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photoshop1234 [79]
3 years ago
8

What phases (solid, liquid, gas) are present in the unlighted candle? In the burning candle? Which phase appears to take part in

the chemical reaction?
Chemistry
2 answers:
PSYCHO15rus [73]3 years ago
8 0
Only the solid phase.
When lit all three phases of matter are present, solid, liquid and gas. The candle itself is still solid. But liquid wax pools under the flame. And the flame itself is releasing gases consisting mostly of water vapor and carbon dioxide. Many consider the flame is in another state of matter, the plasma state. It's a high energy state of matter exhibited by stars, flames and lightning.
Feliz [49]3 years ago
5 0

Answer:

As explained below.

Explanation:

  • In an unlighted, the solid phase can be seen when it's in solid shape and not ignited when it starts to burn it's said to have entered into the liquid phase or wax phase.
  • The liquid phase is to be part of the chemical reaction because the chemical reactions are seen when it is lit with the presence of oxygen and the last phase of gas is given out in the form of flames of fire upon burning as to give out carbon dioxide and ash.

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Which statements are true about Figure I and Figure II below? (Check all that apply)
Scilla [17]

Both figures are mixtures,

Figure II is a heterogenous mixture

Figure I is a homogenous mixture

5 0
3 years ago
EARTH SCIENCE: PLEASE HELP ME !!! Which tools do astronomers use to determine the effects of collisions between galaxies? Select
topjm [15]

In studying effects of collisions between galaxies, astronomers usually make use of space telescopes.

Astronomy deals with the study of outer space. The study focuses on planets and other celestial bodies. Generally telescopes are used to study outer space. The particular type of telescope that is used depends on the need.

In studying effects of collisions between galaxies, astronomers usually make use of space telescopes.

Learn more: brainly.com/question/556195

6 0
2 years ago
Answer these please ASAP need help no idea how to do these
STALIN [3.7K]

Answer:

Explanation:

Cu:

Number of moles = Mass / molar masa

2 mol = mass / 64 g/mol

Mass = 128 g

Mg:

Number of moles = Mass / molar masa

0.5 mol = mass / 24 g/mol

Mass =  g

Cl₂:

Number of moles = Mass / molar masa

Number of moles  = 35.5 g / 24 g/mol

Number of moles = 852 mol

H₂:

Number of moles = Mass / molar mass

8 mol  = Mass / 2 g/mol

Mass =  16 g

P₄:

Number of moles = Mass / molar masa

2 mol  =  mass / 124 g/mol

Mass = 248 g

O₃:

Number of moles = Mass / molar masa

Number of moles  = 1.6 g /48  g/mol

Number of moles = 0.033 mol

H₂O

Number of moles = Mass / molar masa

Number of moles  = 54 g / 18 g/mol

Number of moles = 3 mol

CO₂

Number of moles = Mass / molar masa

2 mol  =  mass / 124 g/mol

Mass = 248 g

NH₃

Number of moles = Mass / molar masa

Number of moles  = 8.5 g / 17 g/mol

Number of moles = 0.5 mol

CaCO₃

Number of moles = Mass / molar masa

Number of moles  = 100 g / 100 g/mol

Number of moles = 1 mol

a)

Given data:

Mass of iron(III)oxide needed = ?

Mass of iron produced = 100 g

Solution:

Chemical equation:

F₂O₃ + 3CO    →    2Fe  + 3CO₂

Number of moles of iron:

Number of moles = mass/ molar mass

Number of moles = 100 g/ 56 g/mol

Number of moles = 1.78 mol

Now we compare the moles of iron with iron oxide.

                        Fe          :           F₂O₃                

                           2          :             1

                          1.78       :        1/2×1.78 = 0.89 mol

Mass of  F₂O₃:

Mass = number of moles × molar mass

Mass = 0.89 mol × 159.69 g/mol

Mass = 142.124 g

100 g of iron is 1.78 moles of Fe, so 0.89 moles of F₂O₃ are needed, or 142.124 g of iron(III) oxide.

b)

Given data:

Number of moles of Al = 0.05 mol

Mass of iodine = 26 g

Limiting reactant = ?

Solution:

Chemical equation:

2Al + 3I₂   →  2AlI₃

Number of moles of iodine = 26 g/ 254 g/mol

Number of moles of iodine = 0.1 mol

Now we will compare the moles of Al and I₂ with AlI₃.

                          Al            :         AlI₃    

                          2             :           2

                         0.05         :        0.05

                           I₂            :         AlI₃

                           3            :          2

                         0.1           :           2/3×0.1 = 0.067

Number of moles of AlI₃ produced by Al are less so it will limiting reactant.

Mass of AlI₃:                            

Mass = number of moles × molar mass

Mass = 0.05 mol × 408 g/mol

Mass = 20.4 g

26 g of iodine is 0.1 moles. From the equation, this will react with 2 moles of Al. So the limiting reactant is Al.

c)

Given data:

Mass of lead = 6.21 g

Mass of lead oxide = 6.85 g

Equation of reaction = ?

Solution:

Chemical equation:

2Pb + O₂   → 2PbO

Number of moles of lead = mass / molar mass

Number of moles = 6.21 g/ 207 g/mol

Number of moles = 0.03 mol

Number of moles of lead oxide = mass / molar mass

Number of moles = 6.85 g/ 223 g/mol

Number of moles = 0.031 mol

Now we will compare the moles of oxygen with lead and lead oxide.

               Pb         :        O₂

                2          :         1

               0.03     :      1/2×0.03 = 0.015 mol

Mass of oxygen:

Mass = number of moles × molar mass

Mass = 0.015 mol × 32 g/mol

Mass =  0.48 g

The mass of oxygen that took part in equation was 0.48 g. which is 0.015 moles of oxygen. The number of moles of Pb in 6.21 g of lead is 0.03 moles. So the balance equation is

2Pb + O₂   → 2PbO

   

6 0
3 years ago
What is the cell potential for the reaction mg(s+fe2+(aq?mg2+(aq+fe(s at 77 ?c when [fe2+]= 3.40 m and [mg2+]= 0.210 m . express
Galina-37 [17]
First, you need to calculate the standard cell potential using standard reduction potential from a textbook or online. Since Mg becomes Mg+2, magnesium is being oxidized because it is losing electrons, you need to flip its potential

Fe+2 + 2e- --> Fe                  potential= -0.44
Mg+2 + 2e- --> Mg                potential= -2.37


Cell potential= (-0.44) + (+2.37)= 1.93 V

Now, you need to use Nernst formula to get the answer. I have attached a PDF with the work.
Download pdf
8 0
3 years ago
Some growth regulators are useful to farmers and gardeners.<br> O True<br> O False
lisov135 [29]

Answer:

True

Explanation:

The first artificial use of a PGR was to stimulate the production of flowers on pineapple plants. They are now used widely in agriculture. Plant hormones are also used in turf management to reduce the need to mow, to suppress seedheads, and to suppress other types of grass.

3 0
3 years ago
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