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LiRa [457]
3 years ago
9

Find the volume of 10g of gasoline if it has a density of 0.7g/mL

Chemistry
1 answer:
Salsk061 [2.6K]3 years ago
5 0

Hey there!

Mass = 10 g

Density = 0.7 g/mL

Volume = ?

Therefore:

D = m/ V

0.7 = 10 / V

V = 10 / 0.7

V = 14.28 mL

Hope that helps!

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5 0
3 years ago
A 45ml of a 4M solution of CaBr2 contains how many grams of CaBr2?
natta225 [31]

The required mass of calcium bromide is 35.98 grams.

<h3>What is molarity?</h3>

Molarity is any solution is define as the number of moles of solute present in per liter of solution as;

M = n/V, where

  • M = molarity = 4M
  • V = volume = 45mL = 0.045L

Moles will be calculated by using the above equation as:

n = (4)(0.045) = 0.18 mole

Relation between the mass and moles of any substance will be represented as:

n = W/M, where

  • W = given mass
  • M = molar mass

Mass of CaBr₂ = (0.18mol)(199.89g/mol) = 35.98g

Hence required mass of CaBr₂ is 35.98 grams.

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6 0
2 years ago
51. The radius of gold is 144 pm and the density is 19.32 g/cm3. Does elemental gold have a face-centered cubic structure or a b
Serjik [45]

Answer:

Elemental gold to have a Face-centered cubic structure.

Explanation:

From the information given:

Radius of gold = 144 pm

Its density = 19.32 g/cm³

Assuming the structure is a face-centered cubic structure, we can determine the density of the crystal by using the following:

a = \sqrt{8} r

a = \sqrt{8} \times 144 pm

a = 407 pm

In a unit cell, Volume (V) = a³

V = (407 pm)³

V = 6.74 × 10⁷ pm³

V = 6.74 × 10⁻²³ cm³

Recall that:

Net no. of an atom in an FCC unit cell = 4

Thus;

density = \dfrac{mass}{volume}

density = \dfrac{ 4 \ atm ( 196.97 \ g/mol) (\dfrac{1 \ mol }{6.022 \times 10^{23} \ atoms})}{6.74 \times 10^{-23} \ cm^3}

density d = 19.41 g/cm³

Similarly; For a  body-centered cubic structure

r = \dfrac{\sqrt{3}}{4}a

where;

r = 144

144 = \dfrac{\sqrt{3}}{4}a

a = \dfrac{144 \times 4}{\sqrt{3}}

a = 332.56 pm

In a unit cell, Volume V = a³

V = (332.56 pm)³

V = 3.68 × 10⁷ pm³

V  3.68 × 10⁻²³   cm³

Recall that:

Net no. of atoms in BCC cell = 2

∴

density = \dfrac{mass}{volume}

density = \dfrac{ 2 \ atm ( 196.97 \ g/mol) (\dfrac{1 \ mol }{6.022 \times 10^{23} \ atoms})}{3.68 \times 10^{-23} \ cm^3}

density =17.78 g/cm³

From the two calculate densities, we will realize that the density in the face-centered cubic structure is closer to the given density.

This makes the elemental gold to have a Face-centered cubic structure.

3 0
3 years ago
Select the compounds from the list below which are insoluble in water
natulia [17]

Answer:

Insoluble in water:

  • BaSO4
  • PbCl2
  • Cu2O
  • AgBr

Explanation:

Water turns out to be a good solvent for ionic substances, or in general, polarized covalent substances. On the other hand, it is not a good solvent for non-polar substances, these being the vast majority of covalent substances.

5 0
3 years ago
Study the image and answer the question.
konstantin123 [22]
The image represents A COMPOUND because the molecules are BONDED CHEMICALLY.
A compound is a substance formed when two or more elements combine together chemically. In the process of chemical combination, the chemical bonds that were present in the participating elements will be broken and new chemical bonds will be formed in the product.
7 0
2 years ago
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