Answer:
H2 P4 O1. Explanation: In order to calculate the Empirical formula , we will assume that we have started with 10 g of the compound.
Explanation:
Answer:
[HClO₄] = 11.7M
Explanation:
First of all we need to know, that a weight percent represents, the mass of solute in 100 g of solution.
Let's convert the mass to moles → 70.5 g . 1mol/100.45 g = 0.702 moles
Now we can apply the density to calculate the volume.
Density always refers to solution → Solution density = Solution mass / Solution volume
1.67 g/mL = 100 g / Solution volume
Solution volume = 100 g / 1.67 g/mL → 59.8 mL
To determine molarity (mol/L) we must convert the mL to L
59.8 mL . 1L/1000mL = 0.0598 L
Molarity → Moles of solute in 1L of solution → 0.702 mol / 0.0598 L = 11.7M
First, we assume that helium behaves as an ideal gas such that the ideal gas law is applicable.
PV = nRT
where P is pressure, V is volume, n is number of moles, R is universal gas constant, and T is temperature. From the equation, if n, R, and T are constant, there is an inverse relationship between P and V. From the given choices, the container with the greatest pressure would be the 50 mL.
Answer:
1.18 V
Explanation:
The given cell is:
Half reactions for the given cell follows:
Oxidation half reaction: 
Reduction half reaction: 
Multiply Oxidation half reaction by 2 and Reduction half reaction by 3
Net reaction: 
Oxidation reaction occurs at anode and reduction reaction occurs at cathode.
To calculate the 
of the reaction, we use the equation:
Putting values in above equation, we get:
To calculate the EMF of the cell, we use the Nernst equation, which is:
![E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[Al^{3+}]^2}{[Fe^{2+}]^3}](https://tex.z-dn.net/?f=E_%7Bcell%7D%3DE%5Eo_%7Bcell%7D-%5Cfrac%7B0.059%7D%7Bn%7D%5Clog%20%5Cfrac%7B%5BAl%5E%7B3%2B%7D%5D%5E2%7D%7B%5BFe%5E%7B2%2B%7D%5D%5E3%7D)
where,
= electrode potential of the cell = ?V
= standard electrode potential of the cell = +1.21 V
n = number of electrons exchanged = 6
Putting values in above equation, we get:
The basis of finding the answer to this problem is to know the electronic configuration of Fluorine. That would be: <span>[He] 2s</span>²<span> 2p</span>⁵. The valence electrons, which are the outermost electrons of the atom, are the ones that participate in bonding. <em>Since the highest orbital for F is 2p, that means the highest energy occupied would be 2.</em>
