Answer:
A balanced equation is an equation for a chemical reaction in which the number of atoms for each element in the reaction and the total charge is the same for both the reactants and the products. In other words, the mass and the charge are balanced on both sides of the reaction.
Explanation:
Answer: PV = nRT
A gas at STP... This means that the temperature is 0°C and pressure is 1 atm.
R is the gas constant which is 0.08206 L*atm/(K*mol)
Rearranging for volume
V = nRT/P
The temperature and number of moles are held constant. This means that this uses Boyle's Law. (The ideal gas law could be manipulated to give us this result when T and n are held constant.)
PV = k
where k is a constant.
This means that
P₁V₁ = k = P₂V₂
P₁V₁ = P₂V₂
(1 atm) * (1 L) = (2 atm) * V₂
V₂ = 0.5 L
The new volume of the gas is 0.5 L.
Explanation:
Hot air rising and cooler air falling.
1) Calculate the number of moles of O2 (g) in 300 cm^3 of gas at 298 k and 1 atm
Ideal gas equation: pV = nRT => n = pV / RT
R = 0.0821 atm*liter/K*mol
V = 300 cm^3 = 0.300 liter
T = 298 K
p = 1 atm
=> n = 1 atm * 0.300 liter / [ (0.0821 atm*liter /K*mol) * 298K] = 0.01226 mol
2) The reaction of a metal with O2(g) to form an ionic compound (with O2- ions) is of the type
X (+) + O2 (g) ---> X2O or
2 X(2+) + O2(g) ----> X2O2 = 2XO or
4X(3+) + 3O2(g) ---> 2X2O3
In the first case, 1 mol of metal react with 1 mol of O2(g); in the second case, 2 moles of metal react with 1 mol of O2(g); in the third, 4 moles of X react with 3 moles of O2(g)
So, lets probe those 3 cases.
3) Case 1: 1 mol of metal X / 1 mol O2(g) = x moles / 0.01226 mol
=> x = 0.01226 moles of metal X
Now you can calculate the atomic mass of the hypotethical metal:
1.15 grams / 0.01226 mol = 93.8 g / mol
That does not correspond to any of the metal with valence 1+
So, now probe the case 2.
4) Case 2:
2moles X metal / 1 mol O2(g) = x / 0.01226 mol
=> x = 2 * 0.01226 = 0.02452 mol
And the atomic mass of the metal is: 1.15 g / 0.02452 mol = 46.9 g/mol
That is similar to the atomic mass of titanium which is 47.9 g / mol and whose valece is 2+.
4) Case 3
4 mol meta X / 3 mol O2 = x / 0.01226 => x = 0.01226 * 4 / 3 = 0.01635
atomic mass = 1.15 g / 0.01635 mol = 70.33 g/mol
That does not correspond to any metal.
Conclusion: the identity of the metallic element could be titanium.
Answer:
P = 1/8
Explanation:
The wave function of a particle in a one-dimensional box is given by:
Hence, the probability of finding the particle in the one-dimensional box is:
Evaluating the above integral from x₁ = 0 to x₂ = L/8 and solving it, we have:
![P = \frac{2}{L} [\frac{L}{16} (1 - 4\frac{sin(\frac{n \pi}{4})}{n \pi})]](https://tex.z-dn.net/?f=%20P%20%3D%20%5Cfrac%7B2%7D%7BL%7D%20%5B%5Cfrac%7BL%7D%7B16%7D%20%281%20-%204%5Cfrac%7Bsin%28%5Cfrac%7Bn%20%5Cpi%7D%7B4%7D%29%7D%7Bn%20%5Cpi%7D%29%5D%20)
Solving for n=4:
I hope it helps you!
