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3 years ago

Antoine Lavoisier burned metals in sealed jars. What is your prediction of his experimental results?

Chemistry

1 answer:

Aliun [14]3 years ago

8 0

Answer:

The mass of the jar and contents remained the same after the metal was burned.

Explanation:

My prediction about the experimental results is that the mass of the jar and contents remained the same after the metal was burned in the jar.

This is compliance with the law of conservation of mass which states that in a chemical reaction, matter is neither created nor destroyed by bonds are rearranged for new compounds to form.

In compliance with this law, it is expected that the mass of the jar and its content will remain the same before and after the reaction.
No new material was added and no material was removed from the jar.

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Answer:

Q = 1267720 J

Explanation:

Qt = QH2O + ΔHv

∴ QH2O = mCpΔT

∴ m H2O = 500 g

∴ Cp H2O = 4.186 J/g°C = 4.183 E-3 KJ/g°C

∴ ΔT = 120 - 50 = 70°C

⇒ QH2O = (500 g)(4.183 E-3 KJ/g°C)(70°C) = 146.51 KJ

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3 years ago

What volume is occupied by 18.4 g oxygen at 28.0°C and a pressure of 0.998 torr? (round to sig figs)

Alexxandr [17]

Answer:

11.0 dm³

Explanation:

From the question,

Applying

PV= nRT............... Equation 1

Where P = pressure of oxygen gas, V = volume of oxygen gas, n = number of moles of oxygen, R = molar constant, T = Temperature.

make V the subeject of the equation

V = nRT/P............. Equation 2

But,

Number of mole (n) = Mass of oxygen(m)/Molar mass of oxygen(m')

n = m/m'....................... Equation 3

Substitute equation 3 into equation 2

V = mRT/Pm'............. Equation 4

Given: T = 28°C = (28+273) = 301 K, P = 0.998 torr = (0.998×0.00131579) = 1.3132 atm, m = 18.4 g

Constant: R = 0.082 atm.dm³/K.mol, m' = 32 g/mol.

Substitute these values into equation 4

V = (301×18.4×0.082)/(32×1.3132)

V = 454.1488/42.0224

V = 10.81 dm³

V = 11.0 dm³

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