The circulating flow of air resulting from temperature differences is called a

Near Cozumel, Mexico, a scuba diver goes down 60 ft to look at the sea creatures. As she descends, the air pressure increases. W

Rudik [331]

Answer: Boyle's Law

Explanation:

Boyle's Law which is also called Mariotte's law was formulated by physicist Robert Boyle in 1662.

It posits that the pressure (p) of a given quantity of gas varies inversely with its volume (v) at constant temperature.

As the Scuba Diver went deeper, the volume of air dropped so the pressure increased.

Please do react or comment to the answer if you have any issues or if you feel like this helped you so we can help others as well. Thank you.

5 0

3 years ago

25 grams of radon-222 remains after 15.28 days. How much radon was in the original sample? Write an equation for the decay of ra

Crazy boy [7]

Answer:

half-life = 3.8 days

total time of decay = 15.2 days

initial amount = 100. g

number of half-lives past: 15.2/3.8 = 4 half-lives

4 half-lives = 1/16 remains

100. g x 1/16 = 6.25 g

8 0

2 years ago

In which process are simple materials chemically combined to form more complex materials?

allsm [11]

Answer:

A) synthesis

I hope this helps!

3 0

2 years ago

Which mixture can be classified as a homogeneous mixture?

Reika [66]

Anything can be homogenous as long as you can only see the same type of liquid
think about it like this
orange juice with pulp is Hetero
orange juice with no pulp is homo

3 0

3 years ago

Read 2 more answers

Ammonia NH3 may react with oxygen to form nitrogen gas and water.4NH3 (aq) + 3O2 (g) \rightarrow 2 N2 (g) + 6H2O (l)If 2.15g of

bagirrra123 [75]

Answer:

NH3 is the limiting reactant

The % yield is 36.1 %

Explanation:

Step 1: Data given

Mass of NH3 = 2.15 grams

Mass of O2 = 3.23 grams

Molar mass of NH3 = 17.03 g/mol

Molar mass of O2 = 32 g/mol

volume of N2 produced = 0.550 L

Temperature = 295 K

Pressure = 1.00 atm

Step 2: The balanced equation:

4NH3 (aq) + 3O2 (g) → 2 N2 (g) + 6H2O (l)

Step 3: Calculate moles of NH3

Moles NH3 = Mass NH3 / Molar Mass NH3

Moles NH3 = 2.15 grams / 17.03 g/mol

Moles NH3 = 0.126 moles

Step 4: Calculate moles of O2

Moles O2 = 3.23 grams / 32 g/mol

Moles O2 = 0.101 moles

Step 5: Calculate the limiting reactant

For 4 moles NH3 consumed, we need 3 moles of O2 to produce, 2 moles of N2 and 6 moles of H2O

NH3 is the limiting reactant. It will completely be consumed ( 0.126 moles).

O2 is in excess, there will be 3/4 * 0.126 = 0.0945 moles consumed

There will remain 0.101 - 0.945 = 0.0065 moles of O2

Step 6: Calculate moles of N2

For 4 moles NH3 consumed, we need 3 moles of O2 to produce, 2 moles of N2 and 6 moles of H2O

For 4 moles NH3 , we'll have 2 moles of N2 produced

For 0.126 moles NH3 consumed, we'll have 0.063 moles of N2 produced.

Step 7: Calculate volume of N2 produced

p*V = n*R*T

⇒ with p = the pressure of the gas = 1.00 atm

⇒ with V = the volume = TO BE DETERMINED

⇒ with n = the number of moles N2 = 0.063 moles

⇒ with R = the gasconstant = 0.08206 L*atm/K*mol

⇒ with T = the temperature = 295

V = (nRT)/p

V = (0.063*0.08206*295)/1

V = 1.525 L = theoretical yield

Step 8: Calculate the % yield

% yield = actual yield / theoretical yield

% yield = (0.550 L / 1.525 L)*100%

% yield = 36.1 %

4 0

3 years ago