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2 years ago

How long would it take for a sound impulse to travel through an aluminum rod 1,500 meters long?

Chemistry

1 answer:

velikii [3]2 years ago

7 0

Answer:

As per Provided Information

Distance travelled by sound impulse through an aluminium rod is 1500 metres .

We have been asked to determine the time taken by the sound impulse to travel through aluminium rod .

As we know

Speed of sound in aluminium rod is 5100m/s .

Using Formulae 

 $\boxed{\bf \: Time \: taken_{(Sound \: impulse)} = \cfrac{distance \: travelled \: by \: sound \: impulse}{speed \: of \: sound \: impluse}}$ 

Substituting the value and we obtain

$\sf \qquad \longrightarrow \: Time \: taken_{(Sound \: impulse)} = \dfrac{1500}{5100} \\ \\ \\ \sf \qquad \longrightarrow \: Time \: taken_{(Sound \: impulse)} = \cancel\dfrac{1500}{5100} \\ \\ \\ \sf \qquad \longrightarrow \: Time \: taken_{(Sound \: impulse)} = 0.29 \: s$

Therefore ,

0.29 second sound impulse takes to travelled.

So, your answer is 0.29 s .

You might be interested in

Which represents a balanced nuclear equation?

storchak [24]

Answer:

The correct option is 2.

Explanation:

In a nuclear reaction balanced we have that:

1. The sum of the mass number (A) of the reactants (r) is equal to the sum of the mass number of the products (p) $\Sigma A_{r} = \Sigma A_{p}$

2. The sum of the atomic number (Z) of the reactants is also equal to the sum of the atomic number of the products $\Sigma Z_{r} = \Sigma A_{p}$

So, let's evaluate each option.

1) $^{23}_{11}Na \rightarrow ^{24}_{11}Mg + ^{1}_{1}H$

The mass number of the reactant is:

$A_{r} = 23$

The sum of the mass number of the products is:

$A_{p} = 24 + 1 = 25$

This is not the correct option because it does not meet the first condition ( $\Sigma A_{r} = \Sigma A_{p}$ ).

2) $^{24}_{11}Na \rightarrow ^{24}_{12}Mg + ^{0}_{-1}e$

The mass number of the reactant and the products is:

$A_{r} = 24$

$A_{p} = 24 + 0 = 24$

Now, the atomic number of the reactants and the products are:

$Z_{r} = 11$

$Z_{p} = 12 + (-1) = 11$

This nuclear reaction is balanced since it does meet the two conditions for a balanced nuclear equation, ( $\Sigma A_{r} = \Sigma A_{p}$ and $\Sigma Z_{r} = \Sigma Z_{p}$ ).

3) $^{24}_{13}Al \rightarrow ^{24}_{12}Mg + ^{0}_{-1}e$

The mass number of the reactant and the products is:

$A_{r} = 24$

$A_{p} = 24 + 0 = 24$

Now, the atomic number of the reactants and the products are:

$Z_{r} = 13$

$Z_{p} = 12 + (-1) = 11$

This reaction does not meet the second condition ( $\Sigma Z_{r} = \Sigma Z_{p}$ ) so this is not a balanced nuclear equation.

4) $^{23}_{12}Mg \rightarrow ^{24}_{12}Mg + ^{1}_{0}n$

The mass number of the reactant and the products is:

$A_{r} = 23$

$A_{p} = 24 + 1 = 25$

This reaction is not a balanced nuclear equation since it does not meet the first condition ( $\Sigma A_{r} = \Sigma A_{p}$ ).

Therefore, the correct option is 2.

I hope it helps you!

4 0

3 years ago

635 mL of a gas is at a pressure of 8.00 atm. What is the volume of the gas at standard pressure (STP)?

Sveta_85 [38]

The new volume at standard temperature and pressure is 5.08 L.

Explanation:

As per the kinetic theory of gases, the volume occupied by gas molecules will be inversely proportional to the pressure of the gas molecules. This is termed as Boyle's law.

So, pressure∝ $\frac{1}{volume}$

Thus, if two pressure and two volumes are given then,

$P_{1} V_{1} = P_{2} V_{2}\\$

Now, we known the values of P₁ = 8 atm, V₁ = 635 mL, P₂ = 1 atm and V₂ we have to determine. We are considering P₂ = 1 atm, because we have to determine V₂ at standard temperature and pressure. And standard pressure is 1 atm.

$8*635*10^{-3} = 1 *V{2} \\\\V_{2} = \frac{8*635*10^{-3} }{1} = 5.08 L$

Thus, the new volume at standard temperature and pressure is 5.08 L.

5 0

3 years ago

307.0 mL of air is at 0 C. What is the volume (L) at 273 C

IgorC [24]

Use Charle's Law because we are dealing with volume and temperature.

$\frac{v}{t} =\frac{v}{t}$

v₁= 3017.0 ml

t₂ = 0

v₂= ???

t₂= 273°C

Change all mL's to L because the question says "What is the volume (L)...".

3017.0 mL = 3.017 L

Change the temperature into Kelvins: Kelvin = Celsius + 273.15

0°C = 273.15 K

273°C = 546.15 K

Now you can plug it in the formula:

$\frac{3.017}{273.15} = \frac{v}{546.15}$

Then, do cross multiplication.

(3.017) (546.15) = (273.15) v₂

1,647.73 = (273.15 K) v₂

Divide 273.15 on both sides

6.0296 L= V₂

8 0

4 years ago

How many moles of ethanol are produced starting with 500.g glucose?

Monica [59]

<h3>Answer:</h3>

5.55 mol C₂H₅OH

<h3>General Formulas and Concepts:</h3>

Math

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Chemistry

Atomic Structure

Reading a Periodic Tables
Moles

Stoichiometry

Using Dimensional Analysis
Analyzing Reactions RxN

<h3>Explanation:</h3>

Step 1: Define

[RxN - Balanced] C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂

[Given] 500. g C₆H₁₂O₆ (Glucose)

[Solve] moles C₂H₅OH (Ethanol)

Step 2: Identify Conversions

[RxN] 1 mol C₆H₁₂O₆ → 2 mol C₂H₅OH

[PT] Molar mass of C - 12.01 g/mol

[PT] Molar Mass of H - 1.01 g/mol

[PT] Molar Mass of O - 16.00 g/mol

Molar Mass of C₆H₁₂O₆ - 6(12.01) + 12(1.01) + 6(16.00) = 180.18 g/mol

Step 3: Stoichiometry

[DA] Set up conversion: $\displaystyle 500 \ g \ C_6H_{12}O_6(\frac{1 \ mol \ C_6H_{12}O_6}{180.18 \ g \ C_6H_{12}O_6})(\frac{2 \ mol \ C_2H_5OH}{1 \ mol \ C_6H_{12}O_6})$
[DA} Multiply/Divide [Cancel out units]: $\displaystyle 5.55001 \ mol \ C_2H_5OH$

Step 4: Check

Follow sig fig rules and round. We are given 3 sig figs.

5.55001 mol C₂H₅OH ≈ 5.55 mol C₂H₅OH

8 0

3 years ago

What are the chemical properties of alcohol and alkene?

Flura [38]

Answer:

they have sweet odour.Alcohol are organic compounds in which a hydrogen atom of an aliphatic carbon is replaced with a hydroxyl group.

Explanation:

7 0

3 years ago