D. All of the above. Developing medicine, analyzing compounds and producing new product such as plastic all have to deal with chemistry.
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>We can use the heat
equation,
Q = mcΔT </span>
<span>Where Q is
the amount of energy transferred (J), m is the mass of the
substance (kg), c is the specific heat (J g</span>⁻¹ °C⁻<span>¹) and ΔT is the temperature
difference (°C).</span>
Density = mass / volume
The density of water = 0.997 g/mL
<span>Hence mass of 1.25 L (1250 mL) of water = 0.997 g/mL x 1250 mL</span>
<span> = 1246.25 g</span>
Specific heat capacity of water = 4.186 J<span>/ g °C.</span>
Let's assume that there is no heat loss to the surrounding and the final temperature is T.
By applying the equation,
5430 J = 1246.25 g x 4.186 J/ g °C x (T - 23) °C
(T - 23) °C = 5430 J / 1246.25 g x 4.186 J/ g °C
(T - 23) °C = 1.04 °C
T = 1.04 °C + 23 °C
T = 24.04 °C
Hence, the final temperature of the water is 24.04 °C.
Answer:
a. Are potentially dangerous and should not be worn.
Answer:
<h3>CH3OH (Methanol)</h3>
MM = 12 + 8 + 16 = 36u
