Hello!
First you need to calculate q
<span>delta U is change in internal energy </span>
<span>delta U = q + w </span>
<span>q is heat and w work done </span>
<span>here work was done by the system means energy leaving the system so w is negative </span>
<span>delta U = q + w </span>
<span>q = delta U - w = 6865 J - (-346 J) = 7211 J = 7.211 KJ </span>
<span>q = m x c x delta T </span>
<span>7211 J = 80.0 g x c x (225-25) °C </span>
<span>c = 0.451 J /g °C
</span>
Hope this Helps! Have A Wonderful Day! :)
8/5lit.. of 12M NaOH
2/5lit.. of 2M NaOH
Answer:
Percent yield = 89.1%
Explanation:
Based on the equation:
Cl₂ + 2KI → 2KCl + I₂
<em>1 mole of Cl₂ reacts with 2 moles of KI to produce to moles of KCl</em>
<em />
To solve this quesiton we must find the moles of each reactant in order to find the limiting reactant. With the limiting reactant we can find the moles of KCl and the mass:
<em>Moles Cl₂:</em>
8x10²⁵ molecules * (1mol / 6.022x10²³ molecules) = 133 moles
<em>Moles KI -Molar mass: 166.0028g/mol-</em>
25g * (1mol / 166.0028g) = 0.15 moles
Here, clarely, the KI is the limiting reactant
As 2 moles of KI produce 2 moles of KCl, the moles of KCl produced are 0.15 moles. The theoretical mass is:
0.15 moles * (74.5513g / mol) =
11.2g KCl
Percent yield is: Actual yield (10.0g) / Theoretical yield (11.2g) * 100
<h3>Percent yield = 89.1%</h3>
I think you mean ionic bond and not conic bond. The pair of elements that form an ionic bond are barium and chlorine.
