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Verdich [7]
3 years ago
7

Compare the way that frogs breathe when they are tadpoles with the way frogs breathe when they are adults.

Chemistry
2 answers:
Ad libitum [116K]3 years ago
7 0

Answer:Frogs breathe through gills underwater to get oxygen and when frogs are adults they breathe through their lungs to get oxygen.

Explanation:

andreev551 [17]3 years ago
6 0
When frogs are tadpoles they, they breathe through their gills, as frogs they have mucus glands in their skin which helps them absorb oxygen from the air, or they can also breathe through their lungs
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A gas at STP has a volume of 1.00 L. If the pressure is doubled and the temperature remains constant, what is the news of the ga
4vir4ik [10]

Answer: PV = nRT

A gas at STP... This means that the temperature is 0°C and pressure is 1 atm.

R is the gas constant which is 0.08206 L*atm/(K*mol)

Rearranging for volume

V = nRT/P

The temperature and number of moles are held constant. This means that this uses Boyle's Law. (The ideal gas law could be manipulated to give us this result when T and n are held constant.)

PV = k

where k is a constant.

This means that

P₁V₁ = k = P₂V₂

P₁V₁ = P₂V₂

(1 atm) * (1 L) = (2 atm) * V₂

V₂ = 0.5 L

The new volume of the gas is 0.5 L.

Explanation:

3 0
2 years ago
What have scientists been able to use to create the Geologic Time Scale?
dem82 [27]
The half-life of the carbon-14 isotope is used in dating fossils in a process called radiocarbon dating.

hope this is adequate. 
3 0
3 years ago
HELP!!! WILL MARK BRAINLIEST!!!
Sophie [7]

Answer:

A , B, C

Explanation: D is a Diamagnetic

8 0
3 years ago
the heat of fusion of acetone is 5.7 kJ/mol. Calculate to two significant figures the entropy change when 6.3 mol of acetone mel
shtirl [24]

<u>Answer:</u> The entropy change of the process is 2.0\times 10^2J/K

<u>Explanation:</u>

To calculate the entropy change for different phase at same temperature, we use the equation:

\Delta S=n\times \frac{\Delta H_{f}}{T}

where,  

\Delta S = Entropy change

n = moles of acetone = 6.3 moles

\Delta H_{f} = enthalpy of fusion = 5.7 kJ/mol = 5700 J/mol    (Conversion factor:  1 kJ = 1000 J)

T = temperature of the system = -94.7^oC=[273-94.7]=178.3K

Putting values in above equation, we get:

\Delta S=\frac{6.3mol\times 5700J/mol}{178.3K}\\\\\Delta S=201.4J/K=2.0\times 10^2J/K

Hence, the entropy change of the process is 2.0\times 10^2J/K

4 0
4 years ago
What is the percent composition for the compound NaBr? (8C)
ehidna [41]

Answer:

the answer is D

Explanation:

percentage composition=  mole of the substance divided by the total molar mass of the compound multiplied by 100.

3 0
3 years ago
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