A species with a positive charge will have a net attraction to a species with a negative charge. Among the choices, N3- is the only one attracted to a positive charge.
Answer: 1. C. polar covalent: electrons shared between silicon and sulfur but attracted more to the sulfur
2. B) 
3. B) Fluorine
Explanation:
1. A polar covalent bond is defined as the bond which is formed when there is a difference of electronegativities between the atoms.
Electronegativity difference = electronegativity of sulphur- electronegativity of silicon = 2.5 -1.8 = 0.7
Thus as electronegativity difference is less than 1.7 , the cond is polar covalent and as electronegativity of sulphur is more , the electrons will be more towards sulphur.
2. A molecular compound is usually composed of two or more nonmetal elements. Example:
Ionic compound is formed by the transfer of electrons from metals to non metals. Example:
,
and 
3. For formation of a neutral ionic compound, the charges on cation and anion must be balanced. The cation is formed by loss of electrons by metals and anions are formed by gain of electrons by non metals.
Here K is having an oxidation state of +1 and as the compound formed is KZ, the oxidation state of non metallic element Z should be -1. Thus the element Z is flourine which exists as diatomic gas 
Answer: 4.15234 m
512 g H2O *
= 0.512 kg H2O
Nitric Acid: HNO3 = 1.008 + 14.007 + 3(15.999) = 63.012 g/mol
H = 1.008 g/mol
N = 14.007 g/mol
O3 = 3*15.999
134 g HNO₃ *
= 2.126 mol
m =
= 4.15234 m
Answer:
9/11 happened on the 11th of September
Answer:
Explanation:
The usefulness of a buffer is its ability to resist changes in pH when small quantities of base or acid are added to it. This ability is the consequence of having both the conjugate base and the weak acid present in solution which will consume the added base or acid.
This capacity is lost if the ratio of the concentration of conjugate base to the concentration of weak acid differ by an order of magnitude. Since buffers having ratios differing by more will have their pH driven by either the weak acid or its conjugate base .
From the Henderson-Hasselbach equation we have that
pH = pKa + log [A⁻]/[HA]
thus
0.1 ≤ [A⁻]/[HA] ≤ 10
Therefore the log of this range is -1 to 1, and the pH will have a useful range of within +/- 1 the pKa of the buffer.
Now we are equipped to answer our question:
pH range = 3.9 +/- 1 = 2.9 through 4.9