Hey There:
First order half life equation :
T 1/2 = ln ( 2 ) /K
T 1/2 = 0.693 / 3.40
T 1/2 = 0.204 s
Answer B
Answer:
Inorganic Phosphorus,phosphate
Answer:
ΔE = 150 J
Explanation:
From first law of thermodynamics, we know that;
ΔE = q + w
Where;
ΔE is change in internal energy
q is total amount of heat energy going in or coming out
w is total amount of work expended or received
From the question, the system receives 575 J of heat. Thus, q = +575 J
Also, we are told that the system delivered 425 J of work. Thus, w = -425 J since work was expended.
Thus;
ΔE = 575 + (-425)
ΔE = 575 - 425
ΔE = 150 J
It seems like the answer is C, you really just need to use the process of elimination.
The pressure of the CO₂ = 0.995 atm
<h3>Further explanation</h3>
The complete question
<em>A student is doing experiments with CO2(g). Originally, a sample of gas is in a rigid container at 299K and 0.70 atm. The student increases the temperature of the CO2(g) in the container to 425K.</em>
<em>Calculate the pressure of the CO₂ (g) in the container at 425 K.</em>
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<em />
Gay Lussac's Law
When the volume is not changed, the gas pressure is proportional to its absolute temperature
P₁=0.7 atm
T₁=299 K
T₂=425 K
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