During a field trip, a student collected a sample of an industry product on analysis the material contains 14.5% Carbon 1.8% Hydrogen 6.4% Cl and 19.4%. reduce the Empirical Formula of the compound.
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Answer:
Each year, 24 students from University geoscience programs across Canada will be invited to Calgary to participate in a two-week intensive petroleum industry field trip known as SIFT (Student Industry Field Trip). Throughout SIFT students will participate in lectures, workshops, core sessions, field trips and a real-life exploration game that expose them to the many aspects of the oil and gas industry within Canada.
2020 marks the 42nd CSPG Student Industry Field Trip. What began in 1978 as a humble four-day field trip to expose students to the geology of the Western Canadian Sedimentary Basin, has now grown to become Canada’s premier geoscience education event. The program is industry sponsored and lead by professionals working within the industry.
Some of the topics that will be covered are:
Core logging and interpretation of clastics and carbonates
Well log interpretation
Prospect generation
Seismic interpretation
Basin analysis
The structural geology of Western Canada
Unconventional reservoirs and oil sands
Field trips to: Dinosaur Provincial Park, Banff National Park, Yoho National Park, Waterton National Park, and more!
Answer:
Given,
The mass of substance A is 2g.
The mass of water is 18 g.
Here, the substance is the solute and water is the solvent. The substance is dissolved in water to form the solution.
In the given question, it is said that we need to find out the mass percent of the solute. The mass percent of the solute is defined as the mass of solute divided by the mass of solute plus mass of the solvent (mass of solution) multiplied by 100.
The formula for calculating the mass percent is shown below.
M%=MsMT×100M%=MsMT×100
Where,
M% is the mass percent of solute.
MSMS is mass of solute
MTMT is the mass of solute and solvent (Total mass of solution)
To find out the mass % of the solute, substitute the given values in the above equation.
⇒M%=2g2g+18g×100⇒M%=2g2g+18g×100
⇒M%=2g20g×100⇒M%=2g20g×100
⇒M%=0.1g×100⇒M%=0.1g×100
⇒M%=10%⇒M%=10%
Therefore, the mass percentage or M% of the solute is 10%.
