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m_a_m_a [10]
3 years ago
15

Show that an aqueous solution with a pH of 5 has a hydroxide ionconcentration of 1×10^9M​

Physics
1 answer:
vekshin13 years ago
5 0

Explanation:

pH + pOH = 14

5 + pOH = 14

pOH = 9

pOH = -log([OH⁻])

9 = -log([OH⁻])

-9 = log([OH⁻])

1×10⁻⁹ = [OH⁻]

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How does a change in resistance affect the conductivity of a material?
max2010maxim [7]

Answer: the answer is B the lower the resistance,the higher the conductivity

Explanation:

Resistance is inversely proportional to conductivity,the lower the resistance the higher the conductivity,the metal silver has a very low resistance,thus it conductivity is high

4 0
4 years ago
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Which graph represents a function with direct variation? A coordinate plane with a line passing through (negative 4, 0) and (0,
fgiga [73]

Answer:

Answer:

A line passing through the points (-1,-4),(0,0) and (1,4)

Step-by-step explanation:

we know that

A relationship between two variables, x, and y, represent a direct variation if it can be expressed in the form y/x=ky/x=k or y=kxy=kx

In a proportional relationship the constant of proportionality k is equal to the slope m of the line and the line passes through the origin

Verify each case

Part 1) A line passing through the points (-4,0) and (0,-2)

This line not represent a direct variation, because the line not passes through the origin.

Part 2) A line passing through the points (-5,4) and (0,3)

This line not represent a direct variation, because the line not passes through the origin.

Part 3) A line passing through the points (-4,-6) and (0,3)

This line not represent a direct variation, because the line not passes through the origin.

Part 4) A line passing through the points (-1,-4),(0,0) and (1,4)

The line passes through the origin

Find out the value of k

k=y/xk=y/x

For the point (-1,-4)

substitute

k=-4/-1=4k=−4/−1=4

For the point (1,4)

substitute

k=4/1=4k=4/1=4

The linear equation is  y=4xy=4x

This line represent a direct variation

7 0
3 years ago
A tube 1.20 m long is closed at one end. A stretched wire is placed near the open end. The wire is 0.350 m long and has a mass o
Ksju [112]

Answer:

71.4583 Hz

67.9064 N

Explanation:

L = Length of tube = 1.2 m

l = Length of wire = 0.35 m

m = Mass of wire = 9.5 g

v = Speed of sound in air = 343 m/s

The fundamental frequency of the tube (closed at one end) is given by

f=\frac{v}{4L}\\\Rightarrow f=\frac{343}{4\times 1.2}\\\Rightarrow f=71.4583\ Hz

The fundamental frequency of the wire and tube is equal so he fundamental frequency of the wire is 71.4583 Hz

The linear density of the wire is

\mu=\frac{m}{l}\\\Rightarrow \mu=\frac{9.5\times 10^{-3}}{0.35}\\\Rightarrow \mu=0.02714\ kg/m

The fundamental frequency of the wire is given by

f=\frac{1}{2l}\sqrt{\frac{T}{\mu}}\\\Rightarrow f^2=\frac{1}{4l^2}\frac{T}{\mu}\\\Rightarrow T=f^2\mu 4l^2\\\Rightarrow T=71.4583^2\times 0.02714\times 4\times 0.35^2\\\Rightarrow T=67.9064\ N

The tension in the wire is 67.9064 N

7 0
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