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Leokris [45]
2 years ago
15

IUPAC NAME FOR [Fe(H2O)6](NO3)3

Chemistry
1 answer:
Kaylis [27]2 years ago
7 0

Answer:

hexaaquairon(III) trinitrate

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5 0
3 years ago
How would you prepare 100 ml of 0.4 M MgSO4 from a stock solution of 2 M MgSO4?
miss Akunina [59]
OK, so to answer this question, you will simply use the molality equation which is as follows:
<span>M1V1 = M2V2 
In the givens you have:
M1 = 2M
V1 is the unknown
M2 = 0.4M
V2 = 100 ml

</span>plug in the givens in the above equation:
<span>2 x V1 = 0.4 x 100 
</span>therefore:
V1 = 20 ml

Based on this: you should take 20 ml of the 2 M solution and make volume exactly 100 ml in a volumetric flask by diluting in water.

7 0
3 years ago
If the solubility of KCl in 100 mL of H₂O is 34 g at 20 °C and 43 g at 50 °C, label each of the following solutions as unsaturat
sertanlavr [38]

Answer:

a) Unsaturated

b) Supersaturated

c) Unsaturated

Explanation:

A saturated  solution contains the <u>maximum amount of a solute that will dissolve in a given  solvent at a specific temperature</u>.

An unsaturated solution contains <u>less solute than it  has the capacity to dissolve. </u>

A supersaturated solution, <u>contains more  solute than is present in a saturated solution</u>. Supersaturated solutions are not very  stable. In time, some of the solute will come out of a supersaturated solution as crystals.

According to these definitions and considering that the solubility of KCl in 100 mL of H₂O at <u>20 °C is 34 g</u>, and at <u>50 °C is 43 g</u> we can label the solutions:

a) 30 g in 100 mL of H₂O at 20 °C  ⇒ unsaturated

b) 65 g in 100 mL of H₂O at 50 °C  ⇒ supersaturated

c) 42 g in 100 mL of H₂O at 50 °C and slowly cooling to 20 °C to give a clear solution <u>with no precipitate</u> ⇒ unsaturated (if it were saturated it would have had precipitate)

8 0
3 years ago
Assume you mixed 5 µl of your cod fish homogenate with 195 µl working solution. The fluorometer displays a measurement of 12.2 µ
11111nata11111 [884]

Answer:

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Explanation:

I took that question

3 0
3 years ago
The student placed 10 mL of PbCl2 (saturated solution) in the test tube and added a pinch of lead acetate. When the test tube wa
STALIN [3.7K]
PbCl₂(s) ⇄ Pb²⁺(aq) + 2Cl⁻(aq)

Pb(CH₃COO)₂(s) → Pb²⁺(aq) + 2CH₃COO⁻(aq)

At lead acetate disolution, concentration of lead cations will increase. According to Le Chatelie's principle equilibrium will be displaced towards formation of solid lead chloride.   
4 0
3 years ago
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