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.
Method 1: gravimetry
advantages: Impurities in the sample can be identified
disadvantages: The process is long, because it goes through several stages
Method 2: titration
advantages: the process is fast, because the titrate and titrant react immediately
disadvantages: Sometimes the determination of the end point of the titration is carried out too fast or too slowly so that the calculations carried out are inaccurate
<u>Answer:</u> The 
for the reaction is -1835 kJ.
<u>Explanation:</u>
Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.
According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.
The given chemical reaction follows:
The intermediate balanced chemical reaction are:
(1) 
( × 4)
(2) 
The expression for enthalpy of the reaction follows:
![\Delta H^o_{rxn}=[4\times (-\Delta H_1)]+[1\times \Delta H_2]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B4%5Ctimes%20%28-%5CDelta%20H_1%29%5D%2B%5B1%5Ctimes%20%5CDelta%20H_2%5D)
Putting values in above equation, we get:
Hence, the for the reaction is -1835 kJ.
Answer:
6.319857 * 10 to the power 7
I believe it's a pure element
