The mass number = protons + neutrons. Bromine has a mass number of 80<span> and 35 protons so </span>80<span>-35 = </span>45<span> neutrons. b) How many electrons does the neutral atom of bromine have? The neutral atom of bromine has 35 electrons because the number of electrons equals the number of protons.</span>
Answer: Nucleus
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Answer:
0.5M
Explanation:
The equation for molarity is:
- M =
; where the "M" stands for molarity, the "mol" stands for moles of solute and the "liters" means the volume in liters of solution.
We are given that there are:
- 1.80 moles of NaCl (the moles of solute)
- 3.60 Liters of solution (the volume in liters of solution)
Now we just plug those numbers into the formula and get our answer:
- M=
= 0.5M
After doing the math and dividing the moles of solute by the liters of solution, we get that the molarity of the solution is 0.5M.
Answer is: near equivalence point indicator should change color, so we must pick indicator who changes color near pH of equivalence point.
Equivalence point is
the point which there is stoichiometrically equivalent amounts of acid and
base. <span>
<span>Chemist can draw pH curve (graph
showing the change in pH of a solution, which is being titrated) for
titration and determine equivalence point.</span></span>
Explanation:
(w/w) % : The percentage mass or fraction of mass of the of solute present in total mass of the solution.

1) 100 g of 0.500% (w/w) NaI
Mass of solution = 100 g
Mass of solute = x
Required w/w % of solution = 0.500%


0.500 grams of solute needed to make 100 g of 0.500% (w/w) NaI.
2) 250 g of 0.500% (w/w) NaBr
Mass of solution = 250 g
Mass of solute = x
Required w/w % of solution = 0.500%


1.25 grams of solute needed to make 250 g of 0.500% (w/w) NaBr
3) 500 g of 1.25% (w/w) glucose
Mass of solution = 500 g
Mass of solute = x
Required w/w % of solution = 1.25%


6.25 grams of solute needed to make 500 g of 1.25% (w/w) (glucose)
4) 750 g of 2.00% (w/w) sulfuric acid.
Mass of solution = 750 g
Mass of solute = x
Required w/w % of solution = 2.00%


15.0 grams of solute needed to make 750 g of 2.00% (w/w) sulfuric acid.