Answer:
2.5 %
Explanation:
Considering:
Or,
Given :
For 
:
Molarity = 0.2850 M
Volume = 63.30 mL
The conversion of mL to L is shown below:
1 mL = 10⁻³ L
Thus, volume = 63.30 × 10⁻³ L
Thus, moles of :
Moles of = 0.0180405 moles
Moles of = Moles of 
Thus, Moles of = 0.0180405 moles
Molar mass of = 35.453 g/mol
Mass = Moles * Molar mass = 0.0180405 moles * 35.453 g/mol = 0.6396 g
Volume of sea water = 25.00 mL
Density = 1.024 g/mL
Density = Mass / Volume
Mass = Density * Volume = 1.024 g/mL * 25.00 mL = 25.6 g
<u>Mass percent of Cl⁻ = 2.5 %</u>
Can you show me a picture or a diagram
The wave will have a frequency that is larger
Answer:
Compare the solubility of silver iodide in each of the following aqueous solutions:
a. 0.10 M AgCH3COO
b. 0.10 M NaI
c. 0.10 M KCH3COO
d. 0.10 M NH4NO3
1. More soluble than in pure water.
2. Similar solubility as in pure water.
3. Less soluble than in pure water.
Explanation:
This can be explained based on common ion effect.
According to common ion effect the solubility of a sparingly soluble salt decreases further in a solution which has a common ion to it.
The solubility of AgI(s) silver iodide in water is shown below:
a. a. 0.10 M AgCH3COO has a common ion Ag+ with AgI.
So, AgI is less soluble than in pure water in this solution.
b. 0.10 M NaI has a common ion I- with AgI.
So, AgI is less soluble than in pure water in this solution.
c. 0.10 M KCH3COO:
This solution has no common ion with AgI.
So, AgI has similar solubility as in pure water.
d. 0.10 M NH4NO3:
In this solution, AgI can be more soluble than in pure water.
Option B is correct
K = Kp /Kr
The given equation indicating, the product containing 6 moles of proton whereas the reactant contains 2 mole of bismuth and 3 mole of hydrogen sulphide.
Hence, in reaction B there are 2 mole of bismuth and 3 mole of hydrogen sulphide reacting to produce 6 moles of proton. whereas the concentration of Bi2S3 is not considered as it is present in solid phase.
