# Butane gas reacts with oxygen gas to form carbon dioxide gas and water gas. if (2023)

Chemical University

see explanation below

explain:

First, let's write and balance the answer:

C₄H₁0 + 13/2O₂ -----> 4CO₂ + 5H₂O

Now that the reaction is at equilibrium, we know that 32.2 g of water was produced. So let's use 42.24 grams of butane and 62.3 grams of oxygen to calculate the theoretical water production.

Butane has a molar mass of 58 g/mol compared to 18 g/mol for water and 32 g/mol for oxygen

The number of moles of each reactant is:

Mole of butane = 42.24 / 58 = 0.73 mole

Line square = 62.3 / 32 = 1.95 line

Let's now calculate the limiting reactants:

1 mol butane ------> 13/2 line cycle

0,73 mol ----------> X

X = 0.73 * 13/2 = 4.75 line square

However, we only have 1.95 moles, so oxygen is the limiting reactant.

Since oxygen is the limiting reactant, this means that the number of moles of water produced will be:

Moles of water = 1.95 moles

And finally, water quality:

m = 1,5 * 18 =¨35,1 g

The yield percentage would be:

% = (32,2 / 35,1) * 100

% = 91,73 %

## Related questions

Hydrogen cyanide (HCN) is produced from ammonia, air and natural gas (CH4) by the following process: Hydrogen cyanide is used to produce sodium cyanide, which is used in part to extract gold from gold-bearing rocks. If the reaction vessel contains 5.90 g of NH3, 11.0 g of O2, and 4.67 g of CH4, what is the maximum mass of hydrogen cyanide that can be produced, assuming the reaction goes to completion as stated?

Answer: The mass of HCN that can be made is 6.183 grams

explain:

To calculate the number of moles, we use the following equation:

.....(1)

• For ammonia:

Given mass of ammonia = 5.90 g

Molar mass of ammonia = 17 g/mol

Substituting the values ​​into equation 1, we get:

• For kisik:

Given mass of oxygen = 11.0 g

Molar mass of oxygen = 32 g/mol

Substituting the values ​​into equation 1, we get:

• For methane:

Methane time fund = 4.67 g

The molar mass of methane = 16 g/mol

Substituting the values ​​into equation 1, we get:

For a specific chemical reaction:

The molar ratio of the reactants is:

Because the number of moles of oxygen is the smallest. It is therefore considered a limiting reagent

Stoichiometry per reaction:

3 moles of oxygen give 2 moles of HCN

Therefore, 0.344 moles of oxygen will produce HCN of =

Now, calculating the mass of HCN according to equation 1, we get:

Molar mass of HCN = 27 g/mol

Molovi HCN = 0.229 moles

Substituting the values ​​into equation 1, we get:

Therefore, the mass of HCN that can be produced is 6.183 grams

Write the net ionic equation for the molecular equation Mg (s) +Snso (aq) Mgso4 (aq) Sn(s) using the lowest possible coefficients. Use the drop-down box to specify the status, such as (aa) or (s). If the box is not needed, leave it blank. )

Mg(s) + Sn²⁺(aq) ⇄ Mg²⁺(aq) + Sn(s)

explain:

Consider the following molecular equation.

Mg(s) + SnSO₄(aq) ⇄ MgSO₄(aq) + Sn(s)

The complete ionic equation includes all ions and species that do not dissociate in water.

Mg(s) + Sn²⁺(aq) + SO₄²⁻(aq) ⇄ Mg²⁺(aq) + SO₄²⁻(aq) + Sn(s)

The net ionic equation includes only reacting ions (not including spectator ions) and species that do not dissociate in water.

Mg(s) + Sn²⁺(aq) ⇄ Mg²⁺(aq) + Sn(s)

In part B, the given conditions are 1.00 mol argon in a 0.500-liter vessel at 27.0 ∘C. You determine that under these conditions the ideal pressure (Pideal) is 49.3 atm, and the actual pressure (Preal) is 47.3 atm. The percentage difference between an ideal gas and a real gas is ______.

Answer: The percentage difference between ideal gas and real gas is 4.06%.

