Why do some metals lose their shine over time but not gold ?

The three alkenes that will form 2-methylbutane upon hydrogenation are 1-pentene, 2-pentene, and 3-pentene.

To find the three alkenes that will form 2-methylbutane upon hydrogenation, we need to consider the number of carbon atoms in the molecule. In this case, the formula is C5H10, which means it has 5 carbon atoms. Since we want to form 2-methylbutane upon hydrogenation, we know that the starting molecule must have a double bond on one of the outer carbon atoms and a methyl group attached to one of the middle carbon atoms.

The three alkenes that satisfy these conditions are:

https://brainly.com/question/40062596

#SPJ3

Answer:

Cl⁻

Explanation:

A potassium ion K+ would bond with chloride ion (Cl⁻) than any other ion or element listed. Because potassium ion (K+) is electro positive and has lost an electron and having a low charge denisty while chloride ion Cl⁻ has a high charge density and is electro negative and has accepted an electron. Magnesium ion (Mg+) and sodium ion (Na+) can't bond with potassium ion (K+) because they have the same charge and has each lost an electron. Same charges or like charges repel while unlike charges or opposite charges attract towards each other.

Oxygen atom (O) can't bond with Potassium ion (K+) because oxygen is in an elemental state while potassium is in an ionic state and hence can't bond with each other. However of oxygen changes into ionic state (O²⁻), it'll definitely bond with potassium ion (K+) to form a compound (K₂O)

The stoichiometric ratio between Fe(NH4)2(SO4)2·6H2O and KMnO4 in the given redox reaction is 5:1. Therefore, for each mole of Fe(NH4)2(SO4)2·6H2O, 0.2 moles of KMnO4 are required.

This problem deals with the stoichiometry of a redox reaction between Fe(NH4)2(SO4)2·6H2O and KMnO4. The balanced redox reaction is:

8H+ + 5Fe+2 + MnO4 - --> Mn+2 + 5Fe+3 + 4H2O

From this balanced chemical equation, we see a stoichiometric ratio of 5:1 between Fe+2 (iron in the compound Fe(NH4)2(SO4)2·6H2O) and MnO4- (manganese in KMnO4). Thus, for each mole of Fe(NH4)2(SO4)2·6H2O, you require 1/5 (or 0.2) moles of KMnO4. In order to fully answer your question, you would need to know the amount (in moles) of Fe(NH4)2(SO4)2·6H2O present.

https://brainly.com/question/29019892

#SPJ12

The equilibrium concentrations of PCl5(g) and PCl3(g) can be calculated using an ICE table and the equilibrium constant Kc for the reaction PCl5 = PCl3 + Cl2 at a given temperature.

To calculate the equilibrium concentrations of PCl5(g) and PCl3(g) after 0.366 mol of PCl5(g) is injected into a 4.45 L vessel held at 250 0C, we use the equilibrium constant (Kc) and perform an ICE (Initial, Change, Equilibrium) table calculation. Since we are not given the Kc for this particular reaction at 250 0C, let's assume we know this value or it is provided in the actual context of the problem.

For example, if the Kc was 0.0211, we would set up the equilibrium as follows: PCl5(g) = PCl3(g) + Cl2(g). With the initial concentration of PCl5 being 0.366 mol / 4.45 L, we would solve for the equilibrium concentrations using the Kc value, and the stoichiometry of the reaction.

At equilibrium, the concentrations are:

[tex]- \( [\text{PCl}_5] \approx 0.0032 \text{ M} \)\\- \( [\text{PCl}_3] \approx 0.079 \text{ M} \)[/tex]

To calculate the equilibrium concentrations of [tex]\( \text{PCl}_5(\text{g}) \)[/tex]and [tex]\( \text{PCl}_3(\text{g}) \)[/tex] in a reaction vessel, we need to consider the equilibrium reaction and the equilibrium constant (if provided). For the decomposition of [tex]\( \text{PCl}_5(\text{g}) \)[/tex]:

[tex]\[ \text{PCl}_5(\text{g}) \rightleftharpoons \text{PCl}_3(\text{g}) + \text{Cl}_2(\text{g}) \][/tex]

Let's denote the equilibrium constant for this reaction as [tex]\( K_c \)[/tex].

