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

alkaline metals are lithium, sodium, rubidium, potassium, cesium and francium.

alkali metals have electronegativity from 0.7 to 1, lowest in periodic table of elements, which means that alkali metals (i group in periodic table) has positive oxidation number in compounds.

electronegativity is a chemical property that describes the tendency of an atom to attract a shared pair of electrons towards itself.

alkaline metals (far left in main group) have lowest ionizations energy and easy remove valence electrons,

for example, potassium is the alkali metal and has a single valence electron in the outer electron shell.

electron configuration of potassium is: ₁₉k 1s²2s²2p⁶3s²3p⁶4s¹.

they can not exist alone in nature, because they are very reactive.

so here we go

atomic mass

239 + 1 = y + 134 + 3 combine like terms

240 = y + 137             subtract 137 from both terms

240 - 137 = y

103 = y

atomic number

94 = z +54 + 0

94 = z + 54               subtract 54 from both sides.

40 = z

answer