Explanation: Given that,

Ideal pressure (true value) = 49.3 atm

Actual pressure (measured) = 47.3 atm

The formula used to calculate the percentage difference is:

Percentage difference =

Percentage difference =

Percentage difference = 4.06 %

Therefore, the percentage difference between the ideal gas and the real gas is 4.06%.

Write the equilibrium equation for the reaction of an aqueous solution of Pb (ClO 3 ) 2 with an aqueous solution of NaI. Turn on the phases. Chemical equation: If you mix 1.50 L of high concentration Pb (ClO 3 ) 2 with 0.200 L of 0.300 M NaI, what mass of precipitate will form

Answer: The mass of the precipitate (lead(II) iodide) that will be formed is 13.83 grams

explain:

To calculate the number of moles for a given molarity, we use the following equation:

Molarity of NaI solution = 0.300 M

Volume of the solution = 0.200 L

Putting the values ​​in the above equation, we get:

The chemical equilibrium equation for the reaction of lead chlorate and sodium iodide is as follows:

Stoichiometry per reaction:

2 moles of NaI give 1 mole of lead(II) iodide

Therefore, 0.06 moles of NaI will produce lead(II) iodide =

To calculate the number of moles, we use the following equation:

Number of moles of lead(II) iodide = 0.03 moles

Molar mass of lead(II) iodide = 461.1 g/mol

Putting the values ​​in the above equation, we get:

Therefore, the mass of the precipitate (lead(II) iodide) that will be formed is 13.83 g

Consider the electron transition from n=3 to n=7 in a hydrogen atom. a) Determine the wavelength of light associated with this transition. (in meters)

b) Will the light be absorbed or emitted?

For a: The wavelength of light is

For b: light is absorbed

explain:

• for one:

To calculate the wavelength of light, we use the Rydberg equation:

Where,

= Rydbergova constant =

= higher energy level = 7

= lower energy level = 3

Putting the values ​​in the above equation, we get:

Therefore, the wavelength of light is

• For B:

There are two ways for electrons to transition between energy levels:

1. Absorption spectrum: This type of spectrum is seen when an electron moves from a lower energy level to a higher energy level. In doing so, energy is absorbed.
2. Emission spectrum: This type of spectrum is seen when an electron moves from a higher energy level to a lower energy level. During this process, energy is released in the form of photons.

Because electrons jump from a lower energy level to a higher energy level. wavelengths are absorbed.

A biochemical engineer isolated a fragment of a bacterial gene and dissolved an 11.3 mg sample of the material in enough water to make a 32.2 milliliter solution. The osmolarity of this solution is 0.340 Torr at 25°C. (a) What is the molar mass of the gene fragment?
(b) If the solution has a density of 0.997 g/mL, what is the freezing point depression of the solution (Kf for water = 1.86 °C/m)?

For a: Molarna masa fragmenta gena je 19182 g/mol

For b: The lowering of the freezing point of the solution is

explain:

• for one:

To calculate the solute concentration, we use the osmotic pressure equation, which reads:

or,

Where,

= Osmolarity of the solution = 0.340 Torr

i = Van't Hoff factor = 1 (for nonelectrolyte)

Given sample mass = 11.3 mg = 0.0113 g (conversion factor: 1 g = 1000 mg)

Molarna massa fragmenta gena = ?

Volume of solution = 32.2 mL

R = gas constant =

T = solution temperature =

Putting the values ​​in the above equation, we get:

Therefore, the molar mass of the gene fragment is 19182 g/mol

• For B:

To calculate the mass of the solution, we use the following equation:

Density of the solution = 0.997 g/mL

Volume of solution = 32.2 mL

Putting the values ​​in the above equation, we get:

Mass of solvent = [32.1034 - 0.0113] = 32.0921 g

The freezing point depression is defined as the difference between the freezing point of the pure solution and the freezing point of the solution.

