A gas occupies a volume of 180 mL at 35 °C and 95.9 kPa. What is the volume of the gas at conditions of STP?

Answers

Answer 1

Answer:

the volume of the gas at conditions of STP = 151.04998 ml

Explanation:

Data given:

V1 = 180 ml

T1 = 35°C or 273.15 + 35 = 308.15 K

P1 = 95.9 KPa

V2  =?

We know that at STP

P2 = 1 atm or 101.3 KPa

T2 = 273.15 K

At STP the pressure is 1 atm and the temperature is 273.15 K

applying Gas Law:

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

putting the values in the equation of Gas Law:

[tex]V_2=\frac{P_1V_1T_2}{T_1P_2}[/tex]

V_2 =[tex]\frac{95.9\times180\times273.15}{308.15\times101.3}[/tex]

V2 = 151.04998

therefore, V2 = 151.04998 ml

Answer 2

Answer:

151 mL is the correct answer to the given question .

Explanation:

We know that

[tex]PV =n RT[/tex]

Where P =pressure ,V=volume and T=Temperature

Given

P=95.9 kPa.

V=[tex]180 * 10 ^{-3}[/tex]

R=25/3

T=273 + 35 =308k

Putting these value into the equation we get

[tex]95.9\ * 180\ *\ 10^{-3} \ =\ n * \frac{25}{3} * 308[/tex]

n=[tex]6.72 * 10^{-3}[/tex]

Now using the equation

[tex]n= \ \frac{V}{22.4}[/tex]

[tex]6.72 * 10^{-3} =\frac{V}{22.4}\\ V\ =\ 150.6mL[/tex]

This can be written as  151mL


Related Questions

What are extensive properties of Oxygen?

Answers

the extensive property of Oxygen is Volume, Enthalpy, Entropy.

The reaction of hydrogen and iodine to produce hydrogen iodide has a Kc of 54.3 at 703 K. Given the initial concentrations of H2 and I2 are 0.453 M, what will the concentration of HI be at equilibrium

Answers

Answer:

[HI] = 0.7126 M

Explanation:

Step 1: Data given

Kc = 54.3

Temperature = 703 K

Initial concentration of H2 and I2 = 0.453 M

Step 2: the balanced equation

H2 + I2 ⇆ 2HI

Step 3: The initial concentration

[H2] = 0.453 M

[I2] = 0.453 M

[HI] = 0 M

Step 4: The concentration at equilibrium

[H2] = 0.453 - X

[I2] = 0.453 - X

[HI] = 2X

Step 5: Calculate Kc

Kc = [Hi]² / [H2][I2]

54.3 = 4x² / (0.453 - X(0.453-X)

X = 0.3563

[H2] = 0.453 - 0.3563 = 0.0967 M

[I2] = 0.453 - 0.3563 = 0.0967 M

[HI] = 2X = 2*0.3563 = 0.7126 M

what is the color of benzene and bromine​

Answers

Bromine is a reddish-brown color.
Benzene is clear or colorless

Explanation:

Benzene is colorless, with a sweet odour.

Color of Bromine is reddish brown .

Hope this helps.

Round off the following measurement to three significant digits: 29.950g

Answers

Answer:

30.0 g.

Explanation:

Hello,

In this case, for us to round the given number off to three significant figures, we firstly realize it has initially five significant figures. Thus, cutting at the third digit, which is the second nine, we will have 29.9 g, nonetheless, as a five is after such nine, we should round the nine to ten, so the result is 30.0 g.

Best regards.

The height of the sun in the sky at noontime is called

Answers

It’s called the solar noon and basically it’s the term for when the sun is at its highest

Answer:

It is called 'solar noon'

Explanation:

It's when the sun is at its highest

And also the moment the sun crosses the meridian.

