Answer:
Explanation:
Approximately how many accidents occurred between 1965 and 1970 inclusive?
about 1800 accidents
about 2500 accidents
about 1200 accidents
about 1500 accidents
Answer:
(c) only Ca2+(aq) and Hg2+(aq)
Explanation:
- In the first step, hydrochloric acid (HCl) is added to the solution. In this case the equilibrium that could take place is:
Ag⁺(aq) + Cl⁻(aq) ↔ AgCl(s)
But no precipitate was formed, so Ag⁺(aq) is absent.
- By adding H₂SO₄(aq) the next equilibrium that could take place is:
Ca⁺²(aq) + SO₄⁻²(aq) ↔ CaSO₄(s)
A white precipitate was formed, so Ca⁺² is present in the solution.
- The following could take place after adding H₂S(aq):
Hg²⁺(aq) + S⁻² ↔ HgS(s)
A black precipitate formed, so Hg⁺² is present as well.
Answer:
The Answer is 2,86 grs Na2CO3
Explanation:
What we have to do is find the mass of Na2CO3 as a pure component or solute. That's because the 11,8 mL are a solution of Na2CO3. This means, the sum between the solute Na2CO3 and water. To find the grams of Na2CO3 as pure component we create a factor series as is shown in the attached file.
Data:
Density of solution (ρ) = 1,10 grs sln Na2CO3/mL sln Na2CO3
Mass Percentage (%) = 22 grs Na2CO3/100 grs sln Na2CO3
The procedure is explained in the attached file
Answer: protons and neutrons.
The nucleus is made up of 3 subatomic particles that are protons,neutrons and electrons.
General notation of an element is 
where, X is the Element, A is the Atomic Mass and Z is the Atomic Number
If we know the number of protons we can easily find out the atomic number of any element because Atomic Number = Number of protons in an element.
And in addition if we know the number of neutrons we can easily find out the atomic mass of an element because
Atomic Mass = (Number of protons) + (Number of neutrons)
If we get to know the atomic number and atomic mass, we can easily tell what element is it by looking from the periodic table.
(133.9) x (595.296)= 79,710.1344
Answer with significant digits: 80,000
1.3 has two significant digits so 79,700 rounds up to 80000.