To calculate the drop in freezing point, we use the following equation:

or,

Where,

Freezing point of pure solution = 0°C

i = Vant Hoff factor = 1 (for nonelectrolyte)

= molar constant of freezing point elevation = 1.86°C/m

= given mass of dissolved substance (gene fragment) = 0.0113 g

= molar mass of dissolved substance (gene fragment) = 19182 g/mol

= mass of solvent (water) = 32.0921 g

Putting the values ​​in the above equation, we get:

Therefore, the depression of the freezing point of this solution

Aspirin can be produced in the laboratory by reacting acetic anhydride (C 4H 6O 3 ) with salicylic acid (C 7H 6O 3 ) to give aspirin (C 9H 8O 4 ) and acetic acid (C 2H 4O 2 ). The balance equation is C 4H 6O 3+C 7H 6O 3?C 9H 8O 4+C 2H 4O 2
In the laboratory synthesis, the student starts with 2.80 mL of acetic anhydride (density = 1.08 g/mL) and 1.24 g of salicylic acid. After the reaction was completed, the student collected 1.24 grams of aspirin.
Determine the limiting reactant for the reaction.
Determine the theoretical yield of aspirin in the reaction.
Determine the percent utilization of aspirin in the reaction.

Answer: The utilization of aspirin is 77.5%.

explain:

To calculate the mass of acetic anhydride, we use the following equation:

Density of acetic anhydride = 1.08 g/mL

Volume of acetic anhydride = 2.80 ml

Putting the values ​​in the above equation, we get:

To calculate the number of moles, we use the following equation:

.....(1)

• For acetic anhydride:

Given mass of acetic anhydride = 3.024 g

Molarna mass of octene kiseline anhydride = 102.1 g/mol

Substituting the values ​​into equation 1, we get:

• For salicylic acid:

Given mass of salicylic acid = 1.24 g

Molar mass of salicylic acid = 138.12 g/mol

Substituting the values ​​into equation 1, we get:

The chemical equation for the combustion of hexane is as follows:

Stoichiometry per reaction:

1 mole of salicylic acid reacts with 1 mole of acetic anhydride

Therefore, 0.0089 moles of salicylic acid will react with = acetic anhydride

Because the amount of acetic anhydride given is greater than necessary. Therefore, it was considered an excess reagent.

Therefore, salicylic acid is considered a limiting reagent because it limits product formation.

Stoichiometry per reaction:

1 mol of salicylic acid gives 1 mol of aspirin

So 0.0089 moles of salicylic acid gives = aspirin

Now, calculating the mass of aspirin according to equation 1, we get:

Molar mass of aspirin = 180.16 g/mol

Moles of aspirin = 0.0089 moles

Substituting the values ​​into equation 1, we get:

• To calculate the percentage of aspirin utilization, we use the following equation:

Experimental use of aspirin = 1.24 g

Theoretical yield of aspirin = 1.60 g

Putting the values ​​in the above equation, we get:

Therefore, the utilization of aspirin is 77.5%.

Consider an ideal gas with N = 1023 particles and internal energy U = 100 J. If the volume of the gas is tripled but its entropy remains the same, what is its internal energy after expansion? Hint: Write the new energy as aU and solve for a.

explain:

This is a question of skill!

Internal energy is the total thermal energy stored in the system. The only factor that affects the internal energy of the system is temperature. If the temperature increases during heating, its internal energy also increases, and when the temperature decreases, it decreases.

This means that tripling the volume of a gas with constant entropy has no effect on the internal energy. The internal energy U of the system is still 100J.

Be sure to answer all sections. To titrate 10.0 mL of 0.250 M acetic acid with 0.200 M sodium hydroxide, to determine the pH: (a) 10.0 mL of base is added. (b) 12.5 mL of base was added. (c) 15.0 mL of base was added.

explain:

Molarity of acetic acid = 0.250 M

Volume of acetic acid solution = 10.0 mL = 0.010 L (1 mL = 0.001 L)

molovi octene kiske;

Molarnost NaOH = 0,200 M

a) Volumen otopine NaOH = 10,0 mL = 0,010 L( 1 mL =0,001L)

Molovi NaOH:

1 mol of NaOH neutralizes 1 mol of acetic acid, then 0.002 mol of NaOH will neutralize 0.002 mol of acetic acid.

Molovi neneutralizane octene kishite = 0.0025 mol - 0.002 = 0.0005 mol

1 mole of acetic acid produces 1 mole of hydrogen ions, then 0.0005 moles of acetic acid will produce 0.0005 moles of hydrogen ions.

