Probably its atomic number (proton number), because this is how they are arranged already.
Other factors are their form (metal/non-metal) etc., which determines how they react with each other and bond.
Another important factor is the number of electrons in their valent (outer) shell. This is because it tells use how the atom will react/bond with other atoms, how 'violent' that reaction will be, and whether it will bond at all (eg. nobel gases)
Answer:
5.25 g NaF
Step-by-step explanation:
We know we will need a balanced equation with masses and molar masses, so let’s gather all the information in one place.
M_r: 41.99
Na₂SiO₃ + 8HF ⟶ H₂SiF₆ + 2 NaF + 3H₂O
n/mol: 8
(1) Calculate the moles of NaF
The molar ratio is 2 mol NaF/8 mol HF.
Moles of F = 0.500 mol HF × (2 mol Na₂SiO₃/8 mol HF)
Moles of Na₂SiO₃ = 0.125 mol Na₂SiO₃
(2) Calculate the mass of NaF
Mass of NaF = 0.125 mol NaF × (41.99 g Na₂SiO₃/1 mol Na₂SiO₃)
Mass of NaF = 5.25 g NaF
Answer:
D - water changes from a
Explanation:
You're correct!
We are asked what happens when the pressure changes from 0 atm to 10 atm at 0C.
If we look at the graph at 0C on the x axis and at 0 atm on the y axis, we can see we are in the yellow area of the graph, so the water starts at a gas.
If we keep the temperature the same, but change the pressure to 10 atm (much higher on the y axis) we are in the blue area, so the water is now a liquid. Therefore, D is correct
Answer:
a. T and P remain the same (T=298 K and P=1 bar)
b. 11.23J/K
Explanation:
a. Since the mixing process of an idea gas doesn't present a change in the enthalpy, we could state that no change in neither temperature and pressure are given.
b. It is not necessary to know enthalpy data, the following formula is enough to compute the entropy change:
Δ![S_{mix}=-n_{N_2}R ln(x_{N_2})-n_{O_2}R ln(x_{O_2})](https://tex.z-dn.net/?f=S_%7Bmix%7D%3D-n_%7BN_2%7DR%20ln%28x_%7BN_2%7D%29-n_%7BO_2%7DR%20ln%28x_%7BO_2%7D%29)
Thus, the molar fractions are equal to 0.5, and the result yields:
Δ![S_{mix}=-(1mol)[(8.314J/(mol*K)]ln(0.5)-(1mol)[(8.314J/(mol*K)]ln(0.5)](https://tex.z-dn.net/?f=S_%7Bmix%7D%3D-%281mol%29%5B%288.314J%2F%28mol%2AK%29%5Dln%280.5%29-%281mol%29%5B%288.314J%2F%28mol%2AK%29%5Dln%280.5%29)
Δ![S_{mix}=11.23J/K](https://tex.z-dn.net/?f=S_%7Bmix%7D%3D11.23J%2FK)
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