Plant cells are affected as they begin to wilt and reduce photosynthesis
PV=nRT
PV= (m/M)RT
PM=(m/v)RT
PM =dRT
d= (PM) ÷(RT)
substitute the the given data to the above equation. d-density, p-pressure, M- molar mass of argon gas, T- temperature. and R is a constant. make sure to substitute its value too.. :)
Answer:
When the volume increased from 2.00 to 5.25L the new temperature is 808.9 K ( =535.75 °C)
Explanation:
Step 1: Data given
The initial volume of the sample = 2.00 L
The initial temperature = 35 °C = 308 K
The increased volume = 5.25 L
Pressure = constant
Step 2: Calculate the new temperature
V1/T1 = V2/T2
⇒ with V1 = the initial volume = 2.00 L
⇒ with T1 = the initial volume = 308 K
⇒ with V2 = the new volume = 5.25 L
⇒ with T2 = the new temperature
2.00 / 308 = 5.25 / T2
0.00649 = 5.25/T2
T2 = 5.25/ 0.00649
T2 = 808.9 K
When the volume increased from 2.00 to 5.25L the new temperature is 808.9 K ( =535.75 °C)
<u>Answer:</u> The pH of the solution is 13.0
<u>Explanation:</u>
To calculate the molarity of solution, we use the equation:

Given mass of KOH = 716. mg = 0.716 g (Conversion factor: 1 g = 1000 mg)
Molar mass of KOH = 56 g/mol
Volume of solution = 130 mL
Putting values in above equation, we get:

1 mole of KOH produces 1 mole of hydroxide ions and 1 mole of potassium ions
- To calculate hydroxide ion concentration of the solution, we use the equation:
![pOH=-\log[OH^-]](https://tex.z-dn.net/?f=pOH%3D-%5Clog%5BOH%5E-%5D)
We are given:
[tex[[OH^-]=0.098M[/tex]
Putting values in above equation, we get:

To calculate the pH of the solution, we use the equation:
pOH + pH = 14
So, pH = 14 - 1.00 = 13.0
Hence, the pH of the solution is 13.0