation over a much longer period of time.

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Amino acids with charged side chains can form ionic interactions with each other, while amino acids with polar but uncharged side chains can form hydrogen bonds with each other. Therefore, the pairs of amino acids that can form ionic interactions and hydrogen bonds between their R-groups are different.

Pairs of amino acids that can form ionic interactions between their R-groups are:

Lysine (K) and glutamic acid (E)

Arginine (R) and aspartic acid (D)

Histidine (H) and glutamic acid (E) or aspartic acid (D)

Pairs of amino acids that can form hydrogen bonds between their R-groups are:

Serine (S) and threonine (T)

Glutamine (Q) and asparagine (N)

Tyrosine (Y) and serine (S) or threonine (T)

Note that some amino acids, such as cysteine (C) and methionine (M), do not form hydrogen bonds or ionic interactions with other amino acids due to the nonpolar nature of their R-groups. Additionally, some amino acids, such as glycine (G), do not have R-groups and cannot form these types of interactions.

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1. The solution with [H+] = 4 × 10^-9 M and [OH-] = 1 × 10^-7 M is basic.
2. The solution with [OH-] = 7 × 10^-13 M is acidic.

To determine whether the solutions with the following concentrations of ions are neutral, acidic, or basic, we will compare the concentrations of [H+] and [OH-] ions in each solution:

1. [H+] = 4 × 10^-9 M
  [OH-] = 1 × 10^-7 M

In this solution, the concentration of [OH-] ions is greater than the concentration of [H+] ions. Therefore, this solution is basic.

2. [OH-] = 7 × 10^-13 M

To find the corresponding [H+] concentration, we can use the ion product of water (Kw), which is:

Kw = [H+] × [OH-]

At 25°C, Kw = 1 × 10^-14

So, we can calculate [H+] as follows:

[H+] = Kw / [OH-]
[H+] = (1 × 10^-14) / (7 × 10^-13)
[H+] ≈ 1.43 × 10^-2 M

In this solution, the concentration of [H+] ions is greater than the concentration of [OH-] ions. Therefore, this solution is acidic.

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The drying agent is removed from an organic solution through a method called solid-liquid separation, which is preferred for its efficiency and ease of use.

Steps for removal of dryoinhg agent :

Step 1: Add the drying agent to the liquid organic solution. The drying agent is a solid substance that has a high affinity for water or other impurities in the solution.

Step 2: Allow the drying agent to interact with the liquid solution for some time. This interaction allows the drying agent to absorb water or other impurities, thereby leaving the organic solution drier.

Step 3: Perform a solid-liquid separation process, such as filtration or decantation. Filtration involves passing the mixture through a filter paper or a porous medium, which traps the solid drying agent and allows the now-dry organic liquid to pass through. Decantation involves carefully pouring off the liquid while leaving the solid drying agent behind.

Step 4: Collect the purified organic solution and dispose of the solid drying agent.

This method of solid-liquid separation is preferred because it is straightforward, efficient, and allows for easy removal of the drying agent, resulting in a purified organic solution.

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Dinitrogen monoxide decomposes into nitrogen and oxygen when heated. the initial rate of the reaction is .022M/s. 0.044 is the  rate of change of the concentration of N[tex]_2[/tex]O. Therefore, the correct option is option C.

Reaction rate is the rate at which the chemical reaction moves forward. It is frequently described in terms of alternatively the amount of a reactant which is consumed in a unit period or the intensity of a product that is generated in a unit of time (amount per unit volume). You may also define it in terms of how much of the reactants are used up or how much of the products are produced in a given amount of time.

2N[tex]_2[/tex]O(g)→ 2N[tex]_2[/tex](g)+O[tex]_2[/tex](g)

-Δ[N[tex]_2[/tex]O]/2ΔT=0.022

-Δ[N[tex]_2[/tex]O]/ΔT=0.022×2

                   = 0.044

Therefore, the correct option is option C.

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The student's claim that ∆S° for the reaction HCNO(g) (fulminic acid) ⇌ HNCO(g) (isocyanic acid) is close to zero is accurate. This is because the two compounds, fulminic acid and isocyanic acid, have very similar structures and therefore have similar molecular complexities.

