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AleksandrR [38]
3 years ago
13

0. Earth's rock is continually cycling through different states and locations. The diagram below shows the rock cycle. The label

s for two processes are missing.
Weathering of
ROCKS at surface
IGNEOUS
ROCK
Uplin
SEDIMENTARY ROCK
METAMORIHIC ROCK
Melting
Which of these processes could be happening at the stage labeled 2?
o folding and mountain building
chemical changes due to heat and pressure
compaction and cementation of weathered sediments
subduction of an oceanic plate beneath a continental plate
Chemistry
2 answers:
cestrela7 [59]3 years ago
7 0

Answer:

Chemical changes due to heat and pressure

Explanation:

Just took the test, hope this helps! :)

Allisa [31]3 years ago
7 0

Answer:

Chemical changes due to heat and pressure

Explanation:

I took this test and got a 100%! :D

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Relative humidity is described as: Group of answer choices a basically unused concept when it comes to weather topics. the amoun
Bad White [126]

Correct answer:

The relative humidity is best described as the amount of water vapor in the air at a given temperature expressed as a percentage of the water vapor capacity of the air.

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5 0
2 years ago
What is a shorter, easier way to show chemical reactions, using symbols instead of words?
Ivanshal [37]

That would be a Chemical Equation.

7 0
3 years ago
Read 2 more answers
The half-life of a first-order reaction is 13 min. If the initial concentration of reactant is 0.13 M, it takes ________ min for
Zigmanuir [339]

Answer:

Therefore it takes 8.0 mins for it to decrease to 0.085 M

Explanation:

First order reaction: The rate of reaction is proportional to the concentration of reactant of power one is called first order reaction.

A→ product

Let the concentration of A = [A]

\textrm{rate of reaction}=-\frac{d[A]}{dt} =k[A]

k=\frac{2.303}{t} log\frac{[A_0]}{[A]}

[A₀] = initial concentration

[A]= final concentration

t= time

k= rate constant

Half life: Half life is time to reduce the concentration of reactant of its half.

t_{\frac{1}{2} }=\frac{0.693}{k}

Here t_{\frac{1}{2} }=0.13 min

k=\frac{0.693}{t_{\frac{1}{2}} }

\Rightarrow k=\frac{0.693}{13 }

To find the time takes for it to decrease to 0.085 we use the below equation

k=\frac{2.303}{t} log\frac{[A_0]}{[A]}

\Rightarrow t=\frac{2.303}{k} log\frac{[A_0]}{[A]}

Here ,   k=\frac{0.693}{13 },  [A₀] = 0.13 m and [ A] = 0.085 M

t=\frac{2.303}{\frac{0.693}{13} } log(\frac{0.13}{0.085})

\Rightarrow t= 7.97\approx 8.0

Therefore it takes 8.0 mins for it to decrease to 0.085 M

7 0
3 years ago
the half life for strontium-90 is 29 years. how many half-lives did the sample go through at the end of 87 year​
Nesterboy [21]
90 divided by 29 x 87 = 290
4 0
2 years ago
Read 2 more answers
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