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Elena-2011 [213]
3 years ago
5

If a weak monoprotic acid, HA, is titrated with strong base, what is NOT true at the half equivalence point?pH = pKapH = 7.0[HA]

= [A-]Buffer capacity is maximized
Mathematics
1 answer:
leonid [27]3 years ago
6 0

Answer:

pH can be different from 7 ( second statement is false)

Step-by-step explanation:

since

pH=pKa + log [A-]/[HA]

at half of the equivalence point

[HA]= [A-] (third statement is true)

and therefore

pH = pKa (first statement is true)

since pH = pKa , pH can be different from 7 ( second statement is false)

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marysya [2.9K]
I'd start by writing an equation for each of the right triangles. (Pythagorean theorem)

y² + 9² = z²
x² + z² = (4+9)²
4² + y² = x²

we want to find z so combine the equations by substituting the other variables x,y out.

substitute y² for (x² - 4²) in 1st equation.

(x² - 4²) + 9² = z²

now by rearranging the 2nd equation we can substitute x² for (13² - z²)

(13² - z²) - 4² + 9² = z²
169 - z² - 16 + 81 = z²
234 - z² = z²
234 = 2z²
234/2 = z²
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√(117) = z
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3 years ago
How much water can be contained by a cylindrical tin can with a radius of 15m and a height of 268m if its 5/6 full?
Law Incorporation [45]
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5 0
2 years ago
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AveGali [126]

The simplified expression of  $4 x \cdot \frac{1}{x^{-5}} \cdot x^{-3} is 4x^3.

Solution:

Given expression is 4 x \cdot \frac{1}{x^{-5}} \cdot x^{-3}.

To simplify the given expression.

$\Rightarrow4 x \cdot \frac{1}{x^{-5}} \cdot x^{-3}

Using exponent rule: \frac{1}{a^{-m}}=a^m

$\Rightarrow4x \cdot x^{5}.x^{-3}

Using another exponent rule: a^m \cdot a^n=a^{m+n}

$\Rightarrow4x \cdot x^{5-3}

$\Rightarrow4x \cdot x^{2}

Using exponent rule: a^m \cdot a^n=a^{m+n}

$\Rightarrow4x^{1+2}

$\Rightarrow4x^3

Hence the simplified expression of  $4 x \cdot \frac{1}{x^{-5}} \cdot x^{-3} is 4x^3.

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3 years ago
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SashulF [63]
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Answer:

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Step-by-step explanation:

Calculate, rewrite, remove the paretheses

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