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IrinaK [193]
3 years ago
9

Given the following equation,

Chemistry
1 answer:
VARVARA [1.3K]3 years ago
7 0

Answer:

5 moles

Explanation:

Given data:

Number of moles of HCl = 5 mol

Number of moles of H₂O produced = ?

Solution:

Chemical equation:

HCl + NaOH     →   NaCl + H₂O

Now we will compare the moles of HCl with H₂O.

                          HCl            :           H₂O

                             1               :             1

                             5               :            5

5 moles of water will be produced.

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The atomic symbol Superscript 206 subscript 82 upper P b. represents lead-206 (Pb-206), an isotope that has 82 protons and 124 n
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Answer:

\left \{ {{y=206} \atop {x=82}}Pb \right.

Explanation:

isotopes are various forms of same elements with different atomic number but different mass number.

Radioactivity is the emission of rays or particles from an atom to produce a new nuclei. There are various forms of radioactive emissions which are

  • Alpha particle emission  \left \{ {{y=4} \atop {x=2}}He \right.
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in the problem the product formed after radiation was Pb-206. isotopes of lead include Pb-204, Pb-206, Pb-207, Pb-208. they all have atomic number 82. which means the radiation cannot be ∝ or β since both radiations will alter the atomic number of the parent nucleus.

Only gamma radiation with \left \{ {{y=0} \atop {x=0}}γ \right. will produce a Pb-206 of atomic number 82 and mass number 206 , since gamma ray have 0 mass and has 0 atomic number.equation is shown below

\left \{ {{y=206} \atop {x=82}}Pb\right ⇒ \left \{ {{y=206} \atop {x=82}}Pb\right +  \left \{ {{y=0} \atop {x=0}}γ\right.

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20 mL of Ba(OH)2 solution with unknown concentration was neutralized by the addition of 43.89 mL of a .1355 M HCl solution. Calc
bezimeni [28]

Answer:

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Moles (n)=Molarity(M)\times Volume (L)

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=1\times 0.029735 mol= 0.029735 mol

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= 20.0 mL + 43.89 mL  = 63.89 mL = 0.06389 L

Concentration of the barium ions =[Ba^{2+}]

[Ba^{2+}]=\frac{0.029735 mol}{0.06389 L}=0.4654 M

Ba(Cl)_2(aq)\rightarrow Ba^{2+}(aq)+2Cl^-(aq)

1 mole of barium chloride gives 1 mole of barium ions and 2 moles of chloride ions in an aqueous solution.

Then concentration of chloride ions will be:

[Cl^-]=2\times [Ba^{2+}]=2\times 0.4654 M=0.9308 M

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