Question

an alloy of iron (53.6%), nickel (45.8%) and manganese (0.6%) has density of 8.17 g/cm. calculate the number of ni atoms present in the alloy of dimensions 10.0cm 20.0cm 15.0cm.

Answer 1

Answer:

The number of Ni atoms present in the alloyis $1.1518\times 10^{26} atoms$.

Explanation:

Density of an alloy = $8.17 g/cm^3$

Volume of an alloy = 10.0 cm × 20.0 cm × 15.0 cm =$3000.000 cm^3$

Mass of an alloy = $Density\times Volume = 8.17 g/cm^3\times 3000.000 cm^3=24,510 g$

Percentage of nickel in alloy = 45.8%

Mass of nickel= $24,510 g\times \frac{45.8}{100}=11,225.58 g$

Moles of nickel = $\frac{112,225.58 g}{58.69 g/mol}=191.2690 mol$

Number of atoms of nickel:

$191.2690\times 6.022\times 10^{23}=1.1518\times 10^{26} atoms$

The number of Ni atoms present in the alloyis $1.1518\times 10^{26} atoms$.

Answer 2

Answer:

answer:-

Explanation:

Volume of the block of alloy

=10cm×20cm×15cm=3000cm^3.

Density of the alloy is 8.17g/cm3.

Mass of the block of alloy .

=density×volume=8.17×3000=24510g.

The amount of iron present in the block of alloy =54.7/100×24510=13407g

The number of iron atoms present in the block =

13407/55.8×6.02×1023=1.442×1026.