<h2>
Hello!</h2>
The answer is:
The empirical formula is the option B. 
<h2>
Why?</h2>
The empirical formula of a compound is the simplest formula that can be written. On the opposite, the molecular formula involves a variant of the same compound, but it can be also simplified to an empirical formula.

We are looking for a formula that cannot be simplified by dividing the number of molecules/atoms that conforms the compound.
Let's discard option by option in order to find which formula is an empirical formula (cannot be simplified)
A. 
It's not an empirical formula, it's a molecular formula since it can be obtained by multiplying the empirical formula of the same compound.

B. 
It's an empirical formula since it cannot be obtained by the multiplication of a whole number and the simplest formula. It's the simplest formula that we can find of the compound.
C. 
It's not an empirical formula, it's a molecular formula since it can be obtained by multiplying the empirical formula of the same compound.

D. 
It's not an empirical formula, it's a molecular formula since it can be obtained by multiplying the empirical formula of the same compound.

Hence, the empirical formula is the option B. 
Have a nice day!
Raised temperature, decreased volume.
Temperature and Pressure are directly related, when volume increases so does the your pressure.
Volume and Pressure are indirectly related. When volume decreases, your pressure will increase.
Answer:
2M
Explanation:
M=mol/L
1. Find moles of CoCl2
mass of substance/molar mass = 130/129.833 = 1.001 mol
3. Substitute in molarity equation
M=(1.001/0.5)
M= around 2M
<span>Answer:
Nothing is balanced in your final equation: not H, not O, not Cr, not I and your charges aren't either.
Start with your 2 half reactions:
I- --> IO3-
Cr2O72- --> 2 Cr3+
Balance O by adding H2O:
I- + 3 H2O --> IO3-
Cr2O72- --> 2 Cr3+ + 7H2O
Balance H by adding H+:
I- + 3 H2O --> IO3- + 6 H+
Cr2O72- + 14 H+ --> 2 Cr3+ + 7H2O
Balance charge by adding e-:
I- + 3 H2O --> IO3- + 6 H+ + 6 e-
Cr2O72- + 14 H+ + 6 e- --> 2 Cr3+ + 7H2O
Since the numbers of electrons in your two half reactions are the same, just add them and simplify to give:
Cr2O72- + I- + 8 H+ --> IO3- + 2 Cr3+ + 4 H2O</span>