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Liono4ka [1.6K]
3 years ago
10

Balance: Fe2(SO4)3 + _K[OH] —> _K2[SO4] + K[SO] + Fe[OH]3

Chemistry
2 answers:
Sindrei [870]3 years ago
8 0

Answer:

Fe2(SO4)3 + 6 KOH → 2 Fe(OH)3 + 3K2SO4

Explanation:

LenKa [72]3 years ago
6 0

Balance the reaction of Fe2(SO4)3 + KOH = K2SO4 + Fe(OH)3 using this chemical equation balancer! ... Fe2(SO4)3 + 6KOH → 3K2SO4 + 2Fe(OH)3 ...

Missing: _K[ ‎| Must include: _K[

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Which of the following statements best describes the current atomic theory
Oksana_A [137]
You did not provide the statements but a possible answer might be that the current atomic theory is sound and that technology that could challenge it does not exist at this time. There is no way to change it because no research can be done about it.
8 0
3 years ago
PLEASE HELP FAST!!!! I'll give brainliest to the best answer!!
MArishka [77]

Answer:

A concentrated acid is an acid which is in either pure form or has a high concentration. Laboratory type sulfuric acid (about 98% by weight) is a concentrated (and strong) acid. A dilute acid is that in which the concentration of the water mixed in the acid is higher than the concentration of the acid itself.

Explanation:

Concentrated acid - Those acids which are pure or have very high concentration in water are called as concentrated acids. For example concentrated Hydrochloric acid (HCl) and concentrated Sulphuric acid are examples of concentrated acids.

7 0
3 years ago
Read 2 more answers
How many bananas are equal to 7.50 moles of bananas?​
Veseljchak [2.6K]

Answer:

4.52×10^24

Explanation:

N = n × Na

where; N = no. of bananas

n = no. of moles

Na = Avogadro's constant

Which is 6.02×10^23

N = 7.5 × 6.02×10^23

N =4.515×10^24

7 0
3 years ago
In the compound CO2, how many lone pairs are on the central atom?
Cloud [144]
There are 4 lone pairs of electrons present in the carbon dioxide molecule 
4 0
3 years ago
Balance the equation :
Sati [7]

The balanced equation is 2 AlI 3 ( a q ) + 3 Cl 2 ( g ) → 2 AlCl 3 ( a q ) + 3 I 2 ( g ) .

<u>Explanation:</u>

  • Aluminum has a typical oxidation condition of  3+  , and that of iodine is  1-  .   Along these lines, three iodides can bond with one aluminum. You get  AlI3.  For comparable reasons, aluminum chloride is  AlCl3.  
  • Chlorine and iodine both exist normally as diatomic components, so they are  Cl2(  g  )  also,  I2(  g  ), individually. In spite of the fact that I would anticipate that iodine should be a strong.  

Balancing the equation, we get:  

            2AlI 3(  aq  )  +  3Cl2 (  g  )  →  2AlCl3 (  aq  )   +  3 I 2  (  g  )  

  • Realizing that there were two chlorines on the left, I simply found the basic numerous of 2 and 3 to be 6, and multiplied the  AlCl  3  on the right.  
  • Normally, presently we have two  Al  on the right, so I multiplied the  AlI  3  on the left. Hence, I have 6  I  on the left, and I needed to significantly increase  I 2  on the right.  
  • We should note, however, that aluminum iodide is viciously receptive in water except if it's a hexahydrate. In this way, it's most likely the anhydrous adaptation broke down in water, and the measure of warmth created may clarify why iodine is a vaporous item, and not a strong.
3 0
4 years ago
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