Use the periodic table to identify the element indicated by each electron configuration by typing in the
1 answer:
Answer:
1s22s22p6: Neon (Ne)
1s22s22p63s23p3: Phosphorous (P)
1s22s22p63s23p64s1: Potassium (K)
1s22s22p63s23p64s2(im not sure what 308 is supposed to be): Calcium (Ca)
1s22s22p63s23p64s23d104p65s24d3: there is no pure element that ends 4d3 that I know of so this can either be Zirconium(Zr) if it ends in 4d2 or Niobium (Nb) if it ends in 4d4
Explanation:
you can look at the periodic table and the trends to find the rough idea of where the electron configuration ends, there are helpful articles and images on these, i attached an image that may help. After that you can look at the atomic number to find the number of electrons for a pure element and use the electron subshell pattern thing to find the exact number
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Answer:
4.8 grams of H₂ will be produced if 175g of HCI are allowed to react completely with sodium
Explanation:
By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction) you can see that the following amounts in moles of each compound react and are produced:
- HCl: 2 moles
- Na: 1 mole
- NaCl: 2 moles
- H₂: 1 mole
You know the following masses of each element:
- H: 1 g/mole
- Cl: 35.45 g/mole
- Na: 23 g/mole
So, the molar mass of each compound participating in the reaction is:
- HCl: 1 g/mole + 35.45 g/mole= 36.45 g/mole
- Na: 23 g/mole
- NaCl: 23 g/mole + 35.45 g/mole= 58.45 g/mole
- H₂: 2* 1 g/mole= 2 g/mole
Then, by stoichiometry of the reaction, the following amounts in grams of each of the compounds participating in the reaction react and are produced:
- HCl: 2 moles* 36.45 g/mole= 72.9 g
- Na: 1 mole* 23 g/mole= 23 g
- NaCl: 2 moles* 58.45 g/mole= 116.9 g
- H₂: 1 mole* 2 g/mole= 2 g
So, a rule of three applies as follows: if by stoichiometry, when reacting 72.9 grams of HCl 2 grams of H₂ are formed, when reacting 175 grams of HCl how much mass of H₂ will be formed?
mass of H₂= 4.8 g
<u><em>4.8 grams of H₂ will be produced if 175g of HCI are allowed to react completely with sodium</em></u>