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vampirchik [111]
3 years ago
6

The difference in the energy exerted by an 8.0 earthquake compared to a 6.0 earthquake?

Chemistry
1 answer:
hammer [34]3 years ago
4 0
<span>The difference in the energy exerted by an 8.0 earthquake compared to a 6.0 earthquake

A magnitude 8.0 earthquake is 100 times bigger and 1000 times stronger (energy released) than a magnitude 6.0 earthquake. 


</span>
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2.What are variables used to describe weather?
Studentka2010 [4]
The answer is, All of the above
4 0
3 years ago
ILL MARK BRAINLIEST :)
DanielleElmas [232]

Explanation:

According to the law of conservation of mass, mass can neither be created nor destroyed but it can simply be transformed from one form to another.

For example, Na^{+} + Cl^{-} \rightarrow NaCl

Mass of Na = 23 g/mol

Mass of Cl = 35.5 g/mol

Sum of mass of reactants = mass of Na + mass of Cl

= 23 + 35.5 g/mol

= 58.5 g/mol

Mass of product formed is as follows.

Mass of NaCl = mass of Na + mass of Cl

= (23 g/mol + 35.5) g/mol

= 58.5 g/mol

As mass reacted is equal to the amount of mass formed. This shows that mass is conserved.

As a result, law of conservation of mass is obeyed.

3 0
3 years ago
A box is 1 m high, 2.5 m long, and 1.5 m wide, its volume is 5 mº.<br> true or false
Maurinko [17]

Answer:

false

Explanation:

Because volume is in m3

5 0
3 years ago
Read 2 more answers
This graph shows two curves pertaining to a hydrogen s orbital.
fgiga [73]

Answer 1) : According to the complete question attached in the answer,

The radial wave function  which is denoted by R_{nl}(r) shown with orange color crosses through zero point. Also, At the the radial nodes, which are spherical shells to some radial distance away from the nucleus there no electron are found.

Also, the radial probability distribution curve denoted as R^{2}_{nl}(r) shown in  blue  color is observed to touch zero, and shows the place of radial node.

Therefore, the total number of nodes will include both the kinds  which has radial and angular nodes which will be represented by <em>'n'</em>.

It is observed that for any atomic orbital, the total number of nodes will be n-1  .


Considering the s orbital of the hydrogen, which has zero angular momentum  (l); (l=0), as it has zero angular nodes.  

Hence, there will be only radial nodes, which is

(n−1  =  total number of radial nodes in s orbitals)

According to the image, there are 4  radial nodes shown, so n  =  5  (as n-1 = 4; therefore, n = 5)

This represents the 5s orbital.


Answer 2) The radial nodes are observed in I'm seeing radial nodes at  

1.9a_{0},  6.4a_{0},13.9a_{0} and  27.0a_{0}.

where  a_{0} represents the  hydorgen bohr atomic radius =  0.0529177 nm


Explanation : It is quite easy to observe the given graph and find out the approximate values of the radial nodes, it does not requires any equation to be solved. Equation can be used to find the radial nodes if it was supplied along with the question. Although by mere speculation one can find out the answer.

3 0
3 years ago
Read 2 more answers
The balanced equation shows how sodium chloride reacts with silver nitrate to form sodium nitrate and silver chloride. NaCl + Ag
Verdich [7]
<span>NaCl First calculate the molar mass of NaCl and AgNO3 by looking up the atomic weights of each element used in either compound Sodium = 22.989769 Chlorine = 35.453 Silver = 107.8682 Nitrogen = 14.0067 Oxygen = 15.999 Now multiply the atomic weight of each element by the number of times that element is in each compound and sum the results For NaCl 22.989769 + 35.453 = 58.44277 For AgNO3 107.8682 + 14.0067 + 3 * 15.999 = 169.8719 Now calculate how many moles of each substance by dividing the total mass by the molar mass For NaCl 4.00 g / 58.44277 g/mol = 0.068443 mol For AgNO3 10.00 g / 169.8719 g/mol = 0.058868 Looking at the balanced equation for the reaction, there is a 1 to 1 ratio in molecules for the reaction. Since there is a smaller number of moles of AgNO3 than there is of NaCl, that means that there will be some NaCl unreacted, so the excess reactant is NaCl</span>
8 0
3 years ago
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