Answer: Solution W and Y solution have more solubility than X and Z
Solutions are homogeneous mixtures of two or more components. By uniform mix we mean that its structure and properties are the same in the whole mix. Generally, the component which is present in the largest quantity is known as solvent. Solvent determines the physical condition in which the solution exists. In addition to the solvent, one or more component present in the solution is called solutes. In this unit we will only consider binary solutions (i.e., with two components)
The structure of the solution can be described by expressing its concentration. The latter can either be expressed qualitatively or quantitatively. For example, in qualitatively we can say that the solution is diluted (i.e., relatively small amounts of solubility) or it is concentrated (i.e., relatively rarely sighs). But in real life such details may be very confusing and thus require a quantitative description of the solution. There are several ways that we can quantitatively describe the concentration of solutions. (i) Mass Percentage (W / W): The mass percentage of a component of the solution is defined as: mass of the component = mass of the component in the solution = 100 Total mass of the solution .For example, if by mass A solution is described by 10% glucose in water, it means that 10 grams of glucose dissolved in 90 grams of water, resulting in 100 grams of solution. The concentration described by a large percentage of the population is usually used in industrial chemical applications. For example, the commercial bleaching solution contains 3.62 mass percentages of sodium hypochlorite in water. (ii) Volume Percentage (V / V): Volume Percentage is defined as: Total Volume of Component Volume 100 (component) Volume% of Component
Explanation:
Air and water have a good day
Answer/Explanation: Speed and direction can change with time. When you throw a ball into the air, it leaves your hand at a certain speed. As the ball rises, it slows down. Then, as the ball falls back toward the ground, it speeds up again. When the ball hits the ground, its direction of motion changes and it bounces back up into the air.
Answer:
ω = 1.83 rad/s clockwise
Explanation:
We are given:
I1 = 3.0kg.m2
ω1 = -5.4rad/s (clockwise being negative)
I2 = 1.3kg.m2
ω2 = 6.4rad/s (counterclockwise being positive)
By conservation of the momentum:
I1 * ω1 + I2 * ω2 = (I1 + I2) * ω
Solving for ω:
Since it is negative, the direction is clockwise.
