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HACTEHA [7]
3 years ago
5

The five boxes below represent a geologic cross-section with five layers that represent the rocks as they would from when earth'

s atmosphere was developing (A is on top and E is on the bottom). Label the cross section with the appropriate rock from the word bank below. Then, explain your answer in relation to the history of the earth's atmosphere.
Chemistry
1 answer:
quester [9]3 years ago
7 0

Answer:

A Limestone Layers

B Banded Iron formations

C Metaphoric rocks with fine grains  

D Red Beds

E Chert Layers

Explanation:

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A certain reaction has an activation energy of 25.88 kJ/mol. At what Kelvin temperature will the reaction proceed 6.00 times fas
Juli2301 [7.4K]

Answer:hi

Explanation:

5 0
2 years ago
An atom of chlorine with a mass number of 37 contains __ protons and __ neutrons.
Natasha2012 [34]

Answer:

Its in the Explanation

Explanation:

Here's what I got.

Aluminium-27 is an isotope of aluminium characterized by the fact that is has a mass number equal to  

27

.

Now, an atom's mass number tells you the total number of protons and of neutrons that atom has in its nucleus. Since you're dealing with an isotope of aluminum, it follows that this atom must have the exact same number of protons in its nucleus.

The number of protons an atom has in its nucleus is given by the atomic number. A quick looks in the periodic table will show that aluminum has an atomic number equal to  

13

.

This means that any atom that is an isotope of aluminum will have  

13

protons in its nucleus.

Since you're dealing with a neutral atom, the number of electrons that surround the nucleus must be equal to the number of protons found in the nucleus.

Therefore, the aluminium-27 isotope will have  

13

electrons surrounding its nucleus.

Finally, use the known mass number to determine how many neutrons you have

mass number

=

no. of protons

+

no. of neutrons

no. of neutrons = 27 − 13 = 14

Your welcome :)

6 0
3 years ago
The term that best describes a 10-gram of KCIO3 per 100 grams of water solution at 30 degrees celsius is:
In-s [12.5K]
The term that best described a 10 gram of KClO3 per 100 grams of water solution at 30 degree Celcius is Saturated. The solubility chart is needed for this work. If the solubility chart is drawn for KClO3, it will be observed that the proportion of KClO3 that is needed to dissolve in 100ml of water to make the solution saturated is 10 grams at 30 degree Celcius.
3 0
3 years ago
Based on solubility rules what ions in water might interfere with the analysis of calcium ions by precipitation of calcium carbo
Pavlova-9 [17]
First, we must know what happens in the precipitation reaction. This type of reaction is a double replacement reactions. It is consists of two reactant compounds which interchange cations and anions to form two products. One of the products is an insoluble solid called a precipitate. For the precipitation of CaCO₃, there are two consecutive reactions involved:

1. Slaking of quicklime, CaO
    CaO + H₂O ⇒ Ca(OH)₂

2. Precipitation
    Ca(OH)₂ + CO₂ ⇒ CaCO₃ + H₂O

The ions that make up the H₂O molecule are H⁺ and OH⁻. According to solubility rules, the cation (positively charged ion) is likely to be attracted to an anion (negatively charged ion). Together, they form an ionic bond. This type of bond is when there is a complete transfer of electrons between the two. The Ca²⁺ cation lacks 2 electrons, while the anion OH⁻ has an excess 1 electron. In order to be stable, 1 Ca²⁺ ion and 2 OH⁻ ions must combine.

Therefore, the answer is OH⁻ ion.
6 0
3 years ago
Read 2 more answers
It takes 495.0 kJ of energy to remove 1 mole of electron from an atom on the surface of sodium metal. How much energy does it ta
Zigmanuir [339]

Answer:

\lambda=241.9\ nm

Explanation:

The work function of the sodium= 495.0 kJ/mol

It means that  

1 mole of electrons can be removed by applying of 495.0 kJ of energy.

Also,  

1 mole = 6.023\times 10^{23}\ electrons

So,  

6.023\times 10^{23} electrons can be removed by applying of 495.0 kJ of energy.

1 electron can be removed by applying of \frac {495.0}{6.023\times 10^{23}}\ kJ of energy.

Energy required = 82.18\times 10^{-23}\ kJ

Also,  

1 kJ = 1000 J

So,  

Energy required = 82.18\times 10^{-20}\ J

Also, E=\frac {h\times c}{\lambda}

Where,  

h is Plank's constant having value 6.626\times 10^{-34}\ Js

c is the speed of light having value 3\times 10^8\ m/s

So,  

79.78\times 10^{-20}=\frac {6.626\times 10^{-34}\times 3\times 10^8}{\lambda}

\lambda=\frac{6.626\times 10^{-34}\times 3\times 10^8}{82.18\times 10^{-20}}

\lambda=\frac{10^{-26}\times \:19.878}{10^{-20}\times \:82.18}

\lambda=\frac{19.878}{10^6\times \:82.18}

\lambda=2.4188\times 10^{-7}\ m

Also,  

1 m = 10⁻⁹ nm

So,  

\lambda=241.9\ nm

6 0
3 years ago
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