Answer:
C-18
Explanation:
Step one follow order of operations
Add and subtract from left to right(-5)+(7)=2-(-4)+(12)
STEP 2
Apply negative Rule -(-4)=+4=2+4+(12)
then add 2+4+12=18
Answer:
Hence, 15.99 g of solid Aluminum Sulfate should be added in 250 mL of Volumetric flask.
Explanation:
To make 0.187 M of Aluminum Sulfate solution in a 250 mL (0.250 L) Volumetric flask
The molar mass of Aluminum Sulfate = 342.15 g/mol
Using the molarity formula:-
Molarity = Number of moles/Volume of solution in a liter
Number of moles = Given weight/ molar mass
Molarity = (Given weight/ molar mass)/Volume of solution in liter
0.187 M = (Given weight/342.15 g/mol)/0.250 L
Given weight = 15.99 g
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Types of Bonds can be predicted by calculating the
difference in electronegativity.
If, Electronegativity difference is,
Less
than 0.4 then it is Non Polar Covalent
Between 0.4 and 1.7 then it is Polar Covalent
Greater than 1.7 then it is Ionic
For Be and F,
E.N of Fluorine = 3.98
E.N of Beryllium = 1.57
________
E.N Difference 2.41 (Ionic Bond)
For H and Cl,
E.N of Chorine = 3.16
E.N of Hydrogen = 2.20
________
E.N Difference 0.96 (Polar Covalent Bond)
For Na and O,
E.N of Oxygen = 3.44
E.N of Sodium = 0.93
________
E.N Difference 2.51 (Ionic Bond)
For F and F,
E.N of Fluorine = 3.98
E.N of Fluorine = 3.98
________
E.N Difference 0.00 (Non-Polar Covalent Bond)
Result:
A polar covalent bond is formed between Hydrogen and Chlorine atoms.
Answer:
2.765amu is the contribution of the X-19 isotope to the weighted average
Explanation:
The average molar mass is defined as the sum of the molar mass of each isotope times its abundance. For the unknown element X that has 2 isotopes the weighted average is defined as:
X = Mass X-19 * Abundance X-19 + MassX-21 * Abundance X-21
The contribution of the X-19 isotope is its mass (19.00 amu) times its abundance (14.55% = 0.1455). That is:
19.00amu * 0.1455 =
2.765amu is the contribution of the X-19 isotope to the weighted average