I think it would be A because hazardous waste would most likely be found in dust, fumes etc. I'm not sure though.
Answer:
Explanation:
destroyerplayz20062
13 hours ago
Chemistry
High School
Predict the Eº values for all (6) combinations of the following: Cu(s) and Cu(NO3)2(aq) Fe(s) and Fe(NO3)3(aq) Zn(s) and Zn(NO3)2(aq) Pb(s) and Pb(NO3)2(aq)
destroyerplayz20062
13 hours ago
Chemistry
High School
Predict the Eº values for all (6) combinations of the following: Cu(s) and Cu(NO3)2(aq) Fe(s) and Fe(NO3)3(aq) Zn(s) and Zn(NO3)2(aq) Pb(s) and Pb(NO3)2(aq)
destroyerplayz20062
13 hours ago
Chemistry
High School
Predict the Eº values for all (6) combinations of the following: Cu(s) and Cu(NO3)2(aq) Fe(s) and Fe(NO3)3(aq) Zn(s) and Zn(NO3)2(aq) Pb(s) and Pb(NO3)2(aq)
destroyerplayz20062
13 hours ago
Chemistry
High School
Predict the Eº values for all (6) combinations of the following: Cu(s) and Cu(NO3)2(aq) Fe(s) and Fe(NO3)3(aq) Zn(s) and Zn(NO3)2(aq) Pb(s) and Pb(NO3)2(aq)
Answer:
First step would be convert to moles
Final Answer: 37.8 g of NaCl
Explanation:
The reaction is:
2Na + Cl₂ → 2NaCI
We convert the mass of each reactant to moles:
18 g . 1mol /23g = 0.783 moles of Na
23g . 1mol / 70.9g = 0.324 moles of chlorine
We use the mole ratio to determine the limiting reactant:
Ratio is 2:1. 2 moles of Na react to 1 mol of chlorine
Then, 0.783 moles of Na, may react to (0.783 . 1)/2 = 0.391 moles.
Excellent!. We need 0.391 moles of Cl₂ and we only have 0.324 moles available. That's why the Cl₂ is our limiting reactant.
We use the mole ratio again, with the product side. (1:2)
1 mol of Cl₂ can produce 2 moles of NaCl
Then, our 0.324 moles of gas, may produce (0.324 . 2)/1 = 0.648 moles
Finally, we convert the moles to grams:
0.648 mol . 58.45g/mol =
