All categories

3 years ago

Is hot asphalt poured onto flat ground cools to from a solid road surface a physical or chemical change?

Chemistry

2 answers:

Phantasy [73]3 years ago

8 0

When hot asphalt is poured onto flat ground cools to form a solid road surface is a chemical change.

Asphalt is a plastic hydrocarbon ,when in liquid condition reacts with oxygen in air to react to become a solid.

aliina [53]3 years ago

7 0

⇒Tar is mainly made up of a collection of very heavy hydrocarbons, with rather high melting temperatures. Commonly, it is mixed with some stones or gravel to make a harder surface for roads. It is possible to heat this up to soften or melt this; you can try it yourself by heating it in a metal can over a stove (don't use a container that you plan to keep clean) or in a wood or coal fire.
it is reversible

You might be interested in

The mass of an object is 32g and its density is 17g /mL. what is its volume

guajiro [1.7K]

Explanation:

here,

Mass (m)= 32 g =0.32 kg

Density(d)= 17 g/ml

volume (v)=?

we know that,

d=m/v

17=0.32/v

v= 0.32/17

v=0.018

3 0

3 years ago

Read 2 more answers

What solution can you add to of the mixture of NH4 (aq) and Ca2 (aq) to precipitate one cation while keeping the other cation in

vekshin1

Answer:H2SO4

Explanation:calcium forms a precipitate in a sulfide compound of so4

7 0

4 years ago

Number of electrons present in 3.6mg of NH₄⁺?

nata0808 [166]

0.0036g of NH4+ x 1mol of NH4+/18.04g of NH4+ x 6.02*10^23 electrons of NH4+/1mol of NH4+

= 1.20 x 10^20 electrons of NH4+

4 0

3 years ago

Question 2 (1 point) Al + S8 = Al2S3 When balanced, what is the coefficient of Al?

gtnhenbr [62]

Answer:

Explanation:

FIRST AND FOREMOST, BALANCE YOUR EQUATION.

Al + S8 ➡️ Al2S3

Numbers of Al=1. ➡️ Numbers of Al = 2

Numbers of S =8. ➡️ Numbers of S. = 3

USE COEFFICIENT TO BALANCE THE EQUATION.

16Al + 3S8 ➡️ 8Al2S3

Now the Numbers of Al and S in both sides of eqn. is balanced

The Answer Is 16

3 0

3 years ago

Read 2 more answers

How many milliliters of 0.100 M NaOH are required to neutralize the following solutions?

gulaghasi [49]

Answers:

a.) 10.0 mL of 0.0500 M HCl: 5.00 ml

b.) 25.0 mL of 0.126 M HNO₃: 31.5 ml

c.) 50.0 mL of 0.215 M H₂SO4: 215. ml

Explanation:

All the reactions are the neutralization of strong acids with the same strong base.

At the neutralization point you have:

number of equivalents of the base = number of equivalent of the acid

And the number of equivalents (#EQ) may be calculated using the normality (N) concentration and the volume (V)

# EQ = N × V

Then, at the neutralization point:

# EQ acid = N acid × V acid

# EQ base = N base × V base

# EQ acid = # EQ base

N acid × V acid = N base × V base

Also, you can use the formula that relates normality with molarity

N = M × number of hydrogen or hydroxide ions

a.) 10.0 mL of 0.0500 M HCl

The number of hydrogen ions for HCl is 1 and the number of hydroxide ions for NaOH is 1.

10.0 ml × 0.0500 M × 1 = V base × 0.100 M × 1

⇒ V base = 10.0 ml ×0.0500 M / 0.100 M = 5.00 ml

b.) 25.0 mL of 0.126 M HNO₃

The number of hydrogen ions for HNO₃ is 1 and the number of hydroxide ions for NaOH is 1.

25.0 ml × 0.126 M × 1 = V base × 0.100 M × 1

⇒ V base = 25.0 ml ×0.126 M / 0.100 M = 31.5 ml

c.) 50.0 mL of 0.215 M H₂SO4

The number of hydrogen ions for H₂SO4 is 2 and the number of hydroxide ions for NaOH is 1.

50.0 ml × 0.215 M × 2 = V base × 0.100 M × 1

⇒ V base = 50.0 ml ×0.215 M × 2 / 0.100 M = 215. ml

7 0

4 years ago