Moles of hydrogen ions = 0.0005 mol

Otopin volume = 0.010 L + 0.010 L = 0.020 L

pH of 10.0 mL of base added to acetic acid solution:

b) Volumen otopine NaOH = 12,0 mL = 0,012 L( 1 mL =0,001L)

Molovi NaOH:

1 mole of NaOH neutralizes 1 mole of acetic acid, then 0.0024 moles of NaOH will neutralize 0.0024 moles of acetic acid.

Molovi neneutralizane octene kishte = 0.0025 mol - 0.0024 = 0.0001 mol

1 mole of acetic acid produces 1 mole of hydrogen ions, then 0.0001 mole of acetic acid will produce 0.0001 mole of hydrogen ions.

Number of moles of hydrogen ions = 0.0001 mole

Otopin volume = 0.010 L + 0.012 L = 0.022 L

pH of 12.0 mL of base added to acetic acid solution:

c) Volume of NaOH solution = 15.0 mL = 0.015 L (1 mL = 0.001 L)

Molovi NaOH:

1 mol of NaOH neutralizes 1 mol of acetic acid, then 0.003 mol of NaOH will neutralize 0.003 mol of acetic acid.

All moles of acetic acid are neutralized by NaOH, leaving unneutralized NaOH.

Molovi neneutraliziranog NaOH = 0.003 mol - 0.0025 = 0.0005 mol

1 mole of NaOH produces 1 mole of hydroxide ions, then 0.0005 moles of NaOH acid will produce 0.0005 moles of hydroxide ions.

Number of moles of hydroxide ions = 0.0005 mol

Otopin volume = 0.010 L + 0.015 L = 0.025 L

pOH of 15.0 mL of base added to acetic acid solution:

pH of 15.0 mL of base added to acetic acid solution:

An observer ion is (check all that apply.) the ionic composition of the reactant that is not changed by the reaction of your eye, b. Carefully observe the reaction process of this experiment, c. Nitrate ions A piece of french fries contaminates the reaction mixture

Correct question:

Spectator Yes (Select all that apply.)

- A piece of fried potatoes contaminates the reaction mixture

- The ionic component of the reactant that does not change during the reaction

- In this experiment, the nitrate ion

- carefully observe the course of the reaction with your eyes

- The ionic component of the reactant that does not change during the reaction

explain:

Spectator ions are ions that exist as reactants and products in chemical equations. Observer ions are ions that exist in the same form on both the reactant side and the product side of a chemical reaction.

Intermediate ions are ions that are present in the solution, but do not participate in the reaction. When reactants dissociate into ions, some ions can combine to form new compounds. Other ions do not participate in this chemical reaction and are therefore called spectator ions.

Therefore, the correct option is an option;

- The ionic component of the reactant that does not change during the reaction

The following reactions were carried out at different temperatures with Kc = 0.055. However, this time the reaction mixture started with only the product [NO]=0.0100 M and no reactants. Determine the equilibrium concentrations of N2, O2 and NO in the equilibrium state. N2(g)+O2(g)⇌2NO(g)

Answer: The equilibrium concentrations are 0.0045 M, 0.0045 M and 0.001 M respectively.

Explanation: Given that,

Equilibrium constant = 0.055

[Ne] = 0.0100 M

The given chemical reaction is:

Initial concentration 0 0 0.0100

in balance. x x (0.0100-2x)

The expression for the equilibrium constant is:

Now putting all default values ​​into this expression, we get:

x = 0,0045 i x = 0,0057

'x' has two values. But let's ignore the value x = 0.0057 because the equilibrium concentration cannot exceed the initial concentration.

Dakle, x = 0,0045M

Concentration at equilibrium = x = 0.0045 M

Concentration at equilibrium = x = 0.0045 M

Equilibrium concentration = (0.0100-2x) = (0.0100-2(0.0045)) = 0.001 M

In the reaction, A → product, the 1/[A] versus time curve is linear with a slope equal to 0.056 M−1 s−1. If the initial concentration of A is 0.80 M, how long does it take for half of the initial amount of A to react?

t = 22.32 seconds

explain:

By plotting experimental concentration versus time data on a single graph, reaction kinetics can be understood graphically.