### Step-by-Step Calculation

1. **Initial conditions and setup:**

  - Initial moles of [tex]\( \text{PCl}_5 \)[/tex] = 0.366 mol

  - Volume of the vessel = 4.45 L

  - Initial concentration of[tex]\( \text{PCl}_5 \)[/tex]:

    [tex]\[ [\text{PCl}_5]_0 = \frac{0.366 \text{ mol}}{4.45 \text{ L}} = 0.0822 \text{ M} \][/tex]

  Initial concentrations of[tex]\( \text{PCl}_3 \) and \( \text{Cl}_2 \)[/tex] are 0 M since they are not present initially.

2. **Change in concentrations:**

  - Let ( x ) be the change in concentration of[tex]\( \text{PCl}_5 \)[/tex]that decomposes.

  - At equilibrium:

   [tex]\[ [\text{PCl}_5] = 0.0822 - x \] \[ [\text{PCl}_3] = x \] \[ [\text{Cl}_2] = x \][/tex]

3. **Expression for the equilibrium constant [tex]\( K_c \)[/tex]:**

  [tex]\[ K_c = \frac{[\text{PCl}_3][\text{Cl}_2]}{[\text{PCl}_5]} = \frac{x \cdot x}{0.0822 - x} = \frac{x^2}{0.0822 - x} \][/tex]

  If[tex]\( K_c \)[/tex] is provided, we can solve for \( x \). If not, let's assume a hypothetical value for[tex]\( K_c \)[/tex]to illustrate the process (e.g.,[tex]\( K_c = 1.80 \)[/tex] at 250 °C, but you should use the actual value given or found from tables).

4. **Solve for ( x ):**

[tex]\[ 1.80 = \frac{x^2}{0.0822 - x} \] \[ 1.80 (0.0822 - x) = x^2 \] \[ 0.14796 - 1.80x = x^2 \] \[ x^2 + 1.80x - 0.14796 = 0 \][/tex]

  Solving this quadratic equation for ( x ):

 [tex]\[ x = \frac{-1.80 \pm \sqrt{(1.80)^2 + 4 \cdot 0.14796}}{2} \] \[ x = \frac{-1.80 \pm \sqrt{3.24 + 0.59184}}{2} \] \[ x = \frac{-1.80 \pm \sqrt{3.83184}}{2} \] \[ x = \frac{-1.80 \pm 1.9574}{2} \][/tex]

  This gives two possible solutions:

[tex]\[ x = \frac{-1.80 + 1.9574}{2} \approx 0.079 \] \[ x = \frac{-1.80 - 1.9574}{2} \approx -1.879 \] (discard this negative value)[/tex]

5. **Equilibrium concentrations:**

  [tex]\[ [\text{PCl}_5] = 0.0822 - 0.079 \approx 0.0032 \text{ M} \] \[ [\text{PCl}_3] = x \approx 0.079 \text{ M} \] \[ [\text{Cl}_2] = x \approx 0.079 \text{ M} \][/tex]

### Conclusion

At equilibrium, the concentrations are:

[tex]- \( [\text{PCl}_5] \approx 0.0032 \text{ M} \)\\- \( [\text{PCl}_3] \approx 0.079 \text{ M} \)[/tex]

The specific values of the concentrations will depend on the actual [tex]\( K_c \)[/tex]value at 250 °C.

The answer is B.) Water molecules at the surface experience fewer hydrogen bonds than water molecules within the liquid.

Hope this helped!

The volume of in milliliters of the HClO₄ needed to neutralize the NaOH solution is 24.9mL

From the question, we are to determine the volume of HClO₄ needed to neutralize the NaOH solution

First, we will write the balanced chemical equation for the reaction

The balanced chemical equation for the reaction is

HClO₄ + NaOH → NaClO₄ + H₂O

This means

1 mole of HClO₄ is required to neutralize 1 mole of NaOH

Now, we will determine the number of moles of NaOH present in the solution

From the given information

Volume of NaOH = 50.00 mL = 0.05 L

Concentration of NaOH = 0.0832 M

Using the formula,

Number of moles = Concentration × Volume

Then,

Number of moles of NaOH present = 0.0832 × 0.05

Number of moles of NaOH present = 0.00416 mole

Now,

Since 1 mole of HClO₄ is required to neutralize 1 mole of NaOH

Then,

0.00416 mole of HClO₄ will be required to neutralize the 0.00416 mole of NaOH

Thus, the number of moles of HClO₄ required is 0.00416 mole

Now, for the volume of HClO₄ required

From the formula,

[tex]Volume = \frac{Number\ of\ moles }{Concentration}[/tex]

Then,

Volume of HClO₄ needed = [tex]\frac{0.00416}{0.167}[/tex]

Volume of HClO₄ needed = 0.0249101 L

Volume of HClO₄ needed = 24.9101 mL

Volume of HClO₄ needed ≅ 24.9 mL

Hence, the volume of in milliliters of the HClO₄ needed to neutralize the NaOH solution is 24.9mL

Learn more on Stoichiometry here: https://brainly.com/question/10382262

Answer:

Container will rupture at temperature of 375 K.

