What is a major product of the reaction in the box?​

Answer:

Molecular compounds are inorganic compounds that take the form of discrete (covalent) molecules. Examples include such familiar substances as water (H2O) and carbon dioxide (CO2).

Answer:

diffusion

Explanation:

Diffusion is the movement of particles from a region of higher concentration to a region of lower concentration in response to a concentration gradient. A concentration gradient simply means a difference in concentration.

Diffusion occurs in solids,liquids and gases. Diffusion is fastest in gases and slowest in solids. Diffusion of solid particles may take very many years while diffusion of gases takes a few milliseconds depending on the mass of the gas.

In materials, atoms and molecules also move from one part of the material to another. This is also refereed to as diffusion.

Answer:

0,

Explanation:

if n was 2, then 1,0,-1

Answer:

[N2O4] = 0.0573M

[NO2] = 0.0163M

Explanation:

The equilibrium of N2O4 is:

N2O4(g)⇌2NO2(g)

Where Kc is defined as:

Kc = 4.64x10⁻³ = [NO2]² / [N2O4]

When you add just N2O4, the reaction will occurs until  [NO2]² / [N2O4] = 4.64x10⁻³. Here, the system reaches equilibrium.

That means if 0.0655M N2O4 begin reaction, in equilibrium we will have:

[N2O4] = 0.0655M - X

[NO2] = 2X

Where X is defined as reaction coordinate

Replacing in Kc:

4.64x10⁻³ = [NO2]² / [N2O4]

4.64x10⁻³ = [2X]² / [0.0655-X]

3.0392x10⁻⁴ - 4.64x10⁻³X = 4X²

3.0392x10⁻⁴ - 4.64x10⁻³X - 4X² = 0

Solving for X:

X = -0.0093 → False solution. there is no negative concentrations

X = 0.008156M → Right solution.

Replacing X, equilibrium concentrations are:

[N2O4] = 0.0655M - X

[NO2] = 2X

Answer:

3.97

Explanation:

pH of buffer solution = pKa+Log(Cb/Ca)

pH of buffer solution = -log(Ka)+log(Cb/Ca)............... Equation 1

Where Ca = concentration of acid, Cb = concentration of base.

Given: Ka = 6.5×10⁻⁵, Ca = 0.25 M, Cb = 0.15 M

Substitute into equation 1

pH of buffer solution = -log(6.5×10⁻⁵)+log(0.15/0.25)

pH of buffer solution = 4.19+(0.22)

pH of buffer solution = 3.97.

Answer:

Option (b) Hydrocyanic acid, 4.9×10^-10

Explanation:

Data obtained from the question include:

Ka of Hydrofluoric acid = 3.5×10^-4

Ka of Hydrocyanic acid = 4.9×10^-10

Ka of Nitrous acid = 4.6×10^-4

To know which acid is least acidic, we shall determine the the pKa value for each acid.

This is illustrated below:

For Hydrofluoric acid

Ka = 3.5×10^-4

pKa =..?

pKa = –Log Ka

pKa = –Log 3.5×10^-4

pKa = 3.5

For Hydrocyanic acid

Ka = 4.9×10^-10

pKa =..?

pKa = –Log Ka

pKa = –Log 4.9×10^-10

pKa = 9.3

For Nitrous acid

Ka = 4.6×10^-4

pKa =..?

pKa = –Log Ka

pKa = –Log 4.6×10^-4

pKa = 3.3

Summary:

Acid >>>>>>>>>>>>> Ka >>>>>>>> pKa

Hydrofluoric acid >> 3.5×10^-4 >> 3.5

Hydrocyanic acid >> 4.9×10^-10 > 9.3

Nitrous acid >>>>>>> 4.6×10^-4 >> 3.3

NB: The smaller the pKa value, the more acidic the compound is and the larger the pKa value, the less acidic the compound will be.

From the above calculations, Hydrocyanic acid has the highest pKa value.

