The least common multiples of 4, 10, and 16 is 80.
Answer:

Explanation:
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In this case, since the normal boiling point of X is 117.80 °C, the boiling point elevation constant is 1.48 °C*kg*mol⁻¹, the mass of X is 100 g and the boiling point of the mixture of X and KBr boils at 119.3 °C, we can use the following formula:

Whereas the Van't Hoff factor of KBr is 2 as it dissociates into potassium cations and bromide ions; it means that we can compute the molality of the solution:

Next, given the mass of solventin kg (0.1 kg from 100 g), we compute the moles KBr:

Finally, considering the molar mass of KBr (119 g/mol) we compute the mass that was dissolved:

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I can’t answer the question sorry. I don’t know what the equation is
It carries at least one hydroxyl functional group ( i.e -OH ) bound to a saturated carbon atom.
Therefore the functional group present is -OH functional group.
Data:
M (Molar Mass) of C2H6
C = 2*12 = 24 amu
H = 6*1 = 6 amu
---------------------
MM C2H6 = 24+6 = 30 g/mol
P (pressure) = 1.6 atm
V (volume) = 12.7 L
<span>R = 0,082 atm .L/mol.K
n (</span>Number of mols) →
![n = \frac{m}{MM} m (mass) = ? T = 24ºC Celsius to Kelvin TK = TºC + 273 TK = 24 + 273 TK = 297 By the equation of state of the gases or equation of Clapeyron, we have: [tex]P*V = n*R*T](https://tex.z-dn.net/?f=n%20%3D%20%5Cfrac%7Bm%7D%7BMM%7D%20m%20%28mass%29%20%3D%20%3F%20T%20%3D%2024%C2%BAC%20%20Celsius%20to%20Kelvin%20TK%20%3D%20T%C2%BAC%20%2B%20273%20TK%20%3D%2024%20%2B%20273%20TK%20%3D%20297%20%20%20%3Cspan%3EBy%20the%20equation%20of%20state%20of%20the%20gases%20or%20equation%20of%20Clapeyron%2C%20we%20have%3A%0A%20%20%20%3C%2Fspan%3E%5Btex%5DP%2AV%20%3D%20n%2AR%2AT%20)
<span>
Since </span>

<span>, we can perform the following substitution in the above Clapeyron equation:
</span>


multiply cross

Solving:




Answer:
25.03 grams of ethane gas