Atoms of a given element will have the same mass. (3 points) True False

An aqueous solution of methylamine (ch3nh2) has a ph of 10.68. how many grams of methylamine are there in 100.0 ml of the soluti

ruslelena [56]

Answer:

3.4 mg of methylamine

Explanation:

To do this, we need to write the overall reaction of the methylamine in solution. This is because all aqueous solution has a pH, and this means that the solutions can be dissociated into it's respective ions. For the case of the methylamine:

CH₃NH₂ + H₂O <-------> CH₃NH₃⁺ + OH⁻ Kb = 3.7x10⁻⁴

Now, we want to know how many grams of methylamine we have in 100 mL of this solution. This is actually pretty easy to solve, we just need to write an ICE chart, and from there, calculate the initial concentration of the methylamine. Then, we can calculate the moles and finally the mass.

First, let's write the ICE chart.

CH₃NH₂ + H₂O <-------> CH₃NH₃⁺ + OH⁻ Kb = 3.7x10⁻⁴

i) x 0 0

e) x - y y y

Now, let's write the expression for the Kb:

Kb = [CH₃NH₃⁺] [OH⁻] / [CH₃NH₂]

We can get the concentrations of the products, because we already know the value of the pH. from there, we calculate the value of pOH and then, the OH⁻:

The pOH:

pOH = 14 - pH

pOH = 14 - 10.68 = 3.32

The [OH⁻]:

[OH⁻] = 10^(-pOH)

[OH⁻] = 10^(-3.32) = 4.79x10⁻⁴ M

With this concentration, we replace it in the expression of Kb, and then, solve for the concentration of methylamine:

3.7*10⁻⁴ = (4.79*10⁻⁴)² / x - 4.79*10⁻⁴

3.7*10⁻⁴(x - 4.79*10⁻⁴) = 2.29*10⁻⁷

3.7*10⁻⁴x - 1.77*10⁻⁷ = 2.29*10⁻⁷

x = 2.29*10⁻⁷ + 1.77*10⁻⁷ / 3.7*10⁻⁴

x = [CH₃NH₂] = 1.097*10⁻³ M

With this concentration, we calculate the moles in 100 mL:

n = 1.097x10⁻³ * 0.100 = 1.097x10⁻⁴ moles

Finally to get the mass, we need to molar mass of methylamine which is 31.05 g/mol so the mass:

m = 1.097x10⁻⁴ * 31.05

<h2>m = 0.0034 g or 3.4 mg of Methylamine</h2>

3 0

3 years ago

Many of water's emergent properties, such as its cohesion, its high specific heat, and its high heat of vaporization, result fro

yarga [219]

Answer:

E) are electrically attracted to each other

Explanation:

Water molecule is polar because there is a difference in electronegativity values between hydrogen and oxygen. The hydrogen side of the molecule has a slight positive charge and the oxygen side is slightly negatively.

Positively and negatively charged ends cause water molecules to attract one another and for this reason water shows the properties mentioned in the question: cohesion, high specific heat, and high heat of vaporization.

3 0

3 years ago

What is the mass, in grams, of 0.125 L of CO2 at STP? Question 4 options: 2.80 g 181 g 0.246 g 4.11 g

Zolol [24]

1) Use the fact that 1 mol of gas at STP occupies 22.4 liter

=> 1 mol / 22.4 l = x / 0.125 l => x = 0.125 l * 1 mol / 22.4 l = 0.00558 mol

2) Now use the molar mass of the gas

molar mass of CO2 ≈ 44 g / mol

Formula: molar mass = mass in grams / number of moles =>

mass in grams = molar mass * number of moles = 44 g/mol * 0.00558 moles

mass = 0.246 g

Answer: 0.246 g

7 0

3 years ago

When 100 mL of 0.200 M NaCl(aq) and 100 mL of 0.200 M AgNO3(aq), both at 21.9 °C, are mixed in a coffee cup calorimeter, the tem

masya89 [10]

Answer:

There is 1.3 kJ heat produced(released)

Explanation:

<u>Step 1:</u> Data given

Volume of a 0.200 M Nacl solution = 100 mL = 0.1 L

Volume of a 0.200 M AgNO3 solution = 100 mL = 0.1 L

Initial temperature = 21.9 °C

Final temperature = 23.5 °C

Solid AgCl will be formed

<u>Step 2</u>: The balanced equation:

AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq)

AgCl(s) + NaNO3(aq) → Na+(aq) + NO3-(aq) + AgCl(s)

<u>Step 3:</u> Define the formula

Pressure is constant. → the heat evolved from the reaction is equivalent to the enthalpy of reaction.

Q=m*c*ΔT

⇒ Q = the heat transfer (in joule)

⇒ m =the mass (in grams)

⇒ c= the heat capacity (J/g°C)

⇒ ΔT = Change in temperature = T2- T1

Step 4: Calculate heat

Let's vonsider the density the same as the density of water (1g/mL)

Mass = volume * density

Mass = 200 mL * 1g/mL

Mass = 200 grams

Q= m*c*ΔT

⇒ m = 200 grams

⇒ c = the heat capacity (let's consider the heat capacity of water) = 4.184 J/g°C

⇒ ΔT = 23.5 -21.9 = 1.6°C

Q = 200 * 4.184 * 1.6 = 1338 .9 J = 1.3 kJ

There is 1.3 kJ heat produced(released)

Therefore, we assumed no heat is absorbed by the calorimeter, no heat is exchanged between the calorimeter and its surroundings, and the specific heat and mass of the solution are the same as those for water (1g/mL and 4.184 J/g°C)

7 0

3 years ago

1. The solubility of lead(II) chloride at some high temperature is 3.1 x 10-2 M. Find the Ksp of PbCl2 at this temperature.

solniwko [45]

Answer:

1) The solubility product of the lead(II) chloride is $1.2\times 10^{-4}$ .