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erma4kov [3.2K]
3 years ago
14

How are amphibians different from reptiles

Chemistry
1 answer:
Brilliant_brown [7]3 years ago
5 0
Amphibians are different from reptiles in that reptiles hatch when they are born and remain basically the same shape throughout their lives. Amphibians, on the other hand, hatch as small, fish-like creatures in water and eventually grow their limbs and live as land creatures afterward.
Hope that helped =)
You might be interested in
The temperature of a sample of water changes from 10°C to 20°C when the water absorbs 100 calories of heat. What is the mass of
Vlad1618 [11]

Answer:

10 g

Explanation:

Right from the start, just by inspecting the values given, you can say that the answer will be  

10 g

.

Now, here's what that is the case.

As you know, a substance's specific heat tells you how much heat is needed to increase the temperature of  

1 g

of that substance by  

1

∘

C

.

Water has a specific heat of approximately  

4.18

J

g

∘

C

. This tells you that in order to increase the temperature of  

1 g

of water by  

1

∘

C

, you need to provide  

4.18 J

of heat.

Now, how much heat would be required to increase the temperature of  

1 g

of water by  

10

∘

C

?

Well, you'd need  

4.18 J

to increase it by  

1

∘

C

, another  

4.18 J

to increase it by another  

1

∘

C

, and so on. This means that you'd need

4.18 J

×

10

=

41.8 J

to increase the temperature of  

1 g

of water by  

10

∘

C

.

Now look at the value given to you. If you need  

41.8 J

to increase the temperature of  

1 g

of water by  

10

∘

C

, what mass of water would require  

10

times as much heat to increase its temperature by  

10

∘

C

?

1 g

×

10

=

10 g

And that's your answer.

Mathematically, you can calculate this by using the equation

q

=

m

⋅

c

⋅

Δ

T

 

, where

q

- heat absorbed/lost

m

- the mass of the sample

c

- the specific heat of the substance

Δ

T

- the change in temperature, defined as final temperature minus initial temperature

Plug in your values to get

418

J

=

m

⋅

4.18

J

g

∘

C

⋅

(

20

−

10

)

∘

C

m

=

418

4.18

⋅

10

=

10 g

5 0
3 years ago
(First to answer gets some good) Order the following components of the universe from largest to smallest: Moon, planet,universe,
cluponka [151]

Answer:

Universe, galaxy, solar system, star, planet, moon and asteroid.

Explanation:

You're welcome!

7 0
4 years ago
The vapor pressure of water is 1.00 atm at 373 K, and the enthalpy of vaporization is 40.68 kJ mol!. Estimate the vapor pressure
Yuki888 [10]

Answer:

The vapor pressure at temperature 363 K is 0.6970 atm

The vapor pressure at 383 K is 1.410 atm

Explanation:

To calculate \Delta H_{vap} of the reaction, we use clausius claypron equation, which is:

\ln(\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}[\frac{1}{T_1}-\frac{1}{T_2}]

where,

P_1 = vapor pressure at temperature T_1

P_2 = vapor pressure at temperature T_2

\Delta H_{vap} = Enthalpy of vaporization  

R = Gas constant = 8.314 J/mol K

1) \Delta H_{vap}=40.68 kJ/mol=40680 J/mol

T_1 = initial temperature =363 K

T_2 = final temperature =373 K

P_2=1 atm, P_1=?

Putting values in above equation, we get:

\ln(\frac{1 atm}{P_1})=\frac{40680 J/mol}{8.314J/mol.K}[\frac{1}{363}-\frac{1}{373}]

P_1=0.69671 atm \approx 0.6970 atm

The vapor pressure at temperature 363 K is 0.6970 atm

2) \Delta H_{vap}=40.68 kJ/mol=40680 J/mol

T_1 = initial temperature =373 K

T_2 = final temperature =383 K

P_1=1 atm, P_2?

Putting values in above equation, we get:

\ln(\frac{P_2}{1 atm})=\frac{40680 J/mol}{8.314J/mol.K}[\frac{1}{373}-\frac{1}{383}]

P_2=1.4084 atm \approx 1.410 atm

The vapor pressure at 383 K is 1.410 atm

8 0
3 years ago
Water is placed in a graduated cylinder and the volume is recorded as 43.5 mL. A homogeneous sample of metal pellets with a mass
Ipatiy [6.2K]

Answer:

1.8g

Explanation:

Initial volume = 43.5ml

Final volume = 49.4ml

Mass = 10.88g

Density = ?

Volume = Final volume - initial volume

= 49.4 - 43.5

= 5.9ml

Density = Mass/volume

Density = 10.88/5.9

= 1.8g/ml

4 0
4 years ago
✞Why is propane stored in household tanks but natural gas is not?♡♡♡​
Klio2033 [76]

Answer:

In order to keep propane a liquid at room temperature (70° F or 21° C), it has to be held in a tank at a pressure of about 850 kPa. ... Household metal tanks cannot withstand this pressure. In short, natural gas is not stored in household tanks because the symmetry of its molecule makes it hard to liquify.

4 0
3 years ago
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