Q = mcΔT = (4.00 g)(0.129 J/g•°C)(40.85 °C - 0.85 °C)
Q = 20.6 J of energy was involved (more specifically, 20.6 J of heat energy was absorbed from the surroundings by the sample of solid gold).
Chemical properties can be determined by heat combustion, how they react with other chemicals, Oxidization (lose electrons, losing hydrogen, gaining oxygen), or toxicity.
Answer:
d = 0.992 g/L
Explanation:
Data Given:
Pressure of nitric oxide (NO) = 0.866 atm
Temperature of a gas = 46.2° C
Convert the temperature to kelvin = 46.2° C + 273
temperature in kelvin = 319.2 K
density of nitric oxide (NO) = ?
Solution:
Density of a gas can be calculated by
d = PM /RT
Where
d = density
P = Pressure
M = molar mass of gas
R = ideal gas constant = 0.0821 L atm mol⁻¹ K⁻¹
T = temperature
So,
Molar mass of NO = 30 g/mol
Put values in the formula:
d = PM /RT
d = 0.866 atm × 30 g/mol / 0.0821 L atm mol⁻¹ K⁻¹ × 319.2 K
d = 25.98 atm. g/mol / 26.2 L atm mol⁻¹
d = 0.992 g/L
Velocity is speed and Direction. It does not include direction