Consider a solution containing 0.100 M fluoride ions and 0.126 M hydrogen fluoride. The concentration of fluoride ions after the addition of 5.00 mL of 0.0100 M HCl to 25.0 mL of this solution is __________ M.

a. 0.0980
b. 0.0817
c. 0.0167
d. 0.0850
e. 0.00253

Answers

Answer:

The answer is "Option b"

Explanation:

In this question first we calculates the moles in F-, HF, and in HCL, which can be defined as follows:

Formula:

[tex]\ Number \ of \ moles\ = \ Molarity \times \ Volume \ in \ litter[/tex]

[tex]\ moles \ in\ F- = 0.100 \ M \times 0.0250 L\\\\[/tex]

                     [tex]=\ 0.0025 \ moles[/tex]

[tex]\ moles \ in \ HF \ = 0.126M \times 0.0250 L[/tex]

                       [tex]= 0.00315 \ moles[/tex]

[tex]\ moles \ in \ HCl = 0.0100M \times 0.00500 L[/tex]

                       [tex]= 0.00005 \ moles[/tex]

[tex]\ Reaction: \\\\F - + H+ \rightarrow HF[/tex]

[tex]\Rightarrow \ moles \ in \ F- = 0.0025 \\\\\Rightarrow \ moles \ in \ H+ = 0.00005 \\\\ \Rightarrow \ moles \ in \ HF = 0.00315\\\\ \ total \ moles = 0.00250 -0.0000500 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.00315 + 0.00005\\\\\ total \ moles =0.00245 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.00245[/tex]

[tex]\ total \ volume \ in \ the \ solution = \ V = \ 0.0300 L\\\\ after \ addition \ of \ HCl \ the \ concentration \ of \ F- \ = 0.00245\ moles \div V[/tex]

                                                                                [tex]=\frac{ 0.00245 \ moles }{0.0300L}\\\\= \frac{245 \times 10^4}{300 \times 10^5} \\\\= \frac{245}{3000} \\\\ = 0.0817 M[/tex]

At 25 °C, what is the hydroxide ion concentration, [OH−] , in an aqueous solution with a hydrogen ion concentration of [H+]=1.9×10−9 M?

Answers

Answer:

The correct answer will be "5.26 × 10⁻⁶".

Explanation:

The given values is:

[tex][H^{+}]=1.9\times 10^{-9} M[/tex]

As we know,

⇒  [tex]pH+pOH=14[/tex]

On taking log, we get

⇒  [tex]-log[H^{+}] + -log[OH^{-}] = 14[/tex]

Now,

Taking "log" as common, we get

⇒  [tex]log[H^{+}][OH^{-}]= -14[/tex]

⇒  [tex][H^{+}][OH^{-}]= 10^{-14}[/tex]

⇒  [tex][OH^{-}]=\frac{10^{-14}}{[H^{+}]}[/tex]

On putting the estimated value of "[tex][H^{+}][/tex]", we get

⇒             [tex]=\frac{10^{-14}}{1.9\times 10^{-9}}[/tex]

⇒             [tex]=5.26\times 10^{-6}[/tex]

WILL GIVE BRAINLIEST!!!!! will give brainliest!!!! ******According to the reaction in question I above, how many grams of solid copper will theoretically be

produced when 14.4 g of aluminum are reacted with 14.4 g of copper (II) sulfate? Which reactant is the

limiting reactant? Show your work. Be sure to include units!

Answers

Answer:

5.73 g Cu

Explanation:

M(CuSO4) = 159.6 g/mol

M(Al) = 27.0 g/mol

M(Cu) = 63.5 g/mol

14.4 g Al * 1 mol/27.0 g = 0.5333 mol Al

14.4 g CuSO4 * 1 mol/159.6 g = 0.0902 mol CuSO4

                                2 Al         +     3CuSO4 ------->  3Cu + Al2(SO4)3

from reaction           2 mol               3 mol

given                     (0.5333 mol )x       0.0902 mol

needed                   0.0601 mol

x= 2*0.0902/3 = 0.0601 mol   Al

Al is excess, CuSO4 is a limiting reactant.

                               2 Al         +     3CuSO4 ------->  3Cu + Al2(SO4)3

from reaction                               3 mol                  3 mol

given                                            0.0902 mol         x mol

x = 0.0902 mol Cu

0.0902 mol Cu * 1 mol/63.5 g Cu = 5.73 g Cu

What are plastic bottles made of?
Polyethylene

halogen

silicon

Alkyl groups

Answers

Answer:

polyethylenes

Explanation:

the plastic bottles used to hold potable water and other drinks are made from polyethylene because, the material is both strong and light.

hope this helped!

