Answer:
185.49 grams of Zinc would react with 454g (1lb) of copper sulfate
Explanation:
Yo know the following balanced reaction:
CuSO₄(aq)+ Zn(s) →Cu(s) + ZnSO₄(aq)
You can see that by stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of reagents and products are part of the reaction:
- CuSO₄: 1 mole
- Zn: 1 mole
- Cu: 1 mole
- ZnSO₄: 1 mole
Being:
- Cu: 63.54 g/mole
- S: 32 g/mole
- O: 16 g/mole
- Zn: 65.37 g/mole
the molar mass of the compounds participating in the reaction is:
- CuSO₄:63.54 g/mole + 32 g/mole + 4*16 g/mole= 159.54 g/mole ≅ 160 g/mole
- Zn: 65.37 g/mole
- Cu: 63.54 g/mole
- ZnSO₄: 65.37 g/mole + 32 g/mole + 4*16 g/mole= 161.37 g/mole
Then, by stoichiometry of the reaction, the following amounts of mass of reagent and product participate in the reaction:
- CuSO₄: 1 moles* 160 g/mole= 160 g
- Zn: 1 mole* 65.37 g/mole= 65.37 g
- Cu: 1 mole* 63.54 g/mole= 63.54 g
- ZnSO₄: 1 mole* 161.37 g/mole= 161.37 g
Now you can apply the following rule of three: if 160 grams of CuSO₄ react with 65.37 grams of Zn by this reaction stoichiometry, 454 grams of CuSO₄ with how much mass of Zn will it react?
mass of Zn= 185.49 grams
<u><em>185.49 grams of Zinc would react with 454g (1lb) of copper sulfate</em></u>
Answer:
Water moves from the ground or oceans into the atmosphere through a process called evaporation. It's a process that happens on a molecular level when the molecules of water are really energized and rise into the air. Now you've got water in the air and water on land. Organisms all over the Earth need water to survive.
Explanation:
Answer:
pH = 1.32
Explanation:
H₂M + KOH ------------------------ HM⁻ + H₂O + K⁺
This problem involves a weak diprotic acid which we can solve by realizing they amount to buffer solutions. In the first deprotonation if all the acid is not consumed we will have an equilibrium of a wak acid and its weak conjugate base. Lets see:
So first calculate the moles reacted and produced:
n H₂M = 0.864 g/mol x 1 mol/ 116.072 g = 0.074 mol H₂M
54 mL x 1L / 1000 mL x 0. 0.276 moles/L = 0.015 mol KOH
it is clear that the maleic acid will not be completely consumed, hence treat it as an equilibrium problem of a buffer solution.
moles H₂M left = 0.074 - 0.015 = 0.059
moles HM⁻ produced = 0.015
Using the Henderson - Hasselbach equation to solve for pH:
ph = pKₐ + log ( HM⁻/ HA) = 1.92 + log ( 0.015 / 0.059) = 1.325
Notes: In the HH equation we used the moles of the species since the volume is the same and they will cancel out in the quotient.
For polyprotic acids the second or third deprotonation contribution to the pH when there is still unreacted acid ( Maleic in this case) unreacted.
Answer:
Below
Explanation:
Balanced form;
1.Benzene + Dioxygen = Carbon Dioxide + Water
2.Tricalcium phosphate +Carbon = Calcium phosphide + carbon monoxide
3.Nitrous acid react with oxygen to produce nitric acid.
4.This means that the carbon dioxide and limewater react to produce calcium carbonate and water.
5.Potassium react with bromine to produce potassium bromide
6. An aqueous solution of ferrous sulphate reacts with aqueous solution of sodium hydroxide to form a precipitate of ferrous hydroxide and sodium sulphate remains in the solution.
