Answer:
HgS
Explanation:
In a 0.350g-sample there are 0.302 g of Hg. Then, there is 0.350 g - 0.302 g = 0.048 g of S. To calculate the empirical formula, we have to follow a series of steps.
Step 1: Calculate the percent composition
Hg: (0.302 g / 0.350 g) × 100% = 86.3%
S: (0.048 g / 0.350 g) × 100% = 13.7%
Step 2: Divide each percentage by the atomic mass of the element
Hg: 86.3 / 200.59 = 0.430
S: 13.7 / 32.07 = 0.427
Step 3: Divide the numbers by the smallest one
Hg: 0.430/0.427 ≈ 1
S: 0.427/0.427 = 1
The empirical formula is HgS.
It's likely what's wanted is
<span><span>Li</span>→<span><span>Li</span><span>2+</span></span>+2<span>e−
</span></span>
The reason is because IEs are usually reported from the neutral atom, that is, IE2 is the energy required to remove two electrons from a neutral Li atom, as above, rather than the additional energy required to remove one more electron from an Li+ cation.
Bonding goes to d sublevel i think??
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Answer:
Explanation:
The equation tells you that for every mol of H2 you need 2 mols of Na to get it.
1 H2 / 4 mols H2 = 2 mols Na / x Cross multiply
1*x = 4 * 2
x = 8 mol Na