The total pressure of a mixture of H2, He, and Ar is 99.3 kPa. The partial pressure of the He is 42.7 kPa, and the partial pressure of Ar is 54.7 kPa. What is the partial pressure of hydrogen

Answer: I think so.. (sorry if i am wrong)

Explanation:  Familiar examples of physical properties include density, color, hardness, melting and boiling points, and electrical conductivity. A physical property is a characteristic of matter that is not associated with a change in its chemical composition.

Answer

Molarity = [tex]1.4mol/L[/tex]

Explanation:

Molarity provides the number of moles of solute per liter of solution (moles/Liter). It is a means by which concentration of solution is measured.

SEE THE ATTACHMENT BELOW FOR STEP BY STEP SOLUTION.

Answer:

1.42 M

Explanation:

In this case have a dilution problem, therefore we need to use the dilution equation:

[tex]C_1*V_1=C_2*V_2[/tex]

What values we have?

[tex]C_1=3.82M[/tex]

[tex]V_1=526mL(0.516L)[/tex]

[tex]C_2=?[/tex]

[tex]V_2=?[/tex]

Now, we can calculate [tex]V_2[/tex] if we add the volumes, so:

[tex]0.516~L+~0.875~L=1.391~L[/tex]

So,  [tex]V_2=1.391~L[/tex]

We can plug the values in the equation:

[tex]3.82~M*0.516~L=C_2*1.391~L[/tex]

[tex]C_2=\frac{3.82~M*0.516~L}{1.391~L}[/tex]

[tex]C_2=1.42~M[/tex]

I hope it helps!

Answer: A

Explanation: Heat is the temperature of an object.

Answer:

Option C

the energy transferred between objects at different temperature.

Answer:

884.8 mL

Explanation:

If we have STP condition (Standard temperature and pressure), we will have the following relationship:

[tex]1~mol=22.4~L[/tex] or [tex]1~mol=22400~mL[/tex]. With this in mind we can do the conversion:

[tex]0.0395~mol\frac{22400~mL}{1~mol}=884.8~mL[/tex]

In the 0.0395 mol of fluorine gas, we will have 884.8 mL of gas at STP conditions.

Answer:

Grams of mercury= 0.06 g of Hg

Note: The question is incomplete. The complete question is as follows:

A compact fluorescent light bulb contains 4 mg of mercury. How many grams of mercury would be contained in 15 compact fluorescent light bulbs?

Explanation:

Since one fluorescent light bulb contains 4 mg of mercury,

15 such bulbs will contain 15 * 4 mg of mercury = 60 mg

1 mg = 0.001 g

Therefore, 60 mg = 0.001 g * 60 = 0.06 g of mercury.

Compact fluorescent lightbulbs (CFLs) are tubes containing mercury and noble gases. When electricity is passed through the bulb, electron-streams flow from a tungsten-coated coil. They collide with mercury atoms, exciting their electrons and creating flashes of ultraviolet light. A phosphor coating on the inside of the tube absorbs this UV light flashes and re-emits it as visible light. The amount of mercury in a fluorescent lamp varies from 3 to 46 mg, depending on lamp size and age.

Total amount of mercury in 15 compact fluorescent light bulbs is 0.06 gram of mercury.

What information do we have?

Number of compact fluorescent light bulbs = 15 bulb

Amount of mercury in each bulb = 4 mg

Total amount of mercury = Number of compact fluorescent light bulbs × Amount of mercury in each bulb

Total amount of mercury = 15 × 4

Total amount of mercury = 60 mg

Total amount of mercury = 60 / 1000

Total amount of mercury = 0.06 gram of mercury

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Answer:

pH = 9.03

Explanation:

The equilibrium of the NH₄Cl / NH₃ buffer in water is:

NH₃ + H₂O ⇄ NH₄⁺ + OH⁻

Initial moles of both NH₃ and NH₄⁺ are:

0.100L ₓ (0.20 mol / L) = 0.0200 moles

The NH₃ reacts with HCl producing NH₄⁺, thus:

NH₃ + HCl → NH₄⁺ + Cl⁻

That means, moles of HCl added to the solution are the same moles are consumed of NH₃ and produced of NH₄⁺

Moles added of HCl were:

0.025L ₓ (0.20mol / L) = 0.0050 moles of HCl. Thus, final moles of NH₃ and NH₄⁺ are:

NH₃: 0.0200 moles - 0.0050 moles = 0.0150 moles

NH₄⁺: 0.0200 moles + 0.0050 moles = 0.0250 moles.

