<span><span>When you write down the electronic configuration of bromine and sodium, you get this
Na:
Br: </span></span>
<span><span />So here we the know the valence electrons for each;</span>
<span><span>Na: (2e)
Br: (7e, you don't count for the d orbitals)
Then, once you know this, you can deduce how many bonds each can do and you discover that bromine can do one bond since he has one electron missing in his p orbital, but that weirdly, since the s orbital of sodium is full and thus, should not make any bond.
However, it is possible for sodium to come in an excited state in wich he will have sent one of its electrons on an higher shell to have this valence configuration:</span></span>
<span><span /></span><span><span>
</span>where here now it has two lonely valence electrons, one on the s and the other on the p, so that it can do a total of two bonds.</span><span>That's why bromine and sodium can form </span>
<span>
</span>
Answer is: cesium sulfide.
Formula for ionic compound is: Cs₂S.
Ionic compound <span>electrically </span>neutral, cesium has charge +1 (2· (+1) = +2) and sulfur has charge -2.
Cesium sulfide is whide powder, soluble in water, pyrophoric in dry air, reductant.
Answer: 0.0508mL
Explanation: Using the basic formula that states: C acid * V acid = C base * V base. we have:0.568 * 17.88 = 20 * C base.
therefore concentration of the base is 1.0156/20 = 0.0508 mL
Answer:
the other variable is also doubled
Explanation:
direct proportion, same thing has to happen to both variables
Answer:
17.55 g of NaCl
Explanation:
The following data were obtained from the question:
Molarity = 3 M
Volume = 100.0 mL
Mass of NaCl =..?
Next, we shall convert 100.0 mL to L. This can be obtained as follow:
1000 mL = 1 L
Therefore,
100 mL = 100/1000
100 mL = 0.1 L
Therefore, 100 mL is equivalent to 0.1 L.
Next, we shall determine the number of mole NaCl in the solution. This can be obtained as follow:
Molarity = 3 M
Volume = 0.1 L
Mole of NaCl =?
Molarity = mole /Volume
3 = mole of NaCl /0.1
Cross multiply
Mole of NaCl = 3 × 0.1
Mole of NaCl = 0.3 mole
Finally, we determine the mass of NaCl required to prepare the solution as follow:
Mole of NaCl = 0.3 mole
Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol
Mass of NaCl =?
Mole = mass /Molar mass
0.3 = mass of NaCl /58.5
Cross multiply
Mass of NaCl = 0.3 × 58.5
Mass of NaCl = 17.55 g
Therefore, 17.55 g of NaCl is needed to prepare the solution.