The two possible units of molarity are

Answers

Answer 1

Answer: The units for molarity are moles/liter.

Similarly, the equation to find molarity is moles divided by liters.

Explanation:  

mol / L is a unit of molar concentration. These are the number of moles of dissolved material per liter of solution. 1 mol / L is also called 1M or 1molar. Mol / m3 is also a unit of molar concentration.

Molarity is expressed in units of moles per liter (mol / L). This is a very common unit, so it has its own symbol, which is the uppercase M. A solution with a concentration of 5 mmol / l is called a 5 M solution or has a concentration value of 5 mol.

The molar concentration of the solution is equal to the number of moles of the solute divided by the mass of the solvent (kilogram), and the molar concentration of the solution is equal to the number of moles of the solute divided by the volume of the solution (liter). increase.


Related Questions

Consider the reaction described by the chemical equation shown.
C2H4(g)+H2O(l)⟶C2H5OH(l)Δ∘rxn=−44.2 kJ

Use the data from the table of thermodynamic properties to calculate the value of Δ∘rxn
at 25.0 ∘C.


ΔS∘rxn= ? J⋅K−1

Calculate Δ∘rxn.

ΔG∘rxn= ? kJ


In which direction is the reaction, as written, spontaneous at 25 ∘C
and standard pressure?
reverse
both
neither
forward

Answers

Answer:

To calculate Δ∘rxn, we can use the following formula:

ΔG∘rxn = ΔH∘rxn - TΔS∘rxn

where ΔH∘rxn is the enthalpy change of the reaction, T is the temperature in Kelvin, and ΔS∘rxn is the entropy change of the reaction.

We know that ΔH∘rxn = -44.2 kJ and we want to find ΔS∘rxn at 25.0 ∘C (298 K). We can use the following formula to calculate ΔS∘rxn:

ΔG∘rxn = -RTlnK

where R is the gas constant (8.314 J/mol K), T is the temperature in Kelvin, and K is the equilibrium constant.

We can find K using the following formula:

ΔG∘rxn = -RTlnK K = e^(-ΔG∘rxn/RT)

We know that ΔG∘rxn = -44.2 kJ/mol and R = 8.314 J/mol K, so we can calculate K:

K = e^(-(-44.2 kJ/mol)/(8.314 J/mol K * 298 K)) K = 1.9 x 10^7

Now we can use K to calculate ΔS∘rxn:

ΔG∘rxn = -RTlnK ΔS∘rxn = -(ΔH∘rxn - ΔG∘rxn)/T ΔS∘rxn = -((-44.2 kJ/mol) - (-8.314 J/mol K * 298 K * ln(1.9 x 10^7)))/(298 K) ΔS∘rxn = -0.143 kJ/K

Therefore, ΔS∘rxn is -0.143 kJ/K.

To determine whether the reaction is spontaneous at 25 ∘C and standard pressure, we can use Gibbs free energy (ΔG). If ΔG < 0, then the reaction is spontaneous in the forward direction; if ΔG > 0, then it is spontaneous in the reverse direction; if ΔG = 0, then it is at equilibrium.

We know that ΔG∘rxn = -44.2 kJ/mol and T = 25 ∘C (298 K). We can use the following formula to calculate ΔG:

ΔG = ΔG∘ + RTlnQ

where Q is the reaction quotient.

At equilibrium, Q = K (the equilibrium constant). Since we calculated K earlier to be 1.9 x 10^7, we can use this value for Q.

ΔG = ΔG∘ + RTlnQ ΔG = (-44.2 kJ/mol) + (8.314 J/mol K * 298 K * ln(1.9 x 10^7)) ΔG = -43.6 kJ/mol

Since ΔG < 0, the reaction is spontaneous in the forward direction at 25 ∘C and standard pressure.

The Ka value for ethanoic acid, CH3COOH is 1.79 x 10-5. What is the pH of an equimolar solution of ethanoic acid and Na+CH3COO-?

Answers

The pH of the solution can be calculated using the following steps:

Write the chemical equation for the dissociation of ethanoic acid:

CH3COOH + H2O ⇌ CH3COO- + H3O+

Write the equilibrium expression for the dissociation of ethanoic acid:

Ka = [CH3COO-][H3O+] / [CH3COOH]

Since the solution is equimolar in CH3COOH and CH3COO-, we can assume that the initial concentrations of CH3COOH and CH3COO- are equal. Let's use the variable x to represent the concentration of CH3COO- and CH3COOH in mol/L.

[CH3COOH] = x mol/L [CH3COO-] = x mol/L

Since CH3COOH is a weak acid, we can assume that only a small fraction of it dissociates in water. Let's use the variable y to represent the concentration of H3O+ ions in mol/L that are produced from the dissociation of CH3COOH. From the dissociation of ethanoic acid, we know that [CH3COO-] = [H3O+].

[CH3COO-] = y mol/L [H3O+] = y mol/L

Use the equilibrium expression to solve for the concentration of H3O+ ions:

Ka = [CH3COO-][H3O+] / [CH3COOH] 1.79 x 10^-5 = y^2 / x

Solving for y in terms of x, we get:

y = sqrt(Ka * x)

Calculate the pH of the solution using the equation:

pH = -log[H3O+]

pH = -log(y)

Substituting in the value of y from Step 5, we get:

pH = -log(sqrt(Ka * x))

Simplifying, we get:

pH = -0.5 * log(Ka * x)

Substituting in the value of Ka, we get:

pH = -0.5 * log(1.79 x 10^-5 * x)

Now we can calculate the pH for the solution by substituting the value of x as it is equimolar.

pH = -0.5 * log(1.79 x 10^-5 * x)

pH = -0.5 * log(1.79 x 10^-5 * 1)

pH = -0.5 * log(1.79 x 10^-5)

pH = 4.74

Therefore, the pH of an equimolar solution of ethanoic acid and Na+CH3COO- is 4.74.

