<u>Answer:</u> The mass percent of lead in lead (IV) carbonate is 63.32 %
<u>Explanation:</u>
The given chemical formula of lead (IV) carbonate is 
To calculate the mass percentage of lead in lead (IV) carbonate, we use the equation:

Mass of lead = (1 × 207.2) = 207.2 g
Mass of lead (IV) carbonate = [(1 × 207.2) + (2 × 12) + (6 × 16)] = 327.2 g
Putting values in above equation, we get:

Hence, the mass percent of lead in lead (IV) carbonate is 63.32 %
Answer:
Phosphorus atoms can bond with oxygen atoms to form ester groups. These can bond with carbon atoms, yielding a large number of organic phosphorus chemicals. These are found in many important biological processes
Answer:
The partial pressure of CO is 5.54x10⁻⁴⁹atm. You shouldn't worry because it is very low pressure
Explanation:
First, the balanced reaction is:
CO + 1/2O₂ → CO₂
The energies of formation are:
ΔG(CO)=-137.168kJ/mol
ΔG(O₂)=0
ΔG(CO₂)=-394.359kJ/mol
The energy of the reaction is:

The expression for calculate the partial pressure of CO is:

<em>Answer:</em>
<h3><em>Answer:</em><em> </em><em>well</em><em> </em><em> </em></h3>
<em>b. a type of gas is evolved ( hydrogen gas</em><em> )</em>
<em> </em>
½H2(g) + ½I2(g) → HI(g) ΔH = +6.2 kcal/mol
or...
½H2(g) + ½I2(g) + 6,2kcal/mole → HI(g)
________
21.0 kcal/mole + C(s) + 2S(s) → CS2(l)
or...
C(s) + 2S(s) → CS2(l) ΔH = +2,1 kcal/mole
_________
ΔH > 0 ----------->>> ENDOTHERMIC REACTIONS