Using H-H equation for bases:

pOH = pKb + log [NH₄⁺] / [NH₃]

Where pKb is -log Kb = 4.745.

Replacing:

pOH = 4.745 + log 0.0250mol / 0.0150mol

pOH = 4.967

As pH = 14- pOH

pH = 9.03

Answer:Fractional distillation of air. About 78 per cent of the air is nitrogen and 21 per cent is oxygen. These two gases can be separated by fractional distillation of liquid air.

Explanation:

Answer:

The molality of this solution is 0.368 molal

Explanation:

Step 1: Data given

Mass of CaCl2 = 10.2 grams

Molar mass of CaCl2 = 110.98 g/mol

Mass of water = 250 grams = 0.250 kg

Step 2: Calculate the number of moles CaCl2

Moles CaCl2 = mass CaCl2 / molar mass CaCl2

Moles CaCl2 = 10.2 grams / 110.98 g/mol

Moles CaCl2 = 0.0919 moles

Step 3: Calculate the molality of the solution

Molality solution = moles CaCl2 / mass water

Molality solution = 0.0919 moles / 0.250 kg

Molality solution = 0.368 mol / kg = 0.368 molal

The molality of this solution is 0.368 molal

Answer:

0.367 m

Explanation:

We have to start with the molality equation:

[tex]m=\frac{mol~of~solute}{Kg~of~solvent}[/tex]

If we want to calculate the molality we need the moles of the solute and the Kg of the solvent. In this case the solute is [tex]CaCl_2[/tex] and the solvent is [tex]H_2O[/tex].

Lets start with the the moles of solute. For this calculation we need the molar mass of [tex]CaCl_2[/tex] ([tex]111~g~CaCl_2[/tex]) and the mass, so:

[tex]10.2~g~CaCl_2\frac{1~mol~CaCl_2}{111~g~CaCl_2}[/tex]

[tex]0.0918~mol~CaCl_2[/tex]

Now, we can calculate the Kg of solvent, if we do the conversion from grams to Kg:

[tex]250~g~CaCl_2\frac{1~Kg}{1000~g}[/tex]

[tex]0.25~Kg[/tex]

With these values we can calculate the "molality":

[tex]m=\frac{0.0918~mol~CaCl_2}{0.25~Kg}[/tex]

[tex]m=0.3672[/tex]

I hope it helps!

Answer:

30.0g/mol

Explanation:

Step 1: Given data

Step 2: Calculate the moles of the gas

We will use the ideal gas equation.

[tex]P \times V = n \times R \times T\\n = \frac{P \times V}{R \times T} = \frac{1atm \times 0.0933L}{\frac{0.0821atm.L}{mol.K} \times 273.15K} = 4.16 \times 10^{-3} mol[/tex]

Step 3: Calculate the molar mass of the gas

4.16 × 10⁻³ moles correspond to a mass of 0.125 g. The molar mass of the gas is:

[tex]\frac{0.125g}{4.16 \times 10^{-3} mol} =30.0g/mol[/tex]

The molar mass of the 0.125 g  of the gas occupies 93.3 mL at STP is 30.0 g/mol.

Number of moles of Gas at STP,

[tex]\bold{n =\dfrac {PV}{RT}}[/tex]

where,

P - pressure

V- volume

R- gas constant

T - temperature

Put the values in the formula,

[tex]\bold{n =\dfrac {1 \times 0.0933} {0.082 \times 273.15 }}\\\\\bold{n =4.16 \timesw 10^-^3}[/tex]

The molar mass of the gas can be calculated using formula,

[tex]\bold {m = \dfrac {w}{n}}\\\\\bold {m = \dfrac {0.125} {4.16 \times 10^-^3}}\\\\\bold {m = 30g/mol}[/tex]

The molar mass of the 0.125 g  of the gas occupies 93.3 mL at STP is 30.0 g/mol.

