Houses and other buildings use circuits that are wired in ___________

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

Calculate the volume of 4.00 molar NaOH solution required to prepare 100 mL of a 0.500 molar solution of NaOH.

Kazeer [188]

Answer:

The answer to your question is V₁ = 12.5 ml

Explanation:

Data

Volume = V₁?

[NaOH] = C₁ = 4.0 M

Volume 2 = V₂ = 100 ml

[NaOH] = C₂ = 0.5 M

Formula of dilution

V₁C₁ = V₂C₂

Solve for V₁ (original solution)

V₁ = $\frac{V2C2}{C1}$

Substitution

V₁ = $\frac{(0.5)(100)}{4}$

Simplification

V₁ = $\frac{50}{4}$

Result

V₁ = 12.5 ml

7 0

3 years ago

A fuel was burned for 5 min, increasing the temperature of 10.0 g of water with a density of 1.00 g/ml by 9.0 oC. The fuel relea

jekas [21]

The fuel released 90 calories of heat.

Let suppose that water experiments an entirely sensible heating. Hence, the heat released by the fuel is equal to the heat absorbed by the water because of principle of energy conservation. The heat released by the fuel is expressed by the following formula:

$Q = m\cdot c \cdot \Delta T$ (1)

Where:

$m$ - Mass of the sample, in grams.
$c$ - Specific heat of water, in calories per gram-degree Celsius.
$\Delta T$ - Temperature change, in degrees Celsius.

If we know that $m = 10\,g$ , $c = 1\,\frac{cal}{g\cdot ^{\circ}C}$ and $\Delta T = 9\,^{\circ}C$ , then the heat released by the fuel is:

$Q = (10\,g)\cdot \left(1\,\frac{cal}{g\cdot ^{\circ}C} \right)\cdot (9\,^{\circ}C)$

The fuel released 90 calories of heat.

We kindly invite to check this question on sensible heat: brainly.com/question/11325154

7 0

2 years ago

How many moles of hydrogen are in the sample? Round your answer to 4 significant digits.

finlep [7]

Answer:

1.56 mol H₂

Explanation:

Mg₃(Si₂O₅)₂(OH)₂

There are 4 Si moles per Mg₃(Si₂O₅)₂(OH)₂ mol. With that in mind we can calculate how many Mg₃(Si₂O₅)₂(OH)₂ moles are there in the sample, using the given number of silicon moles:

3.120 mol Si * $\frac{1molMg_3(Si_2O_5)_2(OH)_2}{4molSi}$ = 0.78 mol Mg₃(Si₂O₅)₂(OH)₂

Then we can convert Mg₃(Si₂O₅)₂(OH)₂ moles into hydrogen moles, keeping in mind that there are 2 hydrogen moles per Mg₃(Si₂O₅)₂(OH)₂ mol:

0.78 mol Mg₃(Si₂O₅)₂(OH)₂ * 2 = 1.56 mol H₂

8 0

3 years ago

Be sure to answer all parts. Calculate the pH during the titration of 30.00 mL of 0.1000 M KOH with 0.1000 M HBr solution after

Fantom [35]

Answer:

(a) pH = 12.73

(b) pH = 10.52

(c) pH = 1.93

Explanation:

The net balanced reaction equation is:

KOH + HBr ⇒ H₂O + KBr

The amount of KOH present is:

n = CV = (0.1000 molL⁻¹)(30.00 mL) = 3.000 mmol

(a) The amount of HBr added in 9.00 mL of 0.1000 M HBr is:

(0.1000 molL⁻¹)(9.00 mL) = 0.900 mmol

This amount of HBr will neutralize an equivalent amount of KOH (0.900 mmol), leaving the following amount of KOH:

(3.000 mmol) - (0.900 mmol) = 2.100 mmol KOH

After the addition of HBr, the volume of the KOH solution is 39.00 mL. The concentration of KOH is calculated as follows:

C = n/V = (2.100 mmol) / (39.00 mL) = 0.0538461 M KOH

The pOH and pH of the solution can then be calculated:

pOH = -log[OH⁻] = -log(0.0538461) = 1.2688

pH = 14 - pOH = 14 - 1.2688 = 12.73

(b) The amount of HBr added in 29.80 mL of 0.1000 M HBr is:

(0.1000 molL⁻¹)(29.80 mL) = 2.980 mmol

This amount of HBr will neutralize an equivalent amount of KOH, leaving the following amount of KOH:

(3.000 mmol) - (2.980 mmol) = 0.0200 mmol KOH

After the addition of HBr, the volume of the KOH solution is 59.80 mL. The concentration of KOH is calculated as follows:

C = n/V = (0.0200 mmol) / (59.80 mL) = 0.0003344 M KOH

The pOH and pH of the solution can then be calculated:

pOH = -log[OH⁻] = -log(0.0003344) = 3.476

pH = 14 - pOH = 14 - 3.476 = 10.52

(c) The amount of HBr added in 38.00 mL of 0.1000 M HBr is:

(0.1000 molL⁻¹)(38.00 mL) = 3.800 mmol

This amount of HBr will neutralize all of the KOH present. The amount of HBr in excess is:

(3.800 mmol) - (3.000 mmol) = 0.800 mmol HBr

After the addition of HBr, the volume of the analyte solution is 68.00 mL. The concentration of HBr is calculated as follows:

C = n/V = (0.800 mmol) / (68.00 mL) = 0.01176 M HBr

The pH of the solution can then be calculated:

pH = -log[H⁺] = -log(0.01176) = 1.93

4 0

3 years ago

Houses and other buildings use circuits that are wired in ____________ ?