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aivan3 [116]
1 year ago
12

Calcium carbide, CaC2, is manufactured by reducing lime with carbon at high temperature. (The carbide is used in turn to make ac

etylene, an industrially important organic chemical.) CaO(s)+3C(s)→CaC2(s)+CO(g) ΔH=464.8kJ Calculate the quantity of energy, in MJ, transferred when a metric ton(1000 kg) of CaC2(s) is manufactured. Ca = 40 g/mol C = 12 g/mol O = 16 g/mol
Chemistry
1 answer:
sweet-ann [11.9K]1 year ago
7 0

The energy of the carbide released is 7262.5MJ.

<h3>What is the energy?</h3>

We know that the reaction between calcium  oxide and carbon occurs in accordance with the reaction; CaO(s)+3C(s)----- > CaC_{2} (s)+CO(g). The reaction is seen to produce 464.8kJ of energy per mole of carbide produced.

Number of moles of CaC_{2} produced = 1000 * 10^3 g/64 g/mol

= 15625 moles of calcium carbide

If 1 mole of CaC_{2} transfers  464.8 * 10^3 J

15625 moles of calcium carbide transfers 15625 moles  * 464.8 * 10^3 J/ 1 mol

= 7262.5MJ

Learn more about reaction enthalpy:brainly.com/question/1657608

#SPJ1

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-BARSIC- [3]

Answer:

The chemist would require to use 43.43 grams.

Explanation:

In order to solve this problem we need to know<u> how much do 0.550 moles of selenium weigh</u>. To do that we use selenium's<em> molar mass </em>and multiply it by the given number of moles:

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4 0
2 years ago
The combustion reaction described in part (b) occurred in a closed room containing 5.56 10g of air
ziro4ka [17]

Answer:

Explanation:

Combustion reaction is given below,

C₂H₅OH(l) + 3O₂(g) ⇒ 2CO₂(g) + 3H₂O(g)

Provided that such a combustion has a normal enthalpy,

ΔH°rxn = -1270 kJ/mol

That would be 1 mol reacting to release of ethanol,

⇒ -1270 kJ of heat

Now,

0.383 Ethanol mol responds to release or unlock,

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Given that :

specific heat c = 1.005 J/(g. °C)

m = 5.56 ×10⁴ g

Using the relation:

q = mcΔT

- 486.34 =  5.56 ×10⁴  × 1.005 × ΔT

ΔT= (486.34 × 1000 )/5.56×10⁴  × 1.005

ΔT= 836.88 °C

ΔT= T₂ - T₁

T₂ =  ΔT +  T₁

T₂ = 836.88 °C + 21.7°C

T₂ = 858.58 °C

Therefore, the final temperature of the air in the room after combustion is 858.58 °C

4 0
3 years ago
How would you prepare 3.5 L of a 0.9M solution of KCl?
Fudgin [204]
Calculate the mass of the solute <span>in the solution :

Molar mass KCl = </span><span>74.55 g/mol

m = Molarity * molar mass * volume

m = 0.9 * 74.55 * 3.5

m = 234.8325 g 

</span><span>To prepare 0.9 M KCl solution, weigh 234.8325 g of salt in an analytical balance, dissolve in a beaker, shortly after transfer with the help of a funnel of transfer to a volumetric flask of 100 cm</span>³<span> and complete with water up to the mark, then cover the balloon and finally shake the solution to mix

hope this helps!</span>
8 0
3 years ago
Identify the reversible chemical reaction below. A. reactants=&gt;products B. Products &lt;=&gt; reactants C. products =&gt; rea
Korvikt [17]

Answer:

Its B

Explanation:

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If the reaction is endothermic, it will be favoured by increase in temperature and equilibrium position will shift to the right ( reactants )

If the reaction is exothermic, its the vice versa

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5 0
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Answer:

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Explanation:

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3 0
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