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mario62 [17]
3 years ago
13

Experiments 11. What 2 factors increase the validity of a scientific experiment?

Chemistry
1 answer:
lorasvet [3.4K]3 years ago
4 0

Answer:

evidince is one of them

Explanation:

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What is the density of a cube of water measuring 2cmX4cmX1cm, with a mass of<br> 8g?
Vinil7 [7]

The actual formula for volume for a cube is the length multiplied by the width and then multiplied by the height. Since all three measurements are the same, the formula results in the measurement of one side cubed. For the example, 5^3 is 125 cm^3. Multiply the volume by the known density, which is the mass per volume.

7 0
3 years ago
5. A shiny, gold-colored bar of metal weighing 57.3 g has a volume of 4.7 cm 3 . Is the bar of metal pure
agasfer [191]

1. The bar of metal is not pure gold.

2. The volume of 50.0 g of air is 42,016 mL.

3. The mass of the pure aluminum is 135 g.

4. The light beam's wavelength in meters is 475 x  10^{-9 m.

<h3>What is density?</h3>

The density of a substance is the ratio of the mass of the substance and its volume.

Density = mass/volume.

For the first problem: mass = 57.3 g, volume = 4.7 cm3

Density of the metal = 57.3/4.7

                                   = 12.19 g/cm3

The density of gold is 19.3 g/cm3. Thus, the bar of metal is not pure gold.

For the second problem: density = 1.19 x 10^{-3 g/ml, mass = 50.0 g

Volume = mass/density

                 = 50/1.19 x 10^{-3

                      = 42,016 ml

The volume of 50.0 g of air is 42,016 mL.

For the third problem: density = 2.70 g/ml, volume = 50.0 ml.

Mass = density x volume

                 = 2.70 x 50

                       = 135 g

The mass of the pure aluminum is 135 g.

Lastly:

nano = 10^{-9

475 nm = 475 x  10^{-9 m

The light beam's wavelength in meters is 475 x  10^{-9 m.

More on density can be found here: brainly.com/question/15164682

#SPJ1

6 0
1 year ago
If an element has 2 fewer protons than boron, what is its electron configuration?
REY [17]
So boron has an electronic configuration of 1S2 2S2 2P1 and the element with an atomic number of 3 is Lithium which has an electronic configuration of 1S2 2S1.

So the answer is 1S2 2S1.
8 0
4 years ago
It is a type of mixture that separates on standing
erma4kov [3.2K]

Answer:

B) Suspensions

<h3>More to know - </h3>

  • In suspension the size of particles are big as compared to solution

  • Colloids shows Tyndall effect (,for ex milk)
6 0
3 years ago
Read 2 more answers
Carbon tetrachloride, CCl4, once used as a cleaning fluid and as a fire extinguisher, is produced by heating methane and chlorin
Elena-2011 [213]

Answer:

1.882 g

Explanation:

Data Given

mass of Cl₂ = 33.4 g

mass of CH₄ = ?

Reaction Given:

                       CH₄+ 4Cl₂ --------→ CCl₄ + HCl

Solution:

First find the mass of CH₄ from the reaction that it combine with how many grams of Chlorin.

Look at the balanced reaction

                    CH₄     +     4Cl₂ --------→ CCl₄ + 4HCl

                    1 mol        4 mol

So 1 mole of CH₄ combine with 4 moles of Cl₂

Now

convert the moles into mass for which we have to know molar mass of CH₄ and Cl₂

Molar mass of Cl₂ = 2 (35.5)

Molar mass of  Cl₂  = 71 g/mol

mass of Cl₂

                mass in grams = no. of moles x molar mass

                mass of Cl₂ = 4 mol x 71 g/mol

                mass of Cl₂  = 284 g

Molar mass of CH₄= 12+ 4(1)

Molar mass of CH₄= 16 g/mol

mass of CH₄

                mass in grams = no. of moles x molar mass

                mass of CH₄= 1 mol x 16 g/mol

                mass of CH₄ = 16 g

So,

284 g of Cl₂  combine with 16 g of methane ( CH₄ ) then how many grams of CH₄ is needed to combine with 33.4 g of Cl₂  

Apply unity Formula

                           284 g of Cl₂  ≅ 16 g of methane ( CH₄ )

                           33.4 g of Cl₂  ≅ X g of methane ( CH₄ )

By cross multiplication

                          X g of methane ( CH₄ ) = 16 g x 33.4 g / 284 g

                          X g of methane ( CH₄ ) = 1.88 g

1.882 g of methane (CH₄) will needed to combine with 33.4 g of Cl₂

So

methane (CH₄) = 1.882 g

5 0
3 years ago
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