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3 years ago

If the contents of a cell have a solute concentration of 0.04 percent which of these solutions would cause it to swell?

2 answers:

4 0

"0.01 percent solute concentration" is the solution among the choices given in the question that <span>would cause it to swell. The correct option among all the options that are given in the question is the fourth option or option "D". I hope that the answer has actually come to your great help.</span>

Akimi4 [234]3 years ago

3 0

Answer : Option D) 0.01 percent solute concentration

Explanation : The contents of a cell if has the solute concentration as 0.04 percent then it should have much lesser concentration that the cell, then only it will allow the cell to swell. Because of the solutions with a lower osmotic pressure than a cell fluid solution concentration, can make it swell.

Hence,the answer will be 0.01 percent of solute concentration.

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3 years ago

The rate of disappearance of HBr in the gas phase reaction2HBr(g) → H2(g) + Br2(g)is 0.301 M s-1 at 150°C. The rate of appearanc

jok3333 [9.3K]

Answer: 0.151

Explanation:

Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

$2HBr(g)\rightarrow H_2(g)+Br_2(g$

The rate in terms of reactants is given as negative as the concentration of reactants is decreasing with time whereas the rate in terms of products is given as positive as the concentration of products is increasing with time.

$Rate=-\frac{1d[HBr]}{2dt}=+\frac{1d[H_2]}{2dt}=+\frac{1d[Br_2]}{dt}$

Given: $-\frac{d[HBr]}{dt}]=0.301$

Putting in the values we get:

$\frac{0.301}{2}=+\frac{1d[Br_2]}{dt}$

$+\frac{1d[Br_2]}{dt}=0.151$

Thus the rate of appearance of $Br_2$ is 0.151

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