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2 years ago

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

2 answers:

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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2 years ago

Read 2 more answers

A 2.00 kg block hangs from a spring balance calibrated in Newtons that is attached to the ceiling of an elevator.(a) What does t

tresset_1 [31]

Answer:

Part a)

$Reading = 2.00 kg$

Part b)

$Reading = 2.00 kg$

Part c)

$Reading = 4.04 kg$

Part d)

from t = 0 to t = 4.9 s

so the reading of the scale will be same as that of weight of the block

Then its speed will reduce to zero in next 3.2 s

from t = 4.9 to t = 8.1 s

The reading of the scale will be less than the actual mass

Explanation:

Part a)

When elevator is ascending with constant speed then we will have

$F_{net} = 0$

$T - mg = 0$

$T = mg$

So it will read same as that of the mass

$Reading = 2.00 kg$

Part b)

When elevator is decending with constant speed then we will have

$F_{net} = 0$

$T - mg = 0$

$T = mg$

So it will read same as that of the mass

$Reading = 2.00 kg$

Part c)

When elevator is ascending with constant speed 39 m/s and acceleration 10 m/s/s then we will have

$F_{net} = ma$

$T - mg = ma$

$T = mg + ma$

Reading is given as

$Reading = \frac{mg + ma}{g}$

$Reading = 2.00\frac{9.81 + 10}{9.81}$

$Reading = 4.04 kg$

Part d)

Here the speed of the elevator is constant initially

from t = 0 to t = 4.9 s

so the reading of the scale will be same as that of weight of the block

Then its speed will reduce to zero in next 3.2 s

from t = 4.9 to t = 8.1 s

The reading of the scale will be less than the actual mass

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3 years ago

This diagram shows how a certain type of precipitation is formed. Water drops are caught in up-drafts and down-drafts, over and

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3 years ago

Calculate the workdone to stretch an elastic string by 40cm if a force of 10 newton produces an extension of 4cm in it

notka56 [123]

Answer:

Workdone = 20 Joules

Explanation:

Given the following data;

Force = 10N

Extension, e = 4cm to meters = 4/100 = 0.04 meters

Workdone extension = 40cm to meters = 40/100 = 0.4 meters

To find the work done;

First of all, we would find the spring constant using the formula;

Force = spring constant * extension

10 = spring constant * 0.04

Spring constant = 10/0.04

Spring constant = 250 N/m

Next, we find the work done;

Workdone = ½ke²

Where;

k is the spring constant.

e is the extension.

Substituting into the formula, we have;

Workdone = ½ * 250 * 0.4²

Workdone = 125 * 0.16

Workdone = 20 Joules

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2 years ago

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

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