V=84.0 mL = 84.0 cm³
m=609.0 g
p=m/v
p=609.0/84.0=7.25 g/cm³
Answer: The given reaction is a neutralization reaction.
Explanation:
When an acid chemically reacts with a base then it leads to the formation of salt and water.
For example, ![3KOH + H_{3}PO_{4} \rightarrow K_{3}PO_{4} + 3H_{2}O](https://tex.z-dn.net/?f=3KOH%20%2B%20H_%7B3%7DPO_%7B4%7D%20%5Crightarrow%20K_%7B3%7DPO_%7B4%7D%20%2B%203H_%7B2%7DO)
Here, KOH is a the base and
is an acid which on chemical reaction with each other leads to the formation of salt
and water
.
Thus, we can conclude that the given reaction is a neutralization reaction.
Answer:
B
Explanation:
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The full ionic equation for the conversion of aqueous lead (II) nitrate to solid lead (II) sulfate and aqueous potassium nitrate is as follows:
<h3>What is
ionic equation ?</h3>
PbSO4(s) + 2 K+ + 2 NO3 = Pb2+(aq) + 2 NO3(aq), 2 K+(aq), and SO42(aq) (aq)
Let's look at the molecular formula that results in solid lead (II) sulfate and aqueous potassium nitrate when aqueous lead (II) nitrate and aqueous potassium sulfate interact. This reaction involves two displacements.
PbSO4(s) + 2 KNO3 = Pb(NO3)2(aq) + K2SO4(aq) (aq)
All ions and molecular species are included in the full ionic equation.
PbSO4(s) + 2 K+ + 2 NO3 = Pb2+(aq) + 2 NO3(aq), 2 K+(aq), and SO42(aq) (aq)
The full ionic equation for the conversion of aqueous lead (II) nitrate to solid lead (II) sulfate and aqueous potassium nitrate is as follows:
PbSO4(s) + 2 K+ + 2 NO3 = Pb2+(aq) + 2 NO3(aq), 2 K+(aq), and SO42(aq) (aq)
To learn more about ionic equation refer to
brainly.com/question/13879496
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Answer:
They represent the valence electrons, so s and p sublevels only! This does not include the d block!