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Nonamiya [84]
3 years ago
5

How is melting simillar evaporation

Chemistry
2 answers:
GrogVix [38]3 years ago
5 0

Answer:

well both disappear over time

Explanation:

kiruha [24]3 years ago
3 0

Answer:

Melting and evaporation both represent changes in matter that involve behavior at the molecular level.

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What is the chemical formula for iron(III) Oxide with explanation, please
OLga [1]
Writing the name for Fe2o3 is a bit more challenging since Fe is a transition metal. this means you need to used parenthesis and Roman Numeral to show the charge on the Fe atoms. in these case each Fe atoms has a charge of +3 so we write Fe2o3 as an iron(III) oxide

4 0
3 years ago
Which states of matter have particles that move independently of one another with very little attraction?
bonufazy [111]

Answer:

Gas

Explanation:

The gaseous state has very loose and unorganized structuring of particles, making them have little attraction and move independently.

4 0
3 years ago
PLEASE HELP!! I have to pass this class! Anything!!
Georgia [21]

Answer:

We'll have 8.0 moles Fe3O4 and 4.0 moles CO2

Explanation:

Step 1: data given

Number of moles Fe2O3 = 12.0 moles

Number of moles CO = 12.0 moles

Step 2: The balanced equation

3Fe2O3 +CO → 2Fe3O4 + CO2

Step 3: Calculate the limiting reactant

For 3 moles Fe2O3 we need 1 mol CO to produce 2 moles Fe3O4 and 1 mol CO2

Fe2O3 is the limiting reactant. It will completely be consumed (12.0 moles).

CO is in excess. There will react 12.0 / 3 = 4.0 moles

There will remain 12.0 - 4.0 = 8.0 moles

Step 4: Calculate moles products

For 3 moles Fe2O3 we need 1 mol CO to produce 2 moles Fe3O4 and 1 mol CO2

For 12.0 moles Fe2O3 we'll have 2/3 * 12.0 = 8.0 moles Fe3O4

For 12.0 moles Fe2O3 we'll have 12.0 / 3 = 4.0 moles CO2

We'll have 8.0 moles Fe3O4 and 4.0 moles CO2

6 0
3 years ago
A lab requires 3.50L of a 2.0M solution of HCl. What volume of a 6.0M solution is needed to make this solution?
Alchen [17]
The answer is 1.167L of 6.0M solution.

4 0
3 years ago
Read 2 more answers
I need chemistry help! How would I set up these problems?
Bond [772]

Answer: -

1) 8.33 minutes

2) 118.39 in/ s

180.43 m/min

10.83 km/ hr

Explanation: -

Speed of light = 3 x 10⁸ m/s

Distance of the earth from the sun= 93 million miles

We know 1 million = 1,000,000

Also 1 mile = 1609 m

Distance of the earth from the sun= 93 million miles

= 93,000,000 miles.

= 1.5 x 10^{11} m

Time taken = \frac{Distance}{Speed}

= \frac{1.5 x [tex] 10^{11} m}{3 x 10⁸ m/s} [/tex]

= 500 s

= 500/ 60

= 8.33 minutes

2) Distance = 1 mile = 63360 inches

Time taken = 8.92 min

= 8.92 x 60

= 535.2 s

Speed = \frac{distance}{time}

= \frac{63360 inches}{535.2 s}

= 118.39 in/ s

Distance = 1 mile = 63360 inches = 63360 x 2.54 cm = 63360 x 2.54 x 10^{-2} m

Time taken = 8.92 min

Speed = \frac{distance}{time}

= \frac{63360 x 2.54 x [tex] 10^{-2} m}{8.92 min} [/tex]

= 180.43 m/ min

1 m = 10⁻³ Km

1 min = 1/60 hour

1 m /min = 10⁻³ km/ \frac{1}{60 hour}

= 60/1000

=0.06 km/hr

180.43 m / min = 180 x 0.06 km / hr

= 10.93 km / hr

4 0
3 years ago
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