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borishaifa [10]
3 years ago
13

What are some synonyms for dissolved

Chemistry
1 answer:
Luba_88 [7]3 years ago
4 0
*Go into solution
*Break down
*Liquefy
*Disintegrate
*Melt away
*Evaporate
*Dissapear

Hope it helps! Brainliest answer, PLZ? :)
You might be interested in
A.it is physical change
PIT_PIT [208]
A physical change is when something don't permanently change
5 0
3 years ago
Hydrocyanic acid, HCN, is a weak acid. (a) Write the chemical equation for the dissociation of HCN in water. (b) Identify the Br
12345 [234]

Answer: a) HCN(aq.)+H_2O\rightleftharoons H_3O^+(aq.)+CN^-(aq.)

b) HCN : acid CN^- :conjugate base.

And, H_2O : base H_3O^+: conjugate acid.

c) HCN(aq.)+NaOH(aq)\rightleftharoons NaCN(aq.)+H_2O(l)

d) NaCN(aq)\rightarrow Na^+(aq.)+CN^-(aq)

e) NaCN(aq)+HCl(aq)\rightarrow NaCl(aq.)+HCN(aq)

Explanation:

a) Weak acid is defined as the acid which does not completely dissociates when dissolved in water. They have high pH. These releases H^+ ions in their aqueous states.

The equation for the dissociation of HCN acid is given by:

HCN(aq.)+H_2O\rightleftharoons H_3O^+(aq.)+CN^-(aq.)

b) According to the Bronsted-Lowry conjugate acid-base theory, an acid is defined as a substance which looses donates protons and thus forming conjugate base and a base is defined as a substance which accepts protons and thus forming conjugate acid.

For the given chemical equation:

HCN is loosing a proton, thus it is considered as an acid and after losing a proton, it forms CN^- which is a conjugate base.

And, H_2O is gaining a proton, thus it is considered as a base and after gaining a proton, it forms H_3O^+ which is a conjugate acid.

c) Neutralization reaction is a reaction in which an acid reacts with base to produce salt and water.

HCN(aq.)+NaOH(aq)\rightarrow NaCN(aq.)+H_2O(l)

d) The chemical equation for dissociation of NaCN in water.

NaCN(aq)\rightarrow Na^+(aq.)+CN^-(aq)

e) The chemical equation for the reaction of NaCN and HCI

NaCN(aq)+HCl(aq)\rightarrow NaCl(aq.)+HCN(aq)

4 0
3 years ago
I need help with this for chemistry. I don’t understand now to do this.
alina1380 [7]

The ipR.O.B.O.T states

 aA+bB⇌ cC+dD  

the equilibrium constant is written as follows:

Kc=[C]c[D]d[A]a[B]b  

The ICE Table

The easiest approach for calculating equilibrium concentrations is to use an ICE Table, which is an organized method to track which quantities are known and which need to be calculated. ICE stands for:

"I" is for the "initial" concentration or the initial amount

"C" is for the "change" in concentration or change in the amount from the initial state to equilibrium

"E" is for the "equilibrium" concentration or amount and represents the expression for the amounts at equilibrium.

For the gaseous hydrogenation reaction below, what is the concentration for each substance at equilibrium?

C2H4(g)+H2(g)⇌C2H6(g)(1)

with  Kc=0.98  characterized from previous experiments and with the following initial concentrations:

[C2H4]0=0.33  

[H2]0=0.53  

SOLUTION

First the equilibrium expression is written for this reaction:

Kc=[C2H6][C2H4][H2]=0.98(2)

ICE Table

The concentrations for the reactants are added to the "Initial" row of the table. The initial amount of  C2H6  is not mentioned, so it is given a value of 0. This amount will change over the course of the reaction.

ICE

C2H4  

H2  

C2H6  

Initial

0.33

0.53

0

Change

Equilibrium

ICE

C2H4  

H2  

C2H6  

Initial

0.33

0.53

0

Change

-x

-x

+x

Equilibrium

Equilibrium is determined by adding "Initial" and "Change together.

ICE

C2H4  

H2  

C2H6  

Initial

0.33

0.53

0

Change

-x

-x

+x

Equilibrium

0.33-x

0.53-x

x

The expressions in the "Equilibrium" row are substituted into the equilibrium constant expression to find calculate the value of x. The equilibrium expression is simplified into a quadratic expression as shown:

0.98=x(0.33−x)(0.53−x)(3)

0.98=xx2−0.86x+0.1749(4)

0.98(x2−0.86x+0.1749)=x(5)

0.98x2−0.8428x+0.171402=x(6)

0.98x2−1.8428x+0.171402=0(7)

The quadratic formula can be used as follows to solve for x:

x=−b±b2−4ac−−−−−−−√2a(8)

x=−0.1572±(−0.1572)2−4(0.98)(0.171402)−−−−−−−−−−−−−−−−−−−−−−−−−√2(0.98)(9)

x=1.78 or0.098(10)

Because there are two possible solutions, each must be checked to determine which is the real solution. They are plugged into the expression in the "Equilibrium" row for  [C2H4]Eq :

[C2H4]Eq=(0.33−1.78)=−1.45(11)

[C2H4]Eq=(0.33−0.098)=0.23(12)

If  x=1.78  then  [C2H4]Eq  is negative, which is impossible, therefore,  x  must equal 0.098.

