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2 years ago

Calculate the mass of O2 produced after 20.0 grams of KClO3 have reacted?

Chemistry

1 answer:

IrinaVladis [17]2 years ago

5 0

Answer:

$RFM \: of \: potassium \: chlorate = (39 + 35.5 + (16 \times 3)) = 122.5 \: g \\ 122.5 \: g \: of \: potassium \: chlorate \: produce \: (16 \times 3) \: g \: of \: oxygen \\ 20.0 \: g \: of \: potassium \: chlorate \: will \: produce \: (\frac{(20.0 \times 16 \times 3)}{122.5} ) \: g \\ = 7.84 \: g \: of \: oxygen$

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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:

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<u>Step 1:</u> Data given

Mass of the piece of metal = 80.0 grams

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<u>Step 2:</u> 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

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

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3 years ago

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Answer:

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