Iron (II) hydroxide, Fe(OH)2, is very insoluble in water. Hence, if it is produced during an aqueous reaction it will precipitate from solution. How many grams of sodium hydroxide (a strong base) must be added to 25.2 mL of a 0.605 M aqueous solution of iron (II) nitrate, in order to precipitate all of the iron as Fe(OH)2

1.2g of NaOH

Explanation:

First we put down the equation of reaction as shown in the image attached. From the equation of reaction, we can calculate the mass of sodium hydroxide that reacts with one mole of iron II ion according to the balanced stoichiometric equation. Using the information provided about the iron II nitrate, we can calculate the amount of Fe2+ that reacted with sodium hydroxide as shown in the image. Lastly, we use this information to obtain the mass of sodium hydroxide that reacted with the stipulated amount of NaOH.

explanation:

venn diagram is pictorial representation with of circles showing the relation between two finite sets or things. the intersection part of two circles conveys the relation present in between those sets or things.

here in the venn diagram of sand and salt two circles represent sand and salt. but the area falling under the intersection of these circle represents the mixture of sand and salt.

so, in a venn diagram the mixture will be placed in a middle that is in between the intersection of two circles.

"you start with 15.67 g mgco3-xh2o, when heated and all the water is removed, you are left with 7.58 g mgco3 only.

therefore, you had 15.67 - 7.58 = 8.09g h2o  

moles of each :

7.58g mgco3 = ? moles = 7.58g / 83.4g/mole = 0.09moles  

8 09g h2o = ? moles = 8.09g / 18g/mole = 0.44moles  

molar ratio between the two by dividing both numbers by the smallest number :

0.09 / 0.09 = 1  

0.44 / 0.09 = 4.8  

mgco3-5h2o"