Given which are missing in your question:
the flask is filled with 1.45 g of argon at 25 C°
So according to this formula (Partial pressure):
PV= nRT
first, we need n, and we can get by substitution by:
n = 1.45/mass weight of argon
= 1.45 / 39.948 = 0.0363 mol of Ar
we have R constant = 0.0821
and T in kelvin = 25 + 273 = 298
and V = 1 L
∴ P * 1 = 0.0363* 0.0821 * 298 = 0.888 atm
Since X is 1 g, therefore O must be 0.1 g. Therefore:
moles O = 0.1 g / (16 g / mol) = 0.00625 mol
We can see that for every 3 moles of O, there are 2 moles
of X, therefore:
moles X = 0.00625 mol O (3 moles X / 2 moles O) =
0.009375 mol
Molar mass X = 1 g / 0.009375 mol
<span>Molar mass X = 106.67 g/mol</span>
Reaction of sodium with water
Sodium metal reacts rapidly with water to form a colourless solution of sodium hydroxide (NaOH) and hydrogen gas (H2). The resulting solution is basic because of the dissolved hydroxide. The reaction is exothermic. During the reaction, the sodium metal may well become so hot that it catches fire and burns with a characteristic orange colour. The reaction is slower than that of potassium (immediately below sodium in the periodic table), but faster than that of lithium (immediately above sodium in the periodic table).
2Na(s) + 2H2O → 2NaOH(aq) + H2(g)
Answer:
C. 26.4 kJ/mol
Explanation:
The Chen's rule for the calculation of heat of vaporization is shown below:
![\Delta H_v=RT_b\left [ \frac{3.974\left ( \frac{T_b}{T_c} \right )-3.958+1.555lnP_c}{1.07-\left ( \frac{T_b}{T_c} \right )} \right ]](https://tex.z-dn.net/?f=%5CDelta%20H_v%3DRT_b%5Cleft%20%5B%20%5Cfrac%7B3.974%5Cleft%20%28%20%5Cfrac%7BT_b%7D%7BT_c%7D%20%5Cright%20%29-3.958%2B1.555lnP_c%7D%7B1.07-%5Cleft%20%28%20%5Cfrac%7BT_b%7D%7BT_c%7D%20%5Cright%20%29%7D%20%5Cright%20%5D)
Where,
is the Heat of vaoprization (J/mol)
is the normal boiling point of the gas (K)
is the Critical temperature of the gas (K)
is the Critical pressure of the gas (bar)
R is the gas constant (8.314 J/Kmol)
For diethyl ether:
Applying the above equation to find heat of vaporization as:
![\Delta H_v=8.314\times307.4 \left [ \frac{3.974\left ( \frac{307.4}{466.7} \right )-3.958+1.555ln36.4}{1.07-\left ( \frac{307.4}{466.7} \right )} \right ]](https://tex.z-dn.net/?f=%5CDelta%20H_v%3D8.314%5Ctimes307.4%20%5Cleft%20%5B%20%5Cfrac%7B3.974%5Cleft%20%28%20%5Cfrac%7B307.4%7D%7B466.7%7D%20%5Cright%20%29-3.958%2B1.555ln36.4%7D%7B1.07-%5Cleft%20%28%20%5Cfrac%7B307.4%7D%7B466.7%7D%20%5Cright%20%29%7D%20%5Cright%20%5D)
The conversion of J into kJ is shown below:
1 J = 10⁻³ kJ
Thus,
<u>Option C is correct</u>
Answer:
Explanation:
Hello!
In this case, considering the Gay-Lussac's law which describes the pressure-temperature behavior as a directly proportional relationship by holding the volume as constant, we write:
Whereas solving for the final temperature T2, we get:
Thus, we plug in the given data (temperature in Kelvins) to obtain:
Best regards!
