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Slav-nsk [51]
2 years ago
10

Limestone (CaCO₃) is the second most abundant mineral on Earth after SiO₂. For many uses, it is first decomposed thermally to qu

icklime (CaO). MgO is prepared similarly from MgCO₃.(a) At what T is each decomposition spontaneous?
Chemistry
1 answer:
steposvetlana [31]2 years ago
5 0

Limestone (CaCO₃) is the second most abundant mineral on Earth after SiO₂. For many uses, it is first decomposed thermally to quicklime (CaO). MgO is prepared similarly from MgCO₃.

AT 871°C CaCO 3 needs about 1 hour for complete decomposition.

<h3>At what temperature Caco3 decompose to Cao?</h3>

At any temperature higher than 835°C, the value of Δ G ∘ will be negative and the decomposition reaction will be spontaneous.

AT 871°C CaCO 3 needs about 1 hour for complete decomposition.

Calcium carbonate decomposes on heating to give calcium oxide and carbon dioxide.

To learn more about Caco3, refer

brainly.com/question/26187973

#SPJ4

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What is the temperature change if 400 J of energy is added to 10 grams of water
Gemiola [76]

The temperature change if 400 J of energy is added to 10 grams of water is 9.57°C.

<h3>How to calculate temperature change?</h3>

The temperature change of a calorimeter can be calculated using the following expression:

E = mc∆T

Where;

  • E = energy in joules
  • m = mass
  • c = specific heat capacity = 4.18J/g°C
  • ∆T = change in temperature

400 = 10 × 4.18 × ∆T

400 = 41.8∆T

∆T = 400/41.8

∆T = 9.57°C

Therefore, the temperature change if 400 J of energy is added to 10 grams of water is 9.57°C.

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3 0
3 years ago
Click on the alignment to view it at a larger size. Scan along the aligned sequences, letter by letter, noting any positions whe
miss Akunina [59]

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8 amino acids

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Read 2 more answers
What is the boiling point of a solution that contains 3 moles of KBr in 2000 g of water?
bija089 [108]

Answer:

C. 101.5°C

Explanation:

The computation of the boiling point of a solution is shown below:

As we know that

\Delta T = i K_bm

where,

i = Factor of Van’t Hoff i.e 2KBr

K_b = Boiling point i.e 0.512° C

m = molality  = \frac{Moles\ of\ solute}{Solvent\ mass}

Now place the values to the above formula

So, the boiling point is

(T - 100^\circ C) = 2\times 0.512^\circ \times \frac{3.00\ ml}{2.00\ kg}

After solving this, the t is

= 100° C + 1.536° C

= 101.536° C

Hence, the correct option is C.

5 0
3 years ago
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