Answer:
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Explanation:
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Answer:
No, it is not enthalpy favored since the chemical system gains energy.
Explanation:
The dissolution of ammonium nitrate in water is an endothermic process.
Endothermic process requires the system to gain energy to can dissolve the particles in water.
So, the reaction is not enthalpy favored.
<span>Let's assume
that the oxygen gas has ideal gas behavior.
Then we can use ideal gas formula,
PV = nRT</span>
Where, P is the pressure of the gas (Pa), V is the volume of the gas
(m³), n is the number of moles of gas (mol), R is the universal gas
constant ( 8.314 J mol⁻¹ K⁻¹) and T is temperature in Kelvin.
<span>
P = 2.2 atm = 222915 Pa
V = 21 L = 21 x 10</span>⁻³ m³
n = ?
R = 8.314 J mol⁻¹ K⁻¹
<span>
T = 87 °C = 360 K
By substitution,
</span>222915 Pa x 21 x 10⁻³ m³ = n x 8.314 J mol⁻¹ K⁻<span>¹ x 360 K
n
= 1.56</span><span> mol</span>
<span>
Hence, 1.56 moles of the oxygen gas are </span><span>
left for you to breath.</span><span>
</span>
A complex, ML₆²⁺, is violet. The same metal forms a complex with another ligand, Q, that creates a weaker field. MQ₆²⁺, be expected to show green color.
<h3>What is spectrochemical series?</h3>
The ligands (attachment to a metal ion) are listed in the spectrochemical series according to the strength of their field. The series has been created by superimposing several sequences discovered through spectroscopic research because it is impossible to produce the full series by examining complexes with a single metal ion. The halides are referred to be weak-field ligands whereas the ligands cyanide and CO are strong-field ligands. Medium field effects are claimed to be produced by ligands like water and ammonia.
To know more about the spectrochemical series, visit:
brainly.com/question/27892620
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Explanation:
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