The empirical formula is K₂O.
The empirical formula is the <em>simplest whole-number ratio</em> of atoms in a compound.
The <em>ratio of atom</em>s is the same as the <em>ratio of moles</em>.
So, our job is to calculate the <em>molar ratio</em> of K to O.
Step 1. Calculate the <em>moles of each element
</em>
Moles of K = 32.1 g K × (1 mol K/(39.10 g K =) = 0.8210 mol K
Moles of O = 6.57 g O × (1 mol O/16.00 g O) = 0.4106 mol 0
Step 2. Calculate the <em>molar ratio of each elemen</em>t
Divide each number by the smallest number of moles and round off to an integer
K:O = 0.8210:0.4106 = 1.999:1 ≈ 2:1
Step 3: Write the <em>empirical formula
</em>
EF = K₂O
Answer: The concentration of the OH-, CB = 0.473 M.
Explanation:
The balanced equation of reaction is:
2HCl + Ca(OH)2 ===> CaCl2 + 2H2O
Using titration equation of formula
CAVA/CBVB = NA/NB
Where NA is the number of mole of acid = 2 (from the balanced equation of reaction)
NB is the number of mole of base = 1 (from the balanced equation of reaction)
CA is the concentration of acid = 1M
CB is the concentration of base = to be calculated
VA is the volume of acid = 23.65 ml
VB is the volume of base = 25mL
Substituting
1×23.65/CB×25 = 2/1
Therefore CB =1×23.65×1/25×2
CB = 0.473 M.
The answer is C, a catalyst. This is due to the fact that catalysts are substances that can be used to speed up a reaction by lowering the activation energy required. However, the catalyst itself is not actually used up during the reaction.
Answer:
K = [H2] [CO] / [HCHO]
Explanation:
HCHO(g) ⇌ H2(g) + CO(g)
We can obtain the expression for the equilibrium constant for the above equation as follow:
Equilibrium constant, K for a given reaction is the ratio of the concentration of the product raised to their coefficient to the concentration of the reactant raised to their coefficient.
Thus, the equilibrium constant, K for the above equation can be written as follow:
K = [H2] [CO] / [HCHO]
