You must use 1880 mL of O₂ to react with 4.03 g Mg.
A_r: 24.305
2Mg + O₂ ⟶ 2MgO
<em>Moles of Mg</em> = 4.03 g Mg × (1 mol Mg/24.305 g Mg) = 0.1658 mol Mg
<em>Moles of O₂</em> = 0.1658 mol Mg × (1 mol O₂/2 mol Mg) = 0.082 90 mol O₂
STP is 25 °C and 1 bar. At STP, 1 mol of an ideal gas has a volume of <em>22.71 L</em>.
<em>Volume of O₂</em> = 0.082 90 mol O₂ × (22.71 L O₂/1 mol O₂) = 1.88 L = 1880 mL
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Answer:- 0.800 moles of the gas were collected.
Solution:- Volume, temperature and pressure is given for the gas and asks to calculate the moles of the gas.
It is an ideal gas law based problem. Ideal gas law equation is used to solve this. The equation is:
PV=nRT
Since it asks to calculate the moles that is n, so let's rearrange this for n:

V = 19.4 L
T = 17 + 273 = 290 K
P = 746 mmHg
we need to convert the pressure from mmHg to atm and for this we divide by 760 since, 1 atm = 760 mmHg

P = 0.982 atm
R = 
Let's plug in the values in the equation to get the moles.

n = 0.800 moles
So, 0.800 moles of the gas were collected.
The answer for this question would be B) False or the second option.
Answer is FALSE: ✅
Answer:
(3R,4R)-4-bromohexan-3-ol
Explanation:
In this case, we have reaction called <u>halohydrin formation</u>. This is a <u>markovnikov reaction</u> with <u>anti configuration</u>. Therefore the halogen in this case "Br" and the "OH" must have <u>different configurations</u>. Additionally, in this molecule both carbons have the <u>same substitution</u>, so the "OH" can go in any carbon.
Finally, in the product we will have <u>chiral carbons</u>, so we have to find the absolute configuration for each carbon. On carbon 3 we will have an "R" configuration on carbon 4 we will have also an "R" configuration. (See figure 1)
I hope it helps!