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kirill115 [55]
3 years ago
10

Since electrons can easily move from one atom to another,

Chemistry
2 answers:
Setler [38]3 years ago
3 0
The correct answer is that last one
horsena [70]3 years ago
3 0
I’m pretty sure it’s B
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7. Suppose each of these isotopes emits a beta particle. Give the iso-
Volgvan

In a beta emission, the mass number of the daughter nucleus remains unchanged while the atomic number of the daughter nucleus increases by one unit. The following are isotopes produced when the following undergo beta emission;

1) potassium-42  ------> Ca - 42

2) iodine-131 ------------> Xe - 131

3)  iron-52 ---------------> Co - 52

4)  sodium-24 -----------> Mg -24

The daughter nucleus formed after beta emission is found one place after its parent in the periodic table.

Regarding the stability of the daughter nuclei, a nucleus is unstable if the neutron-proton ratio is less than 1 or greater than 1.5.

Hence, the following daughter nuclei are stable; Ca - 42, Xe - 131, Mg -24.

Learn more: brainly.com/question/1371390

7 0
2 years ago
Describe a method to calculate the average atomic mass of the sample in the previous question using only the atomic masses of li
alexira [117]

Answer:

Explanation:

To calculate their average atomic masses which is otherwise known as the relative atomic mass, we simply multiply the given abundances of the atoms and the given atomic masses.

The abundace is the proportion or percentage or fraction by which each of the isotopes of an element occurs in nature.

This can be expressed below:

        RAM = Σmₙαₙ

where mₙ is the mass of isotope n

           αₙ is the abundance of isotope n

for this problem:

RAM of Li = m₆α₆ + m₇α₇

       m₆ is mass of isotope Li-6

        α₆ is the abundance of isotope Li-6

       m₇ is mass of isotope Li-7

        α₇ is the abundance of isotope Li-7

3 0
3 years ago
Give the formula of each of the following (don't worry about subscripts or superscripts, example: HCO3- enter as HCO3-): the con
vovangra [49]

Answer:

The relative conjugate acids and bases are listed below:

CH3NH2 → CH3NH3+

H2SO3→ HSO3-

NH3→ NH4+

Explanation:

In a Brønsted-Lowry acid-base reaction, a conjugate acid is the species resulting from a base accepting a proton; likewise, a conjugate base is the species formed after an acid has donated a hydrogen atom (proton).

To this end:

  • HSO3- is the conjugate acid of H2SO3 i.e sulfuric acid has lost a proton (H+)
  • NH4+ is the conjugate acid of NH3 i.e the base ammonia has gained a proton (H+)
  • OH- is the conjugate base of H20
  • CH3NH3+ is the conjugate base of the base CH3NH2 methylamine
3 0
3 years ago
Read 2 more answers
(1) (a) In an equilibrium reaction between gases Q2 and R2 to form QR, the energy content of the reactants is 100KJ and that of
andrezito [222]

Answer:

a is my answer

Explanation:

7 0
2 years ago
Benzoic acid is a natural fungicide that naturally occurs in many fruits and berries. The sodium salt of benzoic acid, sodium be
AlexFokin [52]

Answer:

a. pH = 2.52

b. pH = 8.67

c. pH = 12.83

Explanation:

The equation of the titration between the benzoic acid and NaOH is:

C₆H₅CO₂H + OH⁻ ⇄ C₆H₅CO₂⁻ + H₂O    (1)

a. To find the pH after the addition of 20.0 mL of NaOH we need to find the number of moles of C₆H₅CO₂H and NaOH:

\eta_{NaOH} = C*V = 0.250 M*0.020 L = 5.00 \cdot 10^{-3} moles

\eta_{C_{6}H_{5}CO_{2}H}i = C*V = 0.300 M*0.050 L = 0.015 moles  

From the reaction between the benzoic acid and NaOH we have the following number of moles of benzoic acid remaining:

\eta_{C_{6}H_{5}CO_{2}H} = \eta_{C_{6}H_{5}CO_{2}H}i - \eta_{NaOH} = 0.015 moles - 5.00 \cdot 10^{-3} moles = 0.01 moles

The concentration of benzoic acid is:

C = \frac{\eta}{V} = \frac{0.01 moles}{(0.020 + 0.050) L} = 0.14 M

Now, from the dissociation equilibrium of benzoic acid we have:

C₆H₅CO₂H + H₂O ⇄ C₆H₅CO₂⁻ + H₃O⁺  

0.14 - x                            x                x

Ka = \frac{[C_{6}H_{5}CO_{2}^{-}][H_{3}O^{+}]}{[C_{6}H_{5}CO_{2}H]}

Ka = \frac{x*x}{0.14 - x}

6.5 \cdot 10^{-5}*(0.14 - x) - x^{2} = 0   (2)  

By solving equation (2) for x we have:          

x = 0.0030 = [C₆H₅CO₂⁻] = [H₃O⁺]

Finally, the pH is:

pH = -log([H_{3}O^{+}]) = -log (0.0030) = 2.52

b. At the equivalence point, the benzoic acid has been converted to its conjugate base for the reaction with NaOH so, the equilibrium equation is:

C₆H₅CO₂⁻ + H₂O  ⇄  C₆H₅CO₂H + OH⁻     (3)

The number of moles of C₆H₅CO₂⁻ is:

\eta_{C_{6}H_{5}CO_{2}^{-}} = \eta_{C_{6}H_{5}CO_{2}H}i = 0.015 moles

The volume of NaOH added is:

V = \frac{\eta}{C} = \frac{0.015 moles}{0.250 M} = 0.060 L

The concentration of C₆H₅CO₂⁻ is:

C = \frac{\eta}{V} = \frac{0.015 moles}{(0.060 L + 0.050 L)} = 0.14 M

From the equilibrium of equation (3) we have:

C₆H₅CO₂⁻ + H₂O  ⇄  C₆H₅CO₂H + OH⁻  

0.14 - x                              x               x

Kb = \frac{[C_{6}H_{5}CO_{2}H][OH^{-}]}{[C_{6}H_{5}CO_{2}^{-}]}

(\frac{Kw}{Ka})*(0.14 - x) - x^{2} = 0

(\frac{1.00 \cdot 10^{-14}}{6.5 \cdot 10^{-5}})*(0.14 - x) - x^{2} = 0

By solving the equation above for x, we have:

x = 4.64x10⁻⁶ = [C₆H₅CO₂H] = [OH⁻]

The pH is:

pOH = -log[OH^{-}] = -log(4.64 \cdot 10^{-6}) = 5.33

pH = 14 - pOH = 14 - 5.33 = 8.67

     

c. To find the pH after the addition of 100 mL of NaOH we need to find the number of moles of NaOH:

\eta_{NaOH}i = C*V = 0.250 M*0.100 L = 0.025 moles

From the reaction between the benzoic acid and NaOH we have the following number of moles remaining:                              

\eta_{NaOH} = \eta_{NaOH}i - \eta_{C_{6}H_{5}CO_{2}H} = 0.025 moles - 0.015 moles = 0.010 moles                          

The concentration of NaOH is:

C = \frac{\eta}{V} = \frac{0.010 moles}{0.100 L + 0.050 L} = 0.067 M

Therefore, the pH is given by this excess of NaOH:         

pOH = -log([OH^{-}]) = -log(0.067) = 1.17

pH = 14 - pOH = 12.83

I hope it helps you!    

4 0
3 years ago
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