Explanation:

Initial pressure of the  nitrogen gas =[tex]P_1= 1200 torr = 1.572 atm[/tex]

(1 torr = 0.00131 atm)

Initial temperature of nitrogen gas =[tex]T_1= 250 K[/tex]

Final pressure of the nitrogen gas =[tex]P_2=1800 torr=2.358 atm[/tex]

Final temperature of nitrogen gas =[tex]T_2=?[/tex]

Since, the container is inflexible that is volume remains constant we can apply Gay Lussac's law:

[tex]\frac{P_1}{T_1}=\frac{P_2}{T_2}[/tex]

[tex]\frac{1.572 atm}{250 K}=\frac{2.358 atm}{T_2}[/tex]

[tex]T_2=375 K[/tex]

Container will rupture at temperature of 375 K.

Our planet is formed of elements and the combinations of elements known as compounds. An element is a pure component formed of atoms, which are all of a similar kind. Till now, 116 elements are identified, and of these only 90 occur naturally.  

At the time of universe formation, which is, about 14 billion years ago known as Big Bang, the creation of only the lightest elements took place, that is, hydrogen and helium along with minute concentrations of beryllium and lithium. The rest of the 86 elements found in nature were formed in the nuclear reactions, which took place in the stars and in huge stellar explosions called supernovae.  

A barometer reading of 29.8 inches of mercury is equivalent to 1012.8 millibars of atmospheric pressure.

A barometer displaying a reading of 29.8 inches of mercury would be equivalent to 1012.8 millibars of atmospheric pressure.

To convert inches of mercury to millibars, you can use the conversion factor of 1 inch of mercury = 33.864 millibars. Multiply the given reading by the conversion factor to find the equivalent pressure in millibars.

Therefore, 29.8 inches of mercury = 29.8 inches * 33.864 millibars/inch = 1012.8 millibars.

https://brainly.com/question/33448875

#SPJ2

The name of the compound P₄O₁₀  is Phosphorus pentoxide.

Phosphorus pentoxide can be described as a chemical compound with the molecular formula  P₄O₁₀   and the empirical formula, P₂O₅. This white crystalline solid can be described as the anhydride of phosphoric acid. It acts as a desiccant and dehydrating agent.

Phosphorus pentoxide can be crystallized in at least four forms or polymorphs. The most familiar one, a metastable form comprises molecules of P₄O₁₀. Weak van der Waals forces hold molecules together in a hexagonal lattice.

The structure of the P₄O₁₀  cage is similar to adamantane with a tetrahedral symmetry point group. The density of P₄O₁₀  is 2.30 g/cm³ and boils at 423 °C under atmospheric pressure if heated making it sublimate.

P₄O₁₀ can be made by condensing the vapor of phosphorus pentoxide rapidly and is prepared by burning tetraphosphorus with the supply of oxygen.

Learn more about Phosphorus pentoxide, here:

https://brainly.com/question/14534375

#SPJ6

The question is incomplete, here is the complete question:

The chemistry of fresco painting is the chemistry of limestone and lime plaster. limestone is calcium carbonate, an abundant, naturally occurring mineral. two key reactions, shown below, are involved in the process of converting limestone to lime plaster (calcium hydroxide). classify each of these reactions as one of the four types of reactions listed in this experiment (decomposition, synthesis, single replacement, or double replacement).

(a)  [tex]CaCO_3(s)\rightarrow CaO(s)+CO_2(g)[/tex]

(b)  [tex]CaO(s)+H_2O(l)\rightarrow Ca(OH)_2(aq.)[/tex]

Answer: Equation (a) is a decomposition reaction and equation (b) is a combination reaction.

Explanation:

Decomposition reaction is defined as the reaction in which a single large substance breaks down into two or more smaller substances.

[tex]AB\rightarrow A+B[/tex]

Combination reaction is defined as the reaction in which smaller substances combine to form a larger substance.

[tex]A+B\rightarrow AB[/tex]

Single displacement reaction is defined as the reaction in which more reactive element displaces a less reactive element from its chemical reaction.

