The instrument pictured below is a manometer resistometer aerometer barometer

If an emergency vehicle passes you in traffic, just as it passes you the pitch of

Keith_Richards [23]

If an emergency vehicle passes you in traffic, just as it passes you the pitch of the siren changes as it approaches, the siren’s pitch sounds higher than when it is moving away from you.the frequency increases as the siren is moving toward you, and decreases as it is moving away, is correct

This change is a common physical demonstration of the Doppler effect.

<h3>What is Doppler effect ?</h3>

This effect refers to the increase in the frequency of sound, as the source and observer move toward (or away from) each other.

The effect causes the sudden change in pitch noticeable in a passing siren.

Hence Option (B) i.e, the frequency increases as the siren is moving toward you, and decreases as it is moving away, is correct

Learn more about Doppler' effect here ;

brainly.com/question/27507962

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3 0

2 years ago

Convert 1.3902 hectoliters to centiliters.

hram777 [196]

The answer is 13,902 centiliters.

8 0

4 years ago

Read 2 more answers

Use the Lab Report document as a template. Use the descriptions and examples of each part to help you write your Lab Report. You

GarryVolchara [31]

This link might help you!!

https://www.eiu.edu/biology/bio1500/writing_a_lab_report.pdf

6 0

3 years ago

Fish breathe the dissolved air in water through their gills. Assuming the partial pressures of oxygen and nitrogen in air to be

amid [387]

Answer:

X(O₂) = 0.323

X(N₂) = 0.677

Explanation:

We have the partial pressures of oxygen (O₂) and nitrogen (N₂):

P(O₂) = 0.20 atm

P(N₂) = 0.80 atm

In order to solve the problem, you need the solubilities of each gas in water at 298 K. We can consider 1.3 x 10⁻³ mol/(L atm) for oxygen (O₂) and 6.8 x 10⁻⁴mol/(L atm) for nitrogen (N₂) from the bibliography.

s(O₂) = 1.3 x 10⁻³ mol/(L atm)

s(N₂) = 6.8 x 10⁻⁴mol/(L atm)

So, we calculate the concentration (C) of each gas as the product of its partial pressure (P) and the solubility (s):

C(O₂) = P(O₂) x s(O₂) = 0.20 atm x 1.3 x 10⁻³ mol/(L atm) = 2.6 x 10⁻⁴mol/L

C(N₂) = P(N₂) x s(N₂) = 0.80 atm x 6.8 x 10⁻⁴mol/(L atm) = 5.44 x 10⁻⁴ mol/L

In 1 liter of water, we have the following number of moles (n):

n(O₂) = 2.6 x 10⁻⁴ mol

n(N₂) = 5.44 x 10⁻⁴ mol

Thus, the total number of moles (nt) is calculated as the sum of the number of moles of the gases in the mixture:

nt = n(O₂) + n(N₂) = 2.6 x 10⁻⁴ mol + 5.44 x 10⁻⁴ mol = 8.04 x 10⁻⁴ mol

Finally, the mole fraction of each gas is calculated as the ratio between the number of moles of each gas and the total number of moles:

X(O₂) = n(O₂)/nt = 2.6 x 10⁻⁴ mol/(8.04 x 10⁻⁴ mol) = 0.323

X(N₂) = n(N₂)/nt = 5.44 x 10⁻⁴ mol/(8.04 x 10⁻⁴ mol) = 0.677

5 0

3 years ago

For the following reaction, 13.1 grams of glucose (C6H12O6) are allowed to react with 10.6 grams of oxygen gas. glucose (C6H12O6

mart [117]

Answer:

There will be formed 14.58 grams of CO2

O2 is the limiting reagent

There will remain 3.151 grams of glucose

Explanation:

Step 1: Data given

Mass of glucose = 13.1 grams

molar mass of glucose 180.156 g/mol

Mass of oxygen = 10.6grams

molar mass of oxygen = 32 g/mol

Step 2: The balanced equation

C6H12O6 + 6 O2 → 6H2O + 6CO2

Step 3: Calculate moles of glucose

Moles of glucose = mass glucose / molar mass glucose

Moles of glucose = 13.1 grams / 180.156 g/mol

Moles of glucose = 0.0727 moles

Step 4: Calculate moles of oxygen

Moles O2 = 10.6 grams / 32 g/mol

Moles O2 = 0.33125 moles

Step 5: Calculate the limiting reactant

For 1 mol of glucose , we need 6 moles of O2 to produce 6 moles of H2O and 6 moles of CO2

Oxygen is the limiting reactant. It will completely be consumed ( 0.33125 moles).

Glucose is in excess. There will be consumed 0.33125/6 = 0.05521 moles

There will remain 0.0727 - 0.05521 = 0.01749 moles

This is 0.01749 moles * 180.156 g/mol = 3.151 grams

Step 6: Calculate moles of CO2

For 1 mol of glucose , we need 6 moles of O2 to produce 6 moles of H2O and 6 moles of CO2

For 0.33125 moles of O2 we'll get 0.33125 moles of CO2 produced

Step 7: Calculate mass of CO2

Mass of CO2 = moles CO2 * molar mass CO2

Mass CO2 = 0.33125 moles * 44.01 g/mol

Mass CO2 = 14.58 grams

7 0

3 years ago