As a result, the change in entropy (∆S) between the reactants and products is minimal, which leads to a ∆S° value that is close to zero. Additionally, the reaction involves a simple rearrangement of atoms, which does not significantly affect the randomness or disorder of the system.

Therefore, the entropy change is minimal, and the student's claim is accurate.

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According to ideal gas equation, volume must change by 0.5 factor to keep the pressure and temperature the same.

The ideal gas equation is a equation which is applicable in a hypothetical state of an ideal gas.It is a combination of Boyle's law, Charle's law,Avogadro's law and Gay-Lussac's law . It is given as, PV=nRT where R= gas constant whose value is 8.314.The law has several limitations.

Since there are two conditions before addition and after addition  which is 0.1×RT/V=0.2×RT/V thus volume changes by factor of 0.5 which is pressure.

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An example of two atoms with different electronegativities having no net dipole is when the atoms are arranged in a linear shape, and the molecule is symmetrical.

This means that the bond dipoles cancel each other out, resulting in no net dipole.

One example of this is carbon monoxide (CO). Carbon is less electronegative than oxygen, so there is a partial negative charge on the oxygen atom and a partial positive charge on the carbon atom.

However, because the molecule is linear, and the bond dipoles point in opposite directions, the dipole moments cancel each other out, resulting in a molecule with no net dipole moment.

Another example of two atoms with different electronegativities having no net dipole is when the molecule has a symmetric molecular shape, such as in the case of tetrachloromethane (CCl4).

In this molecule, carbon has a lower electronegativity than chlorine, leading to partial negative charges on the chlorine atoms and a partial positive charge on the carbon atom

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A molecular formula represents the actual number of atoms of each element present in a molecule, while an empirical formula represents the simplest whole number ratio of atoms of each element present in a compound.

The difference between a molecular formula and an empirical formula is that a molecular formula represents the actual number of atoms of each element in a molecule, whereas an empirical formula shows the simplest whole-number ratio of atoms of each element in a compound. For example, the molecular formula for glucose is C6H12O6, which indicates that there are 6 carbon atoms, 12 hydrogen atoms, and 6 oxygen atoms in a glucose molecule. The empirical formula for glucose, on the other hand, is CH2O, which represents the simplest ratio of 1 carbon atom, 2 hydrogen atoms, and 1 oxygen atom.

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The formula for copper (II) sulfate pentahydrate is CuSO₄·5H₂O. The answer is E)

Copper (II) sulfate pentahydrate is a salt compound consisting of copper, sulfur, oxygen, and hydrogen atoms. The formula CuSO₄·5H₂O represents the compound's molecular structure, which consists of one copper (Cu) atom, one sulfur (S) atom, four oxygen (O) atoms, and ten hydrogen (H) atoms.

The "Cu" in the formula represents copper, which has a valency of 2 in this compound, indicated by the Roman numeral II in the name. The "SO₄" in the formula represents the sulfate ion, which consists of one sulfur atom and four oxygen atoms. The sulfate ion has a valency of 2- in this compound, balancing the 2+ charge of the copper ion.

The "5H₂O" in the formula represents the five water molecules that are chemically bound to the copper (II) sulfate molecule. These water molecules are called "water of hydration" and can be removed from the compound by heating it.

Thus, the formula CuSO₄·5H₂O represents copper (II) sulfate pentahydrate, a salt compound with a specific molecular structure.

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The deterioration of metals by an electrochemical process is called corrosion. The metal is oxidized during this process, where electrons are lost and the metal becomes weaker and more prone to further corrosion.

Corrosion is a natural electrochemical process that can occur when metals are exposed to various environmental factors, such as air, water, or certain chemicals. During the process of corrosion, the metal surface begins deterioration due to oxidation, which involves the loss of electrons from the metal atoms. This loss of electrons is accompanied by the formation of metal ions, which can dissolve in water and further accelerate the corrosion process.

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[tex]HClO_{2}[/tex] is a weak acid and partially dissociates in water to form H+ and [tex]ClO_2-[/tex]ions. The correct answer is option: c.

We need to use the equilibrium expression for the dissociation of [tex]HClO_2[/tex] and solve for the concentration of [tex]H_+[/tex] ions.

