Rutherford's gold foil experiment proved that there was a small, dense, positively charged nucleus at the center, which contained most of the mass of the atom. Which contained electrons orbiting the nucleus.
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Hope this helps you take off anything you don't want or anything you don't need
Answer:
A.) 4.0
Explanation:
The general equilibrium expression looks like this:
![K = \frac{[C]^{c} [D]^{d} }{[A]^{a} [B]^{b} }](https://tex.z-dn.net/?f=K%20%3D%20%5Cfrac%7B%5BC%5D%5E%7Bc%7D%20%5BD%5D%5E%7Bd%7D%20%7D%7B%5BA%5D%5E%7Ba%7D%20%5BB%5D%5E%7Bb%7D%20%7D)
In this expression,
-----> K = equilibrium constant
-----> uppercase letters = molarity
-----> lowercase letters = balanced equation coefficients
In this case, the molarity's do not need to be raised to any numbers because the coefficients in the balanced equation are all 1. You can find the constant by plugging the given molarities into the equation and simplifying.
<----- Equilibrium expression
<----- Insert molarities
<----- Multiply
<----- Divide
Given:
Density = 0.7360 g/L.
Pressure = 0.5073 atm.
Step 2
The mathematical expression of an ideal gas is,
Chemistry homework question answer, step 2, image 1
Step 3
Here, R is the universal gas constant (0.0821 L-atm/mol K), T is the temperature in Kelvin, and n is the number of
First, find the number of moles of UF6
Avagadro's number = 6.023 x 10^23
Number of moles = 8.0 x 10^26 / Avagadro's number = 8.0 x 10^26 / 6.023 x 10^23 = 1.328 x 10³ moles
Molecular weight of UF6 = Molecular weight of U (238.02891) + Molecular weight of F6 (6 x 18.9984032) = 238.02891 + 113.9904192 = 352.0193292 g/mol
Therefore mass of 8.0 x 10^26 UF6 molecules = 352.0193292 g/mol x 1.328 x 10³ moles = 467.481669 x 10³ grams