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3 years ago

Explain the factors that affect the absorption and stripping.

Chemistry

1 answer:

RSB [31]3 years ago

7 0

Answer:

Absorption is the process in which one and more than one component which is known as gas stream are removed from the mixture of the gas, which get absorbed into non volatile liquids.

Stripping is the reverse process of absorption in which more than one gaseous state component are removed by vaporized from gas and liquid state into insoluble gas state.

The factors that affect the absorption and stripping are:

Volatility: Low vapor pressure causes low volatility which affect the absorption ans stripping.
Relative solubility: If there is more relative solubility then, it results in higher efficiency in separation.
Viscosity: Viscosity is the important factor which required optimum viscosity.

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Which element probably has the most properties in common with lead (Pb)?

Fed [463]

Answer:

Here is your answer: Tin,

Explanation:

Which of the following elements probably has the most properties in common with lead (Pb)? Tin (Sn).

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3 years ago

To what temperature must a sample of nitrogen at 27°C and 0.625 atm be taken so that it’s pressure becomes 1.125 atm at constant

Karo-lina-s [1.5K]

Answer:

The sample of nitrogen must be taken to 267 $^{0}\textrm{C}$ .

Explanation:

Let's assume nitrogen gas behaves ideally.

Here amount of nitrogen gas in both states remain constant.

So, in accordance with combined gas law for a given amount of an ideal gas in two different states: $\frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}$

where $P_{1}$ and $P_{2}$ are initial and final pressure respectively. $V_{1}$ and $V_{2}$ are initial and final volume respectively. $T_{1}$ and $T_{2}$ are initial and final temperature (in kelvin scale) respectively.

Here $V_{1}=V_{2}$ , $P_{1}=0.625atm$ , $T_{1}=(273+27)K=300K$ and $P_{2}=1.125atm$

So $T_{2}=\frac{P_{2}T_{1}}{P_{1}}=\frac{(1.125atm)\times (300K)}{(0.625atm)}=540K$

540 K = (540-273) $^{0}\textrm{C}$ = 267 $^{0}\textrm{C}$

So the sample of nitrogen must be taken to 267 $^{0}\textrm{C}$ .

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3 years ago

Galileo called the dark, flat parts of the moon______

exis [7]

Answer:

Marias

Explanation:

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3 years ago

Chemistry Lab Question: Filtration

tatyana61 [14]

For the first question, salt is soluble while sand is insoluble or not dissolvable in water. The salt should have vanished or melted, but the sand stayed noticeable or visible, making a dark brown solution probably with some sand particles caught on the walls of the container when the boiling water was put in to the mixture of salt and sand. The solubility of a chemical can be disturbed by temperature, and in the case of salt in water, the hot temperature of the boiling water enhanced the salt's capability to melt in it.

For the second question, the melted or dissolved salt should have easily made its way through the filter paper and into the second container, while the undissolved and muddy sand particles is caught on the filter paper. The size of the pores of the filter paper didn’t change. On the contrary, the size of the salt became smaller because it has been dissolved which is also the reason why it was able to go through the filter paper, while the size of the sand may have doubled or even tripled which made it harder to pass through.

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4 years ago

The gas in an engine cylinder is compressed until it has a volume of 0.045 liter and an absolute pressure of 28 atmospheres. If

Hatshy [7]

$7.7 \times 10^{2} \; \text{K}$ .

<h3>Explanation</h3>

Units of each quantity:

<em>P</em> = 28 atm;
<em>V</em> = 0.045 L;
<em>n</em> = 0.020 mol;

Ideal gas constant:

$R = 0.0820573 \;\text{L} \cdot \text{atm} \cdot \text{K}^{-1} \cdot \text{mol}^{-1}$ .

Apply the ideal gas law:

$T = \dfrac{P \cdot V}{n \cdot R} = \dfrac{{\bf 28} \; \text{atm} \times {\bf 0.045} \;\text{L} }{{\bf 0.020}\; \text{mol} \times {\bf 0.0820573} \;\text{L} \cdot \text{atm} \cdot \text{K}^{-1} \cdot \text{mol}^{-1}} = {\bf 767.76} \; \text{K}$ .

All data given in this question come with two significant figures. Round the value of <em>T</em> to two significant figures:

$T = 767.76 \;\text{K} = 7.7 \times 10^{2} \; \text{K}$ .

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3 years ago

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