[tex]HClO2(aq) + H2O(l)[/tex] ⇌ [tex]H3O+(aq) + ClO2-(aq)[/tex]

Let x be the concentration of [tex]H_+[/tex] ions at equilibrium. Then the equilibrium concentration of[tex]HClO_2[/tex] will be (0.155-x) and the concentration of[tex]ClO_2-[/tex]will also be x.

Using the equilibrium expression for Ka, we get:

[tex]Ka = [H3O+][ClO2-]/[HClO2] = x^2/(0.155-x)[/tex]

Substituting the given value of Ka, we can solve for x:

[tex]0.011 = x^2/(0.155-x) \\x = 0.042 M[/tex]

Thus, the pH of the 0.155 M [tex]HClO_2[/tex] solution can be calculated using the formula:

[tex]pH = -log[H+] = -log(0.042) = 1.38[/tex]

Therefore, the pH of the solution is approximately 1.38, which is closest to option (c).

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Answer:

Gyre

Explanation:

I looked it up, may not be correct

In the Bohr model of the atom, radiation is emitted whenever electrons move to orbits of lower energy. The correct answer is option c.

This model, proposed by Niels Bohr in 1913, is an early description of atomic structure, specifically focusing on hydrogen atoms. It consists of a nucleus with protons and neutrons, and electrons orbiting around it in fixed energy levels called "shells" or "orbitals."

The energy levels in the Bohr model are quantized, meaning that electrons can only occupy specific orbits with distinct energy values. When an electron transitions from a higher energy level (higher orbital) to a lower energy level (lower orbital), it releases energy in the form of electromagnetic radiation, such as light or photons. This energy release is what causes the characteristic emission spectrum observed in hydrogen and other elements.

It's essential to note that while the Bohr model was revolutionary at the time and provided valuable insights into atomic structure and the behavior of electrons, it has since been superseded by the more accurate and comprehensive quantum mechanical model. The modern understanding of atomic structure and electron behavior is rooted in quantum mechanics and wave-particle duality, which explain phenomena that the Bohr model cannot account for.

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The geometry of the hybrid orbitals about a central atom with sp3d hybridization is trigonal bipyramidal.

When an atom undergoes sp3d hybridization, one s orbital, three p orbitals, and one d orbital hybridize to form five hybrid orbitals that are directed towards the five corners of a trigonal bipyramidal geometry.

The hybridization occurs in order to minimize the energy of the system and maximize the overlap of the orbitals, resulting in a stable and energetically favorable configuration.

The trigonal bipyramidal geometry consists of a central atom surrounded by five other atoms or electron pairs. The two axial positions are located along a straight line passing through the central atom, while the three equatorial positions are located in a plane perpendicular to the axial positions.

The five hybrid orbitals occupy these positions, with the two axial hybrid orbitals pointing directly towards the two axial positions and the three equatorial hybrid orbitals pointing towards the three equatorial positions.

Therefore, sp3d hybridization results in five hybrid orbitals that occupy the five positions of a trigonal bipyramidal geometry.

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The negative sign indicates that the age of the rock is before the present time, which means the rock is estimated to be 2.6 billion years old.

The decay of radioactive isotopes is described by the following equation:

N(t) = N0 * (1/2)^(t / t1/2)

where:

N(t) = the amount of remaining radioactive substance after time t

N0 = the initial amount of radioactive substance

t1/2 = the half-life of the radioactive substance

We know that the half-life of potassium-40 is 1.3 billion years. This means that every 1.3 billion years, the amount of potassium-40 in a sample is reduced by half.

If a rock contains only one-fourth of its original potassium-40, this means that the remaining amount of potassium-40 is 1/4 of the initial amount, or N(t) = 1/4 * N0.

Substituting this into the equation, we get:

1/4 * N0 = N0 * (1/2)^(t / t1/2)

Simplifying and solving for t, we get:

t = t1/2 * log2(1/4)t = 1.3 billion years * log2(1/4)t = 1.3 billion years * (-2)t = -2.6 billion years.

