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Lunna [17]
4 years ago
13

When recovering the aqueous and organic layers from the separatory funnel ______. (Select all that apply.) A. the lower layer sh

ould be allowed to drain through the tap B. the stopper must be removed in order to allow the funnel to drain C. the stopper must be kept in place during draining D. the aqueous layer should always be poured from the top, regardless of which layer it is E. magnesium sulfate should be added to the funnel before pouring, to dry the organic layer
Chemistry
1 answer:
ipn [44]4 years ago
5 0

Answer: Option A and C applies

- A. the lower layer should be allowed to drain through the tap

- C. the stopper must be kept in place during draining

Explanation:

Since aqueous and organic substances do not mix, they form a layer between them. Hence, require a separatory funnel with a stopper to recover each fraction effectively.

First, the lower fraction (aqueous) is recovered when the stopper is opened to allow draining through the tap into a well labelled beaker

Second, the stopper must be kept in place to shut the tap once the lower fraction has been collected. So, the organic fraction remains in the seperating funnel

Thus, Option A and C applies

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4. What are the two ways that you can recognize a numeric place changes (1s vs. 10s vs. 100s) on a micropipettor?
Serjik [45]

it's 100,0002 pn a micropipettor

8 0
2 years ago
Calculate the molarity of 4.56 grams of NaCl dissolved in 355 ml of total solution?
alexira [117]

Answer:

0.22 mol / L

Explanation:

Please see the answer in the picture attached below.

Hope this can help you. Have a nice day !

3 0
3 years ago
What volume of .100 M HCl is required to titrate .100 g of sodium bicarbonate to its equivalence point?
Karo-lina-s [1.5K]
The reaction is:  HCl + Na HCO3 → NaCl + H2CO3, which is already balanced.

Then the molar ratios are 1 mol HCL : 1 mol NaHCO3

You know the mass of NaHCO3 is 0.100 g, then you can find the number of moles using the molar mass of NaHCO3

molar mass of NaHCO3 is 23 g/mol + 1 g/mol + 12g/mol + 3*16g/mol = 84 g/mol

And the number of moles is: 0,100 g / 84 g/mol  = 0,0012 mol of NaHCO3.

Then, from the ratio 1:1, you know that, at the titration point, the number of moles of HCl is the same: 0,0012 mol of HCl.

Now from the fomula of molarity you have: M = #of moles / V in liters

=> V in liters = # of moles / M = 0.0012 mol / 0.100 M = 0.012 liters

Answer: 0.012 liters of HCL 
6 0
4 years ago
Read 2 more answers
When ammonium perchlorate, NH4ClO4, is dissolved in water, will it give an acidic, basic, or neutral solution? Ammonium perchlor
ZanzabumX [31]

<u>Answer:</u> The solution will be an acidic solution when ammonium perchlorate is dissolved in water.

<u>Explanation:</u>

Salts are formed when an acid reacts with a base during a neutralization reaction.

  • When a strong acid and a weak base reacts, it leads to the formation of acidic salt.
  • When a strong base and weak acid reacts, it leads to the formation of basic salt.
  • When a strong acid and strong base or weak acid and weak base reacts, it leads to the formation of neutral salts.

Ammonium perchlorate is formed by the combination of perchloric acid (HClO_4) which is a strong acid and ammonium hydroxide (NH_4OH) which is a weak base.

So, the salt formed which is ammonium perchlorate is an acidic salt and its dissolution in water will make the solution acidic in nature.

NH_4ClO_4+H_2O\rightarrow HClO_4+NH_4OH

Hence, the solution will be an acidic solution when ammonium perchlorate is dissolved in water.

4 0
4 years ago
How many grams of hydrogen are needed to produce 1.80g of water according to this equation?
Nastasia [14]

Answer:

0.2g of hydrogen is needed to produce 1.80g of water

Explanation:

$H_{2}+O_{2} \to 2H_{2}O$

No. of moles of water$= \frac{Weight}{Molecular \, weight}$ \\

From given values,

Weight of water $=1.80g$

Molecular weight of water $=18g$

$\therefore$ No. of moles of water$=\frac{1.80}{18}=0.1 $

From equation, it is clear that 2 moles of water can be formed by 2 moles of $H_{2}$

Hence, 0.1 mole of water can be formed from 0.1 mole of $H_{2}$

Now,

No. of moles of $H_{2} = \frac{Weight}{Molecular \, weight}$

$\Rightarrow weight \,of\, H_{2}=No.\,of\,moles \times Molecular\,weight$\\ $=0.1 \times 2=0.2g$

8 0
3 years ago
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