I believe the correct answer from the choices listed above is option A. Fan blades would be an analogy for electron cloud model. Austrian physicist Erwin Schrödinger (1887-1961) developed an “Electron Cloud Model<span>” in 1926. It consisted of a dense nucleus surrounded by a cloud of electrons. Hope this helps.</span>
Seems right, but if it’s just asking for one i would pick B
Answer: 0.172 M
Explanation:
a) To calculate theconcentration of base, we use the equation given by neutralization reaction:

where,
are the n-factor, molarity and volume of acid which is 
are the n-factor, molarity and volume of base which is NaOH.
We are given:

Putting values in above equation, we get:

b) To calculate the concentration of acid, we use the equation given by neutralization reaction:

where,
are the n-factor, molarity and volume of acid which is 
are the n-factor, molarity and volume of base which is NaOH.
We are given:

Putting values in above equation, we get:

The concentration of the phosphoric acid solution is 1.172 M
Explanation:
Fossil fuel is an overall term for covered ignitable geologic stores of natural materials, framed from rotted plants and creatures that have been changed over to unrefined petroleum, coal, flammable gas, or weighty oils by introduction to warmth and weight in the world's outside more than a huge number of years.
The consuming of petroleum products by people is the biggest wellspring of emanations of carbon dioxide, which is one of the ozone depleting substances that permits radiative compelling and adds to an unnatural weather change.
A little bit of hydrocarbon-based powers are biofuels gotten from climatic carbon dioxide, and consequently don't build the net measure of carbon dioxide in the environment.
Answer:
320 g
Step-by-step explanation:
The half-life of Co-63 (5.3 yr) is the time it takes for half of it to decay.
After one half-life, half (50 %) of the original amount will remain.
After a second half-life, half of that amount (25 %) will remain, and so on.
We can construct a table as follows:
No. of Fraction Mass
half-lives t/yr Remaining Remaining/g
0 0 1
1 5.3 ½
2 10.6 ¼
3 15.9 ⅛ 40.0
4 21.2 ¹/₁₆
We see that 40.0 g remain after three half-lives.
This is one-eighth of the original mass.
The mass of the original sample was 8 × 40 g = 320 g