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seropon [69]
2 years ago
6

Number 16 pls answer

Chemistry
1 answer:
barxatty [35]2 years ago
4 0

Answer:it’s D

Explanation:

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A large flask is evacuated and weighed, filled with argon gas, and then reweighed. When reweighed, the flask is found to have ga
Elden [556K]

Answer:

100.52

Explanation:

from the ideal gas equation PV=nRT

for a given container filled with any ideal gas P and V remains constant.So T is also constant.R is as such a constant.

So n i.e no of moles will also be constant.

no of moles of Ar=3.224/40=0.0806

no of moles of unknown gas=0.0806

molecular wt of unknown gas=8.102/0.0806=100.52

8 0
3 years ago
A chemist titrates 130.0mL of a 0.4248 M lidocaine (C14H21NONH) solution with 0.4429 M HBr solution at 25 degree C . Calculate t
jeka57 [31]

Answer:

pH = 3.36

Explanation:

Lidocaine is a weak base to be titrated with the strong acid HBr, therefore at equivalence point we wil have the protonated lidocaine weak conjugate acid of lidocaine which will drive the pH.

Thus to solve the question we will need to calculate the concentration of this weak acid at equivalence point.

Molarity = mol /V ∴ mol = V x M

mol lidocaine = (130 mL/1000 mL/L) x 0.4248 mol/L = 0.0552 mol

The volume of 0.4429 M HBr required to neutralize this 0.0552 mol is

0.0552 mol x  (1L / 0.4429mol) = 0.125 L

Total volume at equivalence is  initial volume lidocaine + volume HBr added

0 .130 L +0.125 L = 0.255L

and the concentration of protonated lidocaine at the end of the titration will be

0.0552 mol / 0.255 L = 0.22M

Now to calculate the pH we setup our customary ICE table for  weak acids for the equilibria:

protonated lidocaine + H₂O   ⇆  lidocaine + H₃O⁺

                      protonated lidocaine          lidocaine        H₃O⁺

Initial(M)               0.22                                       0                  0

Change                   -x                                      +x                 +x

Equilibrium          0.22 - x                                  x                    x

We know for this equilibrium

Ka = [Lidocaine] [H₃O⁺] / [protonaded Lidocaine] =  x² / ( 0.22 - x )

The Ka can be calculated from the given pKb for lidocaine

Kb = antilog( - 7.94 ) = 1.15 x 10⁻⁸

Ka = Kw / Kb = 10⁻¹⁴ / 1.15 x 10⁻⁸  = 8.71 x 10⁻⁷

Since Ka is very small we can make the approximation 0.22  - x  ≈ 0.22

and solve for x. The pH  will then  be the negative log of this value.

8.71 x 10⁻⁷  = x² / 0.22 ⇒ x = √(/ 8.71 X 10⁻⁷ x 0.22) = 4.38 x 10⁻⁴

( Indeed our approximation checks since 4.38 x 10⁻⁴ is just 0.2 % of 0.22 )

pH = - log ( 4.4x 10⁻⁴) = 3.36

3 0
3 years ago
How much does calcite coast
Nikitich [7]

Answer:

It depends on what kind and how much.  Some are about $100, and others are $1,000.

5 0
3 years ago
Describe cold blooded animal.also some five example.​
Vlad [161]

Answer:

Cold-blood animals are the animals that are not capable of regulating their body temperature,Cold blood animals included reptiles,fishes,amphibians,insects and other invertebrates.

3 0
2 years ago
Read 2 more answers
When a 3.00 g 3.00 g sample of KBr KBr is dissolved in water in a calorimeter that has a total heat capacity of 1.36 kJ ⋅ K − 1
cupoosta [38]

Answer:

Molar heat of solution of KBr is 20.0kJ/mol

Explanation:

Molar heat of solution is defined as the energy released (negative) or absorbed (Positive) per mole of solute being dissolved in solvent.

The dissolution of KBr is:

KBr → K⁺ + Br⁻

In the calorimeter, the temperature decreases 0.370K, that means the solution absorbes energy in this process. The energy is:

q = 1.36kJK⁻¹ × 0.370K

q = 0.5032kJ

Moles of KBr in 3.00g are:

3.00g × (1mol / 119g) = 0.0252moles

Thus, molar heat of solution of KBr is:

0.5032kJ / 0.0252moles = <em>20.0kJ/mol</em>

3 0
3 years ago
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