Answer: 234.4K
Explanation:
Given that,
Original volume of gas (V1) = 5.00 L
Original temperature of gas (T1) = 20.0°C
[Convert 20.0°C to Kelvin by adding 273
20.0°C + 273 = 293K]
New volume of gas (V2) = 4.0L
New temperature of gas (T2) = ?
Since volume and temperature are given while pressure is held constant, apply the formula for Charle's law
V1/T1 = V2/T2
5.00L/293K = 4.0L/T2
To get the value of T2, cross multiply
5.00L x T2 = 293K x 4.0L
5.00L•T2 = 1172L•K
Divide both sides by 5.00L
5.00L•T2/5.00L = 1172L•K/5.00L
T2 = 234.4K
Thus, the new temperature of the gas is 234.4 Kelvin
The correct formula of the oxide that forms when X burns in oxygen is X2O3.
Ionization energy is defined as the energy required to remove an electron from an atom. There are as many ionization energies present in an atom as there are electrons in that atom.
However, we can know the ionization energy values that belong to electrons in the outermost shell because they lie close together. If we go further into the inner shells, there is a sudden quantum jump in ionization energy values.
The element X must have three valence electrons because 631 ,1235, 2389 all refer to ionization energies of electrons in the valence shell. As we get into a core shell, there is a sudden jump hence the fourth and fifth ionization energies are 7089 and 8844 respectively.
The correct formula of the oxide that forms when X burns in oxygen is X2O3.
Learn more: brainly.com/question/16243729
It seems that you have missed the necessary options for us to answer this question, but anyway here is the answer. William adds two values, following the rules for using significant figures in computations. He should write the sum of these two numbers by using the same number of decimal places as the least precise value. Hope this helps.
Answer:
Explanation:
Hello,
Clausius Clapeyron equation is suitable in this case, since it allows us to relate the P,T,V behavior along the described melting process and the associated energy change. Such equation is:
As both the enthalpy and volume do not change with neither the temperature nor the pressure for melting processes, its integration turns out:
Solving for the enthalpy of fusion we obtain:
Finally the entropy of fusion is given by:
Best regards.