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Answer:

ΔH = 130.5 kJ

Explanation:

Hello,

In this case, by using the Hess law, we compute the enthalpy of the required reaction:

C(s) + H2O(g) --> CO(g) + H2(g)

Thus, the first step is to keep the following reaction unchanged:

C (s) + O2 (g) → CO2 (g), ΔH = -393.5 kJ

Next, we invert and halve this reaction:

2 CO (g) + O2 (g) → 2 CO2 (g), ΔH= -566.0 kJ

So the enthalpy of reaction is inverted and halved:

CO2 (g) → CO (g) + 1/2 O2 (g) ΔH= 283 kJ

Then, we also invert and halve this reaction:

2 H2 (g) + O2 (g) → 2 H2O ΔH= -483.6 kJ

So the enthalpy of reaction is inverted and halved as well:

H2O → H2 (g) + 1/2 O2 (g) ΔH= 241.8 kJ

Finally, we add the three reactions to obtain the required reaction:

= C (s) + O2 (g) + CO2 (g) + H2O → H2 (g) + 1/2 O2 (g) + CO (g) + 1/2 O2 (g) + CO2 (g)

= C (s) + O2 (g) + CO2 (g) + H2O → H2 (g) + O2 (g) + CO (g) + CO2 (g)

= C (s) + H2O → H2 (g) CO (g)

So enthalpy is computed by:

ΔH = -393.5 kJ + 283 kJ + 241.8 kJ

ΔH = 130.5 kJ

Best regards.

Considering the Hess's Law, the enthalpy change for the reaction is 131.3 kJ.

Hess's Law indicates that the enthalpy change in a chemical reaction will be the same whether it occurs in a single stage or in several stages. That is, the sum of the ∆H of each stage of the reaction will give us a value equal to the ∆H of the reaction when it occurs in a single stage.

In this case you want to calculate the enthalpy change of:

C(s) + H₂O(g) → CO(g) + H₂(g)

You know the following reactions, with their corresponding enthalpies:

Equation 1: C (s) + O₂(g) → CO₂ (g)     ΔH = -393.5 kJ

Equation 2:  2 CO (g) + O₂ (g) → 2 CO₂ (g)     ΔH= -566.0 kJ

Equation 3: 2 H₂ (g) + O₂ (g) → 2 H₂O (g)     ΔH= -483.6 kJ  

First, to obtain the enthalpy of the desired chemical reaction you need one mole of C(s) on reactant side and it is present in first equation so let's write this as such.

Now, 1 mole of CO(g) must be a product and is present in the second equation. Since this equation has 2 moles of CO(g) on the reactant side, it is necessary to locate this component on the product side (invert it) and divide it by 2 to obtain 1 mole of CO(g).

When an equation is inverted, the sign of ΔH also changes.

And since enthalpy is an extensive property, that is, it depends on the amount of matter present, since the equation is divided by 2, the variation of enthalpy also is divided by 2.

Finally, 1 mole of H₂O(g) must be a reactant and is present in the third equation. Since this equation has 2 moles of CO(g) on the product side, it is necessary to locate this component on the product side (invert it) and divide it by 2 to obtain 1 mole of H₂O(g).

So, the sign of ΔH also changes and the variation of enthalpy is divided by 2.

In summary, you know that three equations with their corresponding enthalpies are:

Equation 1: C (s) + O₂(g) → CO₂ (g)     ΔH = -393.5 kJ

Equation 2: CO₂ (g) → CO (g) + [tex]\frac{1}{2}[/tex] O₂ (g)     ΔH= 283 kJ

Equation 3: H₂O (g) → H₂ (g) + [tex]\frac{1}{2}[/tex] O₂ (g)    ΔH= 241.8 kJ  

Adding or canceling the reactants and products as appropriate, and adding the enthalpies algebraically, you obtain:

C(s) + H₂O(g) → CO(g) + H₂(g)  ΔH= 131.3 kJ

Finally, the enthalpy change for the reaction is 131.3 kJ.

Learn more:

Answer:

1-Pentene

Explanation:

If we look at all the options listed, we will notice that the rate of reaction of bromine with each one differs significantly.

For 1-pentene, addition of bromine across the double bond is a relatively fast process. It is usually used as a test for unsaturation. Bromine water is easily decolorized by alkenes.

Cyclohexane, heptane are alkanes. They can only react with chlorine in the presence of sunlight. This is a substitution reaction. It does not occur easily. A certain quantum of light is required for the reaction to occur.

For benzene, bromine can only react with it by electrophilic substitution in which the benzene ring is retained. A Lewis acid is often required for the reaction to occur and it doesn't occur easily.

Answer:

Oxygen and Magnesium.

Based on the difference in electronegativity, potassium and sulfur, chlorine and lithium, and oxygen and magnesium can form ionic compounds, whereas lithium and calcium, sulfur and bromine, and manganese and chlorine cannot.