The reactivity of metal is determined by a series known as reactivity series. The metals lying above in the series are more reactive than the metals which lie below in the series.

[tex]A+BC\rightarrow AC+B[/tex]

Double displacement reactions are defined as the reactions in which exchange of ions takes place.

[tex]AB+CD\rightarrow AD+CB[/tex]

For the given equations:

(a)  [tex]CaCO_3(s)\rightarrow CaO(s)+CO_2(g)[/tex]

This equation is a type of decomposition reaction in which calcium carbonate breaks down into calcium oxide and carbon dioxide

(b)  [tex]CaO(s)+H_2O(l)\rightarrow Ca(OH)_2(aq.)[/tex]

This equation is a type of combination reaction in which calcium oxide reacts with water to form calcium hydroxide.

Answer : The value of [tex]k_a[/tex] for benzoic acid is, [tex]6.4\times 10^{-5}[/tex]

Solution :

The balanced equilibrium reaction will be,

                             [tex]C_6H_5COOH\rightleftharpoons H^++C_6H_5COO^-[/tex]

initial conc.          0.2 M                  0                0

at eqm.  [tex](0.2-0.00355)M[/tex]       [tex]0.00355M[/tex] [tex]0.00355M[/tex]

The expression for dissociation constant for a benzoic acid will be,

[tex]k_a=\frac{[H^+]\times [C_6H_5COO^-]}{[C_6H_5COOH]}[/tex]

Now put all the given values in this formula, we get the value of [tex]k_a[/tex]

[tex]k_a=\frac{(3.55\times 10^{-3})\times (3.55\times 10^{-3})}{(0.2-3.55\times 10^{-3})}=6.4\times 10^{-5}[/tex]

Therefore, the value of [tex]k_a[/tex] for benzoic acid is, [tex]6.4\times 10^{-5}[/tex]

[tex]\boxed{{\text{No precipitate will be formed}}}[/tex] in [tex]2{\text{NaCl}} + {\text{Ba}}{\left( {{\text{OH}}} \right)_2} \to {\text{BaC}}{{\text{l}}_2} + 2{\text{NaOH}}[/tex]

[tex]\boxed{{\text{AgI}}}[/tex] is the precipitate formed in [tex]{\text{AgCl}}{{\text{O}}_3} + {\text{Mg}}{{\text{I}}_2} \to {\text{AgI}} + {\text{Mg}}{\left( {{\text{Cl}}{{\text{O}}_3}} \right)_2}[/tex]

Further Explanation:

Precipitation reaction:

It is the type of reaction in which an insoluble salt is formed by the combination of two solutions containing soluble salts. That insoluble salt is known as precipitate and therefore such reactions are named precipitation reactions. An example of precipitation reaction is,

[tex]{\text{AgN}}{{\text{O}}_3}\left( {aq} \right) + {\text{KBr}}\left( {aq} \right) \to {\text{AgBr}}\left( s \right) + {\text{KN}}{{\text{O}}_3}\left( {aq} \right)[/tex]

Here, AgBr is a precipitate.

The solubility rules to determine the solubility of the compound are as follows:  

1. The common compounds of group 1A are soluble.

2. All the common compounds of ammonium ion and all acetates, chlorides, nitrates, bromides, iodides, and perchlorates are soluble in nature. Only the chlorides, bromides, and iodides of [tex]{\text{A}}{{\text{g}}^ + }[/tex], [tex]{\text{P}}{{\text{b}}^{2 + }}[/tex], [tex]{\text{C}}{{\text{u}}^ + }[/tex] and [tex]{\text{Hg}}_2^{2 + }[/tex] are not soluble.

3. All common fluorides, except for [tex]{\text{Pb}}{{\text{F}}_{\text{2}}}[/tex] and group 2A fluorides, are soluble. Moreover, sulfates except [tex]{\text{CaS}}{{\text{O}}_{\text{4}}}[/tex], [tex]{\text{SrS}}{{\text{O}}_{\text{4}}}[/tex],  [tex]{\text{BaS}}{{\text{O}}_{\text{4}}}[/tex], [tex]{\text{A}}{{\text{g}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}[/tex] and [tex]{\text{PbS}}{{\text{O}}_{\text{4}}}[/tex] are soluble.

4. All common metal hydroxides except [tex]{\text{Ca}}{\left( {{\text{OH}}} \right)_{\text{2}}}[/tex], [tex]{\text{Sr}}{\left( {{\text{OH}}} \right)_{\text{2}}}[/tex], [tex]{\text{Ba}}{\left( {{\text{OH}}} \right)_{\text{2}}}[/tex] and hydroxides of group 1A, are insoluble.

