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Natalka [10]
3 years ago
13

Melvin is exploring the different kinds of changes that can happen to substances. After reading about the kinds of changes that

occur, he saw some rust on a pipe that was going into his house. What kind of change is the formation of rust?
A.reactant
B.physical
C.chemical
Chemistry
2 answers:
Vlada [557]3 years ago
6 0
I think it’s c because when ever it rains it make medal rust
Alexxandr [17]3 years ago
4 0

Answer:

It is B. physical.

Explanation:

The rust on the pipe is something causes by something, which mean it is physical. Hope this helped you out!!

Have a wonderful day! :D

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Which energy changes are associated with a liquid boiling?
Delvig [45]

Answer: Energy is absorbed, and kinetic energy increases

Explanation: Boiling is the phenomenon in which liquid is converted to gaseous state.

Thus in order to break the inter molecular forces of attraction present in a liquid, energy is required and thus the process is endothermic in nature.

As the molecules absorb energy, the kinetic energy of the molecules increases and they start moving faster and more randomly and convert into gaseous state.

7 0
3 years ago
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In electrical work the symbol w is used to indicate what
Lera25 [3.4K]
In electrical work the w represents Watts.
7 0
4 years ago
Which scientist's atomic model is most recent?
Natali [406]

Answer:

B.Rutherford's

Explanation:

I know it

8 0
4 years ago
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Suppose a current of 0.860A flows through a copper wire for 5.0 minutes. Calculate how many moles of electrons travel through th
denpristay [2]

Answer:

2.7×10⁻³ mole

Explanation:

Applying

Q = it.............. Equation 1

Where Q = amount of charge, i = current, t = time

From the question,

Given: i = 0.860 A, t = 5 minutes = (5×60) seconds = 300 seconds

Subsitute these values into equation 1

Q = (0.860×300)

Q = 258 C

If one mole of electron has a charge of 96500 C

Then, x mole of electron will have a charge of 258 C

1 mole ⇒ 96500 C

X moles ⇒ 258 C

Solve for X

X = (258×1)/96500

X = 2.7×10⁻³ mole

7 0
3 years ago
Find the solubility of agi in 2.5 m nh3 [ksp of agi = 8.3 × 10−17; kf of ag(nh3)2+ = 1.7 × 107].
vovangra [49]
When the first reaction equation is:

AgI(S) ↔ Ag+(Aq)  +  I-(Aq)

So, the Ksp expression = [Ag+][I-]

∴Ksp = [Ag+][I-] = 8.3 x 10^-17

Then the second reaction equation is:

 Ag+(aq)  + 2NH3(aq) ↔  Ag(NH3)2+  

So, Kf expression = [Ag(NH3)2+] / [Ag+] [NH3]^2

∴Kf = [Ag(NH3)2+] /[Ag+] [NH3]^2 = 1.7 x 10^7

by combining the two equations and solve for Ag+:

and by using ICE table:

               AgI(aq) + 2NH3  ↔  Ag(NH3)2+  + I-
initial                        2.5                    0            0 

change                    -2X                    +X             +X

Equ                     (2.5-2X)                   X               X

so K = [Ag(NH3)2+] [I-] / [NH3]^2

Kf * Ksp = X^2 / (2.5-2X)

8.3 x 10^-17 * 1.7 x10^7 = X^2 / (2.5-2X) by solving for X

∴ X = 5.9 x 10^-5 

∴ the solubility of AgI = X = 5.9 x 10^-5 M

4 0
3 years ago
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