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katrin [286]
3 years ago
10

Which is the correct terrm that indicated that an oceanic crust is pushed downward by a colliding contintental crust?

Chemistry
1 answer:
NARA [144]3 years ago
3 0
The process by which the oceanic crust is pulled under the continental crust is called subduction.
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how many grams of antifreeze would be required per 500 g of water to prevent the water from feezing at a temperature of -39° C​
andrezito [222]

Answer:

333.7g of antifreeze

Explanation:

Freezing point depression in a solvent (In this case, water) occurs by the addition of a solute. The law is:

ΔT = Kf × m × i

Where:

ΔT is change in temperature (0°C - -20°C = 20°C)

Kf is freezing point depression constant (1.86°C / m)

m is molality of solution (moles solute / 0.5 kg solvent -500g water-)

i is Van't Hoff factor (1, assuming antifreeze is ethylene glycol -C₂H₄(OH)₂)

Replacing:

20°C = 1.86°C / m  × moles solute / 0.5 kg solvent × 1

5.376 = moles solute

As molar mass of ethylene glycol is 62.07g/mol:

5.376 moles × (62.07g / 1mol) = <em>333.7g of antifreeze</em>.

4 0
3 years ago
A 50.00 g sample of an unknown metal is heated to 45.00°C. It is then placed in a coffee-cup calorimeter filled with water. The
AysviL [449]

Answer:- Heat lost by the metal is 279.45 cal.

Solution:- This type of problems are solved by using the concept, heat given = - heat taken

Metal temperature is decreasing from 45.00 degree C to 11.08 degree C. It means the heat is lost by the metal and this heat lost by metal is gained by water and the calorimeter to raise their temperature.

the equation we use is, q=mc\Delta T .

where, q is the heat energy, m is mass, c is specific heat and \Delta T is change in temperature.

Combined mass of calorimeter and water is 250.0 g and the specific heat is \frac{1.035cal}{g.^0C} .

\Delta T  for calorimeter and water (combined) = 11.08 - 10.00 = 1.08 degree C

\Delta T  for metal = 11.08 - 45.00 = -33.92 degree C

let's plug in the values in the above equation and calculate heat gained by combined system.

q=250.0g*\frac{1.035cal}{g.^0C}*1.08^0C

q = 279.45 cal

So, the heat lost by the metal is 279.45 cal.


3 0
4 years ago
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Find the source frequency ω for which the equivalent impedance seen by the voltage source is purely real (i.e., the imaginary pa
Leto [7]
I just took this quiz and the answer is d
8 0
4 years ago
What's the enthalpy of neutralization and explain briefly how it relates to esterification​
Liula [17]

Answer:

The enthalpy of neutralization and its relation to esterification​ is explained below in details.

Explanation:

The conventional enthalpy variation of neutralization is the enthalpy exchange when extracts of an acid and an alkali respond together under conventional circumstances to generate 1 mole of water. Remark that the enthalpy change of neutralization is always estimated per mole of water produced.

8 0
3 years ago
How does the overall charge of oxygen of -16 compare to the overall charge of oxygen -17 and oxygen -18?
nadezda [96]

Answer:

-2

The charge on an oxygen ion is -2. An oxygen atoms contain six electrons in its valence electron shell.

Explanation:

5 0
2 years ago
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