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3 years ago

Which type of variable is kept the same throughout an experiment?

Chemistry

2 answers:

weeeeeb [17]3 years ago

7 0

A Constant variable!

skelet666 [1.2K]3 years ago

6 0

Answer:

Constant variable.

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Which term refers to a variable that isn’t allowed to change during an experiment?

Citrus2011 [14]

D) dependent variable

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3 years ago

What are three ways a fire Extinguisher can put out a fire?

Elis [28]

Answer:

Extinguisher, Fire blanket and water

Explanation:

they all are helpful

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3 years ago

What mixture can be separated by using a filter?

nexus9112 [7]

<h2>Answer: </h2>

Option 2nd - a mixture of solid particles in a liquid.

<h2>Explanation: </h2>

For a Mixture of two immiscible liquids (for example water and oil) it is not possible to filter them with a filter as both of the liquid will pass through it.
For a mixture of a solid and a liquid (for example sand and water) it can be easily separated through a filter as on filtration solid will be left on the filter and the liquid will pass through it.
For a mixture of two miscible liquids (for example milk and water) both of them will form a single identity, and on filtration both of them will pass through the filter. Hence it is not possible to separate them through filter.
For a mixture of liquid and gas (for example drinking soda or Pepsi) filter is not needed for their separation as it will be separated when exposed in air.

Result: Option 2 is the correct answer.

4 0

3 years ago

The graduated cylinder below shows major scale divisions every 5 mL. How should the volume of the liquid be recorded? Use the co

OLga [1]

Explanation:

First, note that the surface of the liquid is curved. This is called the meniscus. This phenomenon is caused by the fact that water molecules are more attracted to glass than to each other (adhesive forces are stronger than cohesive forces). When we read the volume, we read it at the BOTTOM of the meniscus.

The smallest division of this graduated cylinder is 1 mL. Therefore, our reading error will be 0.1 mL or 1/10 of the smallest division. An appropriate reading of the volume is 36.5 0.1 mL. An equally precise value would be 36.6 mL or 36.4 mL.

How many significant figures does our answer have? 3! The "3" and the "6" we know for sure and the "5" we had to estimate a little.

6 0

2 years ago

A 73.6 g sample of aluminum is heated to 95.0°C and dropped into 100.0 g of water at 20.0°C. If the resulting temperature of the

Softa [21]

Answer:

The specific heat of aluminium is 0.875 J/g°C

Explanation:

Step 1: Data given

The mass of the aluminium sample = 73.6 grams

Initial temperature of the sample = 95.0 °C

Mass of water = 100.0 grams

Initial temperature of water = 20.0 °C

Final temperature of water and aluminium = 30.0 °C

The specific heat of water = 4.184 J/g°C

Step 2: Calculate the specific heat of aluminium

Q gained = Q lost

Qwater = -Qaluminium

Q = m*c*ΔT

m(aluminium) * c(aluminium) * ΔT(aluminium) = - m(water) * c(water) * ΔT(aluminium)

⇒ mass of aluminium = 73.6 grams

⇒ c(aluminium) = TO BE DETERMINED

⇒ ΔT(aluminium) = The change of temperature = T2 - T1 = 30 .0 °C - 95.0 °C = -65.0°C

⇒ mass of water = 100.0 grams

⇒ c(water ) = The specific heat of water = 4.184 J/g°C

⇒ ΔT(water) = The change of temperature of water = T2 - T1 = 30.0 - 20.0 = 10.0 °C

73.6g * c(aluminium) * -65.0 °C = 100.0g * 4.184 J/g°C * 10.0°C

-4784 * c(aluminium) = -4184

c(aluminium) = 0.875 J /g°C

The specific heat of aluminium is 0.875 J/g°C

7 0

3 years ago