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3 years ago

What are the benefits of scientists using the metric system worldwide?select all that apply

1 answer:

4 0

There are several scales to determine the weight of material. The chose of the weight scale depends upon the size of the material. Like if we have to measure the molecular or atomic weight of a molecule or atom we use gm. Like that for the bigger molecules like proteins the weight of the molecule is expressed by different scale due to its big size. Like that for big material like iron rods etc we generally use kilograms or ton. There are two types of measuring units one is C.G.S. and another is S.I. The S.I. units are more acceptable worldwide than that the C.G.S. units as the digit comes in SI units are easily handable than that of CGS.

Thus metric is used to determine a large weight of compounds which is a SI unit and handable.

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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

What is the associated deBroglie wavelength of a H2 molecule moving on one direction with kinetic energy of (3/2 kT) at 30 K

andrey2020 [161]

<u>Answer:</u> The de-Broglie's wavelength of a hydrogen molecule is $3.26\AA$

<u>Explanation:</u>

Kinetic energy is the measure of temperature of the system.

The equation used to calculate kinetic energy of a particle follows:

$E=\frac{3}{2}kT$

where,

E = kinetic energy of the particles = ?

k = Boltzmann constant = $1.38\times 10^{-23}J/K$

T = temperature of the particle = 30 K

Putting values in above equation, we get:

$E=\frac{3}{2}\times 1.38\times 10^{-23}J/K\times 30K\\\\E=6.21\times 10^{-22}J$

Calculating the mass of 1 molecule of hydrogen gas:

Conversion factor used: 1 kg = 1000 g

1 mole of hydrogen gas has a mass of 2 grams or $2\times 10^{-3}kg$

According to mole concept:

$6.022\times 10^{23}$ number of molecules occupy 1 mole of a gas.

As, $6.022\times 10^{23}$ number of hydrogen molecules has a mass of $2\times 10^{-3}kg$

So, 1 molecule of hydrogen will have a mass of = $\frac{2\times 10^{-3}kg}{6.022\times 10^{23}}\times 1=3.32\times 10^{-27}kg$

To calculate the wavelength of a particle, we use the equation given by De-Broglie's wavelength, which is:

$\lambda=\frac{h}{\sqrt{2mE_k}}$

where,

$\lambda$ = De-Broglie's wavelength = ?

h = Planck's constant = $6.624\times 10^{-34}Js$

m = mass of 1 hydrogen molecule = $3.32\times 10^{-27}kg$

$E_k$ = kinetic energy of the particle = $6.21\times 10^{-22}J$

Putting values in above equation, we get:

$\lambda=\frac{6.624\times 10^{-34}Js}{\sqrt{2\times 3.32\times 10^{-27}kg\times 6.21\times 10^{-22}J}}$

$\lambda=3.26\times 10^{-10}m=3.26\AA$ (Conversion factor: $1\AA=10^{-10}m$ )

Hence, the de-Broglie's wavelength of a hydrogen molecule is $3.26\AA$

3 0

3 years ago

What is one interaction between the geosphere and the cryosphere

nordsb [41]

The water cycle is one interaction between the geosphere and the cryosphere.

7 0

3 years ago

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Rus_ich [418]

Weeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeee

7 0

3 years ago

Question 4 of 10

frez [133]

Answer: D. Slow down the chain reaction by absorbing free neutrons

Explanation: just got it right on the quiz A P E X

3 0

3 years ago