In antoine lavoisier’s classic experiment, mercuric oxide is heated in a sealed container. the solid red powder is changed into two products: silver liquid mercury and oxygen gas. if lavoisier heated 100 grams of powdered mercuric oxide to produce 93 grams of liquid mercury, how much oxygen would be released?

a). 7 grams
b). 16 grams
c). 32 grams
d). 93 grams

A) 7 grams

Explanation:

The chemical reaction is:

2HgO\rightarrow 2Hg + O2

Amount of HgO heated = 100 g

Molar mass of HgO = 216.59 g/mol

Molar mass of O2 = 32.00 g/mol

Moles of HgO heated =

Based on the reaction stoichiometry:

2 moles of HgO produces 1 mole of O2

Therefore, 0.462 moles of HgO would produce:

=

Molar Mass of O2 = 32 g/mol

Mass of O2 =

pla·gia·rism

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noun

the practice of taking someone else's work or ideas and passing them off as one's own.

explanation:

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol