Answer:
Oxygen, 43kg - Carbon, 16kg - Hydrogen, 7kg - Nitrogen, 1.8kg - Calcium, 1kg - Phosphorus, 0.78kg - Potassium, 0.14kg - Sulfur, 0.14kg - Sodium, 0.10kg - Chlorine, 0.095kg - and Magnesium, 0.019kg.
Answer:
3.24
Explanation:
The dissociation equation for the carboxylic acid can be represented as follows:
RCOOH —-> RCOO- + H+
We can use an ICE table to get the value of the concentration of the hydrogen ion. ICE stands for initial, change and equilibrium.
RCOOH RCOO- H+
Initial 0.2 0.0. 0.0
Change -x +x. +x
Equilibrium 0.2-x. x. x
We can now find the value of x as follows:
Ka = [RCOO-][H+]/[RCOOH]
(1.66* 10^-6) = (x * x)/(0.2-x)
(1.66 * 10^-6) (0.2-x) = x^2
x^2 = (3.32* 10^-7) - (1.66*10^-6)x
x^2 + (1.66 * 10^-6)x - (3.32* 10^-7) = 0
Solving the quadratic equation to get x:
x = 0.0005753650094369094 or - 0.0005753650094369094
As concentration cannot be negative, we discard the negative answer
Hence [H+] = 0.0005753650094369094
By definition, pH = -log[H+]
pH = -log(0.0005753650094369094)
pH = 3.24
In BaF₂ the solubility will decreases on adding NaF
the solubility will increases on adding HCl
<h3>SOLUBILITY OF BARIUM FLUORIDE</h3>
- Increase in temperature will increase the solubility of the solid.
- By adding NaF - decrease the solubility of BaF₂
- By adding HCl - increase the solubility of BaF₂
<h3>BARIUM FLUORIDE</h3>
- It is colourless solid that occur as rare mineral
- It is corroded by moisture
- It is used in window of IR spectroscopy
Hence the barium fluoride the solubility decreases on adding NaF and increases on adding HCl.
Learn more about the solubility on
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<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
