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inn [45]
3 years ago
15

In the waves lab, you changed ______ to see how it would affect wave speed.

Chemistry
2 answers:
Burka [1]3 years ago
5 0

Answer: The correct answer is "medium".

Explanation:

The expression for the relation between the speed of the wave and frequency is as follows;

v=f\lambda

Here, v is the speed of the wave, f is the frequency of the wave and \lambda is the wavelength of the wave.

There is an inverse relationship between the frequency and wavelength. If the frequency of the wave increases then the wavelength of the wave decreases while the speed of the wave remains constant.

The properties of the wave can be described by amplitude, wavelength and frequency but the speed of the wave depends on the medium through which it is moving. The medium may be denser or rarer according to it the speed of the wave will vary.

Therefore, in the waves lab, you changed medium to see how it would affect wave speed.

jasenka [17]3 years ago
4 0
Hi there!

I believe the answer is transversal or transverse.

Have a nice night!
~Brooke
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A 420 mL sample of a 0.100 M formate buffer, pH 3.75, is treated with 7 mL of 1.00 M KOH. What is the pH following this addition
Art [367]

<u>Answer:</u> The pH of the resulting solution will be 3.60

<u>Explanation:</u>

Molarity is calculated by using the equation:

\text{Molarity}=\frac{\text{Moles}}{\text{Volume}} ......(1)

We are given:

Molarity of formic acid = 0.100 M

Molarity of potassium formate = 0.100 M

Volume of solution = 420 mL = 0.420 L (Conversion factor: 1 L = 1000 mL)

Putting values in equation 1, we get:

\text{Moles of formic acid}=(0.100mol/L\times 0.420L)=0.0420mol

\text{Moles of potassium formate}=(0.100mol/L\times 0.420L)=0.042mol

Molarity of KOH = 1.00 M

Volume of solution = 7 mL = 0.007 L

Putting values in equation 1, we get:

\text{Moles of KOH}=(1mol/L\times 0.007L)=0.007mol

The chemical equation for the reaction of formic acid and KOH follows:

                 HCOOH+KOH\rightleftharpoons HCOOK+H_2O

I:                   0.042     0.007       0.042

C:                -0.007    -0.007     +0.007

E:                  0.035         -           0.049

Volume of solution = [420 + 7] = 427 mL = 0.427 L

To calculate the pH of the acidic buffer, the equation for Henderson-Hasselbalch is used:

pH=pK_a+ \log \frac{\text{[conjugate base]}}{\text{[acid]}} .......(2)

Given values:

[HCOOK]=\frac{0.049}{0.427}

[HCOOH]=\frac{0.035}{0.427}

pK_a=3.75

Putting values in equation 2, we get:

pH=3.75-\log \frac{(0.049/0.427)}{(0.035/0.427)}\\\\pH=3.75-0.146\\\\pH=3.60

Hence, the pH of the resulting solution will be 3.60

6 0
3 years ago
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Ivahew [28]

Answer:

<em><u>transmission</u></em>

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When light waves passes straight through an object, it is called <em><u>transmission</u></em>.

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