Answer:B random mutations
Explanation: Because I took the test
Answer:
-476.95 Kj
Explanation:
N2H4(l) + N2O4(g) = 2N2O(g) + 2H20(g)
∆Hrxn = n∆Hf(products) - m∆Hf(reactants)
Where n and m = stoichiometric coefficients of the products and reactants respectively from the balanced chemical equation, ∆Hf = standard enthalpy of formation, ∆Hrxn= standard enthalpy of reaction.
Using the following standard enthalpies of formation ( you did not provide any ):
N2H4(l) = +50.63Kj/mol; N2O4(g) = +9.08Kj/mol; N2O(g) =+33.18Kj/mol; H2O(g) = -241.8Kj/mol
∆Hrxn = [ (2(∆Hf(N2O)) + (2(∆Hf(H2O))] – [(1(∆Hf(N2H4)) + (1(∆Hf(N2O4))]
∆Hrxn = [ 2(+33.18) + 2(-241.8)] – [ (+50.63) + (+9.08)]
∆Hrxn = [ (+66.36)+(-483.6)] – [ +50.63+9.08]
∆Hrxn = [ +66.36-483.6] – [+59.71]
∆Hrxn = -417.24-59.71
∆Hrxn = -476.95 Kj
NOTE: Remember to use the standard enthalpies of formation given to you by your instructor if they differ from the values used herein, and follow the same procedure.
Moles Fe₂O₃ produced : 1.16 moles
<h3>Further explanation</h3>
Given
2.32 moles of iron
Required
moles of Fe2O3
Solution
The reaction coefficient in a chemical equation shows the mole ratio of the reacting compounds (reactants and products)
Reaction
4 Fe + 3 O₂ ⇒ 2 Fe₂O₃
From the equation, mol ratio Fe : mol Fe₂O₃ = 4 : 2, so mol Fe₂O₃ :
= 2/4 x mol Fe
= 2/4 x 2.32
= 1.16 moles
It doesn’t let me see the picture it’s blurry.
<u>Given:</u>
Force acting on the car moving it to the right = F1= 300 N
Force due to friction and air resistance moving it to the left = F2 = 150 N
<u>To determine:</u>
The net force on the car
<u>Explanation:</u>
Force is a vector. The force of friction acts in a direction opposite to the applied force.
If, F1 = 300N, right, then F2 = -150 N, left
Therefore, the net force acting on the car will be:
F = F1+F2 = 300 - 150 = 150 N
<u>Ans</u>: Since the force acting on the right is greater, the car will experience a net force of 150 N which will accelerate it to the right
