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S_A_V [24]
2 years ago
11

The pH of a fruit juice is 5.6. Find the hydronium ion concentration, [H3O+], of the juice. Use the formula pH=-log[H3O+].

Mathematics
1 answer:
Roman55 [17]2 years ago
7 0

The hydroxonium ion concentration of the fruit juice whose pH as given in the task content is; 5.6 can be evaluated as; [H3O+] = 2.51 × 10^-6.

<h3>What is the hydroxonium ion concentration of the fruit juice whose pH is 5.6?</h3>

Since the given pH of the fruit juice is 5.6, the hydroxonium ion concentration of the fruit juice can be evaluated by means of the formula as follows;

pH=-log[H3O+].

5.6 = -log[H3O+].

-5.6 = log[H3O+].

[H3O+] = antilogarithm (-5.6)

[H3O+] = 2.51 × 10^-6.

Read more on logarithm;

brainly.com/question/247340

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∠A = ∠B = 80°

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Which equations represent the line that is perpendicular to the line 5x − 2y = −6 and passes through the point (5, −4)? Check al
Alla [95]

Let's rewrite each equation in the Slope-Intercept Form of the Equation of a Line. First, let's start with the main equation:


\bullet \ 5x-2y=-6 \therefore y=\frac{5}{2}x+3


Then, our options are the following:

A) \ y = -\frac{2}{5}x-2 \\ \\ B) \ 2x+5y=-10 \therefore y=-\frac{2}{5}x-2 \\ \\ C) \ 2x-5y=-10 \therefore y=\frac{2}{5}x+2 \\ \\ D) \ y+4=-\frac{2}{5}(x-5) \therefore y=-\frac{2}{5}x-2 \\ \\ E) \ y-4=\frac{5}{2}(x + 5) \therefore y=\frac{5}{2}x+\frac{33}{2}


For two perpendicular lines it is true that the product of its slopes is:

m_{1}m_{2}=-1


m_{1}m_{2}=-1 \\ \\ m_{1} \ is \ the \ slope \ of \ y=\frac{5}{2}x+3, \ that \ is, \ m_{1}=\frac{5}{2} \\ \\ Then, \ the \ slope \ of \ a \ perpendicular \ line \ is: \\ \\ m_{2}=-\frac{2}{5}


According to this, only A) B) and D) might be the perpendicular lines we are looking for. Notice that these lines are the same. The other condition is that the line must pass through the point (5, -4). By substituting this point in the equation, we have:

y = -\frac{2}{5}x-2 \\ \\ -4=-\frac{2}{5}(5)-2 \\ \\ -4=-2-2 \\ \\ \boxed{-4=-4} \ True!


Finally, the right answer are:

A) \ y = -\frac{2}{5}x-2 \\ \\ B) \ 2x+5y=-10 \\ \\ D) \ y+4=-\frac{2}{5}(x-5)

8 0
3 years ago
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