Answer: 0.67 ml
Explanation:
According to the dilution law,
where,
= concentration of stock solution = 226 mg/ml
= volume of stock solution = ?
= concentration of working solution= 15 mg/ml
= volume of working solution= 10 ml
Putting in the values we get:

Thus volume of stock solution needed to dilute to have 10mL of working solution at the above concentration is 0.67 ml
Answer: D. Use tools to help improve their senses when they make observations.
Observation is the outcome of a scientific procedure, which is followed to answer the questions related to a scientific inquiry. In order to make observation scientists need to use tools that can help in improving their senses like visual, sensation and hearing aids, so as to make observation accurate, and conclusive in a research.
The enthalpy of fusion of a substance is the energy required to change the state of a substance from solid to liquid at a constant temperature. The enthalpy of fusion for iodine, I₂, will be higher. This is because there are stronger intermolecular forces holding the iodine molecules together. The stronger intermolecular forces arise from the fact that iodine is a much larger molecule, so it has much more electrons resulting in higher Van der Waal's forces. This is also visible in the fact that at room temperature, iodine is a solid while nitrogen is a gas.<span />
The question is incomplete, here is the complete question:
A chemist makes 600. mL of magnesium fluoride working solution by adding distilled water to 230. mL of a stock solution of 0.00154 mol/L magnesium fluoride in water. Calculate the concentration of the chemist's working solution. Round your answer to 3 significant digits.
<u>Answer:</u> The concentration of chemist's working solution is 
<u>Explanation:</u>
To calculate the molarity of the diluted solution (chemist's working solution), we use the equation:

where,
are the molarity and volume of the stock magnesium fluoride solution
are the molarity and volume of chemist's magnesium fluoride solution
We are given:

Putting values in above equation, we get:

Hence, the concentration of chemist's working solution is 
Moles XeF6 = 10.0g/ 245.28 g/ mol=0.0408
The ratio between F2 and XeF6 is 3:1
Moles F2 required = 3 x 0.0408=0.122
Mass F2 = 0.122 mol x 37.9968 g/ mol=4.64g