Answer: Polyethylenes

Explanation: I got 100% on the test :)

given a k value of 0.43 for the following aqueous equilibrium suppose sample z is placed into water such that its original concentration is 0.033M assume there was zero initial concentration of either A(aq) or B(ag) once equilibrium has occured what will be the equilibrium concentration of z? K=0.43

Answers

Answer:

Less than 0.033 M:

[tex][Z]_{eq}=2.4x10^{-3}M[/tex]

Explanation:

Hello,

In this case, the described equilibrium is:

[tex]2A+B\rightarrow 2Z[/tex]

Thus, the law of mass action is:

[tex]K=\frac{[Z]^2}{[A]^2[B]}=0.43[/tex]

Nevertheless, given the initial concentration of Z that is 0.033 M, we should invert the equilibrium since the reaction will move leftwards:

[tex]\frac{1}{K}=\frac{[A]^2[B]}{[Z]^2}=\frac{1}{0.43}=2.33[/tex]

Know, by introducing the change [tex]x[/tex] due to the reaction extent, we can write:

[tex]2.33=\frac{(2x)^2*x}{(0.033-2x)^2}[/tex]

Which has the following solution:

[tex]x_1=2.29M\\x_2=0.0181M\\x_3=0.0153M[/tex]

But the correct solution is [tex]x_3=0.0152M[/tex] since the other solutions make the equilibrium concentration of Z negative which is not possible. In such a way, its concentration at equilibrium is:

[tex][Z]_{eq}=0.033M-2(0.0153M)[/tex]

[tex][Z]_{eq}=2.4x10^{-3}M[/tex]

Which is clearly less than 0.033 M since the addition of a product shift the reaction leftwards in order to reestablish equilibrium (Le Chatelier's principle).

Regards.

Now construct a different electrochemical cell. You put a zinc metal anode in contact with a 0.052 M solution of zinc nitrate and a silver cathode in contact with a 0.0042 M solution of silver(I) nitrate. What is the value of the electric potential at the moment the reaction begins

Answers

Answer:

[tex]1.66~V[/tex]

Explanation:

We have to start with the half-reactions for both ions:

[tex]Zn^+^2~+2e^-~->Zn[/tex] V= -0.76

[tex]Ag^+~e^-~->~Ag[/tex] V= +0.80

If we want a spontaneous reaction (galvanic cell) we have to flip the first reaction, so:

[tex]Zn~->~Zn^+^2~+2e^-~[/tex] V= +0.76

[tex]Ag^+~+~e^-~->~Ag[/tex] V= +0.80

If we want to calculate ºE we have to add the two values, so:

ºE=0.76+0.80 = 1.56 V

Now, we have different concentrations. So, if we want to calculate E we have to use the nerts equation:

[tex]E=ºE~+~\frac{0.059}{n}LogQ[/tex]

On this case, Q is equal to:

[tex]Q=\frac{[Zn^+^2]}{[Ag^+]^2}[/tex]

Because the total reaction is:

[tex]Zn~+~2Ag^+~->~Zn^+^2~+~2Ag[/tex]

So, the value of "Q" is:

[tex]Q=\frac{[0.052 M]}{[0.0042]^2}=2947.84[/tex]

Now, we can plug all the values in the equation (n=2, because the amount of electrons transferred is 2). So:

[tex]E=1.56~V~+~\frac{0.059}{2}Log(2947.84)=1.66~V[/tex]

I hope it helps!

For the iodine trichloride molecule: a. Determine the number of valence electrons for each atom in the molecule b. Draw the Lewis Dot structure c. Describe why the molecule is drawn this way (i.e. any extra rules/steps needed?) d. Show the polarity of each bond and for the molecule by drawing in the dipole +à

Answers

Answer:

Explanation:

a. Determine the number of valence electrons for each atom in the molecule

In this case we both atoms are halogens. Therefore we will have 7 electrons for each atom.

b. Draw the Lewis Dot structure

In this case, the formula is [tex]ICl_3[/tex], so the central atom would be "I" and the "Cl" atoms would be placed around "I". See figure 1

c. Describe why the molecule is drawn this way (i.e. any extra rules/steps needed?)

