Answer:
1 = 252g, 2 = 2mL, 3 = 1.5mL, 4 = 3g, 5 = 225g, 6 = 0.92g/mL, 7 = 0.75g/mL, 8 = 0.71g/mL, 9 = 1.9mL, 10= 1.11mL, 11 = 76.9g
Explanation:
This problem is testing how well you can move around the equation D = m/v where D = Density (g/mL), m= mass of sample (g), v = volume of sample (mL).
Five. The measurement 50,600 mg has five significant digits.
I presume that you are using the comma as a decimal separator.
The <em>rules for significant figures</em> are
1. Nonzero digits are always significant.
2. Any zeros between two significant digits are significant.
3. Final or trailing zeros are significant only if they are to the right of a decimal point.
• According to Rule 1, the <em>5 and 6</em> are significant.
• According to Rule 2, the <em>0 between the 5 and 6</em> is significant
• According to Rule 3, the <em>final two zeros</em> are significant.
Thus, there are five significant digits in the measurement 50,600 mg.
Note: If the comma is a thousands separator, the number has only three significant digits.
The valence electron configuration for antimony (Sb) is:
Sb = 5s²5p³5d⁰
In SbCl₅²⁻, antimony has a -2 charge i.e. it has 2 additional electrons
Sb²⁻ = 5s²5p⁵5d⁰
Following a two electron transition from p→d orbital we have:
Sb²⁻ = 5s²5p³5d²
There is a total of 5 unpaired electrons (3 in the p and 2 in the d) which can form five bonds with the 5 Cl atoms.
Thus the hybridisation of Sb in SbCl₅²⁻ is sp³d²
