There are several hazards associated with the chemicals in the qualitative analysis experiment. for each chemical, identify the
associated hazards. (select all that apply.) hazard 0.1 m ag+ 0.1 m ba2+ 0.1 m fe3+ 6 m hcl 6 m h2so4 6 m hno3 7.5 m nh3 corrosive irritating vapors toxic if ingested causes staining on skin flammable none listed
You can determine the hazards of these chemicals by looking at their material data safety sheets (MSDS).
1. 0.1 M Ag⁺: Silver compounds are absorbed by skin causing bluish pigmentation. Thus, it <em>causes </em><span><em>staining on skin</em>.</span> Also, liquid <em>vapor may be irritating</em> to skin and also <em>moderately toxic when ingested</em>.
2. 0.1 M Ba²⁺: This is <em>mildly toxic when ingested</em> causing stomach irritation, muscle weakness, swelling of organs like brain, liver, kidney and heart.
3. 0.1 M Fe³⁺:Iron is <em>corrosive, has irritating vapor especially to the eyes, and toxic if ingested</em>.
4. 6 M HCl: This is a concentrated strong acid, so it is <em>corrosive, has irritating vapors, flammable and toxic when ingested</em>.
5. 6 M H₂SO₄: This is also a concentrated strong acid. Moreover, it is a strong oxidizing agent. So, its hazards include: <span><em>corrosive, has irritating vapors, toxic when ingested and causes staining on skin</em>.
</span>6. 6 M HNO₃: This is a concentrated strong acid, so it is <em>corrosive, has irritating vapors, flammable and toxic when ingested</em>.
7. 7.5 M NH₃: This is a weak base. It is characterized for its pungent odor. This is <em>corrosive, has irritating vapors, toxic if ingested, and flammable</em>.
Carbon is a highly covalent element due to the presence of four valence electrons in its outermost shell. This helps it bond covalently with other elements in various ways.
Carbon is known to be the backbone for most compounds. The carbon backbone vary in length and in the mode of bonding such as having the ability to from straight, branched, rings, double or triple bond which contibutes to the diversity and complexity of organic molecules.
This is equal to the heat lost by the metal, so calculate Cp for the metal, given: q = -419 J (negative because heat was lost) m = 5.00 g ΔT = 15.0°C - 100.0°C = -85.0°C (negative because the temperature decreased)