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2 years ago

Describe how physical and chemical change may have affected by temperature

2 answers:

7 0

The heat changes the physical structure into a chemical structure because some of the cells in the physical structure are removed to make that physical structure a chemical structure

MAXImum [283]2 years ago

5 0

Chemical reaction by temperature can affect the change of matter. If you increase the temp. you increase the reaction rate. Ex. Burning a marshmallow can cause the fire to break down the chemical bonds.

Physical change by temperature can affect depending on pressure. In terms of malleability, a metal being compressed can be shaped into a different object, but the chemical bonds are still the same.

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WILL RATE AND GIVE BRAINLIEST!!

Andrew [12]

A) a disk formed of long trails of stars coiled into a spiral

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2 years ago

Read 2 more answers

What is the molar solubility of nickel(II) sulfide in 0.091 M KCN? For NiS, Ksp = 3.0 × 10 –19; for Ni(CN) 4 2–, Kf = 1.0 × 10 3

Marta_Voda [28]

Answer:

The value is $x = 0.0227 \ M$

Explanation:

From the question we are told that

The concentration of $KCN \ \ i.e \ \ CN^{-}$ is $M_1 = 0.091 \ M$

The solubility product constant for $NiS$ is $K_{sp} = 3.0 *10^{-19}$

The stability constant for $Ni(CN)_4 ^{2-}$ is $K_f = 1.0 *10^{31}$

Generally the dissociation reaction for NiS is

$Ni S \underset{}{\stackrel{}{\rightleftharpoons}} Ni^{2+} + S^{2-}$

Generally the formation reaction for $Ni(CN)_4 ^{2-}$ is

$4CN^- + N_i ^{2+} \underset{}{\stackrel{}{\rightleftharpoons}} \ Ni(CN)^{2-}_{4}$

Combining both reaction we have

$4CN^ - + NiS \ \underset{}{\stackrel{}{\rightleftharpoons}} \ Ni(CN)^{2-}_4 + S^{2-}$

Gnerally the equilibrium constant for this reaction is

$K_c = K_{sp} * K_f$

=> $K_c = 3.0 *10^{-19 } * 1.0 *10^{31}$

=> $K_c = 3.0*10^{12}$

Generally the I C E table for the above reaction is

$4CN^ - \ \ \ + \ \ \ NiS \ \ \ \ \ \ \ \underset{}{\stackrel{}{\rightleftharpoons}} \ \ \ \ \ Ni(CN)^{2-}_4 \ \ \ \ \ \ \ \ \ + \ \ \ \ \ \ \ \ \ S^{2-}$

initial [ I] 0.091 0 0

Change [C] -4x +x + x

Equilibrium [E ] 0.091 - 4x x x

Here is x is the amount in term of concentration that is lost by $CN^-$ and gained by $Ni(CN)_4 ^{2-}$ and $S^{2-}$

Gnerally the equilibrium constant for this reaction is mathematically represented as

$K_c = \frac{[Ni (CN)_4^{2-} ] [S^{2-} ] }{ [CN^{-}]^4}$

=> $3.0*10^{12} = \frac{x * x}{ [0.091 - 4x ]^4}$

=> $3.0*10^{12}* [0.091 - 4x ]^4 = x^2$

=> $[0.091 - 4x ]^4 = \frac{x^2}{3.0*10^{12}}$

=> $[0.091 - 4x ] = \sqrt[4]{ \frac{x^2}{3.0*10^{12}}}$

=> $[0.091 - 4x ] = \frac{\sqrt{x} }{1316}$

=> $119.8 - 5264x =\sqrt{x}$

Square both sides

$(119.8 - 5264x)^2 =x$

=> $14352.04 - 1261255 x + 27709696x^2 = 0$

=> $27709696x^2 - 1261255 x + 14352.04 = 0$

Solving using quadratic equation

The value of x is $x = 0.0227 \ M$

Hence the amount in terms of molarity (concentration) of $Ni(CN)_4 ^{2-}$ and $S^{2-}$ produced at equilibrium is $x = 0.0227 \ M$ it then means that the amount of NiS (nickel(II) sulfide) lost at equilibrium is $x = 0.0227 \ M$

So the molar solubility of nickel(II) sulfide at equilibrium is

$x = 0.0227 \ M$

3 0

2 years ago

The image above shows a type of metamorphic rock called gneiss. Gneiss is characterized by alternating bands of light and dark m

finlep [7]

Answer: first blank is heating second blank is pressure i think but i know that its NOT freezing

Explanation:

I did it and got it right

4 0

2 years ago

What happens to colors in sunlight when droplets of water in the atmosphere act like prisms?

andrew-mc [135]

Dispersion so the correct is A!!!

4 0

2 years ago

Read 2 more answers

A hot air balloon starts with its temperature at 68.7°C and a pressure of 0.987 ATM and volume of 564L at what temperature in de

ICE Princess25 [194]

Answer:

54.7°C is the new temperature

Explanation:

We combine the Ideal Gases Law equation to solve this.

P . V = n. R. T

As moles the balloon does not change and R is a constant, we can think this relation between the two situations:

P₁ . V₁ / T₁ = P₂ . V₂ / T₂

T° is absolute temperature (T°C + 273)

68.7°C + 273 = 341.7K

(0.987 atm . 564L) / 341.7K = (0.852 atm . 625L) / T₂

1.63 atm.L/K = 532.5 atm.L / T₂

T₂ = 532.5 atm.L / 1.63 K/atm.L → 326.7K

T° in C = T°K - 273 → 326.7K + 273 = 54.7°C

3 0

2 years ago