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zaharov [31]
3 years ago
15

A 14.8 L balloon is at 255 K. What will be its volume if the temperature is 304 K?

Chemistry
1 answer:
posledela3 years ago
3 0

Answer: 0.01764392156862745 L

(v1)/(t1)=(v2)/(t2)

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Bess [88]
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C2H6 + Cl2  ---> C2H5Cl + HCl
C2H5Cl+KOH-> C2H5OH+ KCl.
C2H5OH  ----KMnO4 ---> CH3COOH.
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3 years ago
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NaCl is ___ formula of common salt
MrMuchimi

Answer:

None of them, D

Explanation:

The actual answer is chemical formula NaCl, representing a 1:1 ratio of sodium and chloride ions.

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How many moles of ag are formed by the complete reaction of 28.3 mol of pb?
levacccp [35]
<span>Answer: 56.6 moles Explanation: 28.3 moles of Pb would produce twice as much moles as Ag. 28.3 X (2moles Ag/ 1 mol Pb) = 56.6 moles of Ag.</span>
4 0
3 years ago
Which of the following explanations accounts for the fact that the ion-solvent interaction is greater for Li than for K
OLEGan [10]

Li+ has a smaller ionic radius than K+

and smaller molecules have more collisions/interactions between each other

<h3>What is ion-solvent interaction ?</h3>

In the case of ion-solvent interactions, the state in which the interac-tions exist is an obvious one; it is the situation in which ions are inside the solvent.

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  • In the process of solvation, ions are surrounded by a concentric shell of solvent. Solvation is the process of reorganizing solvent and solute molecules into solvation complexes.

Learn more about Ion-solvent interaction here:

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6 0
1 year ago
A mixture of CS2(g) and excess O2(g) is placed in a 10 L reaction vessel at 100.0 ∘C and a pressure of 3.10 atm . A spark causes
ziro4ka [17]

Answer:

PCO2  = 0.6 25 atm

PSO2  = 1.2 75 atm

PO2 = 0.6  atm

Explanation:

Step 1: Data given

Volume = 10.0 L

Temperature = 100.0 °C

Pressure = 3.10 °C

After reaction, the temperature returns to 100.0 ∘C, and the mixture of product gases (CO2, SO2, and unreacted O2) is found to have a pressure of 2.50 atm

Step 2: The balanced equation

CS2(g)+3O2(g)→CO2(g)+2SO2(g)

Step 3: Name the reactants and products

a = CS2

b = O2 before reaction

c = CO2

d = SO2

e = nS O2 after reaction with n = the number of moles

Step 4: Calculate moles before reaction

PV = nRT

n = PV/(RT)

(na + nb) = (3.10atm) * (10.0L) / ((0.08206 Latm/moleK) * (373.15K))

(na + nb) = 1.0124

Step 5: Calculate moles after reaction

PV = nRT

n = PV/(RT)

nc + nd + ne) = PV/(RT) = (2.50 atm)*(10.0L) / ((0.08206 Latm/moleK)*(373.15K))

(nc + nd + ne) = 0.816 moles

Step 6: Calculate mol fraction

For  1 mole CS2 we need 3 moles O2  to produce 1 mole of CO2 and 2 moles of SO2

moles O2 remaining = ne = nb - 3na

moles CO2 produced = nc = na

moles SO2 producted = nd = 2na

(nc + nd + ne) = 0.816 moles = nb - 3na + na + 2na = 0.816

nb = 0.816

. (na + nb) = 1.0124

na = 1.0124 moles - 0.816 moles = 0.208

which leads to  

nc = na = 0.208

nd = 2na = 2*0.208 = 0.416

ne = 0.816 - 3*0.208 = 0.192

mole fraction CO2 = 0.208 / (0.208 + 0.416 + 0.192) = 0.25

mole fraction SO2 = 0.416 / (0.208 + 0.416 + 0.192) = 0.5 1

mole fraction O2 = 0.192 /(0.208 + 0.416 + 0.192) = 0.24

Step 6: Calculate partial pressure

PCO2 = 0.25 * 2.50 atm = 0.6 25 atm

PSO2 = 0.51 * 2.50 atm = 1.2 75 atm

PO2 = 0.24 * 2.50 atm = 0.6  atm

Step 7: Control results

now let's verify a couple of things

PV = nRT

P = nRT/V

before rxn

P = (0.208 + 0.816) * (0.08206 L*atm/mole*K) * (373.15K) / (10.0L) ≈ 3.10 atm

after rxn

P = ((0.208 +0.416+0.192) * (0.08206 L*atm/mole*K) * (373.15K) / (10.0L) ≈ 2.50 atm

8 0
3 years ago
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