So:

[C2H4]Eq=0.23M(13)

[H2]Eq=(0.53−0.0981)=0.43M(14)

[C2H6]Eq=0.098M(15)

Problems

1. Find the concentration of iodine in the following reaction if the equilibrium constant is 3.76 X 103, and 2 mol of iodine are initially placed in a 2 L flask at 100 K.

I2(g)⇌2I−(aq)(16)

2. What is the concentration of silver ions in 1.00 L of solution with 0.020 mol of AgCl and 0.020 mol of Cl- in the following reaction? The equilibrium constant is 1.8 x 10-10.

AgCl(s)⇌Ag+(aq)+Cl−(aq)(17)

3. What are the equilibrium concentrations of the products and reactants for the following equilibrium reaction?

Initial concentrations:   [HSO−4]0=0.4   [H3O+]0=0.01   [SO2−4]0=0.07   K=.012  

HSO−4(aq)+H2O(l)⇌H3O+(aq)+SO2−4(aq)(18)

4. The initial concentration of HCO3 is 0.16 M in the following reaction. What is the H+ concentration at equilibrium? Kc=0.20.

H2CO3⇌H+(aq)+CO2−3(aq)(19)

5.The initial concentration of PCl5 is 0.200 moles per liter and there are no products in the system when the reaction starts. If the equilibrium constant is 0.030, calculate all the concentrations at equilibrium.

Solutions

1.

I2  

I−  

Initial

2mol/2L = 1 M

0

Change

−x  

+2x  

Equilibrium

1−x  

2x  

At equilibrium

Kc=[I−]2[I2]  

3.76×103=(2x)21−x=4x21−x  

cross multiply

4x2+3.76.103x−3.76×103=0  

apply the quadratic formula:

−b±b2−4ac−−−−−−−√2a  

with:  a=4 ,  b=3.76×103   c=−3.76×103 .

The formula gives solutions of of x=0.999 and -940. The latter solution is unphysical (a negative concentration). Therefore, x=0.999 at equilibrium.

[I−]=2x=1.99M(20)

[I2]=1−x=1−.999=0.001M(21)

2.

Ag+  

Cl−  

Initial

0

0.02mol/1.00 L = 0.02 M

Change

+x  

+x  

Equilibrium  

0.02+x  

Kc=[Ag−][Cl−](22)

1.8×10−10=(x)(0.02+x)(23)

x2+0.02x−1.8×1010=0(24)

x=9×10−9(25)

[Ag−]=x=9×10−9(26)

[Cl−]=0.02+x=0.020(27)

3.

H2CO3  

SO2−4  

H3O+  

Initial

0.4

0.01

0.07

Change

−x  

Equilibrium

0.4−x  

0.01+x  

0.07+x  

Kc=[SO2−4][H3O+]H2CO3(28)

0.012=(0.01+x)(0.07+x)0.4−x(29)

cross multiply and get:

x2+0.2x−0.0041=0(30)

apply the quadratic formula

x = 0.0328

[H2CO3]=0.4-x=0.4-0.0328=0.3672

[S042-]=0.01+x=0.01+0.0328=0.0428

[H30]=0.07+x=0.07+0.0328=0.1028

4.

H2CO3

H+  

CO2−3  

Initial

.16

0

Change

-x

Equilibrium

.16-x

apply the quadratic equation

x=0.1049

[H+]=x=0.1049

5. First write out the balanced equation:

PCl5(g)⇌PCl3(g)+Cl2(g)  

PCl5  

PCl3  

Cl2  

Initial

0.2

0

Change

-x

Equilibrium

0.2-x

Kc=[PC3][Cl2][PCl5](31)

0.30=x20.2−x(32)

Cross multiply:

x2+0.03x−0.006=0(33)

Apply the quadratic formula:

x=0.064

[PCl5]=0.2-x=0.136

[PCl3]=0.064

[Cl2]=0.064

Information is verified by Brainly Incorporations.

Do not copy this information without the consent of Brainly Inc.

ipR.O.B.O.T is an international Internet Protocol Recessive Observation Branch Organization Technologies

4 0
3 years ago
Which of the following most likely happens when the volume of a gas increases? (4 points)
Xelga [282]

Answer:

Molecules in the gas move faster due to the fact they have more energy

Explanation:

If you put more heat the higher the volume will be. Less heat means less volume.

(have a great day!)

3 0
3 years ago
Is hydrogen a metal?
Ilia_Sergeevich [38]
No, Hydrogen is a gas.
3 0
3 years ago
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