As per the given chemical formula- Na2CO3.10H2O, one mole of the chemical compound contains 13 moles of oxygen atoms. Hence

Number of moles of oxygen atoms in one mole of Na2CO3.10H2O = 13

number of moles of oxygen atoms in 0.2 moles of Na2CO3.10H2O = 13 X 0.2 = 2.6

Now, one mole of a substance contains 6.022 X 10^23 particles of the substance. Thus

number of atoms of oxygen in one mole of oxygen atom = 6.022 X 10^23

number of moles of oxygen atoms in 2.6 moles of oxygen atoms = 2.6 X 6.022 X 10^23 = 15.657 X 10^23

= 1.566 X 10^24

Thus, there are 1.566 X 10^24 atoms of oxygen in 0.2 moles of Na2CO3.10H2O.
4.7

7 0

2 years ago

A 80.0 g piece of metal at 88.0°C is placed in 125 g of water at 20.0°C contained in a calorimeter. The metal and water come to

grandymaker [24]

Answer:

The specific heat of the metal is 0.485 J/g°C

Explanation:

Step 1: Data given

Mass of the piece of metal = 80.0 grams

Mass of the water = 125 grams

Initial temperature of the metal = 88.0 °C

Initial temperature of water =20.0 °C

Final temperature = 24.7 °C

pecific heat of water is 4.18 J/g*°C

Step 2: Calculate specific heat of the metal

Qgained = -Qlost

Q =m*c*ΔT

Qwater = - Qmetal

m(water)*c(water)*ΔT(water) = -m(metal)*c(metal)*ΔT(metal)

with mass of water = 125 grams

with c( water) = 4.18 J/g°C

with ΔT(water) = T2-T1 = 24.7 - 20 = 4.7°C

with mass of metal = 80.0 grams

with c(metal) = TO BE DETERMINED

with ΔT(metal) = 24.7 - 88.0 = -63.3 °C

125*4.18*4.7 = -80 * C(metal) * -63.3

2455.75 = -80 * C(metal) * -63.3

C(metal) = 2455.75 / (-80*-63.3)

C(metal) = 0.485 J/g°C

The specific heat of the metal is 0.485 J/g°C

3 0

3 years ago