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yKpoI14uk [10]
3 years ago
14

Which is a description of objects found in abundance between Mars and Jupiter?

Chemistry
1 answer:
nirvana33 [79]3 years ago
6 0

I am not sure but rocky bodies ranging from large to small in size

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In which one of the following figures is AB=AC
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Bsbsbsbsbs s s sbsbbsbs
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Which layer of earth atmosphere contains no water vapor, has an atmospheric pressure less than 0.1 atmosphere, and has an air te
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Barium nitrate is combined with sodium phosphate. When the balanced net ionic equation is constructed for this reaction, the sto
alexandr1967 [171]
The complete equation for this reaction is,

Ba(NO3)2 + Na2PO4 = 2NaNO3 + BaPO4

Among the compounds present in the reaction, Barium Nitrate, Sodium Phosphate and Sodium Nitrate are soluble ionic compounds. Hence, they will completely ionize into ions. Only BaSO4 is insoluble which becomes the precipitate. Ionic equation is:

Ba2+ + 2NO3- + 2Na+ + PO42- = 2Na+ +NO3- + BaPO4

Cancel like ions,

Ba2+ + PO42- = BaPO4.

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3 years ago
What is the relationship between radioactive decay and radiometric dating?
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3 years ago
Write the balanced equation for the reaction of aqueous Pb ( ClO 3 ) 2 with aqueous NaI . Include phases. chemical equation: Wha
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<u>Answer:</u> The mass of lead iodide produced is 9.22 grams

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}

Molarity of NaI = 0.200 M

Volume of solution = 0.200 L

Putting values in above equation, we get:

0.200M=\frac{\text{Moles of NaI}}{0.200}\\\\\text{Moles of NaI}=(0.200mol/L\times 0.200L)=0.04moles

The chemical equation for the reaction of NaI and lead chlorate follows:

Pb(ClO_3)_2(aq.)+2NaI(aq.)\rightarrow PbI_2(s)+2NaClO_3(aq.)

By Stoichiometry of the reaction:

2 moles of NaI reacts produces 1 mole of lead iodide

So, 0.04 moles of NaI will react with = \frac{1}{2}\times 0.04=0.02mol of lead iodide

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Molar mass of lead iodide = 461 g/mol

Moles of lead iodide= 0.02 moles

Putting values in above equation, we get:

0.02mol=\frac{\text{Mass of lead iodide}}{461g/mol}\\\\\text{Mass of lead iodide}=(0.02mol\times 461g/mol)=9.22g

Hence, the mass of lead iodide produced is 9.22 grams

6 0
3 years ago
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