Since hydrogen only has 1 electron it shares 1 with flourine to have a pair. It is single bonded.
Answer:
B
Explanation:
The scientist had a wrong hypothesis?
Answer: C2H4
Explanation:
The percentage composition of ethanol ( C2H5OH ) consist of 52.2% Carbon, Hydrogen of 13.0 and 34.8% of Oxygen.
The percentage composition of ethane gas (C2H6) consist of 80.0% carbon and 20.0% hydrogen.
The composition of Ethylene Glycols i.e C2H4(OH)2 is Carbon of 39.7%, 9.7% hydrogen and 51.6% oxygen.
The percent composition of c2h4 is 86% carbon, and 14% hydrogen.
From the information given, the substance with the highest percentage of carbon is C2H4
This is known as Rutherford's gold foil experiment. To align with J.J Thompson's Plum Pudding Model, he expects a beam of alpha particles to just pass through the gold foil undisturbed. However, some were deflected at certain angles. Alpha particles are positive, so it would just go straight through the nucleus, but will deflect if it hits the electrons. <em>Therefore, the answer is: </em><span><em>Particles that struck the nucleus went straight.</em></span>
Explanation:
(a) Formula that shows relation between
and
is as follows.
Here,
= 1
Putting the given values into the above formula as follows.
= 
= 
= 0.01316
(b) As the given reaction equation is as follows.

As there is only one gas so
,
= 1.20
Therefore, pressure of
in the container is 1.20.
(c) Now, expression for
for the given reaction equation is as follows.
![K_{c} = \frac{[CaO][CO_{2}]}{[CaCO_{3}]}](https://tex.z-dn.net/?f=K_%7Bc%7D%20%3D%20%5Cfrac%7B%5BCaO%5D%5BCO_%7B2%7D%5D%7D%7B%5BCaCO_%7B3%7D%5D%7D)
=
= \frac{x^{2}}{(a - x)}[/tex]
where, a = initial conc. of 
=
= 0.023 M
0.0131 =
x = 0.017
Therefore, calculate the percentage of calcium carbonate remained as follows.
% of
remained =
= 75.46%
Thus, the percentage of calcium carbonate remained is 75.46%.