In this specific case, the "I" atom don't follow the octet rule. We will have an expanded octet for iodine (more than 8 electrons).

d. Show the polarity of each bond and for the molecule by drawing in the dipole +d

The negative dipole would be placed in the atom with higher electronegativity, in this case "Cl". The positive dipole would be placed in the atom with low electronegativity, in this case "I".

I hope it helps!

What is the Arrhenius definition of an acid? A substance that increases H3O+ concentration when it is dissolved in water. A substance that increases OH– concentration when it is dissolved in water. A compound that donates protons. A compound that accepts protons.

Answers

Answer:

A substance that increases H3O+ concentration when it is dissolved in water.

Explanation:

Note that H3O+ and H+ are used quite interchangeably in chemistry.

An acid makes the H+ content higher, thereby decreasing the pH.

Answer:

a

A substance that increases H3O+ concentration when it is dissolved in water.

Explanation:

Give two examples (i.e. list 2 elements that are examples) of: a. an atom with a half-filled subshell b. an atom with a completely filled outer shell c. an atom with its outer electrons occupying a half-filled subshell and a filled subshell

Answers

Answer:

an atom with a half-filled subshell - hydrogen

an atom with a completely filled outer shell - argon

an atom with its outer electrons occupying a half-filled subshell and a filled subshell- copper

Explanation:

The outermost shell or the valence shell of the atom is the last shell in the atom. Chemical reactions occur at this outer most shell. The number of electrons on the outermost shell of an atom determines the group to which it belongs in the periodic table as well as its chemical properties.

Hydrogen has a half filled 1s sublevel. Only one electron is present in this sublevel.

Let us consider argon

1s2 2s2 2p6 3s2 3p6

The outermost ns and np levels are completely filled. Thus the outermost shell is completely filled.

In the last case; let us look at the electronic configuration of nitrogen;

1s2 2s2 2p3

The outermost 2p subshell is exactly half filled while the 2s sublevel is fully filled. The outermost shell of nitrogen is made up of 2s2 and 2p3 sublevels.

A compound is known to be Na2CO3, Na2SO4, NaOH, NaCl, NaC2H3O2, or NaNO3. When a barium nitrate solution is added to a solution containing the unknown a white precipitate forms. No precipitate is observed when a magnesium nitrate solution is added to a solution containing the unknown. What is the identity of the unknown compound

Answers

Answer:

Na₂SO₄

Explanation:

Barium nitrate, Ba(NO₃)₂ produce precipitate with SO₄²⁻, CO₃²⁻. That means the precipitate could be obtained from Na₂SO₄ and Na₂CO₃.

Also, magnesium nitrate, Mg(NO₃)₂, produce precipitate just with CO₃²⁻. As the unknown solution produce no precipitate, the unknown compound is:

Na₂SO₄

Identify the Lewis acids and Lewis bases in the following reactions:
1. H+ + OH- <-> H2O Lewis acid: Lewis base:
2. Cl- + BCl3 <-> BCl4- Lewis acid: Lewis base:
3. K+ + 6H2O <-> K(H2O)6+ Lewis acid: Lewis base:

Answers

Answer: 1. [tex]H^++OH^-\rightarrow H_2O[/tex]  Lewis acid : [tex]H^+[/tex], Lewis base : [tex]OH^-[/tex]

2. [tex]Cl^-+BCl_3\rightarrow BCl_4^-[/tex] Lewis acid : [tex]BCl_3[/tex], Lewis base : [tex]Cl^-[/tex]

3. [tex]K^++6H_2O\rightarrow K(H_2O)_6[/tex] Lewis acid : [tex]K^+[/tex], Lewis base : [tex]H_2O[/tex]

Explanation:

According to the Lewis concept, an acid is defined as a substance that accepts electron pairs and base is defined as a substance which donates electron pairs.

