Step 1: Write the unbalanced equation,
C₂H₆ + O₂ → CO₂ + H₂<span>O
There are 2 C at left hand side and 1 carbon at right hand side. So, multiply CO</span>₂ by 2 to balance C atoms at both side. So,
C₂H₆ + O₂ → 2 CO₂ + H₂O
Now, count number of H atoms at both sides. There are 6 H atoms at left hand side and 2 at right hand side. Multiply H₂O by 3 to balance H atoms.
C₂H₆ + O₂ → 2 CO₂ + 3 H₂O
At last, balance O atoms. There are 2 O atoms at left hand side and 3 O atoms at right hand side. Multiply O₂ with 1.5 (i.e. 3/2) to balance O atoms. i.e.
C₂H₆ + 3/2 O₂ → 2 CO₂ + 3 H₂O
Hence, the equation is balanced. If you want to make equation fraction free then multiply all equation with 2. i.e.
( C₂H₆ + 3/2 O₂ → 2 CO₂ + 3 H₂O ) × 2
2 C₂H₆ + 3 O₂ → 4 CO₂ + 6 H₂O
By considering the reaction equation is:
5Br(aq)+BrO3(aq)+6H(aq)= 3Br2(aq)+3H2O(l)
when the average rate of consumption of Br = 1.86x10^-4 m/s
So from the reaction equation
5Br → 3Br2 when we measure the average rate of formation (X) during the same interval So,
∴ 1.86x10^-4/5 = X / 3
∴X = 1.1 x 10^-4 m/s
∴the average rate of formation of Br2 = 1.1x10^-4 m/s
>4Dt. 1950 y fueron. 45Dys Imox x Na motu cues G.91 %10° mole cultes. 3) How many grams are there in 2.3 x 1024 atoms of silver? 2, 3x 10°tatoms x Imol.<
PH = 0.1289<span> for </span>1.50<span> M solution of weak acid with Ka value of </span><span>.73</span>
Answer:
Mn(s)/Mn^2+(aq)//Co^2+(aq)/Co(s)
Explanation:
In writing the cell notation for an electrochemical cell, the anode is written on the left hand side while the cathode is written on the right hand side. The two half cells are separated by two thick lines which represents the salt bridge.
For the cell discussed in the question; the Mn(s)/Mn^2+(aq) is the anode while the Co^2+(aq)/Co(s) half cell is the cathode.
Hence I can write; Mn(s)/Mn^2+(aq)//Co^2+(aq)/Co(s)