Ionic compounds are formed when there is a high electronegativity difference between two elements.

To determine whether the following pairs of elements can form ionic compounds, we need to look at their electronegativity difference:

1. Lithium and calcium: The electronegativity difference between Lithium and calcium is only 0.9, which is less than 1.7. Therefore, they cannot form an ionic compound.

2. Sulfur and bromine: The electronegativity difference between sulfur and bromine is 0.9, which is less than 1.7. Therefore, they cannot form an ionic compound.

3. Manganese and chlorine: The electronegativity difference between manganese and chlorine is 1.5, which is less than 1.7. Therefore, they cannot form an ionic compound.

4. Potassium and sulfur: The electronegativity difference between potassium and sulfur is 2.4, which is greater than 1.7. Therefore, they can form an ionic compound.

5. Chlorine and lithium: The electronegativity difference between chlorine and lithium is 2.8, which is greater than 1.7. Therefore, they can form an ionic compound.

6. Oxygen and magnesium: The electronegativity difference between oxygen and magnesium is 1.7, which is equal to 1.7. Therefore, they can form an ionic compound.

Therefore, based on the electronegativity difference between the elements, potassium and sulfur, chlorine and lithium, and oxygen and magnesium can form ionic compounds, while lithium and calcium, sulfur and bromine, and manganese and chlorine cannot form ionic compounds.

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Answer:

101

Explanation:

Provided that

[tex]S_1 = S_2 = same\ V_{max}[/tex]

And,

[tex]S_1\ k_M = 2.0mM\\S_2\ k_M = 20mM[/tex]

Now we expect the same

{S} (0.1mM)

This determines that [tex]S_1[/tex] generates a  higher rate of product formation as compared to the [tex]S_2[/tex]

So we can easily calculate the [tex]V_{max}[/tex] for either of [tex]S_1[/tex] or [tex]S_2[/tex] as we know that Tube 1 is [tex]S_2[/tex] and tube 2 is [tex]S_1[/tex]

As we know that

[tex]V_0 = V_{max}\ {S} / (K_M + {S})[/tex]

As the rates do not include any kind of units so we do not consider the units for [tex]V_{max}[/tex]

Now the calculation is

[tex]0.5 = V_{max} (0.1\ mM) / (20\ mM + 0.1\ mM)[/tex]

[tex]V_{max} = 0.5 (20.1\ mM) / 0.1\ mM[/tex]

= 100.5

≈ 101

A soluble base is formed when sodium reacts with water.

When sodium reacts with water, it produces strongly alkalic sodium hydroxide which is also called caustic soda and hydrogen gas. In this chemical reaction, energy is absorbed which means it is an exothermic reaction so we can conclude that a soluble base is formed when sodium reacts with water.

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Answer:

A. Reversible reactions.

Explanation:

Reactions that undergo shifts in their equilibrium must be reversible reactions.

A reversible reaction is a chemical reaction that is capable of forming it's reactants back from the resulting formation of products. This simply means that, reversible reactions are chemical reactions that are in equilibrium because the forward and reverse path happens at the same rate.

For example, the reaction of hydrogen gas [tex]H_{2}[/tex] and iodine gas [tex]I_{2}[/tex] to form a chemical compound called hydrogen Iodide [tex]HI[/tex] is a reversible chemical reaction.

Forward reaction:

[tex]H_{2} + I_{2} ----> 2HI[/tex]

Reverse reaction:

[tex]2HI----> H_{2} + I_{2}[/tex]

A reversible reaction is denoted with a double arrow.

Also, reversible chemical reactions are controlled by the Le Chatelier's principle.

Answer: Pollutant

Explanation:

Answer:

Explanation:

The substance that could be found in air, water or soil that is harmful to humans or animals is called pollutant.

Answer:

d) The fraction of collision with total kinetic energy larger than activation energy  increases.

Explanation:

Hello,

In this case, kinetic models explain how the rate of a chemical reaction is affected by several factors. In such a way, specifically for temperature, when it increases, the average velocity of the particles is also increased, for that reason, the collision frequency increases since the molecules are more likely to collide as they move faster and encounter to each other.

Nonetheless, it is the minor reason because the main reason is that the effective collisions increase when the temperature is increased, and they are related with the fraction of collision with total kinetic energy that turns out larger than the activation energy, therefore, answer is d).

Best regards.

