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Kruka [31]
3 years ago
12

Fe,0, is an example of

Chemistry
1 answer:
Amiraneli [1.4K]3 years ago
7 0

Answer:

it is the example of molecules

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PLEASE HELP AND THANKS
Kamila [148]

Answer:

torque =force*distance

4 0
3 years ago
Cobalt-63 has a half-life of 5.3 years. If a pellet that has been in storage for 15.9 years contains 40.0g of Cobalt-63, how muc
Cerrena [4.2K]

Answer:

320 g  

Step-by-step explanation:

The half-life of Co-63 (5.3 yr) is the time it takes for half of it to decay.  

After one half-life, half (50 %) of the original amount will remain.  

After a second half-life, half of that amount (25 %) will remain, and so on.  

We can construct a table as follows:  

  No. of               Fraction         Mass

half-lives   t/yr   Remaining   Remaining/g

      0        0              1

      1         5.3           ½

     2        10.6           ¼

     3        15.9           ⅛                 40.0

     4        21.2           ¹/₁₆

We see that 40.0 g remain after three half-lives.

This is one-eighth of the original mass.

The mass of the original sample was 8 × 40 g = 320 g

5 0
3 years ago
How many moles of solute are contained in 3.0 L of a 1.5 M solution? Helppp pleaseee
Art [367]
4.5 moles! I hope my answer helps
8 0
3 years ago
Convergent boundaries are zones on what plates
Vinil7 [7]
Top surface plates, think of an earthquake.
3 0
3 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 $\large \boxed{\textbf{2.20 g Pb}}$}

5 0
3 years ago
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