Therefore, Hydrocyanic acid is the least acidic compound

Answer:

The molar concentration of HCl in the aqueous solution is 0.0131 mol/dm3

Explanation:

To get the molar concentration of a solution we will use the formula:

Molar concentration = mass of HCl/ molar mass of HCl

Mass of HCl in the aqueous solution will be 40% of the total mass of the solution.

We can extract the mass of the solution from its density which is 1.2g/mL

We will further perform our analysis by considering only 1 ml of this aqueous solution.

The mass of the substance present in this solution is 1.2g.

The mass of HCl Present is 40% of 1.2 = 0.48 g.

The molar mass of HCl can be obtained from standard tables or by adding the masses of Hydrogen (1 g) and Chlorine (35.46 g) = 36.46g/mol

Therefore, the molar concentration of HCl in the aqueous solution is 0.48/36.46 = 0.0131 mol/dm3

Answer: 9.08 L

Explanation:

To calculate the moles :

[tex]\text{Moles of solute}=\frac{\text{given mass}}\times{\text{Molar Mass}}[/tex]    

[tex]\text{Moles of} Al=\frac{14.6g}{27g/mol}=0.54moles[/tex]

[tex]4Al+3O_2\rightarrow 2Al_2O_3[/tex]

According to stoichiometry :

4 moles of [tex]Al[/tex] require  = 3 moles of [tex]O_2[/tex]

Thus 0.54 moles of [tex]Al[/tex] will require=[tex]\frac{3}{4}\times 0.54=0.405moles[/tex]  of [tex]O_2[/tex]

Standard condition of temperature (STP)  is 273 K and atmospheric pressure is 1 atm respectively.

According to the ideal gas equation:

[tex]PV=nRT[/tex]

P = Pressure of the gas = 1 atm

V= Volume of the gas = ?

T= Temperature of the gas = 273 K      

R= Gas constant = 0.0821 atmL/K mol

n=  moles of gas= 0.405

[tex]V=\frac{nRT}{P}=\frac{0.405\times 0.0821\times 273}{1}=9.08L[/tex]

Thus 9.08 L of [tex]O_2[/tex] at STP would be required

Considering the reaction stoichiometry and STP conditions, 9.072 L of O₂ at STP would be required.  

The balanced reaction is:

4 Al + 3 O₂ → 2 Al₂O₃

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

Being 27 g/mole the molar mass of Al, this is the amount of mass that a substance contains in one mole, then if 14.6 grams Al are reacted,   the number of moles of Al that react is calculated as:

[tex]14.6 gramsx\frac{1 mole}{27 grams}= 0.54 moles[/tex]

Then you can apply the following rule of three: if by stoichiometry 4 moles of Al react with 3 moles of O₂, 0.54 moles of Al react with how many moles of O₂?

[tex]amount of moles of O_{2} =\frac{0.54 moles of Alx3 moles of O_{2} }{4 moles of Al}[/tex]

amount of moles of O₂= 0.405 moles

On the other side, the STP conditions refer to the standard temperature and pressure. Pressure values at 1 atmosphere and temperature at 0 ° C are used and are reference values for gases. And in these conditions 1 mole of any gas occupies an approximate volume of 22.4 liters.

Then you can apply the following rule of three: if by definition of STP 1 mole of O₂ occupies 22.4 L, 0.405 moles of O₂, how much volume does it occupy?

[tex]volume=\frac{0.405 moles of O_{2}x22.4 L }{1 mole of O_{2} }[/tex]

volume= 9.072 L

Finally, 9.072 L of O₂ at STP would be required.  

