First row: HCl, ZnCl2, FeCl3, AlCl3, BaCl2, PbCl4
Second row: H3P, Zn3P2, FeP, AlP, Ba3P2, Pb3P4
Third row: HNO3, Zn(NO3)2, Fe(NO3)3, Al(NO3)3, Ba(NO3)2, Pb(NO3)4
Fourth row: ZnO, Fe2O3, Al2O3, BaO, PbO2
Fifth row: HCaF2, Zn(CaF2)2, Fe(CaF2)3, Al(CaF2)3, Ba(CaF2)2, Pb(CaF2)4
Sixth row: H2SO4, ZnSO4, Fe2(SO4)3, Al2(SO4)3, BaSO4, Pb(SO4)2
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>
Ernest Rutherford
J. J Thomson
Explanation:
<u>Ernest Rutherford</u>
In 1911, Ernest Rutherford, a New Zealand chemist performed the gold foil experiment where he gave the modelling of the atom a boost.
Experiment
In his experiment, he bombarded a thin gold foil with alpha particles generated from a radioactive source. He found that most of the alpha particles passed through the gold foil while a few of them were deflected back.
Discovery and reflection on the atomic theory
To account for his observation, Rutherford suggested an atomic model in which an atom has small positively charged center where nearly all the mass is concentrated.
<u>J. J Thomson</u>
Experiment
In 1897 J.J Thomson performed experiments using the gas discharge tube that led to the discovery of the electrons. He called them cathode rays because they originate from the cathode and exits at the anode.
Discovery and reflection on the atomic theory
From his experiment on the gas discharge tube, Thomson was able determine the properties of cathode rays some of which are:
- they move in a straight line
- they possess kinetic energy
- they attract positive charges and repels negative charges
Using his observation, he proposed the plum pudding model of the atom where it is made up of entirely electrons.
learn more:
Rutherford brainly.com/question/1859083
#learnwithBrainly
We are given with the mass of pure iron that reacts with oxygen to form an oxide which has a given mass as well. the mass of oxygen reacted is 8.15-6.25 g or 1.9 grams. THen we convert the mass of the reactants to moles. Iron is equal to 0.1119 moles and oxygen is equal to 0.1188. We divide each number to the less amount. Hence iron is 1 and oxygen is approx 1. The empirical formula hence is FeO or ferrous oxide or Iron (II) oxide.