This problem is requiring the balanced chemical equation that takes place when copper hydroxide and potassium sulfate are produced when reacting potassium hydroxide with copper sulfate.
<h3>Balancing chemical equations:</h3>
In chemistry, balancing chemical equations is based on the law of conservation of mass, which demands us to have equal number of atoms on both sides of the chemical equation. This can be accomplished by inserting coefficients in front of the chemical species.
For this particular case, we have potassium hydroxide with copper sulfate on the reactants side, however, copper can be copper (I) or copper (II) as it has 1+ and 2+ as its possible oxidation numbers. In addition, copper hydroxide and potassium sulfate as the products. Hence, we can assume this is all about copper (II) so we can write:
As we can see, potassium, hydrogen and oxygen have two atoms each on the products side, but just one on the reactants side; drawback we can overcome by putting a 2 in front of KOH so as to balance it:
Learn more about balancing chemical equations: brainly.com/question/8062886
Answer:
Volume of chlorine = 61.943 mL
Explanation:
Given:
Volume of the water in the Pool = 18,000 gal
also,
1 gal = 3785.412 mL
thus,
Volume of water in pool = 18,000 × 3785.412 = 68,137,470 mL
Density of water = 1.00 g/mL
Therefore,
The mass of water in the pool = Volume × Density
or
The mass of water in the pool = 68,137,470 mL × 1.00 g/mL = 68,137,470 g
in terms of million = 
or
= 68.13747 g
also,
1 g of chlorine is present per million grams of water
thus,
chlorine present is 68.13747 g
Now,
volume = 
or
Volume of chlorine = 
or
Volume of chlorine = 61.943 mL
Answer:
electrons
Explanation:
metals do what is called metallic bond
the correct combo is C
We know sunlight is key factor in photosynthesis so A and D are out.
we also know plants get their bulk by breaking up co2 and using the carbon which releases oxygen as a byproduct so B is also out of the question.
Pb(NO3)2 (aq) + 2 NaI (aq) --> PbI2 (s) + 2 NaNO3 (aq)
Starting with with 200.0 grams of Pb(NO3)2 and 120.0 grams of NaI:
A. What is the limiting reagent?
B. How many grams of PbI2 is theoretically formed?
C. How many grams of the excess reactant remains?
D. If 48 grams of NaNO3 actually formed in the reaction, what is the percent yield of this reaction?
