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miss Akunina [59]
3 years ago
9

Calculate the ratio of ch3nh2 to ch3nh3cl required to create a buffer with ph = 10.22.

Chemistry
1 answer:
avanturin [10]3 years ago
6 0
When the  Ka value for CH3NH2 = 2.27 x 10 ^-11

and when we have the Ka value so, we can get Pka from this formula:

Pka = -㏒Ka

∴Pka = -㏒ (2.27 x10^-11)
         = 10.6 

now, to get the base/acid ratio by using Henderson-Hasselblach equation:

PH = Pka + ㏒[base/acid]

so by substitution: 

10.22 = 10.6 + ㏒[base/acid]

∴[base/acid] = 0.417
 
∴[CH3NH2/CH3NH3Cl] = 0.417 
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Which of these plecas of evidence did not help in the development of the theory of plate tectonics
Afina-wow [57]

No mountains in the world are increasing in size

3 0
3 years ago
The image compares the arrangement of electrons in two different neutral atoms.
jekas [21]

Answer:

  • <u>Option b. Atom P has an estimated Zeff of 7 and is therefore to the right of Atom Q, which has a Zeff of 6.</u>

Explanation:

Please, find attached the figures of both atom Q and atom P corresponding to this question.

The <u>features of atom Q are</u>:

  • Each <em>black sphere</em> represents an electron
  • In total this atom has 8 electrons: 2 in the inner shell and 6 in the outermost shell.
  • Since it is assumed that the atom is neutral, it has 8 protons: one positive charge of a proton balances one negative charge of an electron. Thus, the atomic number of this atom is 8.
  • Since only two shells are ocuppied, you can assert that the atom belongs to the period 2 (which is confirmed looking into a periodic table with the atomic number 8).
  • <em>Zeff </em>is the effective nuclear charge of the atom. It accounts for the  net positive charge the valence electrons experience. And may, in a very roughly way, be estimated as the number of protons less the number of electrons in the inner shells. Thus, for this atom, an estimated  Z eff = 8 - 2 = 6.

The <u>features of atom P</u> are:

  • Again, each black sphere represents an electron
  • In total this atom has 9 electrons: 2 in the inner shell and 7 in the outermost shell.
  • Since it is assumed that the atom is neutral, it has 9 protons.
  • The atomic number of this atom is 9.
  • Using the same reasoning used for atom Q, this atom is also in the period 2.
  • Estimated Z eff = 9 - 2 = 7.

Then, since atom P has a greater Z eff than atom Q (an estimated Zeff of 7 for atom P against an estimated Z eff of 6 for atom Q),  and both atoms are in the same period, you can affirm that <em>atom P</em> has a greater atomic number and<em> is therefore to the right of atom Q</em>.

8 0
3 years ago
In a sulphuric acid (h2so4) - sodium hydroxide (naoh) acid-base titration, 17.3 ml of 0.126 m naoh is needed to neutralize 25 ml
katen-ka-za [31]
The balanced equation for the neutralisation reaction is as follows
2NaOH + H₂SO₄ ---> Na₂SO₄ + 2H₂O
stoichiometry of NaOH to H₂SO₄ is 2:1
the number of moles of NaOH reacted  - 0.126 mol/L  x 0.0173 L = 0.00218 mol
if 2 mol of NaOH reacts with 1 mol of H₂SO₄ 
then 0.00218 mol of NaOH reacts with - 0.00218 / 2 = 0.00109 mol of H₂SO₄ 
molarity is the number of moles of solute in 1 L solution
therefore if 25 mL contains - 0.00109 mol 
then 1000 mL contains - 0.00109 mol / 25 mL  x 1000 mL = 0.0436 mol/L
therefore molarity of H₂SO₄ is 0.0436 M
4 0
3 years ago
What mass, in grams, of CO2 and H2O<br> is formed from 2.55 mol of propane?
oksian1 [2.3K]

Answer:

336.6 grams of CO₂ and 183.6 grams of H₂O are formed from 2.55 moles of propane.

Explanation:

In this case, the balanced reaction is:

C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of reactant and product participate in the reaction:

  • C₃H₈: 1 mole
  • O₂: 5 moles
  • CO₂: 3 moles
  • H₂O: 4 moles

Being the molar mass of each compound:

  • C₃H₈: 44 g/mole
  • O₂: 16 g/mole
  • CO₂: 44 g/mole
  • H₂O: 18 g/mole

Then, by stoichiometry, the following quantities of mass participate in the reaction:

  • C₃H₈: 1 mole* 44 g/mole= 44 grams
  • O₂: 5 moles* 16 g/mole= 80 grams
  • CO₂: 3 moles* 44 g/mole= 132 grams
  • H₂O: 4 moles* 18 g/mole= 72 grams

So you can apply the following rules of three:

  • If by stoichiometry 1 mole of C₃H₈ forms 132 grams of CO₂, 2.55 moles of C₃H₈ how much mass of CO₂ will it form?

mass of CO_{2} =\frac{2.55 moles of C_{3} H_{8}*132 gramsof CO_{2} }{ 1 mole of C_{3} H_{8}}

mass of CO₂= 336.6 grams

  • If by stoichiometry 1 mole of C₃H₈ forms 72 grams of H₂O, 2.55 moles of C₃H₈ how much mass of H₂O will it form?

mass of H_{2}O =\frac{2.55 moles of C_{3} H_{8}*72 gramsof H_{2}O }{ 1 mole of C_{3} H_{8}}

mass of H₂O= 183.6 grams

<u><em>336.6 grams of CO₂ and 183.6 grams of H₂O are formed from 2.55 moles of propane.</em></u>

3 0
3 years ago
Dinitrogen tetraoxide, N2O4, decomposes to nitrogen dioxide, NO2, in a first-order process. If k = 1.5 x 103 s-1 at 5 ºC and k =
Arada [10]

Answer:

The activation energy for the decomposition = 33813.28 J/mol

Explanation:

Using the expression,

\ln \dfrac{k_{1}}{k_{2}} =-\dfrac{E_{a}}{R} \left (\dfrac{1}{T_1}-\dfrac{1}{T_2} \right )

Wherem  

k_1\ is\ the\ rate\ constant\ at\ T_1

k_2\ is\ the\ rate\ constant\ at\ T_2

E_a is the activation energy

R is Gas constant having value = 8.314 J / K mol  

Thus, given that, E_a = ?

k_2=4.0\times 10^3s^{-1}

k_1=1.5\times 10^3s^{-1}  

T_1=5\ ^0C  

T_2=25\ ^0C  

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T = (5 + 273.15) K = 278.15 K  

T = (25 + 273.15) K = 298.15 K  

T_1=278.15\ K

T_2=298.15\ K

So,

\ln \frac{1.5\times 10^3}{4.0\times 10^3}\:=-\frac{E_{a}}{8.314}\times \left(\frac{1}{278.15}-\frac{1}{298.15}\right)

E_a=-\ln \frac{1.5\times \:10^3}{4.0\times \:10^3}\:\times \frac{8.314}{\left(\frac{1}{278.15}-\frac{1}{298.15}\right)}

E_a=-\frac{8.314\ln \left(\frac{1.5\times \:10^3}{4\times \:10^3}\right)}{\frac{1}{278.15}-\frac{1}{298.15}}

E_a=-\frac{689483.53266 \ln \left(\frac{1.5}{4}\right)}{20}

E_a=33813.28\ J/mol

<u>The activation energy for the decomposition = 33813.28 J/mol</u>

8 0
3 years ago
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