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podryga [215]
2 years ago
6

3. Name four examples of:(a) mixtures(b) compounds(c) elements​

Physics
2 answers:
Keith_Richards [23]2 years ago
6 0
Examples of mixtures:

Smoke and fog (Smog), Dirt and water (Mud), Sand, water and gravel (Cement), Water and salt (Sea water), Potassium nitrate, sulfur, and carbon (Gunpowder), Oxygen and water (Sea foam), Petroleum, hydrocarbons, and fuel additives (Gasoline)

Examples of compounds:

sodium chloride, sucrose, nitrogen gas, a sample of copper, and water.

Examples of elements: carbon, oxygen, hydrogen, gold, silver and iron.
Anika [276]2 years ago
4 0

Answer:

examples of mixture: smoke,fog,sand, potassium nitrate, sulfur and carbon.

examples of compounds: mithane, ammonia,salt and nitrous oxide.

examples of elements: hydrogen , helium, oxygen and neon.

Explanation:

there are more elements

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Name the type of component that has a greater resistance as the current through it increases​
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3 years ago
How many milliliters of water at 25.0°C with a density of 0.997 g/mL must be mixed with 162 mL of coffee at 94.6°C so that the r
Andre45 [30]

Answer:

The volume of water is 139 mL.

Explanation:

Due to the Law of conservation of energy, the heat lost by coffee is equal to the heat gained by the water, that is, the sum of heats is equal to zero.

Q_{coffee} + Q_{water} = 0\\Q_{coffee} = - Q_{water}

The heat (Q) can be calculated using the following expression:

Q=c \times m \times \Delta T

where,

c is the specific heat of each substance

m is the mass of each substance

ΔT is the difference in temperature for each substance

The mass of coffee is:

162mL.\frac{0.997g}{mL} = 162g

Then,

Q_{coffee} = - Q_{water}\\c_{c} \times m_{c} \times \Delta T_{c} = -c_{w} \times m_{w} \times \Delta T_{w}\\m_{c} \times \Delta T_{c} = -m_{w} \times \Delta T_{w}\\m_{w} = \frac{m_{c} \times \Delta T_{c}}{-\Delta T_{w}} \\m_{w}=\frac{162g \times (62.5 \°C - 94.6 \°C ) }{-(62.5 \°C - 25.0 \°C)} \\m_{w} = 139 g

The volume of water is:

139g.\frac{1mL}{0.997g} =139mL

7 0
3 years ago
The distance recorded for riding a motorcycle on its rear wheel without stopping is more than 320 km! Suppose the rider in this
antiseptic1488 [7]

Answer:

<h3>14.97m/s</h3>

Explanation:

Given

Initial velocity of the car u = 8m/s

Distance travelled by the rider S = 40m

Acceleration a = 2m/s²

Required

rider's velocity after the acceleration v

Using the equation of motion

v² = u²+2as

v² = 8²+2(2)(40)

v² = 64+160

v² = 224

v = √224

v = 14.97m/s

Hence the rider's velocity after the acceleration is 14.97m/s

5 0
3 years ago
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