This is false because it doesn't matter about the mass
Answer:
Polyhydroxyl alcohols
Explanation:
Whenever we have several C-OH bonds, we have a polyhydroxyl alcohol. For example, if we have just one alcohol group, that is, an R-OH group, then the naming is simple, say, we have EtOH, it's ethanol.
The problem becomes more complicated when we have several hydroxyl groups present in the alcohol. Let's say we have an ethane molecule and we replace the hydrogen atoms of carbon 1 and 2 with hydroxyl groups. In that case, we have 1,2-ethanediol. Similarly, we can have triols etc.
That said, we have poly (several) hydroxyl groups and we can generalize this to having polyhydroxyl alcohols.
Answer:
Mass of original sample = 100 g
Explanation:
Half life of cesium-137 = 30.17 years
Where, k is rate constant
So,
The rate constant, k = 0.02297 year⁻¹
Time = 90.6 years
Using integrated rate law for first order kinetics as:
![[A_t]=[A_0]e^{-kt}](https://tex.z-dn.net/?f=%5BA_t%5D%3D%5BA_0%5De%5E%7B-kt%7D)
Where,
![[A_t]](https://tex.z-dn.net/?f=%5BA_t%5D)
is the concentration at time t
![[A_0]](https://tex.z-dn.net/?f=%5BA_0%5D)
is the initial concentration
Initial concentration = ?
Final concentration = 12.5 grams
Applying in the above equation, we get that:-
![12.5\ g=[A_0]e^{-0.02297\times 90.6}](https://tex.z-dn.net/?f=12.5%5C%20g%3D%5BA_0%5De%5E%7B-0.02297%5Ctimes%2090.6%7D)
![[A_0]=\frac{12.5}{e^{-0.02297\times 90.6}}\ g=100\ g](https://tex.z-dn.net/?f=%5BA_0%5D%3D%5Cfrac%7B12.5%7D%7Be%5E%7B-0.02297%5Ctimes%2090.6%7D%7D%5C%20g%3D100%5C%20g)
<u>Mass of original sample = 100 g</u>
17.93 grams of oxygen gas occupy 12.3L of space at 109.4 kPa and 15.4°C. Details about how to calculate mass can be found below.
<h3>How to calculate mass?</h3>
The mass of a given gas can be calculated by multiplying the number of moles of the substance by its molar mass.
However, the number of moles of the gas must be calculated first as follows:
PV = nRT
Where;
- P = pressure = 1.0796941atm
- V = volume = 12.3L
- n = number of moles
- T = temperature = 288.4K
- R = gas law constant = 0.0821 Latm/molK
1.079 × 12.3 = n × 0.0821 × 288.4
13.27 = 23.68n
n = 13.27/23.68
n = 0.56mol
Mass = 0.56 × 32
mass of oxygen gas = 17.93g
Therefore, 17.93 grams of oxygen gas occupy 12.3L of space at 109.4 kPa and 15.4°C.
Learn more about mass at: brainly.com/question/19694949
