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!
Answer:
13.1dm³
Explanation:
Given parameters:
Pressure of gas = 880mmHg
760mmHg = 1 atm
880mmHg will give 1.16atm
Temperature = 303k
Mass of gas = 26.9g
Molar mass of CO₂ = 12 + 2(32) = 44g/mol
Number of moles = mass/molar mass = 26.9/44 = 0.61mole
Unknown:
Volume of the sample = ?
Solution:
To solve this problem, we use the ideal gas equation:
PV = nRT
P is the pressure
V is the volume
n is the number of moles
R is the gas constant = 0.082atmdm³mol⁻¹k⁻¹
T is the temperature
V = 
= 
= 13.1dm³
Answer:
There are exactly 5
Explanation:
IF YOU BREAK IT UP IT WILL BE FIVE
Answer:
38.75 L
Explanation:
From the question,
Applying Boyles Law,
PV = P'V'....................... Equation 1
Where P = Original pressure of the Argon gas, V = Original Volume of Argon gas, P' = Final pressure of Argon gas, V' = Final Volume of Argon gas.
make V the subject of the equation
V = P'V'/P.................... Equation 2
Given: P = 34.6 atm, V' = 456 L, P' = 2.94 atm.
Substitute these values into equation 2
V = (456×2.94)/34.6
V = 38.75 L
Answer:
The pressure changes from 2.13 atm to 1.80 atm.
Explanation:
Given data:
Initial pressure = ?
Final pressure = 1.80 atm
Initial temperature = 86.0°C (86.0 + 273 = 359 K)
Final temperature = 30.0°C (30+273 =303 K)
Solution:
According to Gay-Lussac Law,
The pressure of given amount of a gas is directly proportional to its temperature at constant volume and number of moles.
Mathematical relationship:
P₁/T₁ = P₂/T₂
Now we will put the values in formula:
P₁ = P₂T₁ /T₂
P₁ = 1.80 atm × 359 K / 303 K
P₁ = 646.2 atm. K /303 K
P₁ = 2.13 atm
The pressure changes from 2.13 atm to 1.80 atm.
