3.97×1023 molecules C2H6 1 mol C2H6
------------------------------------------ x ------------------------------------ = 0.66 mol C2H6
6.022 x 1023 molec. C2H6
Answer:
[KCl] = 1.2 M
Explanation:
We need to complete the reaction:
2KCl(aq) + Pb(NO₃)₂(aq) → 2KNO₃(aq) + PbCl₂(s)↓
By stoichiomety we know that 1 mol of chloride needs 1 mol of nitrate to react:
Let's find out the moles of nitrate, we have:
Molarity = mol/volume(L)
We convert the volume → 30 mL . 1L/1000mL = 0.030L
Molarity . volume(L) = moles → 0.400 M . 0.030L = 0.012 moles
Therefore, we can make a rule of three.
1 mol of nitrate reacts with 2 moles of chloride
Then, 0.012 moles of nitrate must react with (0.012 . 2) / 1 = 0.024 moles of KCl
We convert the volume from mL to L → 20 mL . 1L /1000mL = 0.020L
Molarity = mol /volume(L) → 0.024 mol /0.020L = 1.2 M
Answer:
The compound amount is $303.03 .
Explanation:
Formula for compound interest:
Principle amount = $1100
Rate of the interest compounded semiannually :
= R = 9% = 0.09
Number of times interest compounded, n = 
(semi means two times in a year)
Time period = T = 14
The compound amount is $303.03 .
Answer: pH of HCl =5, HNO3 = 1,
NaOH = 9, KOH = 12
Explanation:
pH = -log [H+ ]
1. 1.0 x 10^-5 M HCl
pH = - log (1.0 x 10^-5)
= 5 - log 1 = 5
2. 0.1 M HNO3
pH = - log (1.0 x 10 ^ -1)
pH = 1 - log 1 = 1
3. 1.0 x 10^-5 NaOH
pOH = - log (1.0 x 10^-5)
pOH = 5 - log 1 = 5
pH + pOH = 14
Therefore , pH = 14 - 5 = 9
4. 0.01 M KOH
pOH = - log ( 1.0 x 10^ -2)
= 2 - log 1 = 2
pH + pOH = 14
Therefore, pH = 14 - 2 = 12