5. All carbonates and phosphates, except those formed by group 1A and ammonium ion, are insoluble.

6. All sulfides, except those formed by group 1A, 2A, and ammonium ion are insoluble.

7. Salts that contain [tex]{\text{C}}{{\text{l}}^ - }[/tex], [tex]{\text{B}}{{\text{r}}^ - }[/tex] or [tex]{{\text{I}}^ - }[/tex] are usually soluble except for the halide salts of [tex]{\text{A}}{{\text{g}}^ + }[/tex], [tex]{\text{P}}{{\text{b}}^{2 + }}[/tex] and [tex]{\left( {{\text{H}}{{\text{g}}_2}} \right)^{{\text{2 + }}}}[/tex].

8. The chlorides, bromides, and iodides of all the metals are soluble in water, except for silver, lead, and mercury (II). Mercury (II) iodide is water insoluble. Lead halides are soluble in hot water.

9. The perchlorates of group 1A and group 2A are soluble in nature.

(1) The given reaction is as follows:

 [tex]2{\text{NaCl}} + {\text{Ba}}{\left( {{\text{OH}}} \right)_2} \to {\text{BaC}}{{\text{l}}_2} + 2{\text{NaOH}}[/tex]

This is an example of a double displacement reaction in which two ionic compounds are exchanged with each other and two new compounds are formed. [tex]{\text{BaC}}{{\text{l}}_2}[/tex] and NaOH are soluble salts according to the solubility rules. So no precipitate will be formed in this reaction.

(2) The given reaction is as follows:

 [tex]{\text{AgCl}}{{\text{O}}_3} + {\text{Mg}}{{\text{I}}_2} \to {\text{AgI}} + {\text{Mg}}{\left( {{\text{Cl}}{{\text{O}}_3}} \right)_2}[/tex]

According to the solubility rules, AgI is an insoluble salt. The perchlorates of group 2 are soluble in nature and therefore [tex]{\text{Mg}}{\left( {{\text{Cl}}{{\text{O}}_3}} \right)_2}[/tex] is soluble in water. So AgI forms the precipitate in the above reaction.

Learn more:

1. Balanced chemical equation https://brainly.com/question/1405182

2. The mass of ethylene glycol: https://brainly.com/question/4053884

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Chemical reaction and equation

Keywords: precipitation reaction, precipitate, insoluble, soluble, AgI, AgClO3, Mg(ClO3)2, MgI2, NaCl, Ba(OH)2, BaCl2, NaOH, solubility rules, halides, sulfides.

Answer:   c. potassium nitrate

Among the given substances the solubility of carbon dioxide in liquids is more when compared with potassium nitrate, sugar and sodium chloride.

Explanation:

Increasing pressure doesn’t change the amount of solid or liquid particles dissolved in a solution. Thus pressure doesn’t have an effect on the solubility of solids and liquids. But the case of gases is different.

Increasing pressure above a liquid causes more gas molecules to get dissolved in the liquid. Thus pressure of the system directly affects the solubility of gases in a liquid.

In this question sugar, potassium nitrate and sodium chloride re solids and the pressure cannot change the solubility of these in a liquid.

But carbon dioxide being a gas can be dissolved more in a liquid if the pressure is increased.

The electron pair geometry around the Sb atom in SbF3 is trigonal pyramidal.

The electron pair geometry around the Sb (antimony) atom in SbF3 is trigonal pyramidal. This means that there are three bonding pairs of electrons and one lone pair of electrons around the central atom. The molecular structure of SbF3 is also trigonal pyramidal, which means that the lone pair of electrons forms the apex of the pyramid.

https://brainly.com/question/35160260

#SPJ11

The correct option is C.

The gas in question can most likely identified as SO2 (Sulfur Dioxide), based on a calculated molar mass of approximately 44.12 g/mol.

To find the identity of this gas, we can calculate the molar mass using the ideal gas law equation, which is PV = nRT, where P is the pressure in atm, V is the volume in L, n is the number of moles of gas, R is the ideal gas constant (0.0821 L.atm/K.mol) and T is the temperature in Kelvin.

However, we also know that the density of a gas can be represented as the molar mass divided by the molar volume (22.4 L at STP), so we can rearrange and solve the ideal gas law equation for the molar mass. Doing so gives us molar mass = density * R * T / P.

