CaS is the empirical formula of the compound between calcium and sulphur that has the percent composition 55.6.
When percentages are given, we take a total mass of 100 grams.
Therefore the mass of each element is equal to the percentage given.
Mass of Ca = 55.6 g (given) of
S Mass = 44.4 g (100 - 55.6 = 44.4)
Step 1: Convert the given mass to moles.
moles Ca = given mass Ca / molar mass Ca
moles = 55.6 / 40 = 1.39 moles
mol S = specific mass S / molar mass S
mol = 44.4 / 32 = 1.39 mol
Step 2: Divide the molar ratio of each molar value by the smallest number of moles calculated.
For Ca = 1.39 / 1.39 = 1
For S = 1.39 / 1.39 = 1
The ratio of Ca : S = 1:1
Hence the empirical formula of the given compound will be CaS.
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A good hypothesis describes your ideas and how you think the experiment will conclude. Additionally, hypotheses have scientific information included.
Answer
given,
I is the loudness of sound
I = 10 Log₁₀ r
r is relative intensity
at when relative intensity is 10⁶
I = 60 dB
how much louder when 100 people would be talking together
I = 10 Log₁₀ r
I = 10 Log₁₀ (10⁶ x 100)
I = 10 Log₁₀ (10⁸)
I = 80 dB
hence, the intensity will be increased by (80 dB -60 dB) 20 dB when 100 people start talking together.
Answer:
Equilibrium. • When an object is in equilibrium (either at rest or moving with constant velocity), the net force acting on it zero.
Answer: This type of questions are called probing questions. The correct option is D.
Explanation:
Probing questions are the type of questions that are asked to investigate an ongoing event. It helps the investigator to know more about what is happening and how to obtain conclusive decisions through the personal opinions of the respondent . For example from the question, Liza wanted to know more about the project updates which was held in the weekly meetings. She asked her employees questions like:
- Why were you late meeting your last deadline?
-Were there external factors that delayed your work?
-Did other coworkers get their part of the assignment to you on time?
- Do you need more help from me?".
