Answer:
full moon
Explanation:
....ur welcome..............
Answer:co2
Explanation:because of the oxygen levels
Answer: 72 grams of 
are needed to completely burn 19.7 g 
Explanation:
According to avogadro's law, 1 mole of every substance weighs equal to molecular mass and contains avogadro's number 
of particles.
To calculate the number of moles, we use the equation:
Putting in the values we get:
According to stoichiometry:
1 mole of 
requires 5 moles of oxygen
0.45 moles of require= 
moles of oxygen
Mass of 
72 grams of are needed to completely burn 19.7 g
Answer:
<h2>Lead(II) oxide</h2>
Explanation:
<h3>Lead(II) oxide, also called lead monoxide, is the inorganic compound with the molecular formula PbO. PbO occurs in two polymorphs: litharge having a tetragonal crystal structure, and massicot having an orthorhombic crystal structure. Modern applications for PbO are mostly in lead-based industrial glass and industrial ceramics, including computer components. It is an amphoteric oxide.[3]</h3>
- Other names
- Lead monoxide
- Litharge
- Massicot
- Plumbous oxide
- Galena
<h2> Preparation</h2><h3>PbO may be prepared by heating lead metal in air at approximately 600 °C (1,100 °F). At this temperature it is also the end product of oxidation of other oxides of lead in air:[4]</h3><h3>Thermal decomposition of lead(II) nitrate or lead(II) carbonate also results in the formation of PbO:</h3>
<h3>2 Pb(NO</h3><h3>3)</h3><h3>2 → 2 PbO + 4 NO</h3><h3>2 + O</h3><h3>2</h3><h3>PbCO</h3><h3>3 → PbO + CO2</h3><h3>PbO is produced on a large scale as an intermediate product in refining raw lead ores into metallic lead. The usual lead ore is galena (lead(II) sulfide). At a temperature of around 1,000 °C (1,800 °F) the sulfide is converted to the oxide:[5]</h3>
<h3>2 PbS + 3 O</h3><h3>2 → 2 PbO + 2 SO2</h3><h3>Metallic lead is obtained by reducing PbO with carbon monoxide at around 1,200 °C (2,200 °F):[6]</h3>
<h3>PbO + CO → Pb + CO2</h3>
pls brainlest meh
<span>0.0200 L X 0.06 mol/L HCl = 1.2X10^-3 mol HCl
Moles NaOH added to reach equivalence point = 1.2X10^-3 mol NaOH
Volume NaOH = 1.2X10^-3 mol / 0.0400 M NaOH = 0.030 L NaOH = 30 mL NaOH
</span>
D.There is a rapid change in pH near the equivalence point (pH = 7)
hope it helps
