Answer:
Three
Explanation:
1. Draw a Lewis structure
- Put the least electronegative atom (C) in the centre
- Attach the O and two S atoms to the C.
- Add enough electrons to give each atom an octet.
2. Draw the resonance structures
You should get one of the structures below.
It has a double bond and two single bonds.
The resonance structures will have the double bond in each of the other two positions.
Answer:
101.4pKa
Explanation:
Given parameters:
Initial Volume of helium = 3.20L
Final pressure of helium = 59kPa
Final volume on helium gas = 5.5L
Unknown:
Initial pressure exerted on the balloon = ?
Solution;
To solve this pressure - volume problem, we simply apply the Boyle's law.
It states that "the volume of a fixed mass gas varies inversely as the pressure changes if the temperature is constant".
Mathematically;
P1 V1 = P2 V2
Now, insert parameters and solve;
P1 x 3.2 = 59 x 5.5
P1 = 101.4pKa
Answer:
Oxidizing agent - CrO4^2-
Reducing agent- N2O
Explanation:
Let us look at the equation closely;
CrO4^2- (aq) + 3N2O(g) ------------> Cr^3+ (aq) + 3NO(g) [acidic]
The reduction half equation is;
CrO4^2- (aq) + 3e -------->Cr^3+ (aq)
Oxidation half equation is;
3N2O(g) ------>3 NO(g) +3 e
Note that the oxidizing agent participates in the reduction half equation while the reducing agent participates in the oxidation half equation as seen above.
Answer:
Wavelength, 
Explanation:
We have,
Velocity of the object, 
Mass of the object, 
It is required to find the wavelength of the object. The relation between wavelength and velocity of the object is given by :
h is Planck's constant
So, the wavelength of the object is 
.
Answer:
Explanation:
2Al(s) + Fe₂O₃(s) ⟶ Al₂O₃(s) + 2Fe(s); ΔᵣH = ?
The formula for calculating the enthalpy change of a reaction by using the enthalpies of formation of reactants and products is
2Al(s) + Fe₂O₃(s) ⟶ Al₂O₃(s) + 2Fe(s)
ΔfH°/kJ·mol⁻¹: 0 -824.3 -1675.7 0
![\begin{array}{rcl}\Delta_{\text{r}}H^{\circ} & = & [1(-1675.7) + 2(0)] - [2(0) - 1(-824.3)]\\& = & -1675.7 + 824.3\\& = & \textbf{-851.4 kJ/mol}\\\end{array}\\\text{The enthalpy change is } \large \boxed{\textbf{-851.4 kJ/mol}}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Brcl%7D%5CDelta_%7B%5Ctext%7Br%7D%7DH%5E%7B%5Ccirc%7D%20%26%20%3D%20%26%20%5B1%28-1675.7%29%20%2B%202%280%29%5D%20-%20%5B2%280%29%20-%201%28-824.3%29%5D%5C%5C%26%20%3D%20%26%20-1675.7%20%2B%20824.3%5C%5C%26%20%3D%20%26%20%5Ctextbf%7B-851.4%20kJ%2Fmol%7D%5C%5C%5Cend%7Barray%7D%5C%5C%5Ctext%7BThe%20enthalpy%20change%20is%20%7D%20%5Clarge%20%5Cboxed%7B%5Ctextbf%7B-851.4%20kJ%2Fmol%7D%7D)
