<span>c. Passing electric charge through the reactants Is the answer to you're question.
</span>
Answer:
O H C
Moles in 100g 3.33 6.65 3.33
Ratio 1.00 2.00 1.00
Possible empirical formula = 
I would say the answer is liquids
Answer:
0.296 J/g°C
Explanation:
Step 1:
Data obtained from the question.
Mass (M) =35g
Heat Absorbed (Q) = 1606 J
Initial temperature (T1) = 10°C
Final temperature (T2) = 165°C
Change in temperature (ΔT) = T2 – T1 = 165°C – 10°C = 155°C
Specific heat capacity (C) =..?
Step 2:
Determination of the specific heat capacity of iron.
Q = MCΔT
C = Q/MΔT
C = 1606 / (35 x 155)
C = 0.296 J/g°C
Therefore, the specific heat capacity of iron is 0.296 J/g°C
<u>Given:</u>
Initial concentration of potassium iodate (KIO3) M1 = 0.31 M
Initial volume of KIO3 (stock solution) V1 = 10 ml
Final volume of KIO3 V2 = 100 ml
<u>To determine:</u>
The final concentration of KIO3 i.e. M2
<u>Explanation:</u>
Use the relation-
M1V1 = M2V2
M2 = M1V1/V2 = 0.31 M * 10 ml/100 ml = 0.031 M
Ans: The concentration of KIO3 after dilution is 0.031 M