2. When dinitrogen pentoxide is heated, it decomposes to
nitrogen dioxide and oxygen. How many moles of nitrogen
dioxide can be formed from the decomposition of 1.25 g of
dinitrogen pentoxide?

Answers

0.02314 moles of  NO₂ can be formed from the decomposition of 1.25 g of dinitrogen pentoxide.

The balanced equation for the decomposition of dinitrogen pentoxide is:

2 N₂O₅ → 4 NO₂ + O₂

The molar mass of N₂O₅  is 108.01 g/mol.

To determine the number of moles of N₂O₅  present in 1.25 g, we use the following calculation:

moles N₂O₅  = mass / molar mass

moles N₂O₅ = 1.25 g / 108.01 g/mol

moles N₂O₅ = 0.01157 mol

From the balanced equation, we can see that 2 moles of N₂O₅  decompose to form 4 moles of NO2. Therefore, the number of moles of NO2 produced can be calculated as:

moles  NO₂ = (0.01157 mol N2O5) × (4 mol NO2 / 2 mol N2O5)

moles  NO₂ = 0.02314 mol

Therefore, 0.02314 moles of  NO₂ can be formed from the decomposition of 1.25 g of dinitrogen pentoxide.

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The temperature of a 2.0-liter sample of helium gas at STP is increased to 27C, and the pressure is decreased to 80 kPa. What is the new volume of the helium sample? Round your answer to the nearest tenth of a liter?

Answers

The new volume of the helium sample would be 2.4 L.

Volume of a gas

According to the ideal gas law, PV = nRT, where P is the pressure, V is the volume, n is the number of moles of gas, R is the gas constant, and T is the temperature in kelvins.

At STP (standard temperature and pressure), which is defined as 0°C (273.15 K) and 101.325 kPa, the volume of 2.0 liters of helium gas contains one mole of helium atoms.

To find the new volume of the helium sample when the temperature is increased to 27°C (300.15 K) and the pressure is decreased to 80 kPa, we can use the following equation:

(P1V1)/T1 = (P2V2)/T2

where P1, V1, and T1 are the initial pressure, volume, and temperature, respectively, and P2, V2, and T2 are the final pressure, volume, and temperature, respectively.

Plugging in the values, we get:

(101.325 kPa)(2.0 L)/(273.15 K) = (80 kPa)(V2)/(300.15 K)

Solving for V2, we get:

V2 = (101.325 kPa)(2.0 L)/(273.15 K) * (300.15 K)/(80 kPa) = 2.36 L

Therefore, the new volume of the helium sample is approximately 2.4 L (rounded to the nearest tenth).

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Question 5(Multiple Choice Worth 3 points)
(07.02 LC)

The substances below are listed by increasing specific heat capacity value. Starting at 30.0 °C, they each absorb 100 kJ of thermal energy. Which one do you expect to increase in temperature the least?

a) Cadmium, 0.230 J/(g °C)
b) Sodium, 1.21 J/(g °C)
c) Water, 4.184 J/(g °C)
d) Hydrogen, 14.267 J/(g °C)

Answers

Component form of the vector v is as follows: 4 3 1.5 1 Using the standard basis vectors I and j), express the vector w as follows: 3 two 1 4 pp . 1 3 w 3.5 C. V plus w= d. Determine the vector v's magnitude

What does "vector" mean?

Latin word for "carrier" is "vector." Point A is transported to point B by vectors. The orientation of the vectors AB is the direction in which point A is moved in relation to point B, and the amplitude of the vector is the width of the line connecting the two locations A and B. The terms Euclidean vectors and spatial vectors are also used to refer to vectors.

A vector space is what?

A vector space, also known as a linear space, is a collection of things called vectors that can be added to and multiplied ("scaled") by figures called scalars in the fields of mathematics, physics, and engineering.

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What happens when a solid is dissolved into a liquid?
.

Answers

The solid has broken down into pieces so small that its particles spread all throughout the new mixture

50 points +brainlist (there's going to be 3 more added on my profile with the same points(
which type of process is this?
chemical
physical
nuclear​

Answers

nuclear type of process is this

Is the reaction physical or chemical?

The content of a physical reaction differs from that of a chemical reaction. A chemical reaction changes the makeup of the substances in question; a physical change changes the look, smell, or plain presentation of a sample of matter without changing its content.

Nuclear reactions are not the same as chemical reactions. Atoms become more stable in chemical processes by engaging in electron transfers or by sharing electrons with other atoms.

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In the Periodic Table below, shade all the elements for which the neutral atom has an outer electron configuration of ms2nd2, where n and m are integers, and =m+n1.

Answers

The elements that have an outer electron configuration of ms2nd2 are located in the d-block of the periodic table and include some of the transition metals and lanthanides.

What is the periodic table?

To determine which elements in the periodic table have this outer electron configuration, you can look at the position of the d-block elements in the table. The d-block elements are located in the middle of the table and include the transition metals. These elements have partially filled d orbitals, which can accommodate up to 10 electrons.

Elements in the d-block with an atomic number of 21 through 30 (scandium through zinc) have an outer electron configuration of d10s2 and do not fit the ms2nd2 configuration. However, elements in the d-block with an atomic number of 39 through 48 (yttrium through cadmium) have an outer electron configuration of d10s2p1 and can have the ms2nd2 configuration by removing the single electron in the p orbital. Elements in the d-block with an atomic number of 57 through 80 (lanthanum through mercury) also have the possibility of having an outer electron configuration of ms2nd2.

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