Answer:
[KCl] = 1.2 M
Explanation:
We need to complete the reaction:
2KCl(aq) + Pb(NO₃)₂(aq) → 2KNO₃(aq) + PbCl₂(s)↓
By stoichiomety we know that 1 mol of chloride needs 1 mol of nitrate to react:
Let's find out the moles of nitrate, we have:
Molarity = mol/volume(L)
We convert the volume → 30 mL . 1L/1000mL = 0.030L
Molarity . volume(L) = moles → 0.400 M . 0.030L = 0.012 moles
Therefore, we can make a rule of three.
1 mol of nitrate reacts with 2 moles of chloride
Then, 0.012 moles of nitrate must react with (0.012 . 2) / 1 = 0.024 moles of KCl
We convert the volume from mL to L → 20 mL . 1L /1000mL = 0.020L
Molarity = mol /volume(L) → 0.024 mol /0.020L = 1.2 M
Answer:
a) The theoretical yield is 408.45g of
b) Percent yield =
Explanation:
1. First determine the numer of moles of and .
Molarity is expressed as:
M=
- For the
M=
Therefore there are 1.75 moles of
- For the
M=}{1Lsolution}[/tex]
Therefore there are 2.0 moles of
2. Write the balanced chemical equation for the synthesis of the barium white pigment, :
3. Determine the limiting reagent.
To determine the limiting reagent divide the number of moles by the stoichiometric coefficient of each compound:
- For the :
- For the :
As the is the smalles quantity, this is the limiting reagent.
4. Calculate the mass in grams of the barium white pigment produced from the limiting reagent.
5. The percent yield for your synthesis of the barium white pigment will be calculated using the following equation:
Percent yield =
Percent yield =
The real yield is the quantity of barium white pigment you obtained in the laboratory.
To find the number of neutrons, subtract the number of protons from the mass number. number of neutrons=40−19=21.
Answer:
B is the correct answer hope it helps
When a single compound breaks down into two or more compounds or elements in a chemical reaction then it is known as decomposition reaction.
The chemical symbol for sodium carbonate is .
The decomposition of sodium carbonate is:
The decomposition of sodium bicarbonate, will result in the formation of sodium oxide, and carbon dioxide, .
Hence, carbon dioxide, will produce with sodium oxide, on decomposition of .