answer.
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
oee [108]
2 years ago
6

Consider the following information. The lattice energy of CsCl is Δ H lattice = − 657 kJ/mol. The enthalpy of sublimation of Cs

is Δ H sub = 76.5 kJ/mol. The first ionization energy of Cs is IE 1 = 376 kJ/mol. The electron affinity of Cl is Δ H EA = − 349 kJ/mol. The bond energy of Cl 2 is BE = 243 kJ/mol. Determine the enthalpy of formation, Δ H f , for CsCl ( s ) .
Chemistry
2 answers:
AlladinOne [14]2 years ago
8 0

Answer:

<ERROR>-----------------------------------------------------------------------------<ERROR>

Explanation:

andrew-mc [135]2 years ago
3 0

Answer:

-310.5 kJ/mol

Explanation:

The CsCl(s) will be formed by steps:

  • The metal (Cs) will be made gaseous by a sublimation → ΔHsub = 76.5 kJ/mol;
  • The metal gains energy to ionize → IE = 376 kJ/mol;
  • The nonmetal (Cl) loses energy to ionizes → EA = -349 kJ/mol;
  • The nonmetal is bonded → BE = 243 kJ/mol;
  • The lattice is formed → ΔHlattice = -657 kJ/mol.

The enthanlpy of formation is the sum of all the energies involved in the steps:

ΔHf = 76.5 + 376 -349 + 243 -657

ΔHf = -310.5 kJ/mol

You might be interested in
(A) A chemical reaction takes place in a container of cross-sectional area 100 cm2. As a result of the reaction, a piston is pus
balandron [24]

Answer:

(A) The work done by the system is -101.325J

(B) The workdone by the system is -90.75J

Explanation:

(A) Workdone = -PΔV

Given that A = 100cm2 = 0.01m2

distance d = 10cm = 0.1m

ΔV= Area × distance

ΔV= 0.01 ×0.1

ΔV = 0.001m3

P= external pressure = 1atm = 101325Pa

Workdone = -0.001 × 101325

W= - 101.325Pa m3

1Pam3 = 1J

Therefore W = - 101.325J

The work done on the system is -101.325J

(B) Workdone = -PΔV

Given that A = 50cm2 = 0.005m2

distance d = 15cm = 0.15m

ΔV= Area × distance

ΔV= 0.005×0.15

ΔV = 0.00075m3

P=121kPa = 121000Pa

W= - 121000 × 0.00075

W= -90.75Pa m3

1Pam3 = 1J

W = - 90.75J

The woekdone by the system is -90.75J

5 0
2 years ago
If the actual yield of a reaction is 37.6 g while the theoretical yield is 112.8 g what is the percent yield
Zigmanuir [339]
<h2>Hello!</h2>

The answer is:

The percent yield of the reaction is 32.45%

<h2>Why?</h2>

To calculate the percent yield, we have to consider the theoretical yield and the actual yield. The theoretical yield as its name says is the yield expected, however, many times the difference between the theoretical yield and the actual yield is notorious.

We are given that:

ActualYield=37.6g\\TheoreticalYield=112.8g

Now, to calculate the percent yield, we need to divide the actual yield by the theoretical and multiply it by 100.

So, calculating we have:

PercentYield=\frac{ActualYield}{TheoreticalYield}*100\\\\PercentYield=\frac{37.6g}{112.8g}*100=0.3245*100=32.45(percent)

Hence, we have that the percent yield of the reaction is 32.45%.

Have a nice day!

8 0
2 years ago
Read 2 more answers
Rank the compounds by the ease with which they ionize under sn1 conditions. rank the compounds from easiest to hardest. to rank
marysya [2.9K]
The rate of Formation of Carbocation mainly depends on two factors'

                    1)  Stability of Carbocation:
                                                              The ease of formation of Carbocation mainly depends upon the ionization of substrate. If the forming carbocation id tertiary then it is more stable and hence readily formed as compared to secondary and primary.

                     2) Ease of detaching of Leaving Group:
                                                                                   The more readily and easily the leaving group leaves the more readily the carbocation is formed and vice versa. In given scenario the carbocation formed is tertiary in all three cases, the difference comes in the leaving group. So, among these three substrates the one containing Iodo group will easily dissociate to form tertiary carbocation because due to its large size Iodine easily leaves the substrate, secondly Chlorine is a good leaving group compared to Fluoride. Hence the order of rate of formation of carbocation is,

                                            R-I > R-Cl > R-F

                                               B   >  C  >  A

3 0
1 year ago
A 5.00 L sample of helium expands to 12.0 L at which point the
mina [271]

Answer:

1.73 atm

Explanation:

Given data:

Initial volume of helium = 5.00 L

Final volume of helium = 12.0 L

Final pressure = 0.720 atm

Initial pressure = ?

Solution:

"The volume of given amount of gas is inversely proportional to its pressure by keeping the temperature and number of moles constant"

Mathematical expression:

P₁V₁ = P₂V₂

P₁ = Initial pressure

V₁ = initial volume

P₂ = final pressure

V₂ = final volume  

Now we will put the values in formula,

P₁V₁ = P₂V₂

P₁ × 5.00 L = 0.720 atm × 12.0 L

P₁ = 8.64 atm. L/5 L

P₁ = 1.73 atm

7 0
1 year ago
What is the hydronium ion concentration of a solution whose pH is 7.30
Assoli18 [71]
[ H₃O⁺] = 10 ^ - pH

[ H₃O⁺ ] = 10 ^ - 7.30

[ H₃O⁺ ] = 5.011 x 10⁻⁸ M

hope this helps!
6 0
1 year ago
Other questions:
  • Which statement describes characteristics of a 0.01 M KOH(aq) solution? 1. The solution is acidic with a pH less than 7. 2. The
    11·1 answer
  • Which element in period 4 would have chemical properties similar to magnesium? 7. which metalloids would have chemical propertie
    7·2 answers
  • Solid aluminum metal and diatomic bromine liquid react spontaneously to form a solid product. give the balanced chemical equatio
    5·2 answers
  • A 7.50 liter sealed jar at 18 °c contains 0.125 moles of oxygen and 0.125 moles of nitrogen gas. what is the pressure in the con
    9·2 answers
  • A student has 7.05 g of zinc powder, 1.60 L of a 3.40 M calcium nitrate solution, and 1.50 L of a 1.60 M lead(II) nitrate soluti
    12·1 answer
  • Sodium carbonate (Na2CO3) is available in very pure form and can be used to standardize acid solutions. What is the molarity of
    10·1 answer
  • A 1.00 * 10^-6 -g sample of nobelium, 254/102 No, has a half-life of 55 seconds after it is formed. What is the percentage of 25
    10·1 answer
  • Write down the dissolution equation for rubidium chromate dissolving in water. (Chromate is a polyatomic ion with the formula Cr
    13·1 answer
  • How many total bond are in the Lewis structure for HSIN?
    14·1 answer
  • 7. A gas-filled weather balloon with a volume of 65.0 L is released at sea-level conditions of
    14·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!