Learn more:

Answer:

b) 99 kPa

Explanation:

According to Daltons law of partial pressure, the total pressure of a mixture of two or more non reactive gases is the sum of their individual pressures. Let the total pressure of a mixture of n number of gases be [tex]P_{total}[/tex] and their individual pressure be [tex]P_1,P_2,P_3,\ .\ .\ .\ ,\ P_n[/tex], According to Daltons partial pressure law:

[tex]P_{total}=P_1+P_2+P_3+.\ .\ .+P_n[/tex]

Since A glass cylinder contains 2 gases at a pressure of 106 kPa, therefore n = 2. Also one gas ([tex]P_1[/tex]) is at 7 kPa. Using Daltons partial pressure law:

[tex]P_{total}=P_1+P_2+P_3+.\ .\ .+P_n\\P_{total}=P_1+P_2\\106\ kPa=7\ kPa+P_2\\P_2=106\ kPa-7\ kPa\\P_2=99\ kPa[/tex]

Answer:

8.96g\ cm3

Explanation:

D = ( 89.6g \ 10cm3)

( 89.6\ 10) ( g\ cm3) = 8.96g\cm3

Answer:

(a)  Cu²⁺ +2e⁻ ⇌ Cu

(c) 0.07 V  

Explanation:

(a) Cu half-reaction

Cu²⁺ + 2e⁻ ⇌ Cu

(c) Cell voltage

The standard reduction potentials for the half-reactions are+

                                              E°/V

Cu²⁺ + 2e⁻ ⇌ Cu;                  0.34  

Hg₂Cl₂ + 2e⁻ ⇌ 2Hg + 2Cl⁻; 0.241

The equation for the cell reaction is

                                                                            E°/V

Cu²⁺(0.1 mol·L⁻¹) + 2e⁻ ⇌ Cu;                               0.34  

2Hg + 2Cl⁻ ⇌ Hg₂Cl₂ + 2e⁻;                             -0.241

Cu²⁺(0.1 mol·L⁻¹) + 2Hg + 2Cl⁻ ⇌ Cu + Hg₂Cl₂;   0.10

The concentration is not 1 mol·L⁻¹, so we must use the Nernst equation

(ii) Calculations:  

T = 25 + 273.15 = 298.15 K

[tex]Q = \dfrac{\text{[Cl}^{-}]^{2}}{ \text{[Cu}^{2+}]} = \dfrac{1}{0.1} = 10\\\\E = 0.10 - \left (\dfrac{8.314 \times 298.15 }{2 \times 96485}\right ) \ln(10)\\\\=0.010 -0.01285 \times 2.3 = 0.10 - 0.03 = \textbf{0.07 V}\\\text{The cell potential is }\large\boxed{\textbf{0.07 V}}[/tex]

Answer:

The new volume will be 808 L

Explanation:

Charles's law is a law that says that the volume of gas at constant pressure is directly proportional to its absolute temperature (in degrees Kelvin), that is, when the amount of gas and pressure are kept constant, the quotient between volume and temperature will always have the same value:

[tex]\frac{V}{T}=k[/tex]

Having a certain volume of gas V1 that is at a temperature T1 at the beginning of the experiment, by varying the volume of gas to a new value V2, the temperature will change to T2 and the following will be fulfilled:

[tex]\frac{V1}{T1} =\frac{V2}{T2}[/tex]

In this case:

Replacing:

[tex]\frac{625 L}{273 K} =\frac{V2}{353 K}[/tex]

Solving:

[tex]V2=353 K*\frac{625 L}{273 K}[/tex]

V2= 808 L

The new volume will be 808 L

Answer:

The Highly acidic proton joined to one of the carbon in the ALKYNE bond.

(Kindly Check the attachment for the drawing because the solution will need us to draw).

Explanation:

So, let us start by defining some major key terms in this particular Question given above;

(1). ISOMERIZATION: isomerization can simply be defined as the kind is of chemical rearrangement whichay lead to the breaking and the formation of new bonds.

(2). NaNH2 BASE: Sodium amide is a Chemical compound which has a Molar mass of 39.01 g/mol and Heat capacity (C) of 66.15 J/mol K. It is also known as sodamide. It is a good nucleophile.

(3). ALKYNE BOND: it is a C-C joined together by three bonds.

The chemical reaction given in the Question is given in the attachment too.

Therefore, The Highly acidic proton joined to one of the carbon in the ALKYNE bond  is removed irreversibly by NaNH2 base.

