What kind of substance accepts an electron pair? a. alkali b. acid c. hydroxide ion d. base

b. acid is the correct answer @jayxboof all social media

acid : acids are those which accepts an electron-pairs.

base : bases are those which donates ab electron-pairs.

alkali : it is a base that dissolves in water. it is basically a base and the ph greater than 7.

hydroxide ion : it is a poly-atomic ion which is negatively charged.

to be fair, you have to have a very high iq to understand rick and morty. the humour is extremely subtle, and without a solid grasp of theoretical physics most of the jokes will go over a typical viewer's head. there's also rick's nihilistic outlook, which is deftly woven into his characterisation- his personal philosophy draws heavily from narodnaya volya literature, for instance. the fans understand this stuff; they have the intellectual capacity to truly appreciate the depths of these jokes, to realise that they're not just funny- they say something deep about life. as a consequence people who dislike rick & morty truly are idiots- of course they wouldn't appreciate, for instance, the humour in rick's existential catchphrase "wubba lubba dub dub," which itself is a cryptic reference to turgenev's russian epic fathers and sons. i'm smirking right now just imagining one of those addlepated simpletons scratching their heads in confusion as dan harmon's genius wit unfolds itself on their television screens. what fools.. how i pity them.

and yes, by the way, i do have a rick & morty tattoo. and no, you cannot see it. it's for the ladies' eyes only- and even then they have to demonstrate that they're within 5 iq points of my own (preferably lower) beforehand. nothin personnel kid

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol