Melting
Melting is a change in property of matter from solid to liquid
Matter is anything that occupies space and has mass. Thus, there are fundamentally three types of matter which is solid liquid and gas. But why do gases and liquids diffuse and not solids? It is because of the molecular structure of these components of matter. If we examine the molecular structure of gas the molecules are highly scattered and liquid has also almost the same structure as mediocrely scattered that these particles can easily slip through other substances unlike solid which is entirely intact. <span> </span>
Answer:
D is correct
Explanation:
because
we know that
density of lead is 11.36 g/cm3
and
density of tin is 7.31 g/cm3
so..
density of alloy by mixing 50/50
=(11.36+7.31)/2 g/cm3
=18.67/2 g/cm3
=9.33 g/cm3
A compound can be treated like an element when balancing chemical equations when : A. when there is no change to the atoms in the compound
If they remain together during the reaction, you can treat a compound as a singe atom.
hope this helps
In general, we have this rate law express.:
![\mathrm{Rate} = k \cdot [A]^x [B]^y](https://tex.z-dn.net/?f=%5Cmathrm%7BRate%7D%20%3D%20k%20%5Ccdot%20%5BA%5D%5Ex%20%5BB%5D%5Ey)
we need to find x and y
ignore the given overall chemical reaction equation as we only preduct rate law from mechanism (not given to us).
then we go to compare two experiments in which only one concentration is changed
compare experiments 1 and 4 to find the effect of changing [B]
divide the larger [B] (experiment 4) by the smaller [B] (experiment 1) and call it Δ[B]
Δ[B]= 0.3 / 0.1 = 3
now divide experiment 4 by experient 1 for the given reaction rates, calling it ΔRate:
ΔRate = 1.7 × 10⁻⁵ / 5.5 × 10⁻⁶ = 34/11 = 3.090909...
solve for y in the equation
![\Delta \mathrm{Rate} = \Delta [B]^y](https://tex.z-dn.net/?f=%5CDelta%20%5Cmathrm%7BRate%7D%20%3D%20%5CDelta%20%5BB%5D%5Ey)

To this point,
![\mathrm{Rate} = k \cdot [A]^x [B]^1](https://tex.z-dn.net/?f=%5Cmathrm%7BRate%7D%20%3D%20k%20%5Ccdot%20%5BA%5D%5Ex%20%5BB%5D%5E1%20)
do the same to find x.
choose two experiments in which only the concentration of B is unchanged:
Dividing experiment 3 by experiment 2:
Δ[A] = 0.4 / 0.2 = 2
ΔRate = 8.8 × 10⁻⁵ / 2.2 × 10⁻⁵ = 4
solve for x for
![\Delta \mathrm{Rate} = \Delta [A]^x](https://tex.z-dn.net/?f=%5CDelta%20%5Cmathrm%7BRate%7D%20%3D%20%5CDelta%20%5BA%5D%5Ex)

the rate law is
Rate = k·[A]²[B]