Which of the samples would be classified as a mixture

In a college of exactly 2940 students, exactly 60 % are male. What is the number of female students?

Norma-Jean [14]

Answer:1176

Explanation:

Number of male students: 2940*(60/100)=1764

So the number of female students would be:

2940-1764=1176

5 0

3 years ago

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What kind of electrons are involved in chemical bonding?

mamaluj [8]

Bonding electrons are involved in chemical bonding these electrons have their valnce shell incomplete

6 0

3 years ago

I need 3 facts about each and similarities plz

k0ka [10]

Mitosis- asexual reproduction
Quick
for reproductive purposes

Meiosis-sexual reproduction
Takes longer
For repairing

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

Urea, CO(NH2)2, is manufactured on a large scale for use in producing urea-formaldehyde plastics and as a fertilizer. What is th

Pachacha [2.7K]

Answer:

5.004kg

Explanation:

Combustion of carbon

C+O2=CO2

from the relationship of molar ratio

mass of carbon/molar mass of carbon=volume of CO2 produced\molar vol(22.4 dm3)

mass of carbon =1000kg

atomic mass of carbon =12

volume of CO2 produced=1000×22.4/12

volume of CO2 produced =1866.6dm3

from the combustion reaction equation provided

CO2 (g) + 2NH3 (g) ⟶ CO (NH2 )2 (s) + H2 O(l)

applying the same relationship of molar ratio

no of mole of CO2=no of mole of urea

therefore

vol of CO2\22.4=mass of urea/molar mass of urea

molar mass of urea=60.06g/mol

from the first calculation

vol of CO2=1866.6dm3

mass of urea=1866.6×60.06/22.4

mass of urea=5004.82kg

7 0

3 years ago

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