Answer:
(FeSCN⁺²) = 0.11 mM
Explanation:
Fe ( NO3)3 (aq) [0.200M] + KSCN (aq) [ 0.002M] ⇒ FeSCN+2
M (Fe(NO₃)₃ = 0.200 M
V (Fe(NO₃)₃ = 10.63 mL
n (Fe(NO₃)₃ = 0.200*10.63 = 2.126 mmol
M (KSCN) = 0.00200 M
V (KSCN) = 1.42 mL
n (KSCN) = 0.00200 * 1.42 = 0.00284 mmol
Total volume = V (Fe(NO₃)₃ + V (KSCN)
= 10.63 + 1.42
= 12.05 mL
Limiting reactant = KSCN
So,
FeSCN⁺² = 0.00284 mmol
M (FeSCN⁺²) = 0.00284/12.05
= 0.000236 M
Excess reactant = (Fe(NO₃)₃
n(Fe(NO₃)₃ = 2.126 mmol - 0.00284 mmol
=2.123 mmol
For standard 2:
n (FeSCN⁺²) = 0.000236 * 4.63
=0.00109
V(standard 2) = 4.63 + 5.17
= 9.8 mL
M (FeSCN⁺²) = 0.00109/9.8
= 0.000111 M = 0.11 mM
Therefore, (FeSCN⁺²) = 0.11 mM
Answer: because they take a lot of tie to form
The formula C2H4 can be classified as both a molecular formula
and an empirical formula. The answer is number 4.
Answer:
C. 1.35
Explanation:
2NH3 (g) <--> N2 (g) + 3H2 (g)
Initial concentration 2.2 mol/0.95L 1.1 mol/0.95L 0
change in concentration 2x x 3x
-0.84 M +0.42M +1.26M
Equilibrium 1.4 mol/0.95L=1.47M 1.58 M 1.26 M
concentration
Change in concentration(NH3) = (2.2-1.4)mol/0.95 L = 0.84M
Equilibrium concentration (N2) = 1.1/0.95 +0.42=1.58 M
Equilibrium concentration(NH3) = 1.4/0.95 = 1.47M
K = [N2]*{H2]/[NH3] = 1.58M*1.26M/1.47M = 1.35 M
Answer:
0.1113 mol
Explanation:
Data Given:
no. of atoms of CH₄= 6.70 x 10²² atoms
no. of moles of methane (CH₄) = ?
Solution:
we will find no. of moles of methane (CH₄)
Formula used
no. of moles = no. of atoms / Avogadro's number
Where
Avogadro's number = 6.022 x 10²³
Put values in above equation
no. of moles = 6.70 x 10²² atoms / 6.022 x 10²³ (atoms/mol)
no. of moles = 0.1113 mol
So,
There are 0.1113 moles of methane.
