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Ksivusya [100]
3 years ago
12

The atomic number tells us how many __ or __ an element contains.

Chemistry
1 answer:
CaHeK987 [17]3 years ago
5 0
Answer:-
Electrons or Protons

Explanation:-
Atomic number (Z) is defined as:
>>“The total number of protons in the nucleus of an atom is called atomic number”. All atoms of an element have the same number of protons and electrons.
>>“The total number of protons in the nucleus or electrons revolving around the nucleus of an atom”.
Example:
For instance, Hydrogen has the Atomic number(Z) 1.
Helium has the atomic number 2.
Oxygen has the atomic number 8.
| Note | Atomic number is represented by “Z”. And some ppl don’t consider the 2nd definition it’s upto u to do both or the standard one.

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What is the calculated value of the cell potential at 298K for an
Tpy6a [65]

The question is incomplete, here is the complete question:

What is the calculated value of the cell potential at 298 K for an  electrochemical cell with the following reaction, when the H₂  pressure is 6.56 x 10⁻² atm, the H⁺ concentration is 1.39 M, and  the Sn²⁺ concentration is 9.35 x 10⁻⁴ M?

2H^+(aq)+Sn(s)\rightarrow H_2(g)+Sn^{2+}(aq)

<u>Answer:</u> The cell potential of the given electrochemical cell is 0.273 V

<u>Explanation:</u>

For the given chemical equation:

2H^+(aq)+Sn(s)\rightarrow H_2(g)+Sn^{2+}(aq)

The half cell reactions for the given equation follows:

<u>Oxidation half reaction:</u> Sn(s)\rightarrow Sn^{2+}(aq)+2e^-;E^o_{Sn^{2+}/Sn}=-0.14V

<u>Reduction half reaction:</u> H_2+2e^-\rightarrow H_2(g);E^o_{2H^{+}/H_2}=0.0V

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}=0.0-(-0.14)=0.14V

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{[Sn^{2+}]\times p_{H_2}}{[H^+]^2}

where,

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

E^o_{cell} = standard electrode potential of the cell = +0.14 V

n = number of electrons exchanged = 2

[H^{+}]=1.39M

[Sn^{2+}]=9.35\times 10^{-4}M

p_{H_2}=6.56\times 10^{-2}atm

Putting values in above equation, we get:

E_{cell}=0.14-\frac{0.059}{2}\times \log(\frac{(9.35\times 10^{-4})\times (6.56\times 10^{-2})}{(1.39)^2})\\\\E_{cell}=0.273V

Hence, the cell potential of the given electrochemical cell is 0.273 V

4 0
2 years ago
How many oxygens are used per glucose molecule in glycolysis?
dimulka [17.4K]

Answer:

0

Explanation:

7 0
1 year ago
A 25.0-mL sample of 0.100M Ba(OH)2(aq) is titrated with 0.125 M HCl(aq).
morpeh [17]

Answer:

20.0

Explanation:

NaOH = (25.0) (0.100m) \ 0.125M = 20.0mL

8 0
3 years ago
I need help with 1,2,3, and 4
Schach [20]

Answer:

  • Problem 1: 1.85atm
  • Problem 2: 110mL
  • Problem 3: 290 mL
  • Problem 4: 1.14 atm

Explanation:

Problem 1

<u>1. Data</u>

<u />

a) P₁ = 3.25atm

b) V₁ = 755mL

c) P₂ = ?

d) V₂ = 1325 mL

r) T = 65ºC

<u>2. Formula</u>

Since the temeperature is constant you can use Boyle's law for idial gases:

          PV=constant\\\\P_1V_1=P_2V_2

<u>3. Solution</u>

Solve, substitute and compute:

         P_1V_1=P_2V_2\\\\P_2=P_1V_1/V_2

        P_2=3.25atm\times755mL/1325mL=1.85atm

Problem 2

<u>1. Data</u>

<u />

a) V₁ = 125 mL

b) P₁ = 548mmHg

c) P₁ = 625mmHg

d) V₂ = ?

<u>2. Formula</u>

You assume that the temperature does not change, and then can use Boyl'es law again.

          P_1V_1=P_2V_2

<u>3. Solution</u>

This time, solve for V₂:

           P_1V_1=P_2V_2\\\\V_2=P_1V_1/P_2

Substitute and compute:

        V_2=548mmHg\times 125mL/625mmHg=109.6mL

You must round to 3 significant figures:

        V_2=110mL

Problem 3

<u>1. Data</u>

<u />

a) V₁ = 285mL

b) T₁ = 25ºC

c) V₂ = ?

d) T₂ = 35ºC

<u>2. Formula</u>

At constant pressure, Charle's law states that volume and temperature are inversely related:

         V/T=constant\\\\\\\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}

The temperatures must be in absolute scale.

<u />

<u>3. Solution</u>

a) Convert the temperatures to kelvins:

  • T₁ = 25 + 273.15K = 298.15K

  • T₂ = 35 + 273.15K = 308.15K

b) Substitute in the formula, solve for V₂, and compute:

        \dfrac{V_1}{T_1`}=\dfrac{V_2}{T_2}\\\\\\\\\dfrac{285mL}{298.15K}=\dfrac{V_2}{308.15K}\\\\\\V_2=308.15K\times285mL/298.15K=294.6ml

You must round to two significant figures: 290 ml

Problem 4

<u>1. Data</u>

<u />

a) P = 865mmHg

b) Convert to atm

<u>2. Formula</u>

You must use a conversion factor.

  • 1 atm = 760 mmHg

Divide both sides by 760 mmHg

       \dfrac{1atm}{760mmHg}=\dfrac{760mmHg}{760mmHg}\\\\\\1=\dfrac{1atm}{760mmHg}

<u />

<u>3. Solution</u>

Multiply 865 mmHg by the conversion factor:

    865mmHg\times \dfrac{1atm}{760mmHg}=1.14atm\leftarrow answer

3 0
3 years ago
What are the two fundamental portions of an atom?​
Mice21 [21]
Number of Protons in an Uncharged Atom
The two main components of an atom are the nucleus and the cloud of electrons. The nucleus contains positively charged and neutral subatomic particles, whereas the cloud of electrons contains tiny negatively charged particles.
4 0
2 years ago
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