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Mariulka [41]
1 year ago
11

a parent isotope has atomic number 90 and mass number 232. which are the atomic and mass numbers of the daughter isotope if it e

mits a beta particle?
Chemistry
1 answer:
amm18121 year ago
7 0

The atomic and mass numbers of the daughter isotope, which emits a beta particle 91, 232

Isotopes belong to the same elemental family but have different ratios of neutrons to protons, making them different substances. On the Periodic Table, each element's atomic number is based on how many protons are in its nucleus.

Isotopes are two or more different atom types that share the same atomic number and place in the periodic table but have different quantities of neutrons in their nuclei, resulting in various nucleon numbers. Despite having various atomic weights and physical characteristics, they exhibit roughly the same chemical activity. Isotopes are variations in the mass number, "A," of an element's atoms that share the same number of protons (atomic number, "Z") but a different number of neutrons.

To learn more about isotopes visit here:

brainly.com/question/11680817

#SPJ4

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What is the pH of a solution of RbOH with a concentration of 0.86 M? Answer to 2 decimal places
lubasha [3.4K]

Answer:The pH of the solution is given by pH=−log([H3O+])

Explanation:so you can't use

pH

=

−

log

(

0.150

)

because that's the concentration of the hydroxide anions,

OH

−

, not of the hydronium cations,

H

3

O

+

. In essence, you calculated the

pOH

of the solution, not its

pH

.

Sodium hydroxide is a strong base, which means that it dissociates completely in aqueous solution to produce hydroxide anions in a

1

:

1

mole ratio.

NaOH

(

a

q

)

→

Na

+

(

a

q

)

+

OH

−

(

a

q

)

So your solution has

[

OH

−

]

=

[

NaOH

]

=

0.150 M

Now, the

pOH

of the solution can be calculated by using

pOH

=

−

log

(

[

OH

−

]

)

−−−−−−−−−−−−−−−−−−−−

In your case, you have

pOH

=

−

log

(

0.150

)

=

0.824

Now, an aqueous solution at

25

∘

C

has

pH + pOH

=

14

−−−−−−−−−−−−−−so you can't use

pH

=

−

log

(

0.150

)

because that's the concentration of the hydroxide anions,

OH

−

, not of the hydronium cations,

H

3

O

+

. In essence, you calculated the

pOH

of the solution, not its

pH

.

Sodium hydroxide is a strong base, which means that it dissociates completely in aqueous solution to produce hydroxide anions in a

1

:

1

mole ratio.

NaOH

(

a

q

)

→

Na

+

(

a

q

)

+

OH

−

(

a

q

)

So your solution has

[

OH

−

]

=

[

NaOH

]

=

0.150 M

Now, the

pOH

of the solution can be calculated by using

pOH

=

−

log

(

[

OH

−

]

)

−−−−−−−−−−−−−−−−−−−−

In your case, you have

pOH

=

−

log

(

0.150

)

=

0.824

Now, an aqueous solution at

25

∘

C

has

pH + pOH

=

14

−−−−−−−−−−−−−−

3 0
2 years ago
Help me with these two please​
prohojiy [21]

Answer:

13.A

14.B?

on 14 i don't know

Explanation:

3 0
3 years ago
Read 2 more answers
A gas mixture with 4 mol of Ar, x moles of Ne, and y moles
maks197457 [2]

Answer:

a) \Delta G_{mixing}=\frac{R*T}{12}*[4*ln (1/3) +x*ln (x/12) +(8-x)*ln ((8-x)/12)]

b) x=4

c) \Delta G_{max}=-2721.9 J/mol

Explanation:

Gas mixture:

n_{Ar}= 4 mol

n_{Ne}= x mol

n_{Xe}= y mol

n_{tot}= n_{Ar} + n_{Ne} + n_{Xe}=3*n_{Ar}

n_{Ne} + n_{Xe}=2*n_{Ar}

x + y=8 mol

y=8 mol- x

Mol fractions:

x_{Ar}=\frac{4 mol}{12 mol}=1/3

x_{Ne}=\frac{x mol}{12 mol}=x/12

x_{Xe}=\frac{8 - x mol}{12 mol}=(8-x)/12

Expression of \Delta G_{mixing}

\Delta G_{mixing}=R*T*\sum_{i]*x_i*ln (x_i)

\Delta G_{mixing}=R*T*[1/3*ln (1/3) +x/12*ln (x/12) +(8-x)/12*ln ((8-x)/12)]

\Delta G_{mixing}=\frac{R*T}{12}*[4*ln (1/3) +x*ln (x/12) +(8-x)*ln ((8-x)/12)]

Expression of \Delta G_{max}

\frac{d \Delta G_{mixing}}{dx}=0

\frac{d \Delta G_{mixing}}{dx}=\frac{R*T}{12}*[ln (x/12)+12-ln ((8-x)/12)-12]

0=\frac{R*T}{12}*[ln (x/12)-ln ((8-x)/12)

0=[ln (x/12)-ln ((8-x)/12)

ln (x/12)=ln ((8-x)/12)

x=(8-x)

x=4

\Delta G_{max}=\frac{8.314*298}{12}*[4*ln (1/3) +4*ln (4/12) +(8-4)*ln ((8-4)/12)]

\Delta G_{max}=\frac{8.314*298}{12}*[4*ln (1/3) +4*ln (1/3) +(4)*ln (1/3)]

\Delta G_{max}=\frac{8.314*298}{12}*[12*ln (1/3)]

\Delta G_{max}=-2721.9 J/mol

4 0
3 years ago
What are some factors that affect an element's reactivity?
Alexxandr [17]
From what there’s nothing there? :)
7 0
3 years ago
A dehydration reaction starting with 3.8 g cyclohexanol produces 2.6 g cyclohexene. Calculate the theoretical yield for this rea
diamong [38]

Answer:

Theoretical yield of C6H10 = 3.2 g.

Explanation:

Defining Theoretical yield as the quantity of product obtained from the complete conversion of the limiting reactant in a chemical reaction. It can be expressed as grams or moles.

Equation of the reaction

C6H11OH --> C6H10 + H2O

Moles of C6H11OH:

Molar mass of C6H110H = (12*6) + (1*12) + 16

= 100 g/mol

Mass of C6H10 = 3.8 g

number of moles = mass/molar mass

=3.8/100

= 0.038 mol.

Using stoichoimetry, 1 moles of C6H110H was dehydrated to form 1 mole of C6H10 and 1 mole of water.

Therefore, 0.038 moles of C6H10 was produced.

Mass of C6H10 = molar mass * number of moles

Molar mass of C6H10 = (12*6) + (1*10)

= 82 g/mol.

Mass = 82 * 0.038

= 3.116 g of C6H10.

Theoretical yield of C6H10 = 3.2 g

4 0
3 years ago
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