Our good friend and pseudo scientist, homer simpson, attempts to analyze 300 mg of an unknown compound containing only c and h b

y burning it in excess oxygen. 540 mg of h2o is produced. you may assume that all the hydrogen in the compound ended up as water.

Chemistry

1 answer:

iragen [17]3 years ago

7 0

$C_{x}H_{y} + O_{2} --\ \textgreater \ x CO_{2} + \frac{y}{2} H{2}O

540mg H_{2}O* \frac{1 mmol H_{2}O}{18 mg H_{2}O} =30 mmol H_{2}O

\\ \\30 mmolH_{2}O -have - 60 mmol H
\\ \\ 60 mmol H* \frac{1mg H}{1 mmol H} =60 mg H
\\ \\300mg C_{x}H_{y} - 60 mg H= 240 mg C
\\ \\ 240 mgC* \frac{1mmol}{12mg} =20 mmol C

20 mmol C: 60 mmol H=1 mol C : 3 mol H

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A tank contains 200 gallons of water in which 300 grams of salt is dissolved. A brine solution containing 0.4 kilograms of salt

Nadusha1986 [10]

Answer:

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

Explanation:

given amount of salt at time t is A(t)

initial amount of salt =300 gm =0.3kg

=>A(0)=0.3

rate of salt inflow =5*0.4= 2 kg/min

rate of salt out flow =5*A/(200)=A/40

rate of change of salt at time t , dA/dt= rate of salt inflow- ratew of salt outflow

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

integrating factor

$=e^{\int\limits (1/40) \, dt}$

integrating factor $=e^{(1/40)t}$

multiply on both sides by $=e^{(1/40)t}$

$dAe^{(1/40)t}+(A/40)e^{(1/40)t} dt =2e^{(1/40)t}t\\\\(Ae^{(1/40)t})=2e^{(1/40)t}t$

integrate on both sides

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