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3 years ago

How many atoms are in CeCL3

2 answers:

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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When dissolved sodium hydroxide reacts with hydrogen sulfate, aqueous sodium, sulfate, water, and heat are formed. Write out the

nignag [31]

Answer:

2 NaOH(aq) + H2SO4(aq) ⇒ Na2SO4 (aq) + 2 H2O(l)

Explanation:

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3 years ago

How fast must a 56.5-g tennis ball travel in order to have a de Broglie wavelength that is equal to that of a photon of green li

nikklg [1K]

Answer:vv

$v=2.17\times 10^{-26}\ m/s$

Explanation:

The expression for the deBroglie wavelength is:

$\lambda=\frac {h}{m\times v}$

Where,

$\lambda$ is the deBroglie wavelength

h is Plank's constant having value $6.626\times 10^{-34}\ Js$

m is the mass of = $56.5\ g=0.0565\ kg$

v is the speed.

Wavelength = 5400 Å = $5400\times 10^{-10}\ m$

Applying in the equation as:-

$5400\times 10^{-10}=\frac{6.626\times 10^{-34}}{0.0565\times v}$

$v=\frac{331300000}{10^{34}\times \:1.5255}\ m/s$

$v=2.17\times 10^{-26}\ m/s$

7 0

3 years ago

Whats the answer to number 5?

Hunter-Best [27]

C is the answer because he wants to know if less force is needed to pull on a slippery surface which reduces friction

5 0

3 years ago

Read 2 more answers

How many mililiters are in 0.5000 liters?

balandron [24]

500ml you multiply the volume value by 1000

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2 years ago

I NEED HELP PLEASE, THANKS! :)

krok68 [10]

Answer:

$\large \boxed{\text{2.20 g Pb}}$

Explanation:

They gave us the masses of two reactants and asked us to determine the mass of the product.

This looks like a limiting reactant problem.

1. Assemble the information

We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.

Mᵣ: 239.27 32.00 207.2

2PbS + 3O₂ ⟶ 2Pb + 2SO₃

m/g: 2.54 1.88

2. Calculate the moles of each reactant

$\text{Moles of PbS} = \text{2.54 g PbS } \times \dfrac{\text{1 mol PbS}}{\text{239.27 g PbS}} = \text{0.010 62 mol PbS}\\\\\text{Moles of O}_{2} = \text{1.88 g O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{32.00 g O}_{2}} = \text{0.058 75 mol O}_{2}$

3. Calculate the moles of Pb from each reactant

$\textbf{From PbS:}\\\text{Moles of Pb} = \text{0.010 62 mol PbS} \times \dfrac{\text{2 mol Pb}}{\text{2 mol PbS}} = \text{0.010 62 mol Pb}\\\\\textbf{From O}_{2}:\\\text{Moles of Pb} =\text{0.058 75 mol O}_{2} \times \dfrac{\text{2 mol Pb}}{\text{3 mol O}_{2}}= \text{0.039 17 mol Pb}\\\\\text{PbS is the $\textbf{limiting reactant}$ because it gives fewer moles of Pb}$

4. Calculate the mass of Pb

$\text{ Mass of Pb} = \text{0.010 62 mol Pb} \times \dfrac{\text{207.2 g Pb}}{\text{1 mol Pb}} = \textbf{2.20 g Pb}\\\\\text{The reaction produces $\large \boxed{\textbf{2.20 g Pb}}$}$

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3 years ago