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3 years ago

10

Which characeristic makes omasis different from diffusion.

1 answer:

Ksju [112]3 years ago

5 0

<span>haha I used to think biology was so hard, i find it quite easy now.
Okay, so basically Osmosis is the movement of water molecules from a higher concentration to a lower concentration. Diffusion is generally the movement of a gradient from higher concentration to an area of lower concentration. Osmosis applies to water only, whereas diffusion, you have many types such as Passive transport [ movement of molecules from high- low, NO CELLULAR ENERGY needed! ] then you have faciliated diffusion ( basically uses a channel protein to allow big substances to go through the membrane : NO ENERGY needed]
OSMOSIS, the important thing to remember is that water ALWAYS flow towards the region with the higher concentration of the solute (ex: Salt is solute, water is solvent) solute is the thing that is being dissolved. Solvent is the one doing the dissolving. Hope this helped!</span>

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Agri-Chem's contract with Enerco specified a maximum of 90,000 cu.ft. X 103 per day for its complexes. However, curtailments are

zvonat [6]

Answer:

Caustic soda would be least affected by a gas curtailment

Explanation:

Let,

X1= ammonia

X2= ammonium phosphate

X3= ammonium nitrate

X4 = Urea

X5= hydro-fluoric acid

X6= Chlorine

X7= Caustic soda

X8= Vinyl chloride monomer.

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= (1200 × 8 + 540 × 10 + 490 × 12 + …)

= 85,680,000 cu. ft. per day

When, the curtailment is 20%, availability is

= 0.8 X 85,680

= 68,554,000 cu. ft. per day

Therefore, the gas constant

= 8X1 + 10X2 + 12X3 + 12X4 + 7X5 + 18X6 + 20X7 + 14X8 ≤68,544

When, the curtailment is 40% availability is

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2 years ago

Capacitor C1 is initially charged to V1 and capacitor C2 is initially charged to V2. The capacitors are then connected to each o

o-na [289]

Answer:

<em>20.08 Volts</em>

Explanation:

<u>Parallel Connection of Capacitors</u>

The voltage across any two elements connected in parallel is the same. If the elements are capacitors, then each voltage is

$\displaystyle V_1=\frac{Q_1}{C_1}$

$\displaystyle V_2=\frac{Q_2}{C_2}$

They are both the same after connecting them, thus

$\displaystyle \frac{Q_2}{C_2}=\frac{Q_1}{C_1}$

Or, equivalently

$\displaystyle Q_2=\frac{C_2Q_1}{C_1}$

The total charge of both capacitors is

$\displaystyle Q_t=Q_1\left(1+\frac{C_2}{C_1}\right)$

We can compute the total charge by using the initial conditions where both capacitors were disconnected:

$Q_t=V_{10}C_1+V_{20}C_2=25\cdot 24+13\cdot 11=743\ \mu C$

Now we compute Q1 from the equation above

$\displaystyle Q_1=\frac{Q_t}{\left(1+\frac{C_2}{C_1}\right)}=\frac{743}{\left(1+\frac{13}{24}\right)}=481.95\ \mu C$

The final voltage of any of the capacitors is

$\displaystyle V_1=V_2=\frac{481.95}{24}=20.08\ V$

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2 years ago