Answer:
Explanation:
<u>1) Data:</u>
a) n = ?
b) V = 50.0 dm³ = 50.0 liter
c) p = 100.0 kPa
d) T = 50 °C
<u>2) Physical law:</u>
<u>3) Constants:</u>
- R = 0.08206 atm-liter / K-mol
<u>4) Unit conversions:</u>
- T = 50 + 273.15 K = 323.15 K
- P = 100.0 kPa × (1 amt /101,325 kPa) = 0.9869 atm
<u>5) Solution:</u>
- n = 0.9869 atm × 50.0 liter / (0.08206 atm-liter /K-mol × 323.15 K)
The solubility of Ce(IO3)4, RaSO4, (NH4)2SeO4 in water are 0.123, 2.1x10^-4 and 96 in units of g/100 mL. Therefore, given 100 mL of water 0.123 g of Ce(IO3)3, 2.1x10^-4 g of RaSO4 and 96 g of (NH4)2SeO4 can be dissolve. Hope this helps.
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About 25 percent of energy in an organism is used by the animal that eats it .