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3 years ago

How many fluorine atoms are present in 5.85 g of c2f4?

1 answer:

4 0

The molar mass of C2F4 (tetrafluoroethylene) is equal to 100 g/mol. The number of moles of C2F4 in the given amount,
n = (5.85 g) / (100 g/mol) = 0.0585 mols C2F4
The number of molecules per mol is calculated through the equation,
(0.0585 mols) x (6.022 x 10^23)
Since there are 4 F atoms each molecule then,
(0.0585 mols) x (6.022 x 10^23)(4)
= 1.41 x 10^23 atoms of F

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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

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Initial temperature of water =20.0 °C

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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

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

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

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