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3 years ago

How does the type of medium affect a sound wave? Give an Example

Chemistry

1 answer:

ivanzaharov [21]3 years ago

4 0

Answer:

Sound waves need to travel through a medium such as solids, liquids and gases. The sound waves move through each of these mediums by vibrating the molecules in the matter. The molecules in solids are packed very tightly. Liquids are not packed as tightly.Of the three mediums (gas, liquid, and solid) sound waves travel the slowest through gases, faster through liquids, and fastest through solids. Temperature also affects the speed of sound.Sound waves in air (and any fluid medium) are longitudinal waves because particles of the medium through which the sound is transported vibrate parallel to the direction that the sound wave moves. A vibrating string can create longitudinal waves as depicted in the animation below.

Explanation:

You might be interested in

What percent of I-125 has decayed if there are 37.5g of the original sample left?

garri49 [273]

Answer:

[(x-37.5)/x]*100%

62.5% (assuming the original sample weighs 100.0g)

Explanation:

Let's say that the original sample is x

Mass of I-125 which has decayed: x-37.5

Percentage of decayed mass: [(x-37.5)/x]*100%

Please recheck, for this may not be the correct answer

6 0

3 years ago

What is the concentration of kl solution if 20.68g of solute was added to enough water to form 100ml solution?

rewona [7]

Answer:

1.25 M

Explanation:

Step 1: Given data

Mass of KI (solute): 20.68 g

Volume of the solution: 100 mL (0.100 L)

Step 2: Calculate the moles of solute

The molar mass of KI is 166.00 g/mol.

20.68 g × 1 mol/166.00 g = 0.1246 mol

Step 3: Calculate the molar concentration of KI

Molarity is equal to the moles of solute divided by the liters of solution.

M = 0.1246 mol/0.100 L= 1.25 M

5 0

3 years ago

The electrolysis of water forms H2 and O2. 2H2O 2H2 + O2 What is the percent yield of O2 if 10.2 g of O2 is produced from the de

BigorU [14]

Answer: The percent yield of oxygen gas is 67.53 %.

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

Given mass of water = 17.0 g

Molar mass of water = 18 g/mol

Putting values in equation 1, we get:

$\text{Moles of water}=\frac{17.0g}{18g/mol}=0.944mol$

For the given chemical equation:

$2H_2O\rightarrow 2H_2+O_2$

By Stoichiometry of the reaction:

2 moles of water produces 1 mole of oxygen gas

So, 0.944 moles of water will produce = $\frac{1}{2}\times 0.944=0.472moles$ of oxygen gas

Now, calculating the mass of oxygen gas from equation 1, we get:

Molar mass of oxygen gas = 32 g/mol

Moles of oxygen gas = 0.472 moles

Putting values in equation 1, we get:

$0.472mol=\frac{\text{Mass of oxygen gas}}{32g/mol}\\\\\text{Mass of oxygen gas}=(0.472mol\times 32g/mol)=15.104g$

To calculate the percentage yield of oxygen gas, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of oxygen gas = 10.2 g

Theoretical yield of oxygen gas = 15.104 g

Putting values in above equation, we get:

$\%\text{ yield of oxygen gas}=\frac{10.2g}{15.104g}\times 100\\\\\% \text{yield of oxygen gas}=67.53\%$

Hence, the percent yield of oxygen gas is 67.53 %.

7 0

4 years ago

Read 2 more answers

Work is the only way which energy passes between system and surroundings.

Daniel [21]

Answer:

true :)

Explanation:

6 0

3 years ago

Read 2 more answers

Titration of 0.824 g of potassium hydrogen phthalate required 38.314 g of naoh solution to reach the end point detected by pheno

melamori03 [73]

1.062 mol/kg.

Step 1. Write the balanced equation for the neutralization.

MM = 204.22 40.00

KHC8H4O4 + NaOH → KNaC8H4O4 + H2O

Step 2. Calculate the moles of potassium hydrogen phthalate (KHP)

Moles of KHP = 824 mg KHP × (1 mmol KHP/204.22 mg KHP)

= 4.035 mmol KHP

Step 3. Calculate the moles of NaOH

Moles of NaOH = 4.035 mmol KHP × (1 mmol NaOH/(1 mmol KHP)

= 4.035 mmol NaOH

Step 4. Calculate the mass of the NaOH

Mass of NaOH = 4.035 mmol NaOH × (40.00 mg NaOH/1 mmol NaOH)

= 161 mg NaOH

Step 5. Calculate the mass of the water

Mass of water = mass of solution – mass of NaOH = 38.134 g - 0.161 g

= 37.973 g

Step 6. Calculate the molal concentration of the NaOH

b = moles of NaOH/kg of water = 0.040 35 mol/0.037 973 kg = 1.062 mol/kg

3 0

4 years ago