To make 200 mL of a 10% by volume solution, add 20 mL corn syrup to water.
What is volume solution?Volume percent of a solution is defined as the ratio of the volume of solute present in a solution to the volume of the solution as a whole. It means that the volume of a closed figure determines how much three-dimensional space it can occupy. In terms of numerical value, volume is the amount of three-dimensional space enclosed by a closed surface. For example, a substance's space can be solid, liquid, gas, or plasma, or any other space-occupying shape. The volume percentage of a solution can be calculated by dividing the volume of solute by the volume of solution and multiplying the result by 100. The basic formula for volume is length width height, whereas the basic formula for area of a rectangular shape is length width height. The calculation is unaffected by how you refer to the various dimensions: you can, for example, use 'depth' instead of 'height.'To learn more about volume solution, refer to:
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The substance ammonia has the following properties: normal melting point: 195.4 K normal boiling point: 239.8 K triple point: 5.9×10-2 atm, 195.3 K critical point: 111.5 atm, 405.5 K At temperatures above 405.5 K and pressures above 111.5 atm, NH3 is a _________ . NH3 does not exist as a liquid at pressures below atm. NH3 is a _________ at 5.90×10-2 atm and 249.5 K. NH3 is a _________ at 1.00 atm and 236.0 K. NH3 is a _________ at 24.6 atm and 185.6 K.
Answer:
a) Superficial fluid
b) 5.9*10^-2 atm
c) Gas
d) Liquid
e) Solid
Explanation:
a) At temperatures above 405.5 K and pressures above 111.5 atm, NH3 is a superficial fluid because liquid and gases does not exit at temperature and pressure greater than 405.5 K and 111.5 atm
b) NH3 does not exist as a liquid at pressures below 5.9*10^-2 atm , That is below the triple point there is existence of liquid
c) NH3 is a Gas at 5.90×10^-2 atm and 249.5 K.
d) NH3 is a Liquid at 1.00 atm and 236.0 K. because pressure and temperature ( standard ) is between the given normal melting and boiling point
e) NH3 is a solid at 24.6 atm and 185.6 K . because the pressure here is more than that of triple point while the temperature is lesser
A 500.0 g sample of aluminium, I initially at 25.0 degrees, absorbs heat from its surroundings and reaches a final temperature of 90.7 degrees. How much heat (in KJ) has been absorbed by the sample? To one decimal place
Specific heat= 0.9930j g-1 K-1 for aluminium
A 500.0 g sample of aluminum, initially at 25.0 degrees, absorbs heat from its surroundings and reaches a final temperature of 90.7 degrees. 32.62245 kJ heat has been absorbed by the sample.
What is specific heat?The term specific heat is defined as the amount of heat required to increase the temperature of 1 gram of a substance 1 degree Celsius (°C).
To calculate the amount of heat absorbed by the sample, use the formula:
Q = mcΔT
where Q is the amount of heat absorbed by the sample, m is the mass of the sample, c is the specific heat of aluminum, and ΔT is the change in temperature of the sample.
Substituting the given values into the formula, we get:
Q = 500.0 g × 0.9930 J/g·K × (90.7°C - 25.0°C)
Q = 500.0 g × 0.9930 J/g·K × 65.7 K
Q = 32,622.45 J
To convert the result to kilojoules (kJ), we divide by 1000:
Q = 32.62245 kJ
Thus, the amount of heat absorbed by the sample is 32.6 kJ.
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