Explanation:

First off, we have to understand what we means by oxidation and reduction.

Oxidation is the loss of electrons during a reaction by a molecule, atom or ion. Oxidation occurs when the oxidation state of a molecule, atom or ion is increased.

Reduction on the other hand is the opposite of oxidation. It is the gain of electrons during a reaction by a molecule, atom or ion. Reduction occurs when the oxidation state of a molecule, atom or ion decreases.

Back to the question;

oxidation is what happens when a reactant loses electrons.

Reduction occurs when a reactant gains electrons in the reaction.

Answer:

pH = 7.1581

Explanation:

The equilibrium of NaHSO₃ with Na₂SO₃ is:

HSO₃⁻ ⇄ SO₃²⁻ + H⁺

Where K of equilibrium is the Ka2: 6.5x10⁻⁸

HSO₃⁺ reacts with NaOH, thus:

HSO₃⁻ + NaOH → SO₃²⁻ + H₂O + Na⁺

As the buffer is of 1.0L, initial moles of HSO₃⁻ and SO₃²⁻ are:

HSO₃⁻: 0.252 moles

SO₃²⁻: 0.139 moles

Based on the reaction of NaOH, moles added of NaOH are subtracting moles of HSO₃⁻ and producing SO₃²⁻. The moles added are:

0.0500L ₓ (1mol /L): 0.050 moles of NaOH.

Thus, final moles of both compounds are:

HSO₃⁻: 0.252 moles - 0.050 moles = 0.202 moles

SO₃²⁻: 0.139 moles + 0.050 moles = 0.189 moles

Using H-H equation for the HSO₃⁻ // SO₃²⁻ buffer:

pH = pka + log [SO₃²⁻] / [HSO₃⁻]

Where pKa is - log Ka = 7.187

Replacing:

pH = 7.187 + log [0.189] / [0.202]

Answer:

2.9 M

Explanation:

Step 1: Given data

Moles of barium chloride (solute): 4.4 moles

Volume of solution: 1.5 liters

Step 2: Calculate the molarity of barium chloride in the solution

The molarity is a way to quantitatively express the concentration of a solute in a solution. The molarity is equal to the moles of solute divided by the volume, in liters, of solution.

[tex]M=\frac{4.4mol}{1.5L} =2.9 M[/tex]

Answer:

1. The correct option is;

c. maintains charge balance in the cell

2. The correct option is;

c. +3.272 V

Explanation:

The aqueous solution in a galvanic cell is the electrolyte which is a ionic solution containing that permits the transfer of ions between the separated compartment of the galvanic cell such that the overall system is electrically neutral

Therefore, the aqueous solution maintains the charge balance in the cell

2. Here we have;

B₂ + 2e⁻ → 2B⁻ Ecell = 0.662 V

A⁺ + 1e⁻ → A Ecell = -1.305 V

Hence for the overall reaction, we have;

2A + B₂ → 2AB gives;

(0.662) - 2×(-1.305)  = +3.272 V.

Answer and Explanation:

A chemical formula is composed by chemical symbols (in letters, which indicate the chemical elements) and subscripts (numbers). For example, let see the chemical formula for carbon dioxide:

CO₂

The letters C and O are the chemical symbols for the elements carbon (C) and oxygen (O). The number 2 in subscript indicates that there are 2 atoms of O per molecule. The subscript 1 is generally not indicated in chemical formulae, so it is assumed that the number of atoms of C is 1.

Summarizing, the chemical formula CO₂ indicates us that the molecule if formed by 1 atom of the element carbon (C) and 2 atoms of the element oxygen (O).

Answer:

Explanation:

O₂(g) + 2F₂(g) ↔ 2F₂(g)

Stabdard ΔG values are

[tex]\Delta G_f[F_2O]=41.9kJ/mol =41900J/mol[/tex]

[tex]\Delta G_f[O_2]=0\\\\ \Delta G_f[F_2]=0[/tex]

[tex]\Delta G^0=\sum \Delta G^\circ (products)- \sum \Delta G ^\circ (reactants)[/tex]

[tex]\Delta G^\circ = [2 \times 41900]-0\\\\=83800J/mol[/tex]

Now,

[tex]\Delta G^\circ =-RTInK[/tex]

Given T = 100°C

= 100+ 273.15 = 373.15K

R = 8.314J/k / mol

so,

83800 = -8.314 * 373.15 * InK

InK = -27.0116

K = 1.858 * 10⁻¹²

Answer:

c. I, II and III

Explanation:

The cell is as follows

Cr / Cr⁺²(1M) // H⁺ ( 1 M ) / H₂

Standard reduction potential of hydrogen cell is zero . Standard reduction potential is negative E for Cr⁺² / Cr(1M) half cell

Cell potential = Ecathode - Eanode

= 0 - ( - E)

= E

E is cell potential and also standard cell potential or emf of the cell .