Answer:

B) Ionic

Explanation:

Answer:

See figure 1

Explanation:

In this case, we have to start with the ionization reaction of HBr to produce the hydronium ion ([tex]H^+[/tex]) and the bromide ion ([tex]Br^-[/tex]). Then the double bond in the alkene can attack the hydronium ion to produce a carbocation. The most stable carbocation would be the tertiary one, therefore we have to put the positive charge in the tertiary carbon. Then, the bromide attacks the carbocation to produce the final halide.

See figure 1

I hope it helps!

When a solution is diluted with water, the ratio of the initial to final volumes of solution is equal to the ratio of final to initial molarities. The statement is True.

Concentration refers to the amount of a substance in a defined space. Another definition is that concentration is the ratio of solute in a solution to either solvent or total solution.

There are various methods of expressing the concentration of a solution.

Concentrations are usually expressed in terms of molarity, defined as the number of moles of solute in 1 L of solution.

Solutions of known concentration can be prepared either by dissolving a known mass of solute in a solvent and diluting to a desired final volume or by diluting the appropriate volume of a more concentrated solution (a stock solution) to the desired final volume.

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✔ C5H4 has a molecular molar mass of :

M(C5H4) = 5 x M(C) + 4 x M(H)

✔ The molecular mass of C5H4 is therefore 64 g/mol.

Answer:

C10H8

Explanation:

I clicked on that answer, and it is correct.

Answer:

Kc = 166.7

[Fe³⁺] =  0.18 M

[SCN⁻] = 2×10⁻⁴ M

Explanation:

In the reaction of Fe³⁺ and SCN⁻, it is formed a complex:

Fe³⁺  +  SCN⁻  ⇄  FeSCN²⁺             Kc

Let's make the expression for Kc →  [FeSCN²⁺] / [Fe³⁺] . [SCN⁻]

5.7×10⁻⁵ / 9.5×10⁻⁴. 3.6×10⁻⁴  = 166.7

We determine the mmoles, we add from each reactant:

18 ml . 0.2M = 3.6 mmoles of Fe³⁺

2 ml . 0.002M = 4×10⁻³ mmoles of SCN⁻

General form of the dilution equation is:

Concentrated [C] . Concentrated Volume = Diluted [C] . Diluted Volume

Total volume = 20mL

[Fe³⁺]: 3.6 mmoles /20mL = 0.18 M

[SCN⁻]: 4×10⁻³ mmoles /20mL = 2×10⁻⁴ M

The value should be 1.67 x 10^2

The initial concentration should be 0.18 M and 2.0 x 10^(-4) M

The value is

= 5.7 x 10^(-5)/(9.5 x 10^(-4) x 3.6 x 10^(-4))

= 167

= 1.67 x 10^2

we know that

Initial moles  = volume x concentration

So,

= 18/1000 x 0.200

= 0.0036 mol

Now

Initial moles  = volume x concentration

= 2/1000 x 0.0020

= 4.0 x 10^(-6) mol

So,

Total volume should be

= 18 + 2

= 20 mL

= 0.02 L

Now

Initial concentration   

= moles /total volume

= 0.0036/0.02

= 0.18 M

Now

Initial concentration

= moles  /total volume

= 4.0 x 10^(-6)/0.02

= 2.0 x 10^(-4) M

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Answer:

1.7 × 10⁹ L

Explanation:

Step 1: Write the balanced equation

CO(g) + 2 H₂(g) → CH₃OH(l)

Step 2: Calculate the moles corresponding to 1.0 × 10⁶ kg CH₃OH

The molar mass of CH₃OH is 32.04 g/mol.

[tex]1.0 \times 10^{6} kg \times \frac{10^{3}g }{1kg} \times \frac{1mol}{32.04g} = 3.1 \times 10^{7} mol[/tex]

Step 3: Calculate the theoretical yield of CH₃OH

The real yield of CH₃OH is 3.1 × 10⁷ mol  and the percent yield is 40%. The theoretical yield is:

[tex]3.1 \times 10^{7} mol (R) \times \frac{100mol(T)}{40mol(R)} = 7.8 \times 10^{7}mol(T)[/tex]

Step 4: Calculate the moles of CO required to produce 7.8 × 10⁷ mol of CH₃OH

The molar ratio of CO to CH₃OH is 1:1. The moles of CO required are 1/1 × 7.8 × 10⁷ mol = 7.8 × 10⁷ mol

Step 5: Calculate the volume of 7.8 × 10⁷ mol of CO at STP

The volume of 1 mole of CO at STP is 22.4 L.

