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2 years ago

Fill in the blank.

Chemistry

1 answer:

labwork [276]2 years ago

7 0

Answer:

Kinetic energy, Kelvin temperature, direction, warmer, cooler, Kelvin, Absolute zero

Explanation:

THe answers are in order and when there is a comma move to the next ______

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If 0.50 mol/L of butane is added to the original equilibrium mixture and the system shifts to a new equilibrium position, what i

Ostrovityanka [42]

Consider the isomerization of butane with equilibrium constant is 2.5 .The system is originally at equilibrium with :

[butane]=1.0 M , [isobutane]=2.5 M

If 0.50 mol/L of butane is added to the original equilibrium mixture and the system shifts to a new equilibrium position, what is the equilibrium concentration of each gas?

Answer:

The equilibrium concentration of each gas:

[Butane] = 1.14 M

[isobutane] = 2.86 M

Explanation:

Butane ⇄ Isobutane

At equilibrium

1.0 M 2.5 M

After addition of 0.50 M of butane:

(1.0 + 0.50) M -

After equilibrium reestablishes:

(1.50-x)M (2.5+x)

The equilibrium expression will wriiten as:

$K_c=\frac{[Isobutane]}{[Butane]}$

$2.5=\frac{2.5+x}{(1.50-x)}$

x = 0.36 M

The equilibrium concentration of each gas:

[Butane]= (1.50-x) = 1.50 M - 0.36M = 1.14 M

[isobutane]= (2.5+x) = 2.50 M + 0.36 M = 2.86 M

3 0

3 years ago

3. Indicate the type of reaction represented by each equation:

BartSMP [9]

3. Is double replacement

6 0

4 years ago

For the chemical reaction:

Kamila [148]

Answer:

Volume of ammonia produced = 398.7 dm³

Explanation:

Given data:

Volume of N₂ = 200 dm³

Pressure and temperature = standard

Volume of ammonia produced = ?

Solution:

Chemical equation:

N₂ + 3H₂ → 2NH₃

Number of moles of N₂:

PV = nRT

1 atm× 200 L = n× 0.0821 atm.L/mol.K × 273 K

n = 200 atm.L /22.41 atm.L/mol

n = 8.9 mol

Now we will compare the moles of ammonia and nitrogen.

N₂ : NH₃

1 : 2

8.9 : 2/1×8.9 = 17.8 mol

Volume of ammonia:

1 mole of any gas occupy 22.4 dm³ volume

17.8 mol ×22.4 dm³/1 mol = 398.7 dm³

7 0

3 years ago

Pre-lab questions 1. a concentration gradient affects the direction that solutes diffuse. describe how molecules move with respe

VLD [36.1K]

Just like how heat moves from a region of higher temperature to a region of lower temperature, molecules also tend to move from a region of higher concentration to a region of lower concentration. This is called natural diffusion and is naturally happening to reach stability.

7 0

4 years ago

Hiii pls help me to write out the ionic equation

emmasim [6.3K]

Answer:

STEP I

This is the balanced equation for the given reaction:-

$2KOH_{(aq)} + H_2SO_4{}_{(aq)} \rightarrow K_2SO_4{}_{(aq)} + 2H_2O_{(l)}$

STEP II

The compounds marked with (aq) are soluble ionic compounds. They must be

broken into their respective ions.

see, in the equation KOH, H2SO4, and K2SO4 are marked with (aq).

On breaking them into their respective ions :-

2KOH -> 2K+ + 2OH-
H2SO4 -> 2H+ + (SO4)2-
K2SO4 -> 2K+ + (SO4)2-

STEP III

Rewriting these in the form of equation

$\underline{\pmb{2K^+} }+ 2OH^- + 2H^+ + \pmb{\underline{{SO_4{}^{2-}}} \: \rightarrow \: \underline{\pmb{2K^+}}} + \underline{\pmb{SO_4{}^{2-}}} + 2H_2O$

STEP IV

Canceling spectator ions, the ions that appear the same on either side of the equation

(note: in the above step the ions in bold have gotten canceled.)

$\boxed{ \mathfrak{ \red{ 2OH^-{}_{(aq)} + 2H^+{(aq.)} \rightarrow H_2O{}_{(l)}}}}$

This is the net ionic equation.

____________________________

$\\$

$\mathfrak{\underline{\green{ Why\: KOH \:has\: been\: taken\: as\: aqueous ?}}}$

KOH has been taken as aqueous because the question informs us that we have a solution of KOH. by solution it means that KOH has been dissolved in water before use.

[Alkali metal hydroxides are the only halides soluble in water ]

4 0

3 years ago