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3 years ago

7

What is the hydronium ion (H3O+) concentration in milk if it has a pH of 8.2?

1 answer:

valkas [14]3 years ago

5 0

The hydronium ion is H₃O⁺ and its concentration is equivalent to the H⁺ ion concentration. The H⁺ ion concentration is given by:

pH = -log[H⁺]
[H⁺] = 10^(-8.2)
[H⁺] = 6.31 x 10⁻⁹ mole/dm³

The hydronium ion concentration is 6.31 x 10⁻⁹ mole/dm³

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Sara is mong a call. One of the finie piece she is using is 1 inches 3

tester [92]

Answer:

What are you even asking?

3 0

2 years ago

Find the mass of the lamina that occupies the region D = {(x, y) : 0 ≤ x ≤ 1, 0 ≤ y ≤ 1} with the density function ρ(x, y) = xye

Alona [7]

Answer:

The mass of the lamina is 1

Step-by-step explanation:

Let $\rho(x,y)$ be a continuous density function of a lamina in the plane region D,then the mass of the lamina is given by:

$m=\int\limits \int\limits_D \rho(x,y) \, dA$ .

From the question, the given density function is $\rho (x,y)=xye^{x+y}$ .

Again, the lamina occupies a rectangular region: D={(x, y) : 0 ≤ x ≤ 1, 0 ≤ y ≤ 1}.

The mass of the lamina can be found by evaluating the double integral:

$I=\int\limits^1_0\int\limits^1_0xye^{x+y}dydx$ .

Since D is a rectangular region, we can apply Fubini's Theorem to get:

$I=\int\limits^1_0(\int\limits^1_0xye^{x+y}dy)dx$ .

Let the inner integral be: $I_0=\int\limits^1_0xye^{x+y}dy$ , then

$I=\int\limits^1_0(I_0)dx$ .

The inner integral is evaluated using integration by parts.

Let $u=xy$ , the partial derivative of u wrt y is

$\implies du=xdy$

and

$dv=\int\limits e^{x+y} dy$ , integrating wrt y, we obtain

$v=\int\limits e^{x+y}$

Recall the integration by parts formula: $\int\limits udv=uv- \int\limits vdu$

This implies that:

$\int\limits xye^{x+y}dy=xye^{x+y}-\int\limits e^{x+y}\cdot xdy$

$\int\limits xye^{x+y}dy=xye^{x+y}-xe^{x+y}$

$I_0=\int\limits^1_0 xye^{x+y}dy$

We substitute the limits of integration and evaluate to get:

$I_0=xe^x$

This implies that:

$I=\int\limits^1_0(xe^x)dx$ .

Or

$I=\int\limits^1_0xe^xdx$ .

We again apply integration by parts formula to get:

$\int\limits xe^xdx=e^x(x-1)$ .

$I=\int\limits^1_0xe^xdx=e^1(1-1)-e^0(0-1)$ .

$I=\int\limits^1_0xe^xdx=0-1(0-1)$ .

$I=\int\limits^1_0xe^xdx=0-1(-1)=1$ .

No unit is given, therefore the mass of the lamina is 1.

3 0

3 years ago

If x^(2) - 2 = 2^(2/3) + 2(-2/3) prove that x = 2^(1/3)+2^(-1/3)

dusya [7]

Step-by-step explanation:

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(2^(-1/3))² (It is in the form of a²+2ab+b²)

x² = (2^(1/3) + 2^(-1/3))²

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3 years ago

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aniked [119]

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3 years ago

4over 5 divided by 1 over 10

MAVERICK [17]

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3 years ago