[tex]7.8 \times 10^{7}mol \times \frac{22.4L}{mol} = 1.7 \times 10^{9}L[/tex]

Answer:

[tex]2Al_(_S_) ~+~3NH_4NO_3_(_a_q_) ->3N_2_(_g_) ~+~6H_2O_(_g_) ~+~Al_2O_3_(_S_)[/tex]

[tex]Entalpy=-2861.9~KJ[/tex]

Explanation:

In this case, we have to start with the reagents:

[tex]Al~+~NH_4NO_3[/tex]

The compounds given by the problem are:

-) Nitrogen gas =  [tex]N_2[/tex]

-) Water vapor  =  [tex]H_2O[/tex]

-) Aluminum oxide =  [tex]Al_2O_3[/tex]

Now, we can put the products in the reaction:

[tex]Al_(_S_) ~+~NH_4NO_3_(_aq_) ->N_2_(_g_) ~+~H_2O_(_g_) ~+~Al_2O_3_(_S_) [/tex]

When we balance the reaction we will obtain:

[tex]2Al_(_S_) ~+~3NH_4NO_3_(_a_q_) ->3N_2_(_g_) ~+~6H_2O_(_g_) ~+~Al_2O_3_(_S_)[/tex]

Now, for the enthalpy change, we have to find the standard enthalpy values:

[tex]Al_(_S_)=0~KJ/mol[/tex]

[tex]NH_4NO_3_(_a_q_)=-132.0~KJ/mol[/tex]

[tex]N_2_(_g_)=0~KJ/mol[/tex]

[tex]H_2O_(_g_)=~-~241.8~KJ/mol[/tex]

[tex]Al_2O_3_(_S_)=~-~1675.7~KJ/mol[/tex]

With this in mind, if we multiply the number of moles (in the balanced reaction) by the standard enthalpy value,  we can calculate the energy of the reagents:

[tex](0*2)~+~(-132*3)=~-396~KJ[/tex]

And the products:

[tex](0*3)~+~(-241.8*6)~+~(-1675.7*1)=-3125.9~KJ[/tex]

Finally, for the total enthalpy we have to subtract products by reagents :

[tex](-3125.9~KJ)-(-396~KJ)=-2729.9~KJ[/tex]

I hope it helps!

Answer:

Explanation:

We shall find out volume of air at NTP or at 273 K and 10⁵ Pa ( 1 atm )

Let it be V₂

[tex]\frac{P_1V_1}{T_1} =\frac{P_2V_2}{T_2}[/tex]

[tex]\frac{2\times 10^5\times 5.68}{394} =\frac{10^5\times V_2}{273}[/tex]

V₂ = 7.87 litres

22.4 litres of any gas is equivalent to 1 mole

7.87 litres of air will be equivalent to

7.87 / 22.4 moles

= .35 moles .

Answer:

The peaks are registered from tetramethyl silane (TMS)

Explanation:

Tetramethyl silane (TMS) is used as internal reference in proton nmr (H NMR) spectrometry.

Its peak is usually registered at about a 2.0 chemical shift means that the hydrogen atoms which caused that peak need a magnetic field two millionths less than the field needed by TMS to produce resonance. This is not affected by the chemical shift of the sample analysed.

I hope this helped.

