The answer is Al.
If it is a main group element with 3 electrons in its Lewis dot structure, it must be in group 3A. If it is in the 3p orbital section, then it must be in period 3, since the p orbital is a valence orbital and the number that preceeds it is the principal quantum number. Therefore, your answer is the element in period 3 and group 3A, which is aluminum.
Answer:
XZ2
Explanation:
There are different ways in which compounds can be represented. Broadly, we have three different types of formula;
- Structural formular: This shows how th atoms in te compound are connected to each other.
- Molecular formular: This shows the actual number of atoms of element present in the compound
- Empirical Formular: This is the simplest formular of a compound. It basically shows the number of atoms in simple ration to each other.
This question requires us to input the empirical formular;
X2Z4
The ratio of the elements is; 2 : 4 which can be simplified into 1 : 2
This means the empirical formular is XZ2
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Answer:
5.758 is the density of the metal ingot in grams per cubic centimeter.
Explanation:
1) Mass of pycnometer = M = 27.60 g
Mass of pycnometer with water ,m= 45.65 g
Density of water at 20 °C = d =
1 kg = 1000 g
Mass of water ,m'= m - M = 45.65 g - 27.60 g =18.05 g
Volume of pycnometer = Volume of water present in it = V
2) Mass of metal , water and pycnometer = 56.83 g
Mass of metal,M' = 9.5 g
Mass of water when metal and water are together ,m''= 56.83 g - M'- M
56.83 g - 9.5 g - 27.60 g = 19.7 g
Volume of water when metal and water are together = v
Density of metal = d'
Volume of metal = v' =
Difference in volume will give volume of metal ingot.
v' = v - V
Since volume cannot be in negative .
Density of the metal =d'
=
