I think that it would be <span>either ionic bonds or covalent bonds</span>
Answer:
See image attached
Explanation:
a)
The full reaction mechanism of step 1 was obtained from Bartleby and attached to this answer. The steps involved in the reaction are:
1) Loss of Br- and formation of a carbocation
2) Attack of CH3CN on the carbocation
4) Formation of a quaternary nitrogen intermediate
5) Attack of water on the quaternary nitrogen intermediate
6) Loss of the water molecule
5) Formation of the amide product
b)
i) sodium hydroxide
ii) HCl
Answer:
18.7887 g of NaCl
Explanation:
<em>The question reads - How many grams of NaCl will be produced from 18.0 g of Na and 23.0 g of Cl2?</em>
Let us start by writing out the balanced equation of the reaction:
Na + Cl2 ---> NaCl2
1 mole, each of Na and Cl2 is required to produce 1 mole of NaCl.
mole = mass/molar mass
Therefore
18 g of Na = 18/23 = 0.7826 mole
23 g of Cl2 = 23/71 = 0.3239 mole
In this case, the Na is in excess and the Cl2 becomes the limiting reagent. Hence
0.3239 mole of Cl2 will react with 0.3239 mole of Na to yield 0.3239 mole of NaCl.
mass of 0.3239 mole NaCl = 0.3239 x 58 = 18.7887 g
<u>Hence, 18.7887 grams of NaCl will be produced from 18.0 g of Na and 23.0 g of Cl2.</u>
Answer:
The answer to your question is:
Re (III) has 5 electrons
Sc(III) = has 1 electron
Ru(IV) = has 6 electrons
Hg(II) = has 10 electrons
Explanation:
75 Re(III) = 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p⁶ 6s² 4f¹⁴ 5d⁵
21 Sc(III) = 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹
44 Ru(IV) = 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d⁶
80 Hg(II) = 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p⁶ 6s² 4f¹⁴ 5d¹⁰
