To counter the removal of A the equilibrium change by <u>s</u><em>hifting toward the left</em>
<em> </em><u><em>explanation</em></u>
<u><em> </em></u>If the reaction is at equilibrium and we alter the condition a new equilibrium state is created
<u><em> </em></u>The removal of A led to the shift of equilibrium toward the left since it led to less molecules in reactant side which favor the backward reaction.( equilibrium shift to the left)
The chemical reaction would be written as
2 AsF3<span> + 3 CCl4 = 2 AsCl3 + 3 CCl2F2
</span>
We use the given amounts of the reactants to first find the limiting reactant. Then use the amount of the limiting reactant to proceed to further calculations.
150 g AsF3 ( 1 mol / 131.92 g) = 1.14 mol AsF3
180 g CCl4 (1 mol / 153.82 g) = 1.17 mol CCl4
Therefore, the limiting reactant would be CCl4 since it would be consumed completely. The theoretical yield would be:
1.17 mol CCl4 ( 3 mol CCl2F2 / 3 mol CCl4 ) = 1.17 mol CCl2F2
Answer:
Cl₂O₇
Explanation:
For the reaction:
ClₓOₙ + H₂ → HCl + H₂O
Moles of HCl and moles of H₂O are:
HCl: 0.233g HCl ₓ (1mol / 36.46g) = 6.39x10⁻³ mol HCl
H₂O: 0.403g H₂O ₓ (1mol / 18.02g) = 2.236x10⁻² mol H₂O
As you can see, moles of HCl are equivalent to moles of Cl in the compound and moles of H₂O are equivalent to moles of O in the compound, that means:
6.39x10⁻³ mol Cl
2.236x10⁻² mol O
Empirical formula is the simplest ratio of atoms presents in a molecule. If Cl is <em>1</em>, Oxygen will be:
2.236x10⁻² mol / 6.39x10⁻³ = <em>3.5</em>
As empirical formula must be given in natural numbers, the empirical formula is:
<em>Cl₂O₇</em>
<em></em>
Answer: The concentration of C29H60 in nM per liter is 83,33 nM/liter
Explanation: Let's start from the ppb definition: ppb means parts per billion. In terms of concentracion measuring this means micrograms of solute per liter of solution.
The algebraic expression would be:
<em>ppb [=] micrograms of compound/liter of solution</em>
We can assume that the solvent is water. The solute is dissolved in water and both create the C29H60 solution.
For the exercise we have 34 ppb of C29H60, that means 34 micrograms of C29H60 in one liter of solution. So, since now, we have to convert the units from the initial data to the required answer.
The respective procedure is in a attached file.
Given:
7.20 g sample of Al2(SO4)3
Required:
Mass of oxygen
Solution:
Since you are not given a
chemical reaction, just base your solution to the chemical formula given.
Molar mass of Al2(SO4)3 = 342.15 g/mol
7.20 g Al2(SO4)3 (1 mol/342.15g)(3mol O/2 mol Al)(1 mol O2/1/2 mol
O2)(32g O2/1mol O2) = 4.04 g O2