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jeka94
3 years ago
7

How many atoms are in CeCL3

Chemistry
2 answers:
SOVA2 [1]3 years ago
7 0
Molar mass of CeCl3 = 246.475 g/mol

This compound is also known as Cerium(III) Chloride.

Convert grams CeCl3 to moles or moles CeCl3 to grams

Molecular weight calculation:
140.116 + 35.453*3
kicyunya [14]3 years ago
3 0

Chlorine Cl 35.453 atomic mass 3 atoms Mass percent: 43.152%

Cerium Ce 140.116 atomic mass 1 atom Mass Percent: 56.848%

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Alex777 [14]

a. Emma creates a pressure difference allowing the fluid to flow.

3 0
3 years ago
A slurry of flakes soybeans weighing a total of 100 kg contains 75 kg of inert solids and 25 kg of solution with 10 wt% oil and
lubasha [3.4K]

Answer:

the amounts and compositions of the overflow V1 and underflow L1 leaving the stage are 75kg and 125kg respectively.

Explanation:

Let state the given parameters;

Let A= solvent (hexane)

B= solid(inert soiid)

C= solvent(oil)

F_{solution} = mass of solvent + mass of oil (i.e A+C)

<u>Feed Phase:</u>

Total feed (i.e slurry of flakes soybeans)= 100kg

B= mass of solid =75 kg

F= mass of solvent + mass of oil (i.e A+C)

 = 25kg

Mass ratio of oil to solution Y_{F} =\frac{Mass C}{Mass (A+C)}

mass of oil (C) =25 × 0.1 wt = 2.5kg

mass of hexane  in feed = 25 ×  0.9 =22.5kg + 2.5 =25kg

therefore  Y_{F} = \frac{2.5}{25}

= 0.1

mass ratio of solid to solution Y_{A}  =  \frac{Mass A}{Mass (A+C)}=[tex]\frac{75}{25}

=3

<u>Solvent Phase:</u>

C= Mass of oil= 0(kg)

A= Mass of hexane = 100kg

mass of solutions = A+C = 0+100kg

solvent= 100kg

<u>Underflow:</u>

underflow = L₁ = (unknown) ???

L₁ = E₁ + B

the value of N for the outlet and underflow is 1.5 kg

i.e N₁ = \frac{mass B}{mass(A+C)}

solution in underflow E₁ = Mass (A+C)

<u>Overflow:</u>

Overflow = V₁ = (unknown) ???

solution in overflow V₁ = Mass (A+C)

This is because, B = 0 in overflow

Solid Balance: (since the solid is inert, then is said to be same in feed & underflow).

solid in feed = solid in underflow = 75

75=  E₁ × N₁

75 =  E₁ × 1.5

E₁ = 50kg

Underflow L₁ = E₁ × B

= 50 + 75

=125kg

The Overall Balance: Feed + Solvent = underflow + overflow

100 + 100 = 125 + V₁

V₁ = 75kg

5 0
3 years ago
Humans have three types of cone cells in their eyes, which are responsible for color vision. Each type absorbs a certain part of
harkovskaia [24]

Answer:

The frequency of this light is 6.98\times 10^{14} s^{-1}.

Explanation:

Wavelength of the light = \lambda = 430 nm=4.30\times 10^{-7} m

Speed of the light = c = = 3\times 10^8 m/s

Frequency of the light = \nu

\nu =\frac{c}{\lambda }

\nu =\frac{3\times 10^8 m/s}{4.30\times 10^{-7} m}=6.98\times 10^{14} s^{-1}

The frequency of this light is 6.98\times 10^{14} s^{-1}.

5 0
3 years ago
To how much water should 100 ml of 18m sulfuric acid be added to prepare a 1.5m solution
SashulF [63]

Answer:

Take 100 ml of a 18 molar solution. The total number of moles is (1 liter/1000 ml) 100 ml 18 moles is 1.8 moles.

1.5 moles in 1 liter so If 1.1 liters of water is added, the total volume is 1.2 liters and 1.8 moles are dissolves in it. 1.8 moles/ 1.2 liters is 1.5 moles per liter.

5 0
3 years ago
The half-life of nitrogen-13 is 10.0 minutes. if you begin with 53.3 mg of this isotope, what mass remains after 25.9 minutes ha
zimovet [89]

Hello!

The half-life is the time of half-disintegration, it is the time in which half of the atoms of an isotope disintegrate.

We have the following data:

mo (initial mass) = 53.3 mg

m (final mass after time T) = ? (in mg)

x (number of periods elapsed) = ?

P (Half-life) = 10.0 minutes

T (Elapsed time for sample reduction) = 25.9 minutes

Let's find the number of periods elapsed (x), let us see:

T = x*P

25.9 = x*10.0

25.9 = 10.0\:x

10.0\:x = 25.9

x = \dfrac{25.9}{10.0}

\boxed{x = 2.59}

Now, let's find the final mass (m) of this isotope after the elapsed time, let's see:

m =  \dfrac{m_o}{2^x}

m =  \dfrac{53.3}{2^{2.59}}

m \approx \dfrac{53.3}{6.021}

\boxed{\boxed{m \approx 8.85\:mg}}\end{array}}\qquad\checkmark

I Hope this helps, greetings ... DexteR! =)

3 0
3 years ago
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