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Mars2501 [29]
3 years ago
7

In the cathode ray experiment, the cathode ray is seen to move from the negative disk to the positive disk. What kind of subatom

ic particle will move in. the opposite direction of the cathode ray?
Chemistry
1 answer:
Tomtit [17]3 years ago
8 0

Answer:

Gixt8x75zurx7txr7x7tzzr7zr7z64zz7z84

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I need help on question 3 true or false thxxx quick i’m timed
ruslelena [56]

Answer:

The answer is true.

5 0
3 years ago
Adding O2 to the reaction 6CO2 + 6H2O C6H12O6 + 6O2. WHICH WAY WILL THE REACTION SHIFT?
Oxana [17]

Answer:

The reaction will shift to the left to produce more reactants.

Explanation:

According to the Le- Chatelier principle,

At equilibrium state when stress is applied to the system, the system will behave in such a way to nullify the stress.

The equilibrium can be disturb,

By changing the concentration

By changing the volume

By changing the pressure

By changing the temperature

Consider the following chemical reaction.

Chemical reaction:

6CO₂ +  6H₂O  ⇄  C₆H₁₂O₆ + 6O₂

In this reaction the equilibrium is disturb by increasing the concentration of Product.

When the concentration of product is increased the system will proceed in backward direction in order to regain the equilibrium. Because when product concentration is high it means reaction is not on equilibrium state. As the concentration of O₂  increased the reaction proceed in backward direction to regain the equilibrium state and more reactant is formed.

4 0
3 years ago
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
What type of volcano will most likely form when an eruption takes place with explosive force and layers of ash accumulate? I WIL
Vikki [24]

Answer:

A.) A cinder cone vocano

Explanation:

Cinder cone volcanos give off more ash, rather then an explosion, hense the name "cinder cone".

4 0
2 years ago
Read 2 more answers
A cell consists of a gold wire and a saturated calomel electrode (S.C.E.) in a 0.150 M AuNO 3 solution at 25 °C. The gold wire i
almond37 [142]

The half reaction that occurs at the Au electrode is  1.64

<u>Explanation:</u>

Half cell reaction at 'Au' electrode

We have the equation,

Au + (aq) + e-  ---> Au(s)

Given the concenration of AuNO3=0.150 M

[Au +] =0.150 m

From the equation,

Au + (aq) + e-  ---> Au(s)

Standard electric potential= Eo= 1.69 volt

Solving the problem using the Nerst equation

E cell= E0 cell - 2.303 RT/ nF  log Qc

Where,

T = 298 K

n= no of electron lost or gained

F= faraday's constant= 965000/mole

R= universal constant= 8.314 J/ K/ Mole

Substitue the values we get

E cell = 1.69 volt - 0.05 g/n log (1/0.150M)

E cell = 1.69 volt - 0.05 g/1  0.824

E cell= 1.64

The half reaction that occurs at the Au electrode is  1.64

<u />

6 0
3 years ago
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