The molarity of the lake water is 0.00001 M and the pH of lake water is 5.
The lake water is acidic.
Explanation:
Data given:
molarity of base solution Mbase = 0.1 M
volume of the base solution Vbase = 0.1 ml or 0.0001 litre
volume of lake water Vlake = 1000ml or 1 litre
molarity of the lake water, Mlake = ?
Using the formula for titration:
Mbase X Vbase = Mlake X
Mlake = 
Putting the values in the equation:
Mlake = 
Mlake = 0.00001 M
The pH of the lake water will be calculated by using the following formula:
pH = - 
[
]
pH = - [ 0.00001]
pH = 5
<u>Answer:</u> The boiling point of water in Tibet is 69.9°C
<u>Explanation:</u>
To calculate the boiling point of water in Tibet, we use the Clausius-Clayperon equation, which is:
![\ln(\frac{P_2}{P_1})=\frac{\Delta H}{R}[\frac{1}{T_1}-\frac{1}{T_2}]](https://tex.z-dn.net/?f=%5Cln%28%5Cfrac%7BP_2%7D%7BP_1%7D%29%3D%5Cfrac%7B%5CDelta%20H%7D%7BR%7D%5B%5Cfrac%7B1%7D%7BT_1%7D-%5Cfrac%7B1%7D%7BT_2%7D%5D)
where,
= initial pressure which is the pressure at normal boiling point = 1 atm = 760 mmHg (Conversion factor: 1 atm = 760 mmHg)
= final pressure = 240. mmHg
= Heat of vaporization = 40.7 kJ/mol = 40700 J/mol (Conversion factor: 1 kJ = 1000 J)
R = Gas constant = 8.314 J/mol K
= initial temperature or normal boiling point of water = ![100^oC=[100+273]K=373K](https://tex.z-dn.net/?f=100%5EoC%3D%5B100%2B273%5DK%3D373K)
= final temperature = ?
Putting values in above equation, we get:
![\ln(\frac{240}{760})=\frac{40700J/mol}{8.314J/mol.K}[\frac{1}{373}-\frac{1}{T_2}]\\\\-1.153=4895.36[\frac{T_2-373}{373T_2}]\\\\T_2=342.9K](https://tex.z-dn.net/?f=%5Cln%28%5Cfrac%7B240%7D%7B760%7D%29%3D%5Cfrac%7B40700J%2Fmol%7D%7B8.314J%2Fmol.K%7D%5B%5Cfrac%7B1%7D%7B373%7D-%5Cfrac%7B1%7D%7BT_2%7D%5D%5C%5C%5C%5C-1.153%3D4895.36%5B%5Cfrac%7BT_2-373%7D%7B373T_2%7D%5D%5C%5C%5C%5CT_2%3D342.9K)
Converting the temperature from kelvins to degree Celsius, by using the conversion factor:
Hence, the boiling point of water in Tibet is 69.9°C
The answer I believe would be C hope this helps
Answer:
0.0900 mol/L
Explanation:
<em>A chemist makes 330. mL of nickel(II) chloride working solution by adding distilled water to 220. mL of a 0.135 mol/L stock solution of nickel(II) chloride in water. Calculate the concentration of the chemist's working solution. Round your answer to significant digits.</em>
Step 1: Given data
- Initial concentration (C₁): 0.135 mol/L
- Initial volume (V₁): 220. mL
- Final concentration (C₂): ?
- Final volume (V₂): 330. mL
Step 2: Calculate the concentration of the final solution
We prepare a dilute solution from a concentrated one. We can calculate the concentration of the working solution using the dilution rule.
C₁ × V₁ = C₂ × V₂
C₂ = C₁ × V₁/V₂
C₂ = 0.135 mol/L × 220. mL/330. mL = 0.0900 mol/L
Answer:
for combustion of naphthalene is -5164 kJ/mol
Explanation:
where, C refers heat capacity and 
refers change in temperature.
Here, 
So, 
is generally expressed in terms of per mole unit of reactant. Also, should be negative as it is an exothermic reaction (temperature increases).
Molar mass of naphthalene is 128.17 g/mol
So, 1.025 g of naphthalene = 
of naphthalene
= 0.007997 moles of naphthalene
