Solution :
The reasons for the melting point determines :
- the assignment of the sample's purity
- identification of the unknown sample
- the conversion of solid sample to the liquid completely
In order to prepare the sample in the capillary tube for determining its melting point, we tap the tube gently into the sample with the open end of tube down. We then continue tapping the tube till the sample is couple of millimeters high.
Then the other close end of tube down, we tap the sample down again slowly or then dropping the tube into a longer tube so as to move down the sample faster.
Transporting metals, ions, water-insoluble molecules, and hormones. .... When erythrocytes are removed from circulation,
D) to keep as many things the same as possible across the parts of an experiment.
Answer:
1. 65.1 kJ; 2. 558 g
Step-by-step explanation:
1.
M_r: 34.08
2H₂S+ SO₂ ⟶ 3S + 2H₂O ; ΔH = -56.9 kJ
Treat the heat as if it were a product in the equation. Then use the molar ratio (56.9 kJ/2 mol H₂S) in the usual way.
Moles of H₂S = 78.0 g H₂S × (1 mol H₂S/34.08 g H₂S) = 2.289 mol H₂S
Amount of heat = 2.289 mol H₂S × (56.9 kJ/2 mol H₂S) = 65.1 kJ
The reaction releases 65.1 kJ of energy.
2.
M_r: 169.87
2AgNO₃ + BaCl₂ ⟶ 2AgCl + Ba(NO₃)₂; ΔH = -345 kJ
Moles of AgNO₃ = 567 kJ × (2 mol AgNO₃/345 kJ = 3.287 mol AgNO₃
Mass of AgNO₃ = 3.287 mol AgNO₃ × (169.87 g AgNO₃/1 mol AgNO₃)
= 558 g AgNO₃
You need 558 g of AgNO₃.