Answer: 1
Explanation:
a mole is always equal to 6.02 x 10^23 molecules, or in this case, atoms
so to find the number of moles, divide total atoms / 6.02 x 10^23
total atoms = 6.02 x 10^23
6.02 x 10^23 / 6.02 x 10^23 = 1
so 1 mole
Answer:
The answer to your question is: 0.3 moles of AgNO₃
Explanation:
1.0 L sample
0.1 mol of NaCl
0.1 mol of CaCl₂
AgNO₃ = ? moles
Reactions
NaCl + AgNO₃ ⇒ AgCl + NaNO₃
Then 1 NaCl mol --------------- 1 AgNO₃
0.1 mol -------------- x
x = 0.1 moles of AgNO₃ needed
CaCl₂ + 2 AgNO₃ ⇒ 2 AgCl + Ca(NO₃)₂
Then 1 mol of CaCl₂ ------------- 2 moles of AgNO₃
0.1 mol ------------- x
x = 0.2 moles of AgNO₃
Total moles of AgNO₃ = 0.1 + 0.2 = 0.3
Answer:
to protect astronauts from dieing from no air or presurre
Explanation:
Answer:
The electronengativity values of given elements is as follows.
Fluorine - 4
Chlorine -3
Bromine - 2.9
Iodine- 2.5
Explanation:
Electronegativity =consant (I.E-E.A)
The electron affinity and ionization energy values of the given elements is as follows.
(In attachment)
First we have to find the value of constant by using the fluorine atom to whom the electronengativity taken as "4".
<u>Fluorine:</u>
![4=constant[1678-(-327.8)]](https://tex.z-dn.net/?f=4%3Dconstant%5B1678-%28-327.8%29%5D)

By using this constant values we can find electronegatvity values of remaining elements.
<u>Chlorine:</u>
![Electronegativity=0.0019942168[1255+348.7]=3.1980\sim 3](https://tex.z-dn.net/?f=Electronegativity%3D0.0019942168%5B1255%2B348.7%5D%3D3.1980%5Csim%203)
Therefore, electronegativity of chlorine is 3.
<u>Bromine:</u>
![Electronegativity=0.0019942168[1138+324.5]=2.91\sim 2.9](https://tex.z-dn.net/?f=Electronegativity%3D0.0019942168%5B1138%2B324.5%5D%3D2.91%5Csim%202.9)
Therefore, electronegativity of bromine is 2.9.
<u>Iodine:</u>
![Electronegativity=0.0019942168[1007+295.7]=2.59\sim 2.5](https://tex.z-dn.net/?f=Electronegativity%3D0.0019942168%5B1007%2B295.7%5D%3D2.59%5Csim%202.5)
Therefore, electronegativity of iodine is 2.5.