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konstantin123 [22]

3 years ago

5

Plz help me asap i will die or else :c

2 answers:

Kitty [74]3 years ago

7 0

Answer:

Expow do revolutions occur? In Tunisia, the street vendor Mohamed Bouazizi set himself on fire after his produce cart was confiscated. The injustice of this event provided an emblem for the widespread conditions of poverty, oppression, and humiliation experienced by a majority of the population. In this case, the revolutionary action might be said to have originated in the way that Bouazizi’s act sparked a radicalization in people’s sense of citizenship and power: their internal feelings of individual dignity, rights, and freedom and their capacity to act on them. It was a moment in which, after living through decades of deplorable conditions, people suddenly felt their own power and their own capacity to act. Sociology is interested in studying the conditions of such examples of citizenship and powerlanation:

zhannawk [14.2K]3 years ago

3 0

Answer: give me more of an explaination

Explanation:

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How many moles of Mg are contained in 11.0<br> grams of magnesium?

DanielleElmas [232]

Answer:

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Explanation:

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The isotope 208Tl undergoes Î² decay with a half-life of 3.1 min.

melamori03 [73]

Answer:

a. Pb 208

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c. only a trace amount

Explanation:

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natka813 [3]

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Oxana [17]

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3 years ago

Read 2 more answers

The Ka for formic acid (HCO2H) is 1.8 × 10-4. What is the pH of a 0.35 M aqueous solution of sodium formate (NaHCO2)?

anzhelika [568]

Answer:

9.36

Explanation:

Sodium formate is the conjugate base of formic acid.

Also,

$K_a\times K_b=K_w$

$K_b$ for sodium formate is $K_b=\frac {K_w}{K_a}$

Given that:

$K_a$ of formic acid = $1.8\times 10^{-4}$

And, $K_w=10^{-14}$

So,

$K_b=\frac {10^{-14}}{1.8\times 10^{-4}}$

$K_b=5.5556\times 10^{-11}$

Concentration = 0.35 M

HCOONa ⇒ Na⁺ + HCOO⁻

Consider the ICE take for the formate ion as:

HCOO⁻ + H₂O ⇄ HCOOH + OH⁻

At t=0 0.35 - -

At t =equilibrium (0.35-x) x x

The expression for dissociation constant of sodium formate is:

$K_{b}=\frac {[OH^-][HCOOH]}{[HCOO^-]}$

$5.5556\times 10^{-11}=\frac {x^2}{0.35-x}$

Solving for x, we get:

x = 0.44×10⁻⁵ M

pOH = -log[OH⁻] = -log(0.44×10⁻⁵) = 4.64

pH + pOH = 14

So,

<u>pH = 14 - 4.64 = 9.36</u>

5 0

3 years ago