C6H12O6 + 6 O2 --> 6 CO2 + 6 H2O
Answer
The empirical formula is CrO₂Cl₂
Explanation:
Empirical formula is the simplest whole number ratio of an atom present in a compound.
The compound contain, Chromium=33.6%
Chlorine=45.8%
Oxygen=20.6%
And the molar mass of Chromium(Cr)=51.996 g mol.
Chlorine containing molar mass (Cl)= 35.45 g mol.
Oxygen containing molar mass (O)=15.999 g mol.
Step-1
Then,we will get,
Cr=
mol
Cl=
mol.
O=
mol.
Step-2
Divide the mole value with the smallest number of mole, we will get,
Cr=

Cl=

O=

Then, the empirical formula of the compound is CrO₂Cl₂ (Chromyl chloride)
The equilibrium constant of the reaction is represented by the symbol K. Thus, option C is the correct and accurate statement about the equilibrium constant.
<h3>What is the equilibrium constant?</h3>
The equilibrium constant is a representation of the concentration of the products and the reactants of the reaction that is raised to the powers through their stoichiometry coefficient.
Its value varies and changes at different temperatures and is not always less than 1. The equilibrium constant is the ratio of the coefficient of the products to reactants.
Therefore, option C. equilibrium constant is represented by K is true.
Learn more about the equilibrium constant here:
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First you calculate the pOH of the solution:
pH+ pOH = 14
3.25 + pOH = 14
pOH = 14 - 3.25
pOH = 10.75
<span>Concentration of [OH]</span>⁻<span> in solution:
</span>
[ OH⁻ ] =

[ OH⁻ ] = 10^ - 10.75
[OH⁻] = 1.778 x 10⁻¹¹ Mhope this helps !
Answer: 104 g
Explanation: reaction Cr2O3 + 3 H2 ⇒ 2 Cr + 3 H2O
M(Cr2O3) = 150 g/mol, so n = m/M = 1.0 mol
Number of moles of H2 should be 3.0 moles and
It is much greater (150 g / 2.016 g/mol)
1 mol Cr2O3 produces 2 mol Cr.
Mass m= 2.0 mol· 52g/mol= 104 g