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Novosadov [1.4K]
3 years ago
5

Name 3 common accelerants used in arson

Chemistry
1 answer:
Makovka662 [10]3 years ago
7 0

Answer:

Common ones are Gasoline, Diesel fuel, and Kerosene.

Explanation:

Many accelerants are hydrocarbon-based fuels, sometimes referred to as petroleum distillates: gasoline, diesel fuel, kerosene, turpentine, butane, and various other flammable solvents. These accelerants are also known as ignitable liquids. Ignitable liquids can leave behind tell-tale marks in the fire debris.

Hoped this had helped you :)

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If two oxygen atoms bond covalently, the bond is
PilotLPTM [1.2K]

Answer:

probably b i got nothing but i hop eit is good

Explanation:

4 0
3 years ago
What is the oxidation number of phosphorus (P) in sodium phosphate (Na3PO4)?
joja [24]
Na⁺¹₃P⁺⁵O⁻²₄

+1*3 + (+5) + (-2*4) = 0
8 0
3 years ago
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Which of the following is true for the actual yield of a reaction?
ludmilkaskok [199]

Answer:

It is always less than the theoretical yield

Explanation:

For many chemical reactions, the actual yield is usually less than the theoretical yield. This is due to possible loss in the process or inefficiency of the chemical reaction.

4 0
3 years ago
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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
What is the name of the compound that has the formula Li2SO3?
Oduvanchick [21]
The answer is A lithium sulfite
7 0
4 years ago
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