Standard reduction potential that is for Cr3+/Cr.  is  - E .

Hence statement I , II , III are right . IV th statement is wrong because of sign

Option c is correct.

The branch of chemistry which deals with electricity is called electrochemistry.

The correct answer is C

The cell representation is as follows:-

The standard reduction potential of hydrogen cells is zero. Standard reduction potential is negative E for Cr⁺² / Cr(1M) half cell because it will behave as a cathode.

The formula of the electric cell is as follows :-[tex]Cell potential = E_{cathode} - E_{anode[/tex]

After putting the value the cell potential will be:-

Cell potential = 0 - ( - E)

The standard reduction potential that is for [tex]Cr^{3+}/Cr = - E .[/tex]

Hence, the correct option is C that is I, II, and IV.

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Answer:

(b.) alkene

(i) aldehyde/ketone

(k.) ester

Explanation:

Peaks observed at:

Answer:

Yes, we should invest in renewable energies that decrease the levels of carbon dioxide in the atmosphere, that is, decrease the partial pressure of carbon dioxide in the environment since it is lethal for human life, in turn I consider that the exploitation of sources Oil was widely used but I think they should have an end, since they pollute oceans, oxygen, and fuels that are not necessary for today's technology since automobiles could be a base of electric or solar energy

Explanation:

Environmental pollution is a very serious current problem, since we should stop investing in oil sources and be able to keep those profits or that economy in another way or replace them with another resource, since if this greenhouse effect is not perceived as a current serious problem, we could run the risk of suffering natural catastrophes, systematic diseases, affections in the same life.

Answer:

63.33%

Explanation:

To find the percentage of sugar in the milk chocolate just divide 15 by 9.5. This will give you 0.63333 (recurring). To find the percentage, times it by 100, which will give you 63.33%.

Hope this helped :)

The percentage of sugar present in the milk would be  63.33 %.

In positional notation, significant figures refer to the digits in a number that is trustworthy and required to denote the amount of something, also known as the significant digits, precision, or resolution.

As given in the problem a 15.00 g milk chocolate bar contains 9.500 grams of sugar, and we have calculated the percentage of sugar present,

The percentage of sugar present in the milk = 9.500 ×100/15

                                                                                = 63.33 %

Thus, the percentage of sugar present in the milk would be 63.33 %.

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Answer: Wind current

Explanation:

The wind current is the main driving force along with the water currents when one reaches to the seashore. The wind currents above surface of water generates due to the evaporation of water.

The strong wind currents can pull any animal or human walking or running near the sea shore. These currents bring the animal or human towards the sea.

Answer:

molar mass of the phosphorus chloride = 138.06 g/mol

Explanation:

mass of phosphorus will be the same as mass of CO2, since it is stated that they are of equal amount.

mass = 3.51 g

lets assume that it took the CO2 1 sec to effuse, then the time taken by the phosphorus chloride will be 1.77 sec

From this we can say that

rate of effusion of CO2 = 3.51/1 = 3.51 g/s

rate of effusion of the phosphorus chloride = 3.51/1.77 = 1.98 g/s

From graham's equation of effusion,

[tex]\frac{Rc}{Rp}[/tex] = [tex]\sqrt{\frac{Mp\\}{Mc} }[/tex]

Rc = rate of effusion of CO2 = 3.51 g/s

Rp = rate of effusion of phosphorus chloride = 1.98 g/s

Mc = molar mass of CO2 = 44.01 g/mol

Mp = molar mass of the phosphorus chloride = ?

Imputing values into the equation, we have

[tex]\frac{3.51}{1.98}[/tex] = [tex]\sqrt{\frac{Mp\\}{44.01} }[/tex]

1.77 = [tex]\frac{\sqrt{Mp} }{6.64}[/tex]

11.75 = [tex]\sqrt{Mp}[/tex]

Mp = [tex]11.75^{2}[/tex]

Mp = molar mass of the phosphorus chloride = 138.06 g/mol