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>
The intermolecular force that attracts two nonpolar molecules is London dispersion forces, which are also called induced dipole-induced
By considering the reaction equation is:
5Br(aq)+BrO3(aq)+6H(aq)= 3Br2(aq)+3H2O(l)
when the average rate of consumption of Br = 1.86x10^-4 m/s
So from the reaction equation
5Br → 3Br2 when we measure the average rate of formation (X) during the same interval So,
∴ 1.86x10^-4/5 = X / 3
∴X = 1.1 x 10^-4 m/s
∴the average rate of formation of Br2 = 1.1x10^-4 m/s
The formula that would represent an ionic compound that is composed of calcium and iodide ions is CaI2
Explanation
Ionic compound CaI₂ is formed when Calcium form cation ( <em>a positively charged ion</em>) by losing 2 electrons while two iodine atoms form anion ( <em>a negatively charged ion</em>) by gaining one electron each.
When writing down formula of ionic compound, the formula of cation is written first followed by anion formula. therefore Ca is written first followed by I.
The numeric subscript 2 after I(iodine) indicate that 2 atoms of iodine are involved in bonding.