1. [tex]H^++OH^-\rightarrow H_2O[/tex]

As [tex]H^+[/tex] gained electrons to complete its octet. Thus it acts as lewis acid.[tex]OH^-[/tex] acts as lewis base as it donates lone pair of electrons to electron deficient specie [tex]H^+[/tex].

2. [tex]Cl^-+BCl_3\rightarrow BCl_4^-[/tex]

As [tex]BCl_3[/tex] is short of two electrons to complete its octet. Thus it acts as lewis acid. [tex]Cl^-[/tex] acts as lewis base as it donates lone pair of electrons to electron deficient specie [tex]BCl_3[/tex].

3. [tex]K^++6H_2O\rightarrow K(H_2O)_6[/tex]

As [tex]K^+[/tex] is short of electrons to complete its octet. Thus it acts as lewis acid. [tex]H_2O[/tex] acts as lewis base as it donates lone pair of electrons to electron deficient specie [tex]K^+[/tex].

When you look at an ant up close, using a convex lens, what do you see?

Answers

Answer:

You would be able to see the ants clearly with the unique body parts.

Explanation:

Convex lens is also known as Converging lens. It helps in converging rays of light to become a principal focus which is usually very clear and visible to the eyes. The converging lens when used to view the ant makes the ant appear very visible and the individual is able to see all the unique body parts of the ant. This type of lens is used in individuals who are longsighted.

A 25.0 mLsample of an acetic acid solution is titrated with a 0.175 M NaOH solution. The equivalence point is reached when 37.5 mL of the base is added. The concentration of acetic acid in the sample was:_______,
A) 0.263
B) 0.365
C) 0.175
D) 1.83×10−4
E) 0.119

Answers

Answer:

0.263M of CH₃COOH is the concentration of the solution.

Explanation:

The reaction of acetic acid (CH₃COOH) with NaOH is:

CH₃COOH + NaOH → CH₃COO⁻Na⁺ + H₂O

1 mole of acetic acid reacts per mole of NaOH to produce sodium acetate and water.

In the equivalence point, moles of acetic acid are equal to moles of NaOH and moles of NaOH are:

0.0375L × (0.175 moles / L) = 6.56x10⁻³ moles of NaOH = moles of CH₃COOH.

As the sample of acetic acid had a volume of 25.0mL = 0.025L:

6.56x10⁻³ moles of CH₃COOH / 0.0250L =

0.263M of CH₃COOH is the concentration of the solution

If a pork roast must absorb 1500 kJkJ to fully cook, and if only 14% of the heat produced by the barbeque is actually absorbed by the roast, what mass of CO2CO2 is emitted into the atmosphere during the grilling of the pork roast?

Answers

You need to know the amount of heat generated by the combustion reaction.

Assuming propane as fuel, you can use thiis data:

C3H8(g)+5O2(g)---3CO2(g)+4H2O(g) ΔH= -2217 KJ

So when 3 moles of CO2 is emmitted 2217 kJ of heat is produced.

The molar wegiht of CO2 is 12 g/mol + 2 * 16 g/mol = 44 g/mol.

Then 3 mol * 44 g / mol = 132 g of CO2 are produced with 2217 kJ of heat.

Now you have to calculate how much energy you need to produce if only 12% is abosrbed by the pork

Energy absorbed by the pork = 12% *  total energy =>

total energy = energy absorbed by the pork / 0.12 = 1700 kJ / 0.12 = 14,166.67 kJ.

Now, state the proportion:

132 g CO2 / 2217 kJ = x / 14,166.7 kJ =>

x = 14,166.67 * 132 / 2217 = 843.48 g CO2.

Answer: 843 g of CO2

what is the balanced equation for calcium sulfate?​

Answers

Answer:

CaSO4

Explanation:

Calcium sulfate (or calcium sulphate) is the inorganic compound with the formula CaSO4 and related hydrates.