• The concentration versus time curve for a zero-order reaction is linear with a negative slope.
• A plot of concentration versus time for a first-order reaction is an exponential curve. For first-order kinetics, the graph of the natural logarithm of concentration versus time is a straight line graph with a negative slope.
• A graph of concentration versus time for a secondary reaction is a hyperbola. Also, for second-order kinetics, the inverse concentration versus time plot is a straight-line plot with a positive slope.

Default: - 1/[A] is linear with time, which means it follows second-order dynamics.

therefore,

Assume slope = k = 0.056 M⁻¹ s⁻¹.

The overall rate law for second-order kinetics is:

where the final concentration = half of the initial concentration = 0.80 /2 M = 0.40 M

is the initial concentration = 0.80 M

so,

t = 22.32 seconds

In a specific redox reaction, BrO− is oxidized to BrO−3, and Ag+ is reduced to Ag. The equation for this reaction is completed and balanced in the acidic solution. The stages are optional.
BrO ⁻ + Ag ⁺ → BrO3 ⁻ + Ag

The complete and balanced equation for the redox reaction, BrO− is oxidized to BrO−³ and Ag + is reduced to Ag in acidic solution is -

• 4Ag⁺ + BrO⁻+ 2H₂O ----> 4 Ag + BrO3⁻+ 4H⁺

Steps or phases of this redox or reduction-oxidation reaction:

• First, split the equation into two reactions that take place BrO⁻ ----> BrO3⁻ and Ag+ ----> Ag
• Now balance them one by one

• The first step is always to balance O. We do this by adding water.

BrO⁻ + 2H2O ----> BrO3⁻

• The next step is to balance the H. We do this by adding H+

BrO⁻+ 2H2O ----> BrO3⁻+ 4H⁺

• Then we balance the charge by adding electrons.
1. The charge on the left is only from BrO, so the charge is -1.
2. The charge on the right is -1 from BrO₃ +4 from H+, so +3. To make the charge equal on both sides, we need to add 4 electrons to the side with +3 charge, then it will be -1.

BrO⁻ + 2 H₂0 ----> BrO3 ^- + 4H⁺ + 4e⁻

The equation is now balanced. Now on to silver.

Silver⁺ ---> Silver

• The only thing that changes is the charge, which can be corrected by adding electrons.

Ag⁺ + e⁻ ----> silver

Fees are now balanced.

• Now, when we combine these two half-reactions, we expect the electrons to cancel each other out.
• The reaction requires 4 electrons in the Ag equation to balance the 4e requirement in the BrO equation.
• So we're going to multiply the entire Ag equation by 4

4Ag⁺ + 4e⁻ ------> 4Ag

So finally combine the Ag equation and the BrO equation and cancel out our electrons and you get

- 4Ag⁺ + BrO⁻+ 2H₂O ----> 4 Ag + BrO3⁻+ 4H⁺

Find out more:

pametno.com/question/13978139

4 Ag⁺ + 2 H₂O + BrO⁻ → 4 Ag + BrO3⁻ + 4 H⁺

explain:

We will use the ion-electron method to balance the redox reaction.

Step 1: Identify the two half-reactions

Reduction: Ag⁺ → Ag

Oxidation: BrO⁻ → BrO₃⁻

Step 2: Perform a mass balance, adding H+ and H2O if necessary

silver⁺ → silver

2 H₂O + BrO⁻ → BrO3⁻ + 4 H⁺

Step 3: Perform electrical balancing, adding electrons if necessary

1 e⁻ + Ag⁺ → Ag

2 H₂O + BrO⁻ → BrO3⁻ + 4 H⁺ + 4 e⁻

Step 4: Multiply the two half-reactions by a number that ensures an equal number of electrons gained and lost

4 × (1 e⁻ + Ag⁺ → Ag)

1 × (2 H₂O + BrO⁻ → BrO3⁻ + 4 H⁺ + 4 e⁻)

