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pogonyaev
3 years ago
8

Explain how the mid-ocean ridge supports the theory of sea floor spreading. help plzzz:(

Chemistry
1 answer:
Sedaia [141]3 years ago
4 0
The ridge of mid ocean b is the answer hope this helps
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describe the behavior of the molecules in a liquid. Explain this behavior in terms of intermolecular forces.
tigry1 [53]

Answer:

Molecules in liquids are held to other molecules by intermolecular interactions, which are weaker than the intramolecular interactions that hold the atoms together within molecules and polyatomic ions.

5 0
3 years ago
If 4.168 kJ of heat is added to a calorimeter containing 75.40 g of water, the temperature of the water and the calorimeter incr
Alinara [238K]

Answer:

The value of  the heat capacity of the Calorimeter  C_c = 54.4 \frac{J}{c}

Explanation:

Given data

Heat added Q = 4.168 KJ = 4168 J

Mass of water m_w = 75.40 gm

Temperature change = ΔT = 35.82 - 24.58 = 11.24 ° c

From the given condition

Q = m_w C_w ΔT + C_c ΔT

Put all the values in above equation we get

4168 = 75.70 × 4.18 × 11.24 +  C_c × 11.24

611.37 =  C_c × 11.24

C_c = 54.4 \frac{J}{c}

This is the value of  the heat capacity of the Calorimeter.

7 0
3 years ago
What unit is mass generally measured in
Veronika [31]

Answer:

kilograms

Explanation:

hope this helps, pls mark brainliest :D

6 0
3 years ago
Read 2 more answers
✞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
2 years ago
A 2.40 kg block of ice is heated with 5820 J of heat. The specific heat of water is 4.18 J•g^-1•C^-1. By how much will it’s temp
MrMuchimi

Answer: The temperature rise is 0.53^0C

Explanation:

The quantity of heat required to raise the temperature of a substance by one degree Celsius is called the specific heat capacity.

Q=m\times c\times \Delta T

Q = Heat absorbed by ice = 5280 J

m = mass of ice = 2.40 kg = 2400 g   (1kg=1000g)

c = heat capacity of water = 4.18J/g^0C

Initial temperature  = T_i

Final temperature = T_f  

Change in temperature ,\Delta T=T_f-T_i=?

Putting in the values, we get:

5280J=2400g\times 4.18J/g^0C\times \Delta T

\Delta T=0.53^0C

Thus the temperature rise is 0.53^0C

0 0
3 years ago
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