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The formula weights of the compounds are as follows :

(a) To find the formula weight of KBr, we will add the atomic weights of potassium (K) and bromine (Br):
K: 39.10 amu
Br: 79.90 amu

Formula weight of KBr = 39.10 amu (K) + 79.90 amu (Br) = 119.00 amu.

(b) To find the formula weight of PbCO3, we will add the atomic weights of lead (Pb), carbon (C), and three times the atomic weight of oxygen (O):
Pb: 207.20 amu
C: 12.01 amu
O: 16.00 amu

Formula weight of PbCO3 = 207.20 amu (Pb) + 12.01 amu (C) + 3 * 16.00 amu (O) = 267.21 amu.

So, the formula weights are (a) KBr: 119.00 amu and (b) PbCO3: 267.21 amu.

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The molecule has six bonded atoms and no lone pairs, so its electron domain is six. The ideal bond angle would be 90 degrees. The hybridization is sp3d2. The polarity of the molecule depends on the geometry of the molecule

The molecule with six bonded atoms and no lone pairs has an electron domain of 6. The ideal bond angle for a molecule with six electron domains is octahedral, which corresponds to a bond angle of 90 degrees.

The hybridization of the central atom in this molecule is sp3d2, which involves the hybridization of one 3s, three 3p, and two 3d orbitals.

Whether the molecule is polar or nonpolar depends on the nature and position of the atoms bonded to the central atom. If the atoms are identical and are symmetrically arranged, then the molecule is nonpolar. However, if the atoms are different or if they are arranged asymmetrically, then the molecule is polar.

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To draw the Lewis dot structure of FClO3, we first need to determine the total number of valence electrons in the molecule. Fluorine has 7 valence electrons, chlorine has 7, and oxygen has 6. There are three oxygen atoms, so that's a total of 18 electrons. The total number of valence electrons in FClO3 is:

7 (F) + 7 (Cl) + 18 (3 x O) + 1 (extra electron from F-) = 33

We then arrange the atoms in a way that satisfies the octet rule, meaning that all atoms (except hydrogen) should have 8 valence electrons around them. The central atom in this molecule is chlorine, which forms single bonds with each oxygen atom and one bond with fluorine. The Lewis dot structure of FClO3 with an F-Cl bond looks like this:

Cl: (7 valence electrons)
    |
F - Cl - O
    |
O   O   O
  (6) (6) (6)

Each oxygen atom has a lone pair of electrons, which gives them each 8 valence electrons. Chlorine has 6 electrons around it (2 bonds and 2 lone pairs), so it has a formal charge of +1.

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We need to add approximately 221 ml of solvent to 415 ml of 75% lansoprazole suspension to obtain 415 ml of 35% lansoprazole suspension.

Let x be the volume of solvent (in ml) needed to dilute the solution.

We can use the following formula to calculate the volume of solvent needed for dilution:

Amount of solute in the final solution = Amount of solute in the initial solution

The amount of solute is equal to the concentration multiplied by the volume of the solution. Thus, we have:

0.35 x 415 = 0.75 (415 + x)

Simplifying the equation, we get:

145.25 = 311.25 + 0.75x

0.75x = 145.25 - 311.25

0.75x = -166

x = -166 / 0.75

x = -221.33

Since we cannot have a negative volume of solvent, the result must be due to rounding errors. We round x to the nearest whole ml, which gives:

x ≈ -221 ml ≈ 221 ml

Therefore, we need to add approximately 221 ml of solvent to 415 ml of 75% lansoprazole suspension to obtain 415 ml of 35% lansoprazole suspension.

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The steps for preparing 55 mL of a 1.5 M sodium bicarbonate solution is given below.

The mole idea is a useful way to indicate how much of a substance there is. Any measurement can be divided into two components: the magnitude in numbers and the units in which the magnitude is expressed.

