The answer should be 1 AU as it’s just about close to the number of miles from earth to sun.
Earth to sun in miles is -92.96
1 AU is- 92955807.3
Missing question:
(a) 1s2 2s2 2p6 3s1
(b) 1s2 2s2 2p6 3s2
(c) 1s2 2s2 2p6 3s2 3p1
(d) 1s2 2s2 2p6 3s2 3p4
(e) 1s2 2s2 2p6 3s2 3p5
Answer is: a) 1s²2s²2p⁶3s¹ (sodium).
Sodium have the largest second ionization energy, because when he lost one electron(first ionization energy), he have stable electron configuration of noble gas neon (1s²2s²2p⁶), so sodium do not need to lost second electron, because he will have unstable electron configuration.
Answer:- Molar mass of
.
Solution:- It is a stoichiometry problem. Mass of the grill is 30.0 kg and the mass after burning the grill is also 30.0 kg. It means all the charcol is burned and the gas is given off.
2.0 kg of charcol are converted to grams which is 2000 g. Since charcol is pure solid carbon, the grams are divided by the atomic mass of carbon which is 12.
The combustion equation of charcol is written as:

From this balanced equation, there is 1:1 mol ratio between charcol and carbon dioxide. So, the moles of carbon dioxide gas formed are equivalent to the moles of charcol. To convert the moles of carbon dioxide to grams we multiply the moles by it's molar mass.
Carbon dioxide has one carbon and two oxygen atoms so it's molar mass = 12 + 2(16) = 12 + 32 = 44
So, 44 is the molar mass of carbon dioxide and above calculations clearly shows how and where we get this.
Answer :
(a) The rate of
formed is, 0.066 M/s
(b) The rate of
formed is, 0.033 M/s
Explanation : Given,
= 0.066 M/s
The balanced chemical reaction is,

The rate of disappearance of
= ![-\frac{1}{2}\frac{d[NO]}{dt}](https://tex.z-dn.net/?f=-%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7Bd%5BNO%5D%7D%7Bdt%7D)
The rate of disappearance of
= ![-\frac{d[O_2]}{dt}](https://tex.z-dn.net/?f=-%5Cfrac%7Bd%5BO_2%5D%7D%7Bdt%7D)
The rate of formation of
= ![\frac{1}{2}\frac{d[NO_2]}{dt}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7Bd%5BNO_2%5D%7D%7Bdt%7D)
As we know that,
= 0.066 M/s
(a) Now we have to determine the rate of
formed.
![\frac{1}{2}\frac{d[NO_2]}{dt}=\frac{1}{2}\frac{d[NO]}{dt}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7Bd%5BNO_2%5D%7D%7Bdt%7D%3D%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7Bd%5BNO%5D%7D%7Bdt%7D)
![\frac{d[NO_2]}{dt}=\frac{d[NO]}{dt}=0.066M/s](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5BNO_2%5D%7D%7Bdt%7D%3D%5Cfrac%7Bd%5BNO%5D%7D%7Bdt%7D%3D0.066M%2Fs)
The rate of
formed is, 0.066 M/s
(b) Now we have to determine the rate of molecular oxygen reacting.
![-\frac{d[O_2]}{dt}=-\frac{1}{2}\frac{d[NO]}{dt}](https://tex.z-dn.net/?f=-%5Cfrac%7Bd%5BO_2%5D%7D%7Bdt%7D%3D-%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7Bd%5BNO%5D%7D%7Bdt%7D)
![\frac{d[O_2]}{dt}=\frac{1}{2}\times 0.066M/s=0.033M/s](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5BO_2%5D%7D%7Bdt%7D%3D%5Cfrac%7B1%7D%7B2%7D%5Ctimes%200.066M%2Fs%3D0.033M%2Fs)
The rate of
formed is, 0.033 M/s
Answer:
the answer of your question is below:
Explanation:
decay