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2 years ago

Una solución tiene un valor de pH=5 Cuál es la concentración molar de OH-?

Mathematics

1 answer:

Mnenie [13.5K]2 years ago

5 0

Answer:

Podemos hacer la conversión entre [\text{H}^+][H

]open bracket, start text, H, end text, start superscript, plus, end superscript, close bracket y \text{pH}pHstart text, p, H, end text mediante las siguientes ecuaciones:

\begin{aligned}\text{pH}&=-\log[\text{H}^+]\\ \\ [\text H^+]&=10^{-\text{pH}}\end{aligned}

]

=−log[H

]

=10

−pH

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Can you guys help me to find the measure of each angle tysm.

barxatty [35]

Answer:

∠EBF = 51°

∠DBE = 17°

∠ABF = 141°

∠EBA = 90°

∠DBC = 107°

∠DBF = 68°

Step-by-step explanation:

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2 years ago

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Which logaritnmic equation equivalent to the exponential equation below 5c=125

nataly862011 [7]

I assume you meant to write 5^c = 125. The equivalent logarithmic expression for this would be ㏒₅125=c.

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3 years ago

Using polar coordinates, evaluate the integral which gives the area which lies in the first quadrant below the line y=5 and betw

vfiekz [6]

First, complete the square in the equation for the second circle to determine its center and radius:

x ² - 10x + y ² = 0

x ² - 10x + 25 + y ² = 25

(x - 5)² + y ² = 5²

So the second circle is centered at (5, 0) with radius 5, while the first circle is centered at the origin with radius √100 = 10.

Now convert each equation into polar coordinates, using

x = r cos(θ)

y = r sin(θ)

Then

x ² + y ² = 100 → r ² = 100 → r = 10

x ² - 10x + y ² = 0 → r ² - 10 r cos(θ) = 0 → r = 10 cos(θ)

y = 5 → r sin(θ) = 5 → r = 5 csc(θ)

See the attached graphic for a plot of the circles and line as well as the bounded region between them. The second circle is tangent to the larger one at the point (10, 0), and is also tangent to y = 5 at the point (0, 5).

Split up the region at 3 angles θ₁, θ₂, and θ₃, which denote the angles θ at which the curves intersect. They are

θ₁ = 0 … … … by solving 10 = 10 cos(θ)

θ₂ = π/6 … … by solving 10 = 5 csc(θ)

θ₃ = 5π/6 … the second solution to 10 = 5 csc(θ)

Then the area of the region is given by a sum of integrals:

$\displaystyle \frac12\left(\left\{\int_0^{\frac\pi6}+\int_{\frac{5\pi}6}^{2\pi}\right\}\left(10^2-(10\cos(\theta))^2\right)\,\mathrm d\theta+\int_{\frac\pi6}^{\frac{5\pi}6}\left((5\csc(\theta))^2-(10\cos(\theta))^2\right)\,\mathrm d\theta\right)$

$=\displaystyle 50\left\{\int_0^{\frac\pi6}+\int_{\frac{5\pi}6}^{2\pi}\right\} \sin^2(\theta)\,\mathrm d\theta+\frac12\int_{\frac\pi6}^{\frac{5\pi}6}\left(25\csc^2(\theta) - 100\cos^2(\theta)\right)\,\mathrm d\theta$

To compute the integrals, use the following identities:

sin²(θ) = (1 - cos(2θ)) / 2

cos²(θ) = (1 + cos(2θ)) / 2

and recall that

d(cot(θ))/dθ = -csc²(θ)

You should end up with an area of

$=\displaystyle25\left(\left\{\int_0^{\frac\pi6}+\int_{\frac{5\pi}6}^{2\pi}\right\}(1-\cos(2\theta))\,\mathrm d\theta-\int_{\frac\pi6}^{\frac{5\pi}6}(1+\cos(2\theta))\,\mathrm d\theta\right)+\frac{25}2\int_{\frac\pi6}^{\frac{5\pi}6}\csc^2(\theta)\,\mathrm d\theta$

$=\boxed{25\sqrt3+\dfrac{125\pi}3}$

We can verify this geometrically:

• the area of the larger circle is 100π

• the area of the smaller circle is 25π

• the area of the circular segment, i.e. the part of the larger circle that is bounded below by the line y = 5, has area 100π/3 - 25√3

Hence the area of the region of interest is

100π - 25π - (100π/3 - 25√3) = 125π/3 + 25√3

as expected.

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2 years ago

If Will gave Molly $9, he would now have the same amount of money as her. If Molly would have given Will $9, the ratio of the mo

Rufina [12.5K]

Answer:

What is the wI-FI PASSWORD

Step-by-step explanation:

Friend me on

(2x+1)-1=0

Divide

2=0

Reitalizie

2=0

Reatomize

⊕∴∵

Reunatomize

2=0

It is not true

2=0

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2 years ago

Hello, please help thank youu!!

uranmaximum [27]

Answer:

36 $\frac{1}{2}$

Step-by-step explanation:

(5 $\frac{1}{2}$ x 3) + (10 ÷ $\frac{1}{2}$ )

(16 $\frac{1}{2}$ ) + (10 ÷ $\frac{1}{2}$ )

(16 $\frac{1}{2}$ ) + (20)

36 $\frac{1}{2}$

7 0

3 years ago

Read 2 more answers

Una solución tiene un valor de pH=5 Cuál es la concentración molar de OH-?​

Una solución tiene un valor de pH=5 Cuál es la concentración molar de OH-?