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Answer:
239.7 g
Explanation:
Step 1: Write the balanced equation
2 LiBr + I₂ → 2 LiI + Br₂
Step 2: Convert the molecules of iodine to moles
We have 9.033 × 10²³ particles (molecules) of iodine. In order to convert molecules to moles, we will use the <em>Avogadro's number</em>: there are 6.022 × 10²³ molecules of iodine in 1 mole of iodine.

Step 3: Calculate the moles of bromine produced
The <em>molar ratio of I₂ to Br₂</em> is 1:1. Then, the moles of bromine produced are 1.500 moles.
Step 4: Calculate the mass of bromine
The <em>molar mass of bromine</em> is 159.81 g/mol. The mass corresponding to 1.500 moles is:

Is by turning a few seconds before
Answer:
The pOH of HNO₃ solution that ha OH⁻ concentration 9.50 ×10⁻⁹M is 8.
Explanation:
Given data:
[OH⁻] = 9.50 ×10⁻⁹M
pOH = ?
Solution:
pOH = -log[OH⁻]
Now we will put the value of OH⁻ concentration.
pOH = -log[9.50 ×10⁻⁹M]
pOH = 8
Thus the pOH of HNO₃ solution that ha OH⁻ concentration 9.50 ×10⁻⁹M is 8.