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Blizzard [7]
2 years ago
10

One brand of vinegar has a pH of 4.5. Another brand has a pH of 5.0. The equation for the pH of a substance is pH = –log[H+], wh

ere H+ is the concentration of hydrogen ions. What is the approximate difference in the concentration of hydrogen ions between the two brands of vinegar?
Mathematics
2 answers:
likoan [24]2 years ago
6 0

Answer:

The vinegar with pH 4.5 has 2.2*10^-5 more hydrogen ions than the vinegar with pH 5.0.

Explanation:

To identify the difference, we must first find the H+ of both. We can do this by using the equation [H+]=10^-pH. After we do that we know that the vinegar with a pH of 5.0 has [H+]=1.0*10^-5 and the vinegar with a pH of 4.5 has [H+]=3.2*10^-5. The difference is found by subtraction.

(3.2*10^-5)-(1.0*10^-5)=2.2*10^-5

kiruha [24]2 years ago
4 0

Answer:

2.16 * (10^-5).

Step-by-step explanation:

5 = -log H+

5 = log (1 / H+)

1/ H+ = 10^5

H+ = 10^-5

In a similar fashion  the other brand has H+ of 10^-4.5.

So the difference in hydrogen ion concentration =  10^(-4.5) - 10^(-5)

= 2.16 * (10^-5).

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Find the distance from (4, −7, 6) to each of the following.
LenKa [72]

Answer:

(a) 6 units

(b) 4 units

(c) 7 units

(d) 9.22 units

(e) 7.21 units

(f) 8.06 units

Step-by-step explanation:

The distance d from one point (x₁, y₁, z₁) to another point (x₂, y₂, z₂) is given by;

d = √[(x₂ - x₁)² + (y₂ - y₁)² + (z₂ - z₁)²]

Now from the question;

<em>(a) The distance from (4, -7, 6) to the xy-plane</em>

The xy-plane is the point where z is 0. i.e

xy-plane = (4, -7, 0).

Therefore, the distance d is from <em>(4, -7, 6) </em> to <em>(4, -7, 0)</em>

d = √[(4 - 4)² + (-7 - (-7))² + (0 - 6)²]

d = √[(0)² + (0)² + (-6)²]

d = √(-6)²

d = √36

d = 6

Hence, the distance to the xy plane is 6 units

<em>(b) The distance from (4, -7, 6) to the yz-plane</em>

The yz-plane is the point where x is 0. i.e

yz-plane = (0, -7, 6).

Therefore, the distance d is from <em>(4, -7, 6) </em> to <em>(0, -7, 6)</em>

d = √[(4 - 0)² + (-7 - (-7))² + (6 - 6)²]

d = √[(4)² + (0)² + (0)²]

d = √(4)²

d = √16

d = 4

Hence, the distance to the yz plane is 4 units

<em>(c) The distance from (4, -7, 6) to the xz-plane</em>

The xz-plane is the point where y is 0. i.e

xz-plane = (4, 0, 6).

Therefore, the distance d is from <em>(4, -7, 6) </em> to <em>(4, 0, 6)</em>

d = √[(4 - 4)² + (-7 - 0)² + (6 - 6)²]

d = √[(0)² + (-7)² + (0)²]

d = √[(-7)²]

d = √49

d = 7

Hence, the distance to the xz plane is 7 units

<em>(d) The distance from (4, -7, 6) to the x axis</em>

The x axis is the point where y and z are 0. i.e

x-axis = (4, 0, 0).

Therefore, the distance d is from <em>(4, -7, 6) </em> to <em>(4, 0, 0)</em>

d = √[(4 - 4)² + (-7 - 0)² + (6 - 0)²]

d = √[(0)² + (-7)² + (6)²]

d = √[(-7)² + (6)²]

d = √[(49 + 36)]

d = √(85)

d = 9.22

Hence, the distance to the x axis is 9.22 units

<em>(e) The distance from (4, -7, 6) to the y axis</em>

The x axis is the point where x and z are 0. i.e

y-axis = (0, -7, 0).

Therefore, the distance d is from <em>(4, -7, 6) </em> to <em>(0, -7, 0)</em>

d = √[(4 - 0)² + (-7 - (-7))² + (6 - 0)²]

d = √[(4)² + (0)² + (6)²]

d = √[(4)² + (6)²]

d = √[(16 + 36)]

d = √(52)

d = 7.22

Hence, the distance to the y axis is 7.21 units

<em>(f) The distance from (4, -7, 6) to the z axis</em>

The z axis is the point where x and y are 0. i.e

z-axis = (0, 0, 6).

Therefore, the distance d is from <em>(4, -7, 6) </em> to <em>(0, 0 6)</em>

d = √[(4 - 0)² + (-7 - (0))² + (6 - 6)²]

d = √[(4)² + (-7)² + (0)²]

d = √[(4)² + (-7)²]

d = √[(16 + 49)]

d = √(65)

d = 8.06

Hence, the distance to the z axis is 8.06 units

5 0
1 year ago
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