The only list containing molecules is list #4
NaCl, N2, and H2O are all made up of more than one element, making them molecules. The other lists contain an element. Mg, He, and Fe are elements, not molecules.
Hope this helps!
Answer: Alkali metals, group 1
Explanation: Group 1 means 1 valence electron. This group is called alkali metals, which indicates that it is a metal. They are highly reactive because only 1 electron needs to move to bond with something (since it only has 1 valence electron)
Answer:
See explanation
Explanation:
What I have written in the image attached is called a nuclear equation. It differs from a chemical reaction equation in the sense that it involves transformations that occur in the nucleus of atoms.
The nuclear equation must be balanced. This means that the mass and charge on both sides of the reaction equation must be the same.
On the left hand side the U-235 interacts with a neutron. The total mass on the left hand side is 236 while the total charge is 92. If we sum up the masses and charges of Ba and Kr, we also get a total of 236 mass units and a charge of 92.
Hence the other nucleus is barium-141
4.20 mol Al would react completely with 4.20 x (1/2) = 2.10 mol Fe2O3, but there is not that much Fe2O3 present, so Fe2O3 is the limiting reactant. (1.75 mol Fe2O3) x (2/1) x ( 55.8452 g Fe/mol) = 195 g Fe 3 MgO + 2 H3PO4 → Mg3(PO4)2 + 3 H2O (15.0 g MgO) / (40.3045 g MgO/mol) = 0.37217 mol MgO (18.5 g H3PO4) / (97.9953 g H3PO4/mol) = 0.18878 mol H3PO4 0.18878 mol H3PO4 would react completely with 0.18878 x (3/2) = 0.28317 mole of MgO, but there is more MgO present than that, so MgO is in excess and H3PO4 is the limiting reactant. Now we must consider why the problem tells us "17.6g of Mg3(PO4)2 is obtained". The first possibility is that it's just there for the sake of confusion -- in which case ignore it and proceed this way: ((0.37217 mol MgO initially) - (0.28317 mole MgO reacted)) x (40.3045 g MgO/mol) = 3.59 g MgO left over However, if the amount of magnesium phosphate obtained is given because the reaction was stopped before it was complete, the amount obtained governs the amount reacted and the amount left over, so proceed this way: (17.6g Mg3(PO4)2) / (262.8581 g Mg3(PO4)2/mol) x (3/1) = 0.20087 mol MgO reacted ((0.37217 mol MgO initially) - (0.20087 mole MgO reacted)) x (40.3045 g MgO/mol) = 6.90 MgO left over
A(C)+A(F)+A(Br)+A(O)=12.01+19.00+79.90+16.00=126.91
n=381.01/126.91= 3
C₃F₃Br₃O₃
