Answer:
14.533 grams of solid precipitate of mercury(II) dichromate will form.
Explanation:
Moles of mercury(II) acetate = 
Moles of sodium dichromate = 
According to reaction , 1 mole of sodium dichromate reacts with 1 mole of mercury(II) acetate , then 0.045906 moles of sodium dichromate will recat with :
of mercury(II) acetate
This means that mercury(II) acetate is present in an excess amount and sodium dichromate is present in limiting amount.So, amount of precipitate will depend upon moles of sodium dichromate.
According to reaction , 1 mole of sodium dichromate gives 1 mole of mercury(II) dichromate , then 0.045906 moles of sodium dichromate will give :
of mercury(II) dichromate
Mass of 0.045906 moles of mercury(II) dichromate:
0.045906 mol × 316.59 g/mol = 14.533 g
14.533 grams of solid precipitate of mercury(II) dichromate will form.
Answer:
E_a = 103.626 × 10³ KJ/mol
Explanation:
Formula to solve this is given by;
Log(k2/k1) = (E_a/2.303R)((1/T1) - (1/T2))
Where;
k2 is rate constant at second temperature
k1 is rate constant at first temperature
R is universal gas constant
T1 is first temperature
T2 is second temperature
We are given;
k1 = 2.8 × 10^(-3) /s
k2 = 4.8 × 10^(-4) /s
R = 8.314 J/mol.k
T1 = 60°C = 333.15 K
T2 = 45°C = 318.15 K
Thus;
Log((4.8 × 10^(-4))/(2.8 × 10^(-3))) = (E_a/(2.303 × 8.314))((1/333.15) - (1/318.15))
We now have;
-0.76592 = -0.00000739121E_a
E_a = -0.76592/-0.00000739121
E_a = 103.626 × 10³ KJ/mol
The Correct choice is D. Fe + S = FeS
because, in nature Iron (Fe) and Sulphur (S) are found in mono - Atomic form. so, There will be 1 Fe atom reacting with 1 S atom to form FeS (Iron sulphide)
Answer:
The answer to your question is letter B.
Explanation:
Process
1.- Calculate the molecular mass of Ferric oxide
Fe = 56g x 2 = 112 g
O = 16 x 3 = 48 g
Mass = 112 + 48 = 160 g
2.- Calculate the mass of 0.7891 moles of Ferric oxide
160 g of Fe₂O₃ ------------- 1 mol
x ------------- 0.7891 mol
x = (160 x 0.7891) / 1
x = 126.2 g
Activation energy
Hope this helps
