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Paraphin [41]
3 years ago
13

What tests can you do to identify gases, based on their chemical properties?

Chemistry
1 answer:
GuDViN [60]3 years ago
5 0
Tests for gases
Hydrogen, oxygen, carbon dioxide, ammonia and chlorine can be identified using different tests.
Hydrogen. A lighted wooden splint makes a popping sound in a test tube of hydrogen.
Oxygen. A glowing wooden splint relights in a test tube of oxygen.
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A gas sample at stp contains 1.15 g oxygen gas and 1.55 g nitrogen gas.what is the volume of the gas sample?
ss7ja [257]
O2=32 g/ mol
1.15/32=0.035
N2=28 g/mol
1.55/28=0.055
in STP every 22.4 litters is 1 mol

4 0
3 years ago
Read 2 more answers
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
When one of the atoms involved in a covalent bond has a much higher electronegativity than the other atom, what type of bond res
ICE Princess25 [194]
If electronegativity difference between atoms is much higher then they form ionic bond.

Hope this helps!
5 0
3 years ago
All of the following conditions of STP are true except A. 101.3 kPa B.3.81kPa.L/mol.K C. 24.2 L D. 273.15 K
Tasya [4]
The answer is c 24.2l
6 0
2 years ago
Read 2 more answers
Consider the balanced chemical reaction below and determine the percent yield of sodium bromide if 2.36 moles of iron(iii) bromi
hammer [34]
        First   the  theoretical yield   of Nabr
  by use  of  mole  ratio   between  FeBr3  and  NaBr  which  is  2:6   the   theoretical yield  

  =2.36  x6/2= 7.08  moles

the  %  yield  =  actual  yield/  theoretical  yield  x  100

that  is    6.14/7.08  x100=  86.72%
8 0
3 years ago
Read 2 more answers
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