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TiliK225 [7]
3 years ago
11

What happens to the oceanic crust at a deep ocean floor

Chemistry
1 answer:
erica [24]3 years ago
3 0
Mid-ocean ridges and seafloor spreading can also influence sea levels. As oceanic crust moves away from the shallow mid-ocean ridges, it cools and sinks as it becomes more dense.
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The solubility of nacl at 30°c is 36.3 g/100 g of water. what mass of nacl is contained in a saturated solution that contains 30
sleet_krkn [62]
Because a solubility is calculated for saturated solutions we can write:

36.3 g NaCl ---100g water
x g NaCl ---- 300.0 g water

x=36.3*300/100=108.9 g NaCl
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4 years ago
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Which of these is not an example of a molecule?
r-ruslan [8.4K]

Answer:

D

Explanation:

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A galvanic cell generates a cell potential of 0.32V when operated under standard conditions according to the reaction above. Whi
Ugo [173]

The complete question is shown in the image attached to this answer.

Answer:

C

Explanation:

Let us quickly remember that the EMF of a cell under non standard conditions in given by the Nernst equation.

This equation states that;

E = E°cell - 0.592/n log Q

Where

E = EMF under non standard conditions

E°cell= standard EMF of the cell

n = number of electrons transferred

Q = reaction quotient

If the reaction quotient is greater than 1 then cell potential is less than the standard cell potential.

The cell that generates the lowest cell potential is the cell depicted in option C because Q has the greatest positive value(Q<1).

6 0
3 years ago
What is the mass of 25 Liters of nitrogen dioxide gas?
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Answer:

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7 0
3 years ago
A horizontal cylinder equipped with a frictionless piston contains 785 cm3 of steam at 400 K and 125 kPa pressure. A total of 83
guapka [62]

Answer:

a. 478.69 K

b. 939.43 cm^{3}

c. 19.30 J

d. 64.5J

Explanation:

From the question, we can identify the following;

V_{o} = 785cm^{3} = 0.000785 m^{3}

T_{o} = 400K

P_{o} = 125 Kpa =  125 000 Pa

Using the ideal gas equation,

PV = nRT

where R is the molar gas constant = 8.314 m^{3}⋅Pa⋅K^{-1}⋅mol^{-1}

Thus, n = PV/RT = (125000 × 0.000785)/(8.314 × 400) = 0.03 mol

a. Steam temperature in K

To calculate this, we use the constant pressure process;

q = nΔH

Where q is 83.8J according to the question

Thus;

83.8 = 0.03 × [34980 + 35.5T_{1} - (34980 + 35.5T_{o})]

83.8 = (0.03 × 35.5) (T_{1} - 400K)

83.8 = 1.065 (T_{1}  - 400)

78.69 = (T_{1}  - 400)

T_{1} = 400 + 78.69

T_{1}  = 478.69 K

b. Final cylinder volume

To calculate this, we make use of the Charles' law(Temperature and pressure are directly proportional)

V_{1}/T_{1} = V_{o}/T_{o}

V_{1}  =  V_{o}T_{1}/T_{o}

V_{1}   = (785 × 478.69)/400

V_{1}   = 939.43 cm^{3}

c. Work done by the system

Mathematically, the work done by the system is calculated as follows;

w = P(V_{1}- V_{o}) = 125 KPa ( 939.43 - 785) = 19.30 J

d. Change in internal energy of the steam in J

ΔU = q - w = 83.8 - 19.3 = 64.5J

6 0
3 years ago
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