It is two or more objects and different things that can be removed from each other.
Answer:
∆H° rxn = - 93 kJ
Explanation:
Recall that a change in standard in enthalpy, ∆H°, can be calculated from the inventory of the energies, H, of the bonds broken minus bonds formed (H according to Hess Law.
We need to find in an appropiate reference table the bond energies for all the species in the reactions and then compute the result.
N₂ (g) + 3H₂ (g) ⇒ 2NH₃ (g)
1 N≡N = 1(945 kJ/mol) 3 H-H = 3 (432 kJ/mol) 6 N-H = 6 ( 389 kJ/mol)
∆H° rxn = ∑ H bonds broken - ∑ H bonds formed
∆H° rxn = [ 1(945 kJ) + 3 (432 kJ) ] - [ 6 (389 k J]
∆H° rxn = 2,241 kJ -2334 kJ = -93 kJ
be careful when reading values from the reference table since you will find listed N-N bond energy (single bond), but we have instead a triple bond, N≡N, we have to use this one .
For stainless steel different kinds of compositions are used. Based on that different series of stainless steel has been coined.
1. Series 200 - Iron alloyed with <span>chromium, nickel and manganese.
2. Series 300 - It has
a. Stainless Steel 304 - it has composition of 18% chromium and 8% Nickel
b. </span>Stainless Steel 316 - This has 18% chromium and 10% Nickel
Each kind of stainless steel is of different cost and has different applications.
You need evidence to support a conclustion
Answer:
- <u><em>1.7 × 10³ kg of ore.</em></u>
Explanation:
Call X the amount of aluminum ore mined to produce 1.0 × 10³ kg the aluminum metal.
Then, taking into account the yield of the reaction (82 % = 0.82) and the percent of aluminun in the ore (71% = 0.71), you can write the following equation:
- X × 71% × 82% = 1.0 × 10³ kg
↑ ↑ ↑ ↑
(mass of ore) (% of Al in the ore) (yield) ( Al metal to obtain)
You must just simplify, solve and compute:
- X = 1,000 / (0.71 × 0.82) = 1,000 / 0.5822 = 1,717.6 Kg
Round to two significant figures; 1,700 kg = 1.7 × 10³ kg of ore ← answer.
