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xxMikexx [17]
3 years ago
11

Given that the nucleophilic substitution reaction used 5.0 mL of t-pentyl alcohol and 12.0 mL of conc. hydrochloric acid to prod

uce t-pentyl chloride, what is the theoretical yield of t-pentyl chloride, in grams
Chemistry
1 answer:
Vladimir [108]3 years ago
8 0

Answer:

4.90 g

Explanation:

Given that:

volume of t-pentyl alcohol = 5 mL

the standard density of t-pentyl alcohol = 0.805 g/mL

Recall that:

density = mass(in wt) /volume

mass = density × volume

mass = 0.805 g/mL × 5 mL

mass = 4.03 g

Volume of HCl used = 12 mL

The reaction for this equation is shown in the image attached below.

From the reaction,

88.15 g of t-pentyl alcohol reacts with concentrated HCl to yield 106.59 g pf t-pentyl chloride.

4.03 g of t-pentyl alcohol forms,

= \dfrac{106.59 \ g \times 4.03 \ g}{88.15 \ g} of t-pentyl chloride.

Therefore,

Theoretical yield of t-pentyl chloride = 4.90 g

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Answer:

<h3>Therefore, after long period of time 80kg of salt will remain in tank</h3>

Explanation:

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dA/dt=2-(A/40)\\\\dA=2dt-(A/40)dt\\\\dA+(A/40)dt=2dt

integrating factor

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\int\limits(Ae^{(1/40)t})=\int\limits2e^{(1/40)t}dt\\\\(Ae^{(1/40)t})=2*40e^{(1/40)t}+C\\\\A=80+(C/e^{(1/40)t})\\\\A(0)=0.3\\\\0.3=80+(C/e^{(1/40)t}^*^0)\\\\0.3=80+(C/1)\\\\C=0.3-80\\\\C=-79.7\\\\A(t)=80-(79.7/e^{(1/40)t})

b)

after long period of time means t - > ∞

{t \to \infty}\\\\ \lim_{t \to \infty} A_t \\\\ \lim_{t \to \infty} (80)-(79/{e^{(1/40)t}}\\\\=80-(0)\\\\=80

<h3>Therefore, after long period of time 80kg of salt will remain in tank</h3>
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