Alkyl derivatives of mercury are highly toxic and can cause mercury poisoning in humans. Dimethylmercury is one of the strongest known neurotoxins. Although it is said to have a slightly sweet smell, inhaling enough to discern this would be hazardous.
Give the empirical formula of dimethylmercury.

Answers

Answer:

The empirical formula of dimethylmercury is C2H6Hg

Explanation:

Dimethylmercury, as it says in the name, presents not only the mercury metal in its structure (Hg) but also two radical groups called methyl, which is why its name begins with the prefix DI, referring to the fact that there are two methyl.

how many moles of helium gas occupy 22.4 L at 0 degreeC at 1 atm pressure

Answers

Answer:

1 mole of the gas occupies 22. 4L at O0C at 1atm pressure. Hence, the correct option is C.

Explanation:

Interpret the following equation for a chemical reaction using the coefficients given: CO(g) Cl2(g) COCl2(g) On the particulate level: _________ of CO(g) reacts with _________ of Cl2(g) to form _________ of COCl2(g). On the molar level: _________ of CO(g) reacts with _________ of Cl2(g) to form _________ of COCl2(g).

Answers

Answer:

On the particulate level: 6.02 * 10²³ particles of CO(g) reacts with 6.02 * 10²³ particles of Cl₂(g) to form 6.02 * 10²³ particles of COCl2(g).

On the molar level: 1 mole of CO(g) reacts with 1 mole of Cl2(g) to form 1 mole of COCl₂(g).

Explanation:

The particulate level refers to the microscopic or atomic level of substances. It also involves the ions, protons, neutrons and molecules present in substances.

The molar level refers to the quantitative measure of substances in terms of the mole, where a mole represents the amount of substances containing the Avogadro number of particles which is equal to 6.02 * 10³ particles.

Equation of the reaction: CO(g) + Cl₂(g) ----> COCl₂(g)

From the equation above, I mole of CO gas reacts with 1 mole of Cl₂ gas to produce 1 mole of COCl₂ gas.

Since 1 mole of a substance contains 6.02 * 10²³ particles, on a particulate level, 6.02 * 10²³ particles of CO gas reacts with 6.02 * 10²³ particles of Cl₂ gas to produce 6.02 * 10²³ particles of COCl₂ gas.

Identify whether each species functions as a Brønsted-Lowry acid or a Brønsted-Lowry base in this net ionic equation. HF (aq) + SO32- ⇌ F- + HSO3- Brønsted-Lowry _____ Brønsted-Lowry _____ Brønsted-Lowry _____ Brønsted-Lowry _____ In this reaction: The formula for the conjugate _____ of HF is The formula for the conjugate _____ of SO32- is

Answers

Explanation:

A Bronsted-Lowry base is a substance that accepts a proton in the form of a hydrogen (H) atom.

On the other hand;

Bronsted-Lowry acid is the substance that donates the proton.

HF (aq) + SO32- ⇌ F- + HSO3-

In the forward reaction;

Bronsted-Lowry acid : HF

Bronsted-Lowry base: SO32-

In the backward reaction;

Bronsted-Lowry acid : HSO3-

Bronsted-Lowry base: F-

The conjugate base of HF is F-

The conjugate acid of SO32- is HSO3-

AHP for the formation of rust (Fe2O3) is -826 kJ/mol. How much energy is
involved in the formation of 5.00 grams of rust?
A 25.9 kJ
B 25.9 J
C 66.0 kJ
D 66.0)

Answers

Answer:

A- 25.9 kJ

Explanation:

ΔH of formation is defined as the amount of energy that is involved in the formation of 1 mole of substance.

ΔH of rust is -826kJ/mol, that means when 1 mole of rust is formed, there are released -826kJ.

Moles of 5.00g of Fe₂O₃ (Molar mass: 159.69g/mol) are:

5.00g ₓ (1 mole / 159.69g) = 0.0313 moles of Fe₂O₃.

If 1 mole release -826kJ, 0.0313 moles release:

0.0313 moles ₓ (-826kJ / 1 mole) = -25.9kJ

Thus, heat involved is:

A- 25.9 kJ

How many moles of CO2 are produced when 84 0 mol O2 completely react?


