1. What volume is required to contain 100.0 mol O2 at STP?
1 answer:
You might be interested in
Data:
m (<span>Sample Mass) = ?
n (</span><span>Number of moles) = 0.714 mol
MM (Molar Mass) of </span>Mercury (I) Chloride (
)
Hg = 2*200.59 = 401.18 amu
Cl = 2*35.453 = 70.906 amu
----------------------------------------
Molar Mass
= 401.18 + 70.906 = 472.086 ≈ 472.09<span> amu or 472.09 g/mol
</span>
Formula:
Solving:
Answer:
By approximation would be letter D) <span>337.2 g</span>
m (<span>Sample Mass) = ?
n (</span><span>Number of moles) = 0.714 mol
MM (Molar Mass) of </span>Mercury (I) Chloride (
Hg = 2*200.59 = 401.18 amu
Cl = 2*35.453 = 70.906 amu
----------------------------------------
Molar Mass
</span>
Formula:
Solving:
Answer:
By approximation would be letter D) <span>337.2 g</span>
Answer:
Explanation:
Hello there!
In this case, according to the described chemical reaction, Cl2 replaces iodine in NaI in order to produce I2 and NaCl:
It is possible to realize how chlorine replaces iodine in agreement with the single displacement reaction. Moreover, since chlorine and iodine atoms are not correctly balanced, we add a 2 in front of both NaI and NaCl in order to do so:
Best regards!