1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Airida [17]
2 years ago
14

Which mineral might scratch the mineral fluorite, but would not scratch the mineral amphibole? 1 brucite 2. magnesite 3. carnall

ite 4. olivine​
Chemistry
1 answer:
NARA [144]2 years ago
4 0

Answer:

olivine i think

Explanation:

You might be interested in
Some magnesium powder was mixed with some copper( II)oxide and heated strongly. there was a victorious reaction producing a lot
-BARSIC- [3]

Reaction:

\mathrm{Mg \: + \: CuO \rightarrow Cu\: + \: MgO}

Magnesium is a stronger reducing agent than copper and is thus able to reduce copper(II) oxide.

Products of the reaction: Magnesium oxide and metallic copper.

4 0
3 years ago
Read 2 more answers
11.0 L of hydrogen and 5.52 L of oxygen are exploded together in a reaction tube. What volume of water vapor was formed, at STP?
Marysya12 [62]

Answer:

11.0 L

Explanation:

The equation for this reaction is given as;

2H2  +  O2  -->  2H2O

2 mol of H2 reacts with 1 mol of O2 to form 2 mol of H2O

At STP;

1 mol = 22.4 L

This means;

44.8 L of H2 reacts with 22.4 L of O2 to form 44.8 L of H2O

In this reaction, the limiting reactant is H2 as O2 is in excess.

The relationship between H2 and H2O;

44.8 L = 44.8 L

11.0 L would produce x

Solving for x;

x = 11 * 44.8 / 44.8

x = 11.0 L

4 0
2 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
How can an atom that has seven valence electrons complete its outermost level
GaryK [48]
It can form a covalent bond with a hydrogen bond that has one valence electron to have eight valence electrons and become stable.
3 0
3 years ago
Read 2 more answers
The activation energy of an uncatalyzed reaction is 70 kJ/mol. When a catalyst is added, the activation energy (at 20 °C) is 42
denis23 [38]

Answer:

T = 215.33 °C

Explanation:

The activation energy is given by the Arrhenius equation:

k = Ae^{\frac{-Ea}{RT}}

<u>Where:</u>

k: is the rate constant

A: is the frequency factor    

Ea: is the activation energy

R: is the gas constant = 8.314 J/(K*mol)

T: is the temperature

We have for the uncatalyzed reaction:

Ea₁ = 70 kJ/mol

And for the catalyzed reaction:

Ea₂ = 42 kJ/mol

T₂ = 20 °C = 293 K

The frequency factor A is constant and the initial concentrations are the same.

Since the rate of the uncatalyzed reaction (k₁) is equal to the rate of the catalyzed reaction (k₂), we have:

k_{1} = k_{2}

Ae^{\frac{-Ea_{1}}{RT_{1}}} = Ae^{\frac{-Ea_{2}}{RT_{2}}}   (1)

By solving equation (1) for T₁ we have:

T_{1} = \frac{T_{2}*Ea_{1}}{Ea_{2}} = \frac{293 K*70 kJ/mol}{42 kJ/mol} = 488. 33 K = 215.33 ^\circ C  

Therefore, we need to heat the solution at 215.33 °C so that the rate of the uncatalyzed reaction is equal to the rate of the catalyzed reaction.

I hope it helps you!      

4 0
3 years ago
Other questions:
  • Why does opening the air valve of a tire at a constant temperature decrease the pressure
    8·1 answer
  • A decrease in height of a column in a mercury barometer means that
    5·1 answer
  • A local meteorologist reports the days weather. Which information is qualitative? Check all that apply.
    15·1 answer
  • the air pressure from the atmosphere measures 0.5atm at an altitude of 18,000 ft. How much pressure is this in pounds per square
    7·1 answer
  • Can someone find the independent variable and the dependent variable ? :)))
    8·1 answer
  • Which of the following interactions are generally considered to be unfavorable? Group of answer choices a)Interaction between Ly
    9·1 answer
  • Describe the size, content,<br> and the density of an atom’s nucleus.
    14·1 answer
  • Which type of bond occurs when atoms complete their outermost electron shells by
    14·1 answer
  • Write the balanced chemical equation for the dissociation reaction of thiosulfuric acid.
    15·1 answer
  • 14. Which of the following groups is most at risk for developing this disease?
    9·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!