Answer: False
Explanation:
Molecular formula is the chemical formula which depicts the actual number of atoms of each element present in the compound.
Empirical formula is the simplest chemical formula which depicts the whole number of atoms of each element present in the compound.
Example: 
has similar molecular formula and empirical formula as the elements are already present in simplest of the ratios.
has molecular formula of but 
as the empirical formula.
The crust is the outside layer and also the thinnest.
<span>Answer: 0.00649M
The question is incomplete,
</span>
<span>You are told that the first ionization of the sulfuric acid is complete and the second ionization of the sulfuric acid has a constant Ka₂ = 0.012
</span>
<span>
With that you can solve the question following these steps"
</span>
<span>1) First ionization:
</span>
<span>
H₂SO₄(aq) --> H⁺ (aq) + HSO₄⁻ (aq)
Under the fully ionization assumption the concentration of HSO4- is the same of the acid = 0.01 M
2) Second ionization
</span>
<span>HSO₄⁻ (aq) ⇄ H⁺ + SO₄²⁻ with a Ka₂ = 0.012
</span>
<span>Do the mass balance:
</span>
<span><span> HSO₄⁻ (aq) H⁺ SO₄²⁻</span>
</span>
<span /><span /><span> 0.01 M - x x x
</span><span>Ka₂ = [H⁺] [SO₄²⁻] / [HSO₄⁻]</span>
<span /><span>
=> Ka₂ = (x²) / (0.01 - x) = 0.012
</span><span />
<span>3) Solve the equation:
</span><span>x² = 0.012(0.01 - x) = 0.00012 - 0.012x</span>
<span /><span>
x² + 0.012x - 0.0012 = 0
</span><span />
<span>Using the quadratic formula: x = 0.00649
</span><span />
<span>So, the requested concentratioN is [SO₄²⁻] = 0.00649M</span>
Zero order are reactions in which concentration of reactant has NO effect on RATE OF REACTION.
2. First order are reactions in which concentration of one reactant is proportionate to the RATE OF REACTION.
Exp: That means when you increase the concentration of the one reactant, then the rate of reaction will increase by the same degree of extent.
3. Second order are reactions in which concentration of two reactant has an effect on the RATE OF REACTION.
Formula:
1. Zero order Rate = k
2. First order Rate = k(A)^m
3. Second order Rate = k(A)^m(B)^n
where () represents concentration
and equation is mA + nB -> Product.
Answer: C) 0.020 m
Explanation:
Molality of a solution is defined as the number of moles of solute dissolved per kg of the solvent.
where,
n = moles of solute
= weight of solvent in g
Mole fraction of 
is = 
i.e. moles of is present in 1 mole of solution.
Moles of solute 
=
moles of solvent (water) = 1 - = 0.99
weight of solvent =
Molality =
Thus approximate molality of in this solution is 0.020 m
