Answer:
0.1g (Gallon) of chlorine
Explanation:
<u>Formula</u>
1 gallon = 3.7L; the density of water is 1.0g/ml
<u>Given</u>
2g (gallon) of chlorine to sanitize = 1,000,000g (gallon) of water
<u>Solve</u>
If 2g (gallon) chlorine = 1,000,000g (gallon)
∴, ? chlorine = 40,000
The First step; set up an equation
1000000/2 = 40000/?
The Next step; divide 1 million to 2
1000000 ÷ 2 = 500000
Then, divide the result by 40000
40000 ÷ 500000 = 0.08
In the nearest unit that is 0.1
Therefore, it will take 0.1g (gallon) of chlorine to sanitize a 40,000-gallon pool.
Answer:
<h3>5.06282 × 10²⁴ molecules</h3>
Explanation:
The number of molecules of Ca2(SO3) can be found by using the formula
<h3>N = n × L</h3>
where n is the number of moles
N is the number of entities
L is the Avogadro's constant which is
6.02 × 10²³ entities
From the question we have
N = 8.41 × 6.02 × 10²³
We have the final answer as
<h3>5.06282 × 10²⁴ molecules</h3>
Hope this helps you
Answer:
Explanation:
Atoms are held together by covalent bonds when they share electrons between themselves.
Covalent bonds are bonds that are formed between non-metals usually with a low electronegative difference between them. In this bond type, two non-metals donate electrons which are shared between the combining atoms and this makes them both like the corresponding noble gases. The shared electrons is what forms the covalent bonds.
An example of covalent bond is HCl, H₂S, SO₂, CO₂, O₂ etc
Answer:
Explanation:
there are a certain amount of atoms
Answer:
- The limiting reactant is lead(II) nitrate.
- 7.20 g of precipitate are formed.
- 1.9 g of the excess reactant remain.
Explanation:
The reaction that takes place is:
- Pb(NO₃)₂(aq) + 2KCl(aq) → PbCl₂(s) + 2KNO₃(aq)
With a percent yield of 87.5%.
To determine the limiting reactant, first we <u>convert the masses of each reactant to moles</u>, using their molar mass:
- 9.82 g Pb(NO₃)₂ ÷ 331.2 g/mol = 0.0296 mol Pb(NO₃)₂
- 5.76 g KCl ÷ 74.55 g/mol = 0.0773 mol KCl
Looking at the stoichiometric coefficients, we see that 1 mol of Pb(NO₃)₂ would react completely with 2 moles of KCl. Following that logic, 0.0296 mol Pb(NO₃)₂ would react completely with (2x0.0296) 0.0592 mol of KCl. We have more than that amount of KCl, this means KCl is the reactant in excess and Pb(NO₃)₂ is the limiting reactant.
To calculate the mass of precipitate (PbCl₂) formed, we <u>use the moles of the limiting reactant</u>:
- 0.0296 mol Pb(NO₃)₂
* 
* 87.5/100 = 7.20 g PbCl₂
- Keeping in mind the reaction yield, the moles of Pb(NO₃)₂ that would react are:
- 0.0296 mol Pb(NO₃)₂ * 87.5/100 = 0.0259 mol Pb(NO₃)₂
Now we <u>convert that amount to moles of KCl and finally into grams of KCl</u>:
- 0.0259 mol Pb(NO₃)₂
* 
= 3.86 g KCl
3.86 g of KCl would react, so the amount remaining would be:
