6 km/h west would be your answer
Fe + NiSO4 ----> FeSO4 + No
NaF + HI ---> NaI + HF
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:-
0.229 L
Explanation:-
Molar mass of AgBr = 107.87 x 1 + 79.9 x 1
=187.77 grams mol-1
Mass of AgBr = 150 grams
Number of moles of AgBr = 150 grams / 187.77 gram mol-1
= 0.8 mol
The balanced chemical equation is
NaBr (aq) + AgNO3 (aq)--> AgBr(s) + NaNO3(aq)
From the equation we can see that
1 mol of AgBr is produced from 1 mol of AgNO3.
∴ 0.8 mol of AgBr is produced from 1 x 0.8 / 1 = 0.8 mol of AgNO3.
Strength of AgNO3 = 3.5 M
Volume of AgNO3 required = Number of moles / strength
= 0.8 moles / 3.5
=0.229 L
Answer : The value of
for the reaction is -959.1 kJ
Explanation :
The given balanced chemical reaction is,

First we have to calculate the enthalpy of reaction
.

![\Delta H^o=[n_{H_2O}\times \Delta H_f^0_{(H_2O)}+n_{SO_2}\times \Delta H_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta H_f^0_{(H_2S)}+n_{O_2}\times \Delta H_f^0_{(O_2)}]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo%3D%5Bn_%7BH_2O%7D%5Ctimes%20%5CDelta%20H_f%5E0_%7B%28H_2O%29%7D%2Bn_%7BSO_2%7D%5Ctimes%20%5CDelta%20H_f%5E0_%7B%28SO_2%29%7D%5D-%5Bn_%7BH_2S%7D%5Ctimes%20%5CDelta%20H_f%5E0_%7B%28H_2S%29%7D%2Bn_%7BO_2%7D%5Ctimes%20%5CDelta%20H_f%5E0_%7B%28O_2%29%7D%5D)
where,
= enthalpy of reaction = ?
n = number of moles
= standard enthalpy of formation
Now put all the given values in this expression, we get:
![\Delta H^o=[2mole\times (-242kJ/mol)+2mole\times (-296.8kJ/mol)}]-[2mole\times (-21kJ/mol)+3mole\times (0kJ/mol)]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo%3D%5B2mole%5Ctimes%20%28-242kJ%2Fmol%29%2B2mole%5Ctimes%20%28-296.8kJ%2Fmol%29%7D%5D-%5B2mole%5Ctimes%20%28-21kJ%2Fmol%29%2B3mole%5Ctimes%20%280kJ%2Fmol%29%5D)

conversion used : (1 kJ = 1000 J)
Now we have to calculate the entropy of reaction
.

![\Delta S^o=[n_{H_2O}\times \Delta S_f^0_{(H_2O)}+n_{SO_2}\times \Delta S_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta S_f^0_{(H_2S)}+n_{O_2}\times \Delta S_f^0_{(O_2)}]](https://tex.z-dn.net/?f=%5CDelta%20S%5Eo%3D%5Bn_%7BH_2O%7D%5Ctimes%20%5CDelta%20S_f%5E0_%7B%28H_2O%29%7D%2Bn_%7BSO_2%7D%5Ctimes%20%5CDelta%20S_f%5E0_%7B%28SO_2%29%7D%5D-%5Bn_%7BH_2S%7D%5Ctimes%20%5CDelta%20S_f%5E0_%7B%28H_2S%29%7D%2Bn_%7BO_2%7D%5Ctimes%20%5CDelta%20S_f%5E0_%7B%28O_2%29%7D%5D)
where,
= entropy of reaction = ?
n = number of moles
= standard entropy of formation
Now put all the given values in this expression, we get:
![\Delta S^o=[2mole\times (189J/K.mol)+2mole\times (248J/K.mol)}]-[2mole\times (206J/K.mol)+3mole\times (205J/K.mol)]](https://tex.z-dn.net/?f=%5CDelta%20S%5Eo%3D%5B2mole%5Ctimes%20%28189J%2FK.mol%29%2B2mole%5Ctimes%20%28248J%2FK.mol%29%7D%5D-%5B2mole%5Ctimes%20%28206J%2FK.mol%29%2B3mole%5Ctimes%20%28205J%2FK.mol%29%5D)

Now we have to calculate the Gibbs free energy of reaction
.
As we know that,

At room temperature, the temperature is 500 K.


Therefore, the value of
for the reaction is -959.1 kJ