Answer: The change in volume will be 9.30 L
Explanation:
To calculate the new pressure, we use the equation given by Boyle's law. This law states that pressure is directly proportional to the volume of the gas at constant temperature.
The equation given by this law is:
![P_1V_1=P_2V_2](https://tex.z-dn.net/?f=P_1V_1%3DP_2V_2)
where,
are initial pressure and volume.
are final pressure and volume.
We are given:
![P_1=0.950atm\\V_1=35.0L\\P_2=0.750atm\\V_2=?L](https://tex.z-dn.net/?f=P_1%3D0.950atm%5C%5CV_1%3D35.0L%5C%5CP_2%3D0.750atm%5C%5CV_2%3D%3FL)
Putting values in above equation, we get:
![0.950\times 35.0L=0.750\times V_2\\\\V_2=44.3L](https://tex.z-dn.net/?f=0.950%5Ctimes%2035.0L%3D0.750%5Ctimes%20V_2%5C%5C%5C%5CV_2%3D44.3L)
The change in volume will be (44.3-35.0)L = 9.30 L
Answer:
Always carry the microscope with two hands. One on the arm and one underneath the base of the microscope.
Answer:
<u>Option B is correct</u>
Explanation:
Step 1: Define volatility
In chemistry, the term volatility, is a way to describe how readily a substance transitions from a liquid phase to a gas phase, also called evaporating.
At a given temperature and pressure, a substance with high volatility is more likely to evaporate more quickly , while a substance with a lower volatility is more likely to be a liquid or solid, so not to evaporate or slower.
The higher the volatility, the higher the (vapor) pressure of a compound. Increasing temperature means the vapor pressure will also increase,
Step 2: In this case:
⇒ O<u>ption A is false</u> because the pressure will be higher when volatility is higher.
<u>⇒ Option B is correct</u> because higher volatility means evaporating more quickly
<u>⇒ Option C is false</u> because higher volatility means higher pressure. When pressure increases, the surface tension decreases.
<u>⇒ Option D is false</u> because when the volatility is higher, the liquid/gas escape the container, easier, so there will be less resistance.
Answer:
The molarity of this KOH solutions 3.21 mol / L (option C is correct)
Explanation:
Step 1: Data given
Mass of KOH = 45.0 grams
Molar mass of KOH = 56.1 g/mol
Volume = 250.0 mL = 0.250 L
Step 2: Calculate moles KOH
Moles KOH = mass KOH / molar mass KOH
Moles KOH = 45.0 grams / 56.1 g/mol
Moles KOH = 0.802 moles
Step 3: Calculate molarity
Molarity KOH solution = moles KOH / volume water
Molarity KOH solution = 0.802 moles / 0.250 L
Molarity KOH solution = 3.21 mol / L
The molarity of this KOH solutions 3.21 mol / L (option C is correct)
Answer:
The jewelry is 2896.54_Kg/m^3 less dense than pure silver
Explanation:
Density of jewellery = (mass of jewellery) ÷ (volume of jewellery)
=3.25g ÷ 0.428mL = 0.00325Kg÷0.000000428m^3 = 7583.46Kg/m^3
The density of silver is 10490_Kg/m^3 which is (10490 - 7583.46) 2896.54_Kg/m^3 more dense than the jewellery
The density of Silver [Ag]
The weight of Silver per cubic centimeter is 10.49 grams or the weight of silver per cubic meter is 10490 kilograms, that is the density of silver is 10490 kg/m³; at 20°C (68°F or 293.15K) at a pressure of one atmospheres.