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:
I am typing this to get more answers for one of my tests good luck dawg
Answer:
N2I6 = 789 g
N2I6 = 8.2x1022 molecules N2I6 x 1 mole/6.02x1023 molecules = 1.36x10-1 moles = 0.136 moles
N2I6=0.136molesx789g/mole=107g=110g
----------------------
Hope this helps, have a BLESSED AND WONDERFUL DAY!
- Cutiepatutie ☺❀❤
Answer:
255.51cm3
Explanation:
Data obtained from the question include:
V1 (initial volume) =?
T1 (initial temperature) = 50°C = 50 + 273 = 323K
T2 (final temperature) = - 5°C = - 5 + 237 = 268K
V2 (final volume) = 212cm3
Using the Charles' law equation V1/T1 = V2/T2, the initial volume of the gas can be obtained as follow:
V1/T1 = V2/T2
V1/323 = 212/268
Cross multiply to express in linear form
V1 x 268 = 323 x 212
Divide both side by 268
V1 = (323 x 212)/268
V1 = 255.51cm3
Therefore, the initial volume of the gas is 255.51cm3
