Answer:- D. 1.8 moles of Fe and 
.
Solution:- The balanced equation is:
let's first figure out the limiting reactant using the given moles and mol ratio:
= 5.4 mol CO
From calculations, 5.4 moles of CO are required to react completely with 1.8 moles of Iron(III)oxide but only 2.7 moles of CO are available. It means CO is limiting reactant.
Products moles depends on limiting reactant. Let's calculate the moles of each reactant formed for given 2.7 moles of CO.
= 1.8 mol Fe
=
So, the correct choice is D. 1.8 moles of Fe and are formed.
They all have 2 elections in the outer S orbital, and 4 electrons in the P orbitals.
The answer is C : 15.7 m/s
Use the idea of : momentum before collision = momentum after collision
Before collision;
For car:mass=1.1×10^3, velocity=22
For truck:mass=2.3×10^3, velocity=0
After collision;
For car:mass=2.3×10^3, velocity=-11
For truck:mass=2.3×10^3, velocity=V
(1.1×10^3 × 22) + (2.3×10^3 × 0) = (1.1×10^3 × -11) + (2.3×10^3 × V)
24200 = -12100 + 2.3×10^3V
2.3×10^3V = 36300
V = 15.7 m/s
To answer the question above, multiply the given number of moles by the molar masses.
(A) (0.20 mole) x (32 g / 1 mole) = 6.4 grams O2
(B) (0.75 mole) x (62 g / 1 mole) = 46.5 grams H2CO3
(C) (3.42 moles) x (28 g / 1 mole) = 95.7 grams CO
(D) (4.1 moles) x (29.88 g / 1 mole) = 122.508 g Li2O
The answer to the question above is letter D.
Au^2S^3+ 3H^2 = 2Au + 3H^2S
