26. of of (aq) by (aq) is . if 50.00 ml of 1.05 m is to 25.00 ml of 1.86 m ,at be? ( no is toina of aof) , h.. (p. ). . .

T₂ = 33.2⁰C

Explanation:

Needed in problem text is the temperature of the acid and base solutions before reaction and the accepted (published) molar heat of neutralization (strong acids) for HCl by NaOH. => Assuming solution temperature is 25⁰C* for both acid and base solutions before reaction and the molar heat of neutralization for HCl by NaOH is 55.7Kj/mole,**  then …

* Standard Thermodynamic conditions => 25⁰C (298K) & 1.00 Atm.

**(https://chemdemos.uoregon.edu/demos/Heat-of-Neutralization-HClaq-NaOHaq)  

NaOH + HCl => NaCl + H₂O

=> 50ml(1.05M NaOH) + 25ml(1.86M HCl)  

=> 0.05(1.05)mole NaOH + 0.025(1.86)mole HCl

=> 0.0525mole NaOH + 0.0465mole HCl

=> (0.0525 – 0.0465)mole NaOH excess + 0.0465mole NaCl + H₂O + Heat

=> 0.0060mole NaOH in excess + 0.0465mole NaCl + H₂O + Heat

Note => NaOH neutralizes 0.0465mole HCl (Limiting Reactant) and produces 0.0465mole NaCl  & H₂O + Heat of Neutralization.

Heat flow (Q) = Heat received by solvent water from the NaOH + HCl reaction  

=> Q = mcΔT = mc(T₂ - T₁) = specific heat produced by 0.0465mole HCl

=> Q(m) = Molar Heat of Neutralization = Q/mole = mcΔT/n

-m = mass of solvent water receiving heat = (50ml + 25ml)1g/ml = 75g

-c = specific heat of water = 4.184j/g⁰C

-T₂ - T₁ = T₂ - 25⁰C

-Q(m) = 55,700 joules/mole (published heat of neutralization)

-n = moles of HCl neutralized = 0.0465mole HCl

=>Q(m) = mcΔT/n = 55,700j/mole = (75g)(4.184j/g⁰C)(T₂ - 25⁰C) /0.0465mole

Solving for T₂ => T₂ = 33.2⁰C

sorry i was late, but i don't know what assignment your talking about, i'm in flvs as well, im in earth space science. but u are behind pace bc every class should be in the 3's or 4'