Answer: The average atomic mass of oxygen is 15.999 amu

Explanation:

Mass of isotope O-16 = 15.995 amu

% abundance of isotope O-16= 99.759 % = [tex]\frac{99.759}{100}=0.99759[/tex]

Mass of isotope O-17 = 16.995 amu

% abundance of isotope O-17 = 0.037% = [tex]\frac{0.037}{100}=0.00037[/tex]

Mass of isotope O-18 = 17.999 amu

% abundance of isotope O-18 = 0.204% = [tex]\frac{0.204}{100}=0.00204[/tex]

Formula used for average atomic mass of an element :

[tex]\text{ Average atomic mass of an element}=\sum(\text{atomic mass of an isotopes}\times {{\text { fractional abundance}})[/tex]

[tex]A=\sum[(15.995\times 0.99759)+(16.995\times 0.00037)+(17.999 \times 0.00204)][/tex]

[tex]A=15.999[/tex]

Thus the average atomic mass of oxygen is 15.999 amu

Answer:

Converting the percent abundance into decimal form, we get:

O-16: 99.759% = 99.759/100 = 0.9975

O-17: 0.037% = 0.037/100 = 0.00037

O-18: 0.204% = 0.204/100 = 0.0020

Average atomic mass of oxygen is:

(15.995) x (0.9975) + (16.995) x (0.00037) + (17.999) x (0.0020)

= 15.955 + 0.0062 + 0.0359

= 15.997 amu

Explanation:

From PLATO

Answer:

a. (i) the distance of R from P is 13 Km to the nearest  kilometer

(ii) the bearing of R from P = 360 - (38 + 25) = 297° to the nearest degree

b. the distance of m from P is 11 Km to the nearest kilometre

Explanation:

a) A triangle PQR is formed. Q = 90°,  p = 8 km; r = 10 km; distance of R from P is q is to be found.

(i) Using the cosine rule: q² = p² + r² - 2prCosQ

q² = 8² + 10² - 2 * 8 * 10 * Cos90

q² = 64 + 100 + 0

q² = 164

q = 13 Km  to the nearest kilometre

Therefore, the distance of R from P is 13 Km to the nearest  kilometer

(ii) the bearing of R from P

The angle at P is found using the formula Cos P = (q² + r² - p²)/2qr

Cos P = 13² + 10² - 8²/2 * 13 *10

Cos P = 0.7884

P = Cos⁻¹ 0.7884

P = 38°

Therefore, the bearing of R from P = 360 - (38 + 25) = 297° to the nearest degree

Note : 25° is alternate (Northwest) to 65°at P

b) A right-angled triangle  QMP is formed

using the trigonometrical ratios; cos Θ = adjacent/hypotenuse

where the hypotenuse side = 10 km, adjacent side = distance of M from P, x

cos P = x/10

x = cos 38 * 10

x = 11 Km to the nearest kilometre

Therefore, the distance of m from P is 11 Km to the nearest kilometre

Answer: E = - 19.611×[tex]10^{-18}[/tex] J

Explanation: The lowest possible energy can be calculated using the formula:

[tex]E_{n} = - Z^{2}.\frac{k}{n^{2}}[/tex]

where:

Z is atomic number of the atom;

k is a constant which contains other constants and is 2.179×[tex]10^{-18}[/tex] J

n is a layer;

For the lowest possible, n=1.

Atom of Lithium has atomic number of Z=3

Substituing:

[tex]E_{1} = - 3^{2}.\frac{2.179.10^{-18}}{1}[/tex]

[tex]E_{1} =[/tex] [tex]-19.611.10^{-18}[/tex] J

The energy for the electron in the [tex]Li^{+2}[/tex] ion is - 19.611 joules

The lowest possible energy, in Joules, for the electron in the [tex]Li^{2+}[/tex] ion is equal to [tex]1.96\times 10^{-17}\; Joules[/tex]

To determine the lowest possible energy, in Joules, for the electron in the [tex]Li^{2+}[/tex] ion, we would use the Bohr model:

Mathematically, Bohr's model is given by the equation:

[tex]Energy = -Z^2 \frac{k}{n^2}[/tex]

Where:

We know that the atomic number of lithium (Li) is equal to 3.

Also, at the lowest possible energy, n = 1.

