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3 years ago

Why do rocks have different color stripes

Chemistry

1 answer:

Gekata [30.6K]3 years ago

4 0

For example Existing rock is weathered into sediment, then compacted, cemented and finally Lithified (Lithification)

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What happens to a nitrogen atom when heated

KATRIN_1 [288]

Answer: Nitrogen normally exists as two-atom molecules in the form of a gas at room temperature.

Explanation:

4 0

4 years ago

A student has an unknown organism under a microscope. The organism was found in a local pond. He observes that the sample organi

timurjin [86]

Answer:

Eubacteria

Explanation:

The two prokaryotic kingdoms are Eubacteria and Archaea. A prokaryote is a relatively simple single-celled organism;

5 0

3 years ago

What fraction of a Sr-90 sample remains unchanged after 87.3 years

jolli1 [7]

The answer is 1/8.

Half-life is the time required for the amount of a sample to half its value.
To calculate this, we will use the following formulas:
1. $(1/2)^{n} = x$ ,
where:
n - a number of half-lives
x - a remained fraction of a sample

2. $t_{1/2} = \frac{t}{n}$
where:
 $t_{1/2}$ - half-life
t - total time elapsed
n - a number of half-lives

The half-life of Sr-90 is 28.8 years.
So, we know:
t = 87.3 years
 $t_{1/2}$ = 28.8 years

We need:
n = ?
x = ?

We could first use the second equation, to calculate n:
If:
$t_{1/2} = \frac{t}{n}$ ,
Then:
$n = \frac{t}{ t_{1/2} }$
⇒ $n = \frac{87.3 years}{28.8 years}$
⇒ $n=3.03$
⇒ n ≈ 3

Now we can use the first equation to calculate the remained amount of the sample.
 $(1/2)^{n} = x$
⇒ $x=(1/2)^3$
⇒ $x= \frac{1}{8}$

8 0

4 years ago

Please someone help meeeeee!!!

Lorico [155]

Answer:

1 = Q = 7315 j

2 =Q = -21937.5 j

Explanation:

Given data:

Mass of water = 50 g

Initial temperature = 20°C

Final temperature = 55°C

Energy required to change the temperature = ?

Solution:

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Specific heat capacity of water is 4.18 j/g.°C.

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = T2 - T1

ΔT = 55°C - 20°C

ΔT = 35°C

Q = 50 g× 4.18 j/g.°C×35°C

Q = 7315 j

Q 2:

Given data:

Mass of metal = 100 g

Initial temperature = 1000°C

Final temperature = 25°C

Energy released = ?

Specific heat capacity = 0.225 j/g.°C

Solution:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = T2 - T1

ΔT = 25°C - 1000°C

ΔT = -975°C

Now we will put the values in formula.

Q = 100 g × 0.225 j/g.°C × -975°C

Q = -21937.5 j

Negative sign show that energy is released.

7 0

3 years ago

AlCl3 is lewis acid. Explain how?

professor190 [17]

For AlCl3 to be a Lewis acid, it would have to react in such a way that it accepted a lone pair from some other atom or molecule from a Lewis base.AlCl3 has an electron-deficient aluminum atom. It has only six electrons in its valence shell. It readily accepts electrons from other atoms, in an attempt to get a full valence shell of eight electrons.

3 0

4 years ago