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Convert 25°c to kelvin scale.

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4 parts
11

10%\text{10\%}

12

298.15 K\text{298.15 K}

15

meter (m)\text{meter (m)}

16

-2 m/s2\text{-2 m/s}^2

Step-by-step solution

7 steps

Here are the solutions to the questions.

Part 11· 3 steps

A solution is prepared by dissolving 50 gm of sugar in 450 gm of water. Express the concentration of the solution in mass/mass percentage.

  1. Identify the mass of the solute and solvent.

    Mass of solute (sugar) = 50 gm
    Mass of solvent (water) = 450 gm

  2. Calculate the total mass of the solution.

    Mass of solution = Mass of solute + Mass of solvent

    Mass of solution=50 gm+450 gm=500 gm\text{Mass of solution} = 50 \text{ gm} + 450 \text{ gm} = 500 \text{ gm}

  3. Calculate the mass percentage concentration.

    Mass percentage concentration = Mass of soluteMass of solution×100%\frac{\text{Mass of solute}}{\text{Mass of solution}} \times 100\%

    Mass percentage concentration=50 gm500 gm×100%\text{Mass percentage concentration} = \frac{50 \text{ gm}}{500 \text{ gm}} \times 100\% Mass percentage concentration=110×100%=10%\text{Mass percentage concentration} = \frac{1}{10} \times 100\% = 10\%

    The concentration of the solution is \boxed{\text{10%}}.

Part 12

298.15 K\text{298.15 K}

Convert 25°C to kelvin scale.

  1. Use the conversion formula from Celsius to Kelvin.

    The formula to convert temperature from Celsius to Kelvin is:

    K=°C+273.15K = \text{°C} + 273.15

  2. Substitute the given Celsius temperature into the formula.

    K=25+273.15K = 25 + 273.15 K=298.15K = 298.15

    The temperature in Kelvin is 298.15 K\boxed{\text{298.15 K}}.

Part 13

A diver is able to cut through water in a swimming pool. Which property of matter does this observation show?

This observation shows that the particles of water have intermolecular spaces between them. It also demonstrates that the force of attraction between water particles is not very strong, allowing the diver to overcome these forces and move through the water.

Part 14

Name two organelles you have studied that contain genetic material?

Two organelles that contain genetic material are:

  • Nucleus (contains DNA)
  • Mitochondria (contains its own DNA)
  • Chloroplasts (in plant cells, contain their own DNA)

Part 15

meter (m)\text{meter (m)}

Area under a velocity time Graph represents a physical quantity. Write the S.I. unit of the physical quantity.

The area under a velocity-time graph represents displacement.
The S.I. unit of displacement is meter (m)\boxed{\text{meter (m)}}.

Part 16

-2 m/s2\text{-2 m/s}^2

A bus decreases its speed from 72 km/hr to 36 km/hr in 5 seconds. Find the acceleration of the bus.

  1. Convert initial and final velocities from km/hr to m/s.

    Initial velocity, u=72 km/hru = 72 \text{ km/hr}

    u=72×1000 m3600 s=72×518 m/s=4×5 m/s=20 m/su = 72 \times \frac{1000 \text{ m}}{3600 \text{ s}} = 72 \times \frac{5}{18} \text{ m/s} = 4 \times 5 \text{ m/s} = 20 \text{ m/s}

    Final velocity, v=36 km/hrv = 36 \text{ km/hr}

    v=36×1000 m3600 s=36×518 m/s=2×5 m/s=10 m/sv = 36 \times \frac{1000 \text{ m}}{3600 \text{ s}} = 36 \times \frac{5}{18} \text{ m/s} = 2 \times 5 \text{ m/s} = 10 \text{ m/s}

    Time, t=5 st = 5 \text{ s}

  2. Calculate the acceleration using the formula a=v−uta = \frac{v - u}{t}.

    a=10 m/s−20 m/s5 sa = \frac{10 \text{ m/s} - 20 \text{ m/s}}{5 \text{ s}} a=−10 m/s5 sa = \frac{-10 \text{ m/s}}{5 \text{ s}} a=−2 m/s2a = -2 \text{ m/s}^2

    The acceleration of the bus is -2 m/s2\boxed{\text{-2 m/s}^2}. (The negative sign indicates deceleration).

Part 17

How would you confirm that a colourless liquid given to you is pure water?

To confirm that a colourless liquid is pure water, you can measure its boiling point and freezing point. Pure water boils at 100∘C100^\circ\text{C} (at standard atmospheric pressure) and freezes at 0∘C0^\circ\text{C}. If the liquid exhibits these precise boiling and freezing points, it is pure water.

Part 18

Smoke and fog both are aerosols. In what way are they different?

Both smoke and fog are aerosols, meaning they are colloidal systems where solid or liquid particles are dispersed in a gas.

  • Smoke is a dispersion of solid particles (like carbon) in a gas (usually air), typically formed by combustion.
  • Fog is a dispersion of liquid droplets (water) in a gas (air), formed when water vapor condenses into tiny droplets near the ground.
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