Rydberg constant = [tex]2.179 \times 10^{-18}[/tex]

Substituting the parameters into the equation, we have;

[tex]E_1 = -3^2 \times \frac{2.179 \times 10^{-18}}{1^2} \\\\E_1 =9 \times 2.179 \times 10^{-18}\\\\E_1 =1.96\times 10^{-17}\; Joules[/tex]

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Answer:

The final temperature of sulfur dioxide gas is 215.43 C

Explanation:

Gay Lussac's Law establishes the relationship between the temperature and the pressure of a gas when the volume is constant. This law says that if the temperature increases the pressure increases, while if the temperature decreases the pressure decreases. In other words, the pressure and temperature are directly proportional quantities.

Mathematically, the Gay-Lussac law states that, when a gas undergoes a transformation at constant volume, the quotient of the pressure exerted by the temperature of the gas remains constant:

[tex]\frac{P}{T}=k[/tex]

Assuming you have a gas that is at a pressure P1 and at a temperature T1 at the beginning of the experiment, by varying the temperature to a new value T2, then the pressure will change to P2, and it will be true:

[tex]\frac{P1}{T1} =\frac{P2}{T2}[/tex]

The reference temperature is the absolute temperature (in degrees Kelvin)

In this case:

Replacing:

[tex]\frac{0.450atm}{293.15 K} =\frac{0.750 atm}{T2}[/tex]

Solving:

[tex]T2 =\frac{0.750 atm}{\frac{0.450atm}{293.15 K} }[/tex]

[tex]T2=\frac{0.750 atm}{0.450 atm} *293.15K[/tex]

T2=488.58 K

Being 273.15 K= 0 C, then 488.58 K= 215.43 C

The final temperature of sulfur dioxide gas is 215.43 C

Answer:

5.81 g/L

Explanation:

Let's apply the Ideal Gases Law to determine this:

P . V = n . R . T

Pressure = 1.6 atm

Volume = ?

Mol = 1 mol

Temperature = 96 K

In order to find the density, we should know the volume of the atmosphere which is a mixture of gases so, we consider all the atmosphere as a unique ideal gas.

1. 6 atm . V = 1 mol . 0.082 L.atm/mol.K . 96K

V = (1 mol . 0.082 L.atm/mol.K . 96K) / 1.6 atm

V = 4.92 L → As this is the volume for the whole atmosphere and the mass of 1 mol is 28.6 g, density should be:

28.6 g / 4.92L = 5.81 g/L

Density → mass / volume

Answer:

35%

Explanation:

Percentage yield = actual yield / theoretical yield × 100.

Given:

Actual yield = 63.07g

Theoretical yield = ?

Mole ratio of C7H6O3 to C4H6O3 = 1 : 1

1 mole of C7H6O3 - 138.1g

Which implies that only 1 mole s[tex]\frac{63.07}{180.2} * 100[/tex]hould be used up in the reaction, yielding 180.2 g of C9H8O4. ⇒ Theoretical yield = 180.2g

∴ % Yield = [tex]\frac{63.07}{180.2} * 100[/tex]

= 35% yield.

Let me know if you found this easy to understand.

Answer:

C₄H₁₀(g) + O₂(g) ⇒ CO₂(g) + H₂O(g)

2 C₄H₁₀(g) + 13 O₂(g) ⇒ 8 CO₂(g) + 10 H₂O(g)

Explanation:

Butane gas (C₄H₁₀) burns in oxygen gas to produce carbon dioxide gas and water vapor. The unbalanced equation is:

C₄H₁₀(g) + O₂(g) ⇒ CO₂(g) + H₂O(g)

First, we will balance carbon and hydrogen which are in just one compound on each side.

C₄H₁₀(g) + O₂(g) ⇒ 4 CO₂(g) + 5 H₂O(g)

Finally, we will balance the oxygen atoms.

C₄H₁₀(g) + 6.5 O₂(g) ⇒ 4 CO₂(g) + 5 H₂O(g)

In order to have integers, we will multiply everý compound by 2.

2 C₄H₁₀(g) + 13 O₂(g) ⇒ 8 CO₂(g) + 10 H₂O(g)