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3 years ago

An isotope has 15 protons, 16 neutrons, and 15 electrons. What is the isotopic notation of the isotope

Chemistry

1 answer:

VikaD [51]3 years ago

3 0

Answer:

$^{31}_{15}P$ is the isotopic notation of the atom

Explanation:

The isotope notation is:

$^a_bX$

Where a is the mass number = Number of protons + Number of neutrons

b is atomic number = Number of protons

The atomic number define the nature of the atom, the element with atomic number = 15 is phosphorus, P:

$^a_bP$

a = 15 protons + 16 neutrons = 31

b = 15

$^{31}_{15}P$ is the isotopic notation of the atom.

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A reaction yields 6.26 grams of a CuCl2. What is the percent yield of CuCl2 if the theoretical yield is 18.81g?

Lilit [14]

Answer:

33% yield

Explanation:

6.26/18.81 =0.33280170122 = 33%

6 0

3 years ago

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Sholpan [36]

Answer:

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5 0

3 years ago

A metal ion (X) with a charge of 2+ is attracted to a nonmetal ion (Z) with a charge of 3-.Which formula represents the resultin

Viefleur [7K]

Answer:

X₃Z₂

Explanation:

A metal ion (X) with a charge of 2+ = X²⁺

nonmetal ion (Z) with a charge of 3- = Z³⁻

In a neutral compound, the charges but balance each other to remain neutral. However in this case, the charges are not balanced.

A simple way to make the charges balance is by tweaking the number of the ions present in the compound.

We do this by multiplying the metal ion with 3 and the non metal ion by 2.

That way we have;

Metal ion = 3X2⁺ (Total charge - 6⁺)

Non metal ion = 2Z3⁻ (Total charge - 6⁻)

The charges are now balanced. The formular of the compound is;

X₃Z₂

5 0

3 years ago

Read 2 more answers

The standard reduction potentials for the Ag+|Ag(s) and Zn2+| Zn(s) half-cell reactions are +0.799 V and -0.762 V, respectively.

mihalych1998 [28]

Answer: The potential of the given cell is 1.551 V

Explanation:

The given chemical cell follows:

$Zn(s)|Zn^{2+}(0.125M)||Ag^{+}(0.240M)|Ag(s)$

Oxidation half reaction: $Zn(s)\rightarrow Zn^{2+}(0.125M)+2e^-;E^o_{Zn^{2+}/Zn}=-0.762V$

Reduction half reaction: $Ag^{+}(0.240M)+e^-\rightarrow Ag(s);E^o_{Ag^{+}/Ag}=0.799V$ ( × 2)

Net cell reaction: $Zn(s)+2Ag^{+}(0.240M)\rightarrow Zn^{2+}(0.125M)+2Ag(s)$

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

Putting values in above equation, we get:

$E^o_{cell}=0.799-(-0.762)=1.561V$

To calculate the EMF of the cell, we use the Nernst equation, which is:

$E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[Zn^{2+}]}{[Ag^{+}]^2}$

where,

$E_{cell}$ = electrode potential of the cell = ? V

$E^o_{cell}$ = standard electrode potential of the cell = +1.561 V

n = number of electrons exchanged = 2

$[Zn^{2+}]=0.125M$

$[Ag^{+}]=0.240M$

Putting values in above equation, we get:

$E_{cell}=1.561-\frac{0.059}{2}\times \log(\frac{(0.125)}{(0.240)^2})$

$E_{cell}=1.551V$

Hence, the potential of the given cell is 1.551 V

6 0

4 years ago

A 0.130 mole quantity of NiCl 2 is added to a liter of 1.20 M NH 3 solution. What is the concentration of Ni 2 + ions at equilib

rosijanka [135]

This is an incomplete question, here is a complete question.

A 0.130 mole quantity of NiCl₂ is added to a liter of 1.20 M NH₃ solution. What is the concentration of Ni²⁺ ions at equilibrium? Assume the formation constant of Ni(NH₃)₆²⁺ is 5.5 × 10⁸

Answer : The concentration of $Ni^{2+}$ ions at equilibrium is, $4.31\times 10^{-8}$

Explanation : Given,

Moles of $NiCl_2$ = 0.130 mol

Volume of solution = 1 L

$Concentration=\frac{Moles }{Volume}$

Concentration of $NiCl_2$ = Concentration of $Ni^{2+}$ = 0.130 M

Concentration of $NH_3$ = 1.20 M

$K_f=5.5\times 10^8$

The equilibrium reaction will be:

$Ni^{2+}(aq)+6NH_3(aq)\rightarrow [Ni(NH_3)_6]^{2+}$

Initial conc. 0.130 1.20 0

At eqm. x [1.20-6(0.130)] 0.130

= 0.42

The expression for equilibrium constant is:

$K_f=\frac{[Ni(NH_3)_6^{2+}]}{[Ni^{2+}][NH_3]^6}$

Now put all the given values in this expression, we get:

$5.5\times 10^8=\frac{(0.130)}{(x)\times (0.42)^6}$

$x=4.31\times 10^{-8}$

Thus, the concentration of $Ni^{2+}$ ions at equilibrium is, $4.31\times 10^{-8}$

7 0

4 years ago