1. To prepare 55 mL of a 1.5 M sodium bicarbonate solution, we need to dissolve a certain amount of sodium bicarbonate in water. Here are the steps to follow along with the supporting calculations:

2. Determine the number of moles of sodium bicarbonate needed:

Moles = Molarity x Volume (in liters)

Moles = 1.5 M x 0.055 L

Moles = 0.0825 moles

3. Calculate the mass of sodium bicarbonate needed using its molar mass:

Mass = Moles x Molar Mass

Mass = 0.0825 moles x 84.01 g/mol (molar mass of sodium bicarbonate)

Mass = 6.93 grams

4. Add the calculated amount of sodium bicarbonate to a volumetric flask containing a small amount of water.

5. Add distilled water to the flask until the volume reaches 55 mL.

6. Mix the solution thoroughly until the sodium bicarbonate is completely dissolved.

7. The 1.5 M sodium bicarbonate solution is now ready for use.

Note: It's essential to use a volumetric flask to prepare the solution accurately and to ensure that the final volume is precisely 55 mL. Also, make sure to wear appropriate PPE such as gloves, safety glasses, and a lab coat while handling the chemicals.

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To find Kb, we really want to initially decide the centralization of hydroxide particles (Gracious ) in the arrangement, which we can do utilizing the pH.

pH = -log[H+]

11.9 = -log[H+]

[H+] = 10[tex]^(-11.9)[/tex]

[H+] = 7.94 x 10[tex]^-12[/tex]

Since CH₃NH₂ is a weak base, we can use the equilibrium expression for weak bases to determine Kb because we can assume that it does not completely dissociate in water:

CH₃NH₂ + H₂O ⇌ CH₃NH₃+ + OH-

Kb = [CH₃NH₃+][OH-]/[CH₃NH₂]

We know that [OH-] = 7.94 x 10[tex]^-12[/tex] from the pH computation, and we can accept that [CH₃NH₃+] is immaterial contrasted with [ CH₃NH₂], so we can set [CH₃NH₂] ≈ [CH₃NH₂] initial = 1.0 M.

Kb = [OH-][CH₃NH₃+]/[CH₃NH₃]

Kb = (7.94 x 10[tex]^-12)(x)/(1.0-x)[/tex]

where x is the concentration of CH₃NH₃+ that is formed.

We can simplify the equation by assuming that x is less than 1.0 M because CH₃NH₃ is a weak base:

Kb = (7.94 x 10[tex]^-12)(x)/(1.0)[/tex]

Kb = 7.94 x 10[tex]^-12 x[/tex]

Now, we need to find x. We can use the equilibrium constant expression for weak bases:

Kb = [CH₃NH₃+][OH-]/[CH₃NH₂]

Kb = (x)(7.94 x 10[tex]^-12)/(1.0)[/tex]

Solving for x:

x = Kb(1.0)/(7.94 x 10[tex]^-12)[/tex]

x = Kb/7.94 x 10[tex]^-12[/tex]

Subbing this worth of x back into the improved articulation for Kb:

Kb = 7.94 x 10[tex]^-12 x[/tex]

Kb = 7.94 x 10[tex]^-12[/tex] (Kb/7.94 x 10[tex]^-12)[/tex]

Kb = Kb

Therefore, the value of Kb CH₃NH₂ is 7.94 x 10[tex]^-12.[/tex]

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The angles phi and psi are commonly used in biochemistry and molecular biology to describe the backbone torsion angles of amino acids in a protein chain.

These angles describe the rotation around the peptide bond between the amino group and the carbonyl group of adjacent amino acids. The phi angle refers to the rotation around the N-Cα bond, while the psi angle refers to the rotation around the Cα-C bond. The values of phi and psi angles determine the conformation of the protein backbone and play a crucial role in protein folding and stability.

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Answer:Activation Energy

Explanation:

The number of moles of TiCl4 present in 12.5g of titanium (IV) chloride is approximately 0.066 moles.

To find the number of moles of TiCl4 in 12.5g of titanium (IV) chloride, we have to follow these steps:

1. Determine the molar mass of TiCl4.
The molar mass of titanium (Ti) is 47.87 g/mol, and the molar mass of chlorine (Cl) is 35.45 g/mol. Since there are 4 chlorine atoms in TiCl4:

Molar mass of TiCl4 = 47.87 + (4 × 35.45) = 47.87 + 141.8 = 189.67 g/mol

2. Calculate the number of moles.
Use the formula: moles = mass / molar mass

Number of moles of TiCl4 = 12.5 g / 189.67 g/mol ≈ 0.066 moles

So, there are approximately 0.066 moles of TiCl4 in 12.5g of titanium (IV) chloride.

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