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! :)

Explanation:
Sodium hydroxide completely ionizes in water to produce sodium ions and hydroxide ions. Hydroxide ions are in excess and neutralize all acetic acid added by the following ionic equation:

The mixture would contain
if
undergoes no hydrolysis; the solution is of volume
after the mixing. The two species would thus be of concentration
and
, respectively.
Construct a RICE table for the hydrolysis of
under a basic aqueous environment (with a negligible hydronium concentration.)

The question supplied the <em>acid</em> dissociation constant
for acetic acid
; however, calculating the hydrolysis equilibrium taking place in this basic mixture requires the <em>base</em> dissociation constant
for its conjugate base,
. The following relationship relates the two quantities:

... where the water self-ionization constant
under standard conditions. Thus
. By the definition of
:
![[\text{HAc} (aq)] \cdot [\text{OH}^{-} (aq)] / [\text{Ac}^{-} (aq) ] = K_b = 10^{-pK_{b}}](https://tex.z-dn.net/?f=%20%5B%5Ctext%7BHAc%7D%20%28aq%29%5D%20%5Ccdot%20%5B%5Ctext%7BOH%7D%5E%7B-%7D%20%28aq%29%5D%20%2F%20%5B%5Ctext%7BAc%7D%5E%7B-%7D%20%28aq%29%20%5D%20%3D%20K_b%20%3D%20%2010%5E%7B-pK_%7Bb%7D%7D%20)


![[\text{OH}^{-}] = 0.30 +x \approx 0.30 \; \text{M}](https://tex.z-dn.net/?f=%20%5B%5Ctext%7BOH%7D%5E%7B-%7D%5D%20%3D%200.30%20%2Bx%20%5Capprox%200.30%20%5C%3B%20%5Ctext%7BM%7D%20)
![pH = pK_{w} - pOH = 14 + \text{log}_{10}[\text{OH}^{-}] = 14 + \text{log}_{10}{0.30} = 13.5](https://tex.z-dn.net/?f=%20pH%20%3D%20pK_%7Bw%7D%20-%20pOH%20%3D%2014%20%2B%20%5Ctext%7Blog%7D_%7B10%7D%5B%5Ctext%7BOH%7D%5E%7B-%7D%5D%20%3D%2014%20%2B%20%5Ctext%7Blog%7D_%7B10%7D%7B0.30%7D%20%3D%2013.5%20)
There are three perfect squares in a standard die; 1, 2, 4. If there is two standard dies, then the probability of getting a perfect square is 1/3 x 1/3 = 1/9.
There are 4 numbers less than 5 in a standard die, making it 1/4 x 1/4=1/16.
Answer:
3
Explanation:
Because in product 3CuSO4 is present it means 3 SO4 sulphate ions are present
The four ionic species initially in solution are Na⁺, PO₄³⁻, Cr³⁺, and Cl⁻. Since the precipitate is composed of Cr³⁺ and PO₄³⁻ ions, the spectator ions must be Na⁺ and Cl⁻.
The complete ionic equation is 3Na⁺(aq) + PO₄³⁻(aq) + Cr₃⁺(aq) + 3Cl⁻(aq) → 3Na⁺(aq) + 3Cl⁻(aq) + CrPO₄(s).
So the balanced <u>net ionic equation</u> for this reaction would be Cr³⁺(aq) + PO₄³⁻(aq) → CrPO₄(s).