Answers

Answer:

Explanation:boom

What is the limiting reactant if 8 g of Ba reacts with 2.8 g of Al2(SO4)3?

Answers

Answer:

Al2(SO4)3 is the limiting reactant

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

3Ba + Al2(SO4)3 → 2Al + 3BaSO4

Next, we shall determine the mass of Ba and the mass of Al2(SO4)3 that reacted from the balanced equation. This is illustrated below:

Molar mass of Ba = 137g/mol

Mass of Ba from the balanced equation = 3 x 137 = 411g

Molar mass of Al2(SO4)3 = 2x27 + 3[32 + (16x4)]

= 54 + 3[32 + 64]

= 54 + 3[96]

= 54 + 288 = 342g/mol

Mass of Al2(SO4)3 from the balanced equation = 1 x 342 = 342g

Summary:

From the balanced equation above,

411g of Ba reacted with 342g of Al2(SO4)3.

Finally, we shall determine the limiting reactant as follow:

From the balanced equation above,

411g of Ba reacted with 342g of Al2(SO4)3.

Therefore, 8g of Ba will react with

= (8 x 342/411 = 6.66g of Al2(SO4)3.

From the calculations made above, we can see that it will take a higher mass of Al2(SO4)3 i.e 6.66g than what was given i.e 2.8g to react completely with 8g of Ba.

Therefore, Al2(SO4)3 is the limiting reactant and Ba is the excess reactant.

Determine the amount of heat (in kJ) associated with the production of 5.71 × 104 g of ammonia according to the following equation. N2(g) + 3H2(g) 2NH3ΔH°rxn = −92.6 kJ Assume that the reaction takes place under standard-state conditions at 25°C.

Answers

Answer:

[tex]Q=-3.11x10^5kJ[/tex]

Explanation:

Hello,

In this case, for the given reaction, we are given the standard enthalpy of reaction per mole of ammonia that is -92.6 kJ, it means, that forming one mole of ammonia will release 92.6 kJ of energy. In such a way, for the formation of 5.71x10⁴ g of ammonia, the following amount of heat will be released:

[tex]Q=5.71x10^4gNH_3*\frac{1molNH_3}{17gNH_3}*-92.6\frac{kJ}{molNH_3}\\ \\Q=-3.11x10^5kJ[/tex]

Best regards.

The amount of the heat associated with the production of 5.71×10⁴ g of ammonia, NH₃ is –311026.732 KJ

We'll begin by calculating the number of mole in 5.71×10⁴ g of NH₃

Mass of NH₃ = 5.71×10⁴ g

Molar mass of NH₃ = 14 + (3×1) = 17 g/mol

Mole of NH₃ =?

Mole = mass / molar mass

Mole of NH₃ = 5.71×10⁴ / 17

Mole of NH₃ = 3358.82 moles

Finally, we shall determine the heat required to produce 3358.82 moles (i.e 5.71×10⁴ g) of NH₃. This can be obtained as follow:

N₂(g) + 3H₂(g) —> 2NH₃(g) ΔH°rxn = −92.6 kJ

Since reaction took place at standard conditions, it means:

1 moles of NH₃ required −92.6 kJ

Therefore,

3358.82 moles of NH₃ will require = 3358.82 × –92.6 = –311026.732 KJ

Thus, the amount of the heat associated with the production of 5.71×10⁴ g of ammonia, NH₃ is –311026.732 KJ

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A compound has an empirical formula of CHN. What is the molecular formula, if it’s molar mass is 135.13 g/mol? (C=12.01 amu, H=1.008 amu, N= 14.01)

Answers

Answer:

well the MF is 224.78 g/mol

Explanation:

just times them all by the molor mass and divide it by 3

Which diagram represents this molecule?

Answers

Answer:

  C

Explanation:

The molecule has 8 carbon atoms joined by 7 C-C bonds.

The first two diagrams show 6 carbon atoms, not 8.

The last two diagrams show line segments representing C-C bonds. Only choice C shows 7 such segments.

The appropriate choice is C.

Answer:

C.

Explanation:

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