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1 year ago

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Nahco3 can be used to neutralize hydrogen ion, h+. if nahco3 can be purchased at a cost of $9.20/kg, how much will it cost to ne

utralize 1.0 mole of h+?

2 answers:

tresset_1 [31]1 year ago

8 0

NaHCO₃ + H⁺ = Na⁺ + H₂O + CO₂

n(NaHCO₃)=n(H⁺)

m(NaHCO₃)=M(NaHCO₃)n(NaHCO₃)=M(NaHCO₃)n(H⁺)

p=$m(NaHCO₃)=$M(NaHCO₃)n(H⁺)

p=9.20$/kg×84.0kg/kmol×10⁻³kmol=0.7728$

p≈0.77$

jok3333 [9.3K]1 year ago

6 0

Answer:

$Cost=\$0.77$

Explanation:

Hello,

At first, the ionic neutralization chemical reaction turns out into:

$NaHCO_3+H^+-->Na^++H_2O+CO_2$

In such a way, we develop the stoichiometry calculation to obtain the kilograms of sodium bicarbonate:

$m_{NaHCO_3}=1.0 molH^+*\frac{1molNaHCO_3}{1molH^+}*\frac{84gNaHCO_3}{1molNaHCO_3} *\frac{1kgNaHCO_3}{1000gNaHCO_3} \\m_{NaHCO_3}=0.084kgNaHCO_3$

Now, we compute the cost by considering the neutralized kilograms of sodium bicarbonate as shown below:

$Cost=\$9.20/kg*0.084kg\\Cost=\$0.77$

Best regards.

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Explanation :

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The neutron capture equation is represented as,

$_Z^A\textrm{X}+_{0}^1\textrm{n}\rightarrow _{Z}^{A+1}\textrm{X}+\gamma$

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The beta minus decay equation is represented as,

$_Z^A\textrm{X}\rightarrow _{Z+1}^A\textrm{Y}+_{-1}^0e$

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As per question, the cobalt-60 is produced by a three reaction process involving neutron capture, beta-emission, and neutron capture.

Process 1 : Neutron capture.

$_{26}^{58}\textrm{Fe}+_{0}^1\textrm{n}\rightarrow _{26}^{59}\textrm{Fe}+\gamma$

Process 2 : Beta emission.

$_{26}^{59}\textrm{Fe}\rightarrow _{27}^{59}\textrm{Co}+_{-1}^0e$

Process 3 : Neutron capture.

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now we have to determine P₁ and n₂ /n₁.

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$\eta = \frac{T_{2} - T_{1}}{T_{2}}$

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It is known that for $NH_{3}$ the value of $C_{v}$ = 0.028 kJ/mol.

Therefore, putting the given values into the above formula as follows.

$\Delta S = C_{v} ln \frac{T_{2}}{T_{1}} + R ln \frac{P_{1}}{P_{2}}$

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