Answer:
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Answer: The empirical formula for the given compound is 
Explanation : Given,
Percentage of C = 38.8 %
Percentage of H = 16.2 %
Percentage of N = 45.1 %
Let the mass of compound be 100 g. So, percentages given are taken as mass.
Mass of C = 38.8 g
Mass of H = 16.2 g
Mass of N = 45.4 g
To formulate the empirical formula, we need to follow some steps:
Step 1: Converting the given masses into moles.
Moles of Carbon =
Moles of Hydrogen = 
Moles of Nitrogen = 
Step 2: Calculating the mole ratio of the given elements.
For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 3.23 moles.
For Carbon = 
For Hydrogen = 
For Oxygen = 
Step 3: Taking the mole ratio as their subscripts.
The ratio of C : H : N = 1 : 5 : 1
Hence, the empirical formula for the given compound is 
Pressure has little effect on the solubility of liquids and solids because they are almost incompressible True.
Liquids and solids show little change in solubility with changes in pressure. As expected, gases increase in solubility with increasing pressure. Henry's Law states that the solubility of a gas in a liquid is directly proportional to the pressure of that gas above the surface of the solution.
External pressure has little effect on liquid and solid solubility. In contrast, the solubility of a gas increases as the partial pressure of the gas above the solution increases.
Solubility is a measure of the concentration of dissolved gas particles in a liquid and is a function of gas pressure. Increasing the gas pressure increases the number of collisions and increases the solubility, and decreasing the pressure decreases the solubility.
Learn more about pressure here : brainly.com/question/28012687
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Answer:
22.27 °C = ΔT
Explanation:
Specific heat capacity:
It is the amount of heat required to raise the temperature of one gram of substance by one degree.
Formula:
Q = m × c × ΔT
Given data:
mass = 28 g
heat absorbed = 58 cal
specific heat of copper = 0.093 cal/g .°C
temperature change =ΔT= ?
Solution:
Q = m × c × ΔT
58 cal = 28 g × 0.093 cal /g.°C × ΔT
58 cal = 2.604 cal.°C × ΔT
58 cal / 2.604 cal .°C = ΔT
22.27 °C = ΔT