Please need this ASAP. Calculate the mass of lime, CaO, that would be produced from 250 tonnes of limestone, CaCO3.
Relative atomic masses: C 12; O 16; Ca 40.

1.4×10⁸ g of CaO

Explanation:

We'll begin by converting 250 tonnes to grams (g). This can be obtained as follow:

1 tonne = 1×10⁶ g

Therefore,

250 tonne = 250 × 1×10⁶

250 tonne = 2.5×10⁸ g

Next, the balanced equation for the reaction.

CaCO₃ —> CaO + CO₂

Next, we shall determine the mass of CaCO₃ that decomposed and the mass CaO produced from the balanced equation. This can be obtained as follow:

Molar mass of CaCO₃ = 40 + 12 + (16×3)

= 40 + 12 + 48

= 100 g/mol

Mass of CaCO₃ from the balanced equation = 1 × 100 = 100 g

Molar mass of CaO = 40 + 16

= 56 g/mol

Mass of CaO from the balanced equation = 1 × 56 = 56 g

SUMMARY:

From the balanced equation above,

100 g of CaCO₃ decomposed to produce 56 g of CaO.

Finally, we shall determine the mass of CaO produced by the decomposition of 250 tonnes (i.e 2.5×10⁸ g) of CaCO₃. This can be obtained as follow:

From the balanced equation above,

100 g of CaCO₃ decomposed to produce 56 g of CaO.

Therefore, 2.5×10⁸ g of CaCO₃ will decompose to produce =

(2.5×10⁸ × 56)/100 = 1.4×10⁸ g of CaO.

Thus, 1.4×10⁸ g of CaO will be obtained from 250 tonnes (i.e 2.5×10⁸ g) of CaCO₃.

the density of gold = 19.3 g/cm3

the relation between density and mass and volume is

density = mass / volume

given :

mass of gold = 13 g

density = 13g / volume = 19.3

volume = 13 /19.3 = 0.67 cm^3

thus volume of gold nugget = 0.67 cm^3