Name the following ionic compound KBr

Question 2 (2 points)

inysia [295]

Answer:

v = 46.5 m/s

Explanation:

Given data:

Mass of car = 1210 kg

Momentum of car = 56250 kg m/s

Velocity of car = ?

Solution:

Formula:

p = mv

p = momentum

m = mass

v = velocity

Now we will put values in formula:

56250 kg m/s = 1210 kg × v

v = 56250 kg m/s / 1210 kg

v = 46.5 m/s

So a car having mass of 1210 kg with momentum 56250 kg m/s having 46.5 m/s velocity.

5 0

3 years ago

The standard reduction potentials of the following half-reactions are given in Appendix E in the textbook:

german

Answer:

For 1: The largest positive cell potential is of cell having 1st and 4th half reactions.

For 2: The standard electrode potential of the cell is 1.539 V

For 3: The smallest positive cell potential is of cell having 3rd and 4th half reactions. The standard electrode potential of the cell is 0.46 V

Explanation:

The substance having highest positive $E^o$ potential will always get reduced and will undergo reduction reaction.

We are given:

$Ag^++(aq.)+e^-\rightarrow Ag(s);E^o_{Ag^+/Ag}=0.799V\\\\Cu^{2+}+(aq.)+2e^-\rightarrow Cu(s);E^o_{Cu^{2+}/Cu}=0.337V\\\\Ni^{2+}(aq.)+2e^-\rightarrow Ni(s);E^o_{Ni^{2+}/Ni}=-0.28V\\\\Cr^{3+}(aq.)+3e^-\rightarrow Cr(s);E^o_{Cr^{3+}/Cr}=-0.74V$

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

Cell having 1st and 2nd half reactions:

Silver has higher electrode potential. So, this will undergo reduction reaction and act as anode. Copper will undergo oxidation reaction and act as cathode.

$E^o_{cell}=0.799-0.337=0.462V$

Cell having 1st and 3rd half reactions:

Silver has higher electrode potential. So, this will undergo reduction reaction and act as anode. Nickel will undergo oxidation reaction and act as cathode.

$E^o_{cell}=0.799-(-0.28)=1.079V$

Cell having 1st and 4th half reactions:

Silver has higher electrode potential. So, this will undergo reduction reaction and act as anode. Chromium will undergo oxidation reaction and act as cathode.

$E^o_{cell}=0.799-(-0.74)=1.539V$

Cell having 2nd and 3rd half reactions:

Copper has higher electrode potential. So, this will undergo reduction reaction and act as anode. Nickel will undergo oxidation reaction and act as cathode.

$E^o_{cell}=0.337-(-0.28)=0.617V$

Cell having 3rd and 4th half reactions:

Nickel has higher electrode potential. So, this will undergo reduction reaction and act as anode. Chromium will undergo oxidation reaction and act as cathode.

$E^o_{cell}=-0.28-(-0.74)=0.46V$

Hence,

For 1: The largest positive cell potential is of cell having 1st and 4th half reactions.

For 2: The standard electrode potential of the cell is 1.539 V

For 3: The smallest positive cell potential is of cell having 3rd and 4th half reactions. The standard electrode potential of the cell is 0.46 V

8 0

3 years ago

The reaction of C4H8 and Br2 to yield C4H2Br2 represents

natima [27]

Answer:

A Synthesis Reaction

Explanation:

A + B --> AB

8 0

3 years ago

Based on the balanced equation 2C2H2 + 5O2 → 4CO2 + 2H2O calculate the number of excess reagent units remaining when 52 C2H2 mol

erica [24]

Answer:

20 molecules of oxygen gas remains after the reaction.

Explanation:

$2C_2H_2 + 5O_2\rightarrow 4CO_2 + 2H_2O$

Molecules of ethyne = 52

Molecules of oxygen gas = 150

According to reaction, 2 molecules of ethyne reacts with 5 molecules of oxygen gas.

Then 52 molecules of ethyne will react with:

$\frac{5}{2}\times 52=130 molecules$ of oxygen gas.

As we can see that we have 150 molecules of oxygen gas, but 52 molecules of ethyne will react with 130 molecules of oxygen gas. So, this means that ethyne is a limiting reagent and oxygen gas is an excessive reagent.

Remaining molecules of recessive reagent = 150 - 130 = 20

20 molecules of oxygen gas remains after the reaction.

5 0

3 years ago

Plot your values of ln(Ksp) vs. 1/T and find the slope and y-intercept of the best fit line. Use the equation for the best fit l

labwork [276]

Answer:

a) The slope of the line of best fit plot = -12629.507

b) ΔH∘ = 105 kJ

c) Intercept of the line of best fit plot = 39.099

d) ΔS∘ = 325.1 J/K

e) Option A is correct.

Solubility will increase as temperature increases, because as T increases the (−ΔH∘/RT) term becomes smaller therefore K will get larger.

f) Option D is correct. All of the options are correct.

Explanation:

The complete question is presented in the first attached image to this question. This complete question has the data readings required to plot the graph.

The second attached image has the plotted graph and the regression analysis to obtain the line of best fit.

The equation of the line of best fit obtained is

y = -12629.507x + 39.099

Comparing the given expression for the question with the equation of a straight line

ln (K) = (−ΔH∘/RT) + (ΔS∘/R)

y = mx + c

y = In K

Slope = m = (−ΔH∘/R)

x = (1/T)

Intercept = c = (ΔS∘/R)

So, to answer the question now

a) The slope of the line of best fit plot = -12629.507

b) Slope = (−ΔH∘/R)

(−ΔH∘/R) = -12629.507

But R = molar gas constant = 8.314 J/mol.K

ΔH∘ = 12629.507 × 8.314 = 105,001.721198 J = 105,002 J = 105 kJ

c) Intercept of the line of best fit plot = 39.099

d) Intercept = (ΔS∘/R)

(ΔS∘/R) = 39.099

ΔS∘ = 39.099 × 8.314 = 325.069086 J/K = 325.1 J/K

e) Do you expect the solubility of Borax to increase or decrease as temperature increases?

Solubility will increase as temperature increases, because as T increases the (−ΔH∘/RT) term becomes smaller therefore K will get larger.

f) Why was it necessary to make sure that some solid was present in the main solution before taking the samples to measure Ksp? Select the option that best explains why.

A. To make sure no more sodium borate would dissolve in solution.

B. To ensure the dissolution process was at equilibrium.

C. To make sure the solution was saturated with sodium and borate ions.

D. All of the above

Hope this Helps!!!

5 0

3 years ago