Step 5: Add the two half-reactions

4 e⁻ + 4 Ag⁺ + 2 H₂O + BrO⁻ → 4 Ag + BrO3⁻ + 4 H⁺ + 4 e⁻

4 Ag⁺ + 2 H₂O + BrO⁻ → 4 Ag + BrO3⁻ + 4 H⁺

The gas phase reaction A2 + B2 ---->2AB takes place by bimolecular collisions between A2 and B2 c,o(\'e'9 molecules. The rate law is rate = k[A2][B2] if the concentrations of A2 and B2 Doubling both, the reaction rate will change by a factor

The speed of the reaction is quadrupled.

explain:

According to the Mass Action Act:-

The reaction rate is proportional to the active concentration of reactants, and each concentration of reactants increases by an experimentally determined coefficient, which is called order. The rate is determined by the slowest step in the reaction mechanism.

The order in the law of mass action is the coefficient of increase in the active concentration of reactants. It is determined experimentally and can be zero, plus or minus or a decimal.

The order of the overall reaction is the sum of the series of each reactant raised to its power in the rate law.

therefore,

Given that: - The rate law is: -

now, and

so,

The speed of the reaction is quadrupled.

Radioactive gold 198 is used to diagnose liver problems. The half-life of this isotope is 2.7 days. If you start with a sample of 8.1 mg of the isotope, how much of the sample is left after 2.6 days?

see explanation below

explain:

To fix this, we need to use an expression for the half-life concentration (or mass) decay as follows:

m = m₀e^-kt (1)

In this case k will be a constant rate for that element. This is calculated using the following expression:

k = ln2/t₁/₂ (2)

First we calculate the value of k:

k = ln2/2,7 = 0.2567 d⁻¹

Now we can use expression (1) to calculate the residual mass:

m = 8,1 * e^(-0,2567 * 2,6)

m = 8,1 * e^(-0,6674)

Meters = 8.1 * 0.51303

m = 4.16 mg remaining

In the reaction, A -----> Products, the rate constant is 3.6 x 10^−4 s−1. If the initial concentration of A is 0.548 M, what is the concentration of A (in M) at t = 99.2 s? Fill in the value with only three valid digits for the answer. Do not enter units (M).

0,529

explain:

Let's look at the reaction A → product

Since the units of the rate constant are s-1, this is a first-order reaction with respect to A.

We can find the concentration of A at a certain time t() using the following expression.

Where,

[A]₀: initial concentration of A

k: rate constant

On another planet, titanium isotopes have the following natural abundances. a. Share of isotope 46Ti 70.900% by mass (amu) 45.95263
b. Share of isotope 48Ti 10,000% by mass (amu) 47.94795
c. Isotope fraction 50Ti 19.100% by mass (amu) 49.94479
d. What is the average atomic mass of titanium on this planet?
e. I have 46.9 amu, but that is wrong.

Average atomic mass = 46.91466 amu

explain:

Step 1: given data

Isotope titana

46Ti = 70.900% ⇒ 45.95263 amu

48Ti = 10,000 % ⇒ 47,94795 amu

50Ti = 19,100 % ⇒ 49,94479 amu

Step Two: Calculate the average atomic mass of titanium

Average atomic mass = 0.7090 * 45.95263 + 0.10 * 47.94795 + 0.1910 * 49.94479

Average atomic mass = 46.91466 amu

Acetonitrile (CH3CN) is a polar organic solvent that dissolves a large number of solutes, including many salts. A solution of 1.80 M LiBr in acetonitrile has a density of 0.824 g/cm^3 (cubic) a.) Calculate the molarity of the solution.
b.) Calculate the concentration of the solution as a mole fraction of lithium bromide
c.) Calculate the concentration of the solution as a mass percentage of CH3CN

According to the calculation:

• The molarity of the solution is 2.70 mol/kg
• The mole fraction of lithium bromide is 0.1
• The mass percentage of the solution is 81%

How to calculate molarity

The molarity of the solution is calculated as follows:

