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Allisa [31]
3 years ago
15

Sodium phosphate dodecahydrate, na3po4(12h2o), is crystalline salt that is water soluble. describe what is present in an aqueous

solution of sodium phosphate dodecahydrate.
Chemistry
1 answer:
Elza [17]3 years ago
7 0
Hey there!:

Sodium cations and phosphate anions .

hope this helps!
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For the aqueous solution containing 75 mg of compound C in 1.0 mL of water, what will be the total amount of the solute C that w
Sindrei [870]

Answer:

75 mg

Explanation:

We can write the extraction formula as

x = m/[1 + (1/K)(Vaq/Vo)], where

x = mass extracted

m = total mass of solute

K = distribution coefficient

Vo = volume of organic layer

Vaq = volume of aqueous layer

Data:

m = 75 mg

K = 1.8

Vo = 0.90 mL

Vaq = 1.00 mL

Calculations:

For each extraction,

1 + (1/K)(Vaq/Vo) = 1  + (1/1.8)(1.00/0.90) = 1 + 0.62 = 1.62  

x = m/1.62 = 0.618m

So, 61.8 % of the solute is extracted in each step.

In other words, 38.2 % of the solute remains.

Let r = the amount remaining after n extractions. Then  

r = m(0.382)^n.

If n = 7,

r = 75(0.382)^7 = 75 × 0.001 18 = 0.088 mg

m = 75 - 0.088 = 75 mg

After seven extractions, 75 mg (99.999 %) of the solute will be extracted.

5 0
3 years ago
True of false metals like copper are sometimes used to fill cavities in teeth
ElenaW [278]
I'd say false because copper is what they used to drink out of in the middle ages, and it caused them to get very sick, so I doubt they would use copper in your mouth.
7 0
3 years ago
Read 2 more answers
If a substance is soluble in water, it will _______in water. A. settle B. float C. dissolve D. all of these
MrRa [10]

C.Dissolve because its another word for soluble

6 0
3 years ago
Read 2 more answers
Separate a mixture of sand, common salt, copper pieces and iron fillings. Also measure the amount of common salt in the mixture.
Neporo4naja [7]

Answer:

Explanation:

This question seeks to test the knowledge of separation techniques.

From the narration in the question, the first separation to be done is the removal of Iron fillings by the use of magnet (magnetic separation). Since Iron is magnetic, the iron fillings will be attracted by the magnet hence removing the iron fillings from the mixture.

The second constituent to be removed will be the copper pieces by the use of a sieve (sieving). Copper pieces have relatively larger sizes than sand and common salt, hence a sieve (which separates particles based on size) can be used to remove the copper pieces from the mixture.

What will be left in the mixture after the processes above will be salt and water. This mixture will have to be dissolved in water; the salt will dissolve in water while the sand will not. After which, filtration will be done to remove the sand which will be collected on the filter paper as filtride and the salt solution will pass through the filter paper as filtrate.

The salt solution can then be evaporated to dryness to retrieve the solid salt from the solution.

The amount of salt in the mixture can then be measured using a weighing balance.

Some of safety measures to be taken during the course of this experiment includes performing the experiment in an airtight and controlled environment. Lab coat and hand gloves should be worn during the course of the experiment. The evaporation to dryness should not be done close to an inflammable material/substance

5 0
3 years ago
Chlorine gas is added to a large flask to a pressure of 1.85 atm, at a temperature of 322 K. Phosphorus is added, and a reaction
AVprozaik [17]

Answer:

The volume of the flask is<u> 20.245 litres .</u>

Explanation:

We are given with following information-

PV=nRT -------- 1

where R =0.0821L.atom/mole.K

Molar mass of PCl_5=208.22g/mole

The given chemical equation is -

2P _(_s_)+5Cl_2 _(_g_)\rightarrow2PCl_5  _(_s_) --------- 2

Now , calculation -

Mass of PCl_5 formed = 118g

Molar mass of PCl_5 = 208.22g/mole

      Mole = \frac{mass (g)}{molar mass}

Therefore , moles of PCl_5 formed = \frac{118}{208.22}

     From equation 2 , we get to know that ,

 2mole PCl_5 formed from 5 mole Cl_2 _(_g_)

Therefore , \frac{118}{208.22} mole PCl_5 formed from \frac{5}{2} \times\frac{118}{208.22} mole Cl_2 _(_g_)

Moles of Cl_2 _(_g_) used =\frac{5\times118}{2\times208.22} mole

R= 0.0821L.atom/mole.K

Pressure (P)= 1.85atm

Temperature (T)= 322K

Moles of  Cl_2 _(_g_) (n)= \frac{5}{2} \times\frac{118}{208.22} moles

Applying  the formula above in 1 equation , that is

PV = nRT

putting the given values -

1.85 \times V=\frac{5}{2} \times\frac{118}{208.22}\times0.0821\times322

 V = 20.245 litres.

Hence , the volume of the flask is <u>20.245 litres . </u>

 

8 0
3 years ago
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