Substituting the given values into this equation we get molar mass = 1.4975g/L * 0.0821 L.atm/K.mol * 349.6 K / 2.384 atm = 44.12 g/mol.

Looking at the options you've provided, the molar mass closest to our calculated molar mass is SO2 or Sulfur Dioxide, which has a molar mass of 64.07 g/mol.

https://brainly.com/question/27922808

#SPJ3

The complete question is given below:

What is the identity of a gas that has a density of 1.4975 g/L and a volume of 8.64 L at a pressure of 2.384 atm and with a temperature of 349.6 Kelvin?

A) H2O B) N2 C) SO2 D) NO3 E) Cl2

Final answer:

To find the molecular formula of a compound with an empirical formula of C5H7N and a molar mass of 208.17 g/mol, divide the molar mass by the molar mass of the empirical formula and multiply the empirical formula by the resulting whole number to obtain the molecular formula, which is C15H21N3.

Explanation:

The student's question pertains to finding the molecular formula of a compound given its empirical formula and molar mass. The empirical formula is C5H7N, which has a molar mass of 81.13 g/mol. To find the molecular formula, we divide the given molar mass of the compound, 208.17 g/mol, by the empirical formula mass of 81.13 g/mol to find the multiple (n). This multiple, which should be a whole number, will be used to scale the empirical formula up to the molecular formula.

Performing this calculation, we find that n = 208.17 / 81.13, which simplifies to approximately 2.57. Since n must be a whole number, and this value is close to 2.5, we can deduce that the empirical formula needs to be multiplied by 3 to obtain the molecular formula. Therefore, the molecular formula of the compound is C15H21N3.

Answer:

295.3744 grams of NaBr will produce 244 grams of [tex]NaNO_3[/tex].

Explanation:

[tex]Pb(NO_3)_2(aq)+2 NaBr(aq)\rightarrow PbBr_2(s)+2 NaNO_3(aq)[/tex]

Moles of sodium nitarte= [tex]\frac{244 g}{85 g/mol}=2.8705 mol[/tex]

According to reaction, 2 moles of sodium nitrate is obtained from 2 moles of sodium bromide.

Then 2.8705 mol of sodium nitrate will be obtained from :

[tex]\frac{2}{2}\times 2.8705mol=2.8705 mol[/tex] of sodium nitrate

Mass of 2.8705 moles of sodium nitrate:

[tex]2.8705 mol\times 102.9 g/mol=295.3744 g[/tex]

295.3744 grams of NaBr will produce 244 grams of [tex]NaNO_3[/tex].

Answer:

1. 1/2

2. 36

3. 18

‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ 

The ratio of the central angle to the entire circle measure is 1:2. The area of the sector is equal to the area of the entire circle.

The ratio of the measure of the central angle MNL to the entire circle measure is 1:2. This is because the measure of the central angle is given as π radians, which is half the measure of the entire circle, which is 2π radians.

The area of the entire circle is given as π units², and the area of the sector (or the sector is the region enclosed by the central angle MNL in the circle) is also given as π units². This means that the sector occupies the entire area of the circle.

Answer:

The molar mass of the unknown gas is 200 g/mol. R: [tex]2*10^2 g/mol[/tex]

Explanation:

We have an effusion experiment with oxygen and another unknown substance.

Oxygen effuses (v1) through a tiny hole 2.5 times faster than unknown substances (v2). It means

v1 = 2.5v2.

Molar mas of Oxygen is M1 = 32.0 g/mol.

This process can be studied by using Graham law.

[tex]\frac{v1}{v2} = \sqrt[2]{\frac{M2}{M1} }[/tex]

Where M2 is the unknown molecular mass, all the other data are given in the problem. Replacing and isolating M2. we can fin its value:

[tex]\frac{v1}{v2} = \sqrt[2]{\frac{M2}{M1} } \\v1 = 2.5 v2\\\frac{2.5v2}{v2} = \sqrt[2]{\frac{M2}{M1} } \\\\2.5 =  \sqrt[2]{\frac{M2}{M1} } \\\frac{M2}{M1} = (2.5)^2\\M2 = (2.5)^2 M1 = (2.5)^2*32.0 \frac{g}{mol}\\ M2 = 6.25* 32.0 \frac{g}{mol} = 200 \frac{g}{mol}\\M2 = 200 \frac{g}{mol}[/tex]

The molar mass of the unknown gas is 200 g/mol. R: [tex]2*10^2 g/mol[/tex]