• Mass molarity = moles of solute/mass of solvent

The concentration of the solute is 1.80 M

According to; moles of solute = 1.80 moles in 1 liter

Mass of solvent = mass of solution - mass of solute

Mass of solution = volume × density

Volume of solution = 1 L = 1000 cm^3

Mass of the solution = 1000 × 0.824

Mass of the solution = 824 g

Mass of solute = number of moles x molar mass

Molar mass of solute = 86.84 g/mol

Mass of solute = 1.80 × 86.84

Mass of solute = 156.3 g

Solvent mass = 824 - 156.3

Mass otapala = 667,7 g = 0,6677 kg

Molarity of the solution = 1.80/0.6677

Molarity of the solution = 2.70 mol/kg

molni udio

• Mole fraction = moles of solute/total moles
• Total moles = moles of solute + moles of solvent

Molovi otapala = masa/molarna masa

Molar mass of solvent = 41 g/mol

Number of moles of solvent = 667.7/41

Molovi otapala = 16.3

Total number of moles = 16.3 + 1.8 = 18.1 moles

Mole fraction of lithium bromide = 1.8/18.1

Molni udio LiBr = 0.1

Mass percent of solvent

• Mass percentage of solvent = mass of solvent/mass of solution × 100%

Mass percentage of solvent = 667.7/ 824 × 100 %

Mass percentage of solvent = 81 %

So, according to the calculation:

• The molarity of the solution is 2.70 mol/kg
• The mole fraction of lithium bromide is 0.1
• The mass percentage of the solution is 81%

For more information on molar concentrations, mole fractions, and mass percents of substances, visit: brainly.com/question/14594475

a. [LiBr] = 2,70 m

b. Xm za LiBr = 0,1

c. 81 % mase CH3CN

explain:

Solvent → acetonitrile (CH3CN)

Dissolved substance → lithium bromide, lithium bromide

Moles of dissolved substances are converted into mass → 1.80 mol. 86.84 grams/1 mole = 156.3 grams

This mass of solute is found in 1 L of solution

1 L = 1000 mL → 1 mL = 1 cm³

The mass of the solution is determined by density

Density of solution = mass of solution/volume of solution

density of the solution. volume of solution = mass of solution

0.824 g/cm3. 1000 cubic centimeters = 824 grams

Mass of the solution = 824 g (solvent + solute)

Mass of solute = 156.3 g

Masa otapala = 824 g - 156.3 g = 667.7 g

molar concentration → moles of solute in 1 kg of solvent

We convert the mass of the solvent from g to kg → 667.7 g. 1kg/1000g=0.667kg

Mol/kg → 1.80 mol / 0.667 kg = 2.70 m → molar concentration

Mole fraction → moles of solute / moles of total (moles of solute + moles of solvent)

Molovi otapala → 667.7 g. 1 mol/ 41 g = 16.3 mol

Ukupni molovi = 16.3 + 1.8 = 18.1

Molni udio Li Br → 1.80 mol / 18.1 mol = 0.1

mass percentage → (mass of solvent, in this case / total mass). 100

We were asked to provide acetonitrile → (667.7 g / 824 g). 100 = 81%

The ideal gas law (PV=nRT) describes the relationship between pressure P, volume V, temperature T and molar mass n. When n and V are fixed, the equations can be rearranged as follows, where k is a constant: Boyle's law PV=nRT=kn and T; Charles's law VT=nRP=kn and P; Avogadro's law Vn= RTP = kT and P.
A quantity of chlorine gas is placed in a cylinder with a movable piston at one end. The initial volume is 3.00 L, and the initial pressure of chlorine gas is 1.80 atm. Push the plunger down to change the volume to 1.00 L.
If the temperature and moles of chlorine are held constant, calculate the final pressure of the gas. Express the answer in the appropriate units.

The new pressure is 5.40 atm

explain:

When T° and number of moles are kept constant, we should use these relationships to determine pressure or volume

P₁。 V1 = P₂。 V2

1,80 atmospheric 3L = P2. 1 liter

(1,80 atm. 3L) / 1L = P2

5.40 atmosfera = P2

When aqueous solutions of manganese(II) iodide and sodium phosphate are mixed, a solid solution of manganese(II) phosphate and sodium iodide is formed. Give the net ionic equation for this reaction.

Net ionic equation for the given reaction:

explain:

...

..

...

Replace, NaI i in s usig i 

By removing the common ions present on both sides, we get the net ionic equation for the given reaction :

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