Answer:
The answer to your question is P2 = 2676.6 kPa
Explanation:
Data
Volume 1 = V1 = 12.8 L Volume 2 = V2 = 855 ml
Temperature 1 = T1 = -108°C Temperature 2 = 22°C
Pressure 1 = P1 = 100 kPa Pressure 2 = P2 = ?
Process
- To solve this problem use the Combined gas law.
P1V1/T1 = P2V2/T2
-Solve for P2
P2 = P1V1T2 / T1V2
- Convert temperature to °K
T1 = -108 + 273 = 165°K
T2 = 22 + 273 = 295°K
- Convert volume 2 to liters
1000 ml -------------------- 1 l
855 ml -------------------- x
x = (855 x 1) / 1000
x = 0.855 l
-Substitution
P2 = (12.8 x 100 x 295) / (165 x 0.855)
-Simplification
P2 = 377600 / 141.075
-Result
P2 = 2676.6 kPa
Hypoventilation can cause oxygen levels to fall too low, a condition called Hypoxia and carbon dioxide levels may rise too high, a condition called Hypercapnia.
Hypoxia is a state in which there is insufficient oxygen reaching the tissues of the body or a specific area of the body.
Generalized hypoxia, which affects the entire body, and local hypoxia, which affects a specific area of the body, are the two types of hypoxia.
Although fluctuations in arterial oxygen concentrations are frequently associated with clinical conditions, they can also occur naturally during severe physical activity or hypoventilation training.
A rise in carbon dioxide partial pressure (PaCO2) above 45 mmHg is referred to as hypercapnia.
The body produces carbon dioxide as a metabolic byproduct of its numerous cellular functions, and it has a number of physiological systems at its disposal to control its levels.
Learn more about Hypoxia here brainly.com/question/13870938
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Question is incomplete, complete question is;
A 34.8 mL solution of 
(aq) of an unknown concentration was titrated with 0.15 M of NaOH(aq).
If it takes 20.4 mL of NaOH(aq) to reach the equivalence point of the titration, what is the molarity of 
? For your answer, only type in the numerical value with two significant figures. Do NOT include the unit.
Answer:
0.044 M is the molarity of (aq).
Explanation:
The reaction taking place here is in between acid and base which means that it is a neutralization reaction .
To calculate the concentration of acid, we use the equation given by neutralization reaction:
where,
are the n-factor, molarity and volume of acid which is
are the n-factor, molarity and volume of base which is NaOH.
We are given:
Putting values in above equation, we get:
0.044 M is the molarity of (aq).
Answer: The correct option is, (C) 0.53
Explanation:
The given chemical reaction is:
The rate of the reaction for disappearance of A and formation of C is given as:
![\text{Rate of disappearance of }A=-\frac{1}{9}\times \frac{\Delta [A]}{\Delta t}](https://tex.z-dn.net/?f=%5Ctext%7BRate%20of%20disappearance%20of%20%7DA%3D-%5Cfrac%7B1%7D%7B9%7D%5Ctimes%20%5Cfrac%7B%5CDelta%20%5BA%5D%7D%7B%5CDelta%20t%7D)
Or,
![\text{Rate of formation of }C=+\frac{1}{5}\times \frac{\Delta [C]}{\Delta t}](https://tex.z-dn.net/?f=%5Ctext%7BRate%20of%20formation%20of%20%7DC%3D%2B%5Cfrac%7B1%7D%7B5%7D%5Ctimes%20%5Cfrac%7B%5CDelta%20%5BC%5D%7D%7B%5CDelta%20t%7D)
where,
= change in concentration of C = 1.33 M
= change in time = 4.5 min
Putting values in above equation, we get:
![\frac{1}{9}\times \frac{\Delta [A]}{\Delta t}=\frac{1}{5}\times \frac{\Delta [C]}{\Delta t}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B9%7D%5Ctimes%20%5Cfrac%7B%5CDelta%20%5BA%5D%7D%7B%5CDelta%20t%7D%3D%5Cfrac%7B1%7D%7B5%7D%5Ctimes%20%5Cfrac%7B%5CDelta%20%5BC%5D%7D%7B%5CDelta%20t%7D)
![\frac{\Delta [A]}{\Delta t}=\frac{9}{5}\times \frac{\Delta [C]}{\Delta t}](https://tex.z-dn.net/?f=%5Cfrac%7B%5CDelta%20%5BA%5D%7D%7B%5CDelta%20t%7D%3D%5Cfrac%7B9%7D%7B5%7D%5Ctimes%20%5Cfrac%7B%5CDelta%20%5BC%5D%7D%7B%5CDelta%20t%7D)
![\frac{\Delta [A]}{\Delta t}=\frac{9}{5}\times \frac{1.33M}{4.5min}](https://tex.z-dn.net/?f=%5Cfrac%7B%5CDelta%20%5BA%5D%7D%7B%5CDelta%20t%7D%3D%5Cfrac%7B9%7D%7B5%7D%5Ctimes%20%5Cfrac%7B1.33M%7D%7B4.5min%7D)
![\frac{\Delta [A]}{\Delta t}=0.53M/min](https://tex.z-dn.net/?f=%5Cfrac%7B%5CDelta%20%5BA%5D%7D%7B%5CDelta%20t%7D%3D0.53M%2Fmin)
Thus, the decrease in A during this time interval is, 0.53
