Answer:
AsF3:C2CI6
4:3
1.3618 moles: 1.02135 moles(1.3618÷4×3)
C2CI6 is the limting reagent
So the number of moles for AsCI3 is 0.817 moles( number of moles of the limting reagant) ÷3 ×4 (according to ratio by balancing chemical equation)=1.09 moles(3 s.f.)
or
Balanced equation
4AsF3 + 3C2Cl6 → 4AsCl3 + 3C2Cl2F4
Use stoichiometry to calculate the moles of AsCl3 that can be produced by each reactant.
Multiply the moles of each reactant by the mole ratio between it and AsCl3 in the balanced equation, so that the moles of the reactant cancel, leaving moles of AsCl3.
Explanation:
Answer:
<h2>2.70 g/cm³</h2>
Explanation:
The density of a substance can be found by using the formula
From the question we have
We have the final answer as
<h3>2.70 g/cm³</h3>
Hope this helps you
Just do it and believe in you self come on man you got it
I believe it’s c because if you really read and close read it will make more sense
