All categories

4 years ago

Ocean lithosphere is lighter than continental lithosphere

Chemistry

2 answers:

lesantik [10]4 years ago

8 0

Answer:

False.

Explanation:

Ocean lithosphere is about 60km thick and very dense, whereas continental lithosphere is about 120km and less dense. That is why when they collide, ocean plates subduct under continental ones.

Nady [450]4 years ago

6 0

if this is is a True or False question I think it would be (False). Because the continental crust is lighter than the oceanic crust, and the continental crust cannot subduct.

You might be interested in

Definition of volume

AysviL [449]

Answer:

The amount of space an object occupies.

Explanation:

3 0

3 years ago

How many licks does it take to get to the end of a tootsie pop

worty [1.4K]

3 lol i was thinking about that from the ad

4 0

3 years ago

Read 2 more answers

On the basis of the information above, what is the approximate percent ionization of HNO2 in a 1.0 M HNO2 (aq) solution?

enot [183]

Answer:

The answer is "2%"

Explanation:

Equation:

$HNO_2\ (aq) \leftrightharpoons H^{+} \ (aq) + NO_2^{-}\ (aq) \\\\\ K_a = 4.0\times \ 10^{-4}$

$H^{+}=?$

Formula:

$Ka = \frac{[H^{+}][NO_2^{-}]}{[HNO_2]}$

Let

$[H^{+}] = [NO_2^{-}] = x$ at equilibrium

$x^2 = (4.0\times 10^{-4})\times 1.0\\\\x = ((4.0\times 10^{-4})\times 1.0)^{0.5} = 2.0 \times 10^{-2} \ M\\\\$

therefore,

$[H^{+}] = 2.0\times 10^{-2} \ M = 0.02 \ M$

Calculating the % ionization:

$= \frac{([H^{+}]}{[HNO_2])} \times 100 \\\\= \frac{0.02}{1}\times 100 \\\\= 2\%\\\\$

6 0

3 years ago

Which of the following electron configurations for neutral atoms is correct?

geniusboy [140]

Argon: 1s22s22p63s23p6 is right answer,

8 0

4 years ago

Read 2 more answers

Given the value of the equilibrium constant (Kc) for the equation (a), calculate the equilibrium constant for equation (b)

matrenka [14]

Answer: The value of equilibrium constant for new reaction is $1.92\times 10^{-25}$

Explanation:

The given chemical equation follows:

$O_2(g)\rightarrow \frac{2}{3}O_3(g)+\frac{1}{2}O_2(g)$

The equilibrium constant for the above equation is $5.77\times 10^{-9}$

We need to calculate the equilibrium constant for the equation of 3 times of the above chemical equation, which is:

$3O_2(g)\rightarrow 2O_3(g)$

The equilibrium constant for this reaction will be the cube of the initial reaction.

If the equation is multiplied by a factor of '3', the equilibrium constant of the new reaction will be the cube of the equilibrium constant of initial reaction.

The value of equilibrium constant for reverse reaction is:

$K_{eq}'=(5.77\times 10^{-9})^3=1.92\times 10^{-25}$

Hence, the value of equilibrium constant for new reaction is $1.92\times 10^{-25}$

5 0

3 years ago