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3 years ago

Sublimation and boiling both happen at the surface of the substance true or false

Chemistry

2 answers:

dolphi86 [110]3 years ago

6 0

The answer to your question is false, boiling happens below the surface of a substance.

Snowcat [4.5K]3 years ago

5 0

I think the answer to this question is

False

Hope It Helps

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For the reaction A(g)

zubka84 [21]

Answer:

Yes

Explanation:

By definition, the equilibrium constanct, Kc, for the reaction A ⇒ 2B is

= [A]^1 / [B]^2

Substitute [A] = 4 and [B] = 2 in the equation,

[A]^1 / [B]^2

= 4^1 / 2^2

= 1

= Kc

So yes the reaction is at equilibrium.

6 0

2 years ago

Which equation describes a reduction?

ira [324]

The answer choice is going to be B.

7 0

3 years ago

Can someone help me? It needs to have a diagram that has arrows.

daser333 [38]

Answer: The enthalpy change for formation of butane is -125 kJ/mol

Explanation:

The balanced chemical reaction is,

$C_4H_{10}(g)+\frac{13}{2}O_2(g)\rightarrow 4CO_2(g)+5H_2O(l)$

The expression for enthalpy change is,

$\Delta H=[n\times H_f{products}]-[n\times H_f{reactants}]$

Putting the values we get :

$\Delta H=[4\times H_f_{CO_2}+5\times H_f_{H_2O}]-[1\times H_f_{C_4H_{10}}+\frac{13}{2}\times H_f_{O_2}]$

$-2877=[(4\times -393)+(5\times -286)]-[1\times H_f_{C_4H_{10}}+\frac{13}{2}\times 0]$

$H_f_{C_4H_{10}=-125kJ/mol$

Thus enthalpy change for formation of butane is -125 kJ/mol

5 0

3 years ago

Solid magnesium has mass of 1300g and a volume of 743cm . what's the density of magnesium?

Lilit [14]

D = m / V

d = 1300 g / 743 cm³

d = 1.749 g/cm³

3 0

3 years ago

A 30.0-g piece of metal at 203oC is dropped into 400.0 g of water at 25.0 oC. The water temperature rises to 29.0 oC. Calculate

oee [108]

The specific heat of the metal is 2.4733 J/g°C.

Given the following data:

Initial temperature of water = 25.0°C

Final temperature of water = 29.0°C

Mass of water = 400.0 g

Mass of metal = 30.0 g

Temperature of metal = 203.0°C

We know that the specific heat capacity of water is 4.184 J/g°C.

To find the specific heat of the metal (J/g°C):

Heat lost by metal = Heat gained by water.

$Q_{metal} = Q_{water}$

Mathematically, heat capacity or quantity of heat is given by the formula;

$Q = mc\theta$

<u>Where:</u>

Q is the heat capacity or quantity of heat.

m is the mass of an object.

c represents the specific heat capacity.

∅ represents the change in temperature.

Substituting the values into the formula, we have:

$30(203)c = 400(4.184)(29 \; -\; 20)\\\\6090c = 1673.6(9)\\\\6090c = 15062.4\\\\c = \frac{15062.4}{6090}$

Specific heat capacity of metal, c = 2.4733 J/g°C

Therefore, the specific heat of the metal is 2.4733 J/g°C.

8 0

2 years ago