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2 years ago

Type in the correct values to correctly represent the valence electron configuration of oxygen: AsB2pC

Chemistry

2 answers:

valkas [14]2 years ago

7 0

Answer:

2s²2p⁴

Explanation:

Oxygen is an element on the periodic table with a total of 8 electrons. It's electronic configuration is given as 2,6.

Using the orbital notation we write as 1s²2s²2p⁴

Also, the valence electrons are the electrons in the outermost shell of an atom. These electrons mostly determine the chemical properties of an atom.

Oxygen has a total of 6 electrons in its outermost shell and it is given as 2s²2p⁴

PilotLPTM [1.2K]2 years ago

3 0

Answer:

2s²2p⁴

Explanation:

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Standard Thermodynamic Quantities for Selected Substances at 25 ∘C Reactant or product ΔHf∘(kJ/mol) Al(s) 0.0 MnO2(s) −520.0 Al2

Svet_ta [14]

Answer:

-1815.4 kJ/mol

Explanation:

Starting with standard enthalpies of formation you can calculate the standard enthalpy for the reaction doing this simple calculation:

∑ n *ΔH formation (products) - ∑ n *ΔH formation (reagents)

This is possible because enthalpy is state function meaning it only deppends on the initial and final state of the system (That's why is also possible to "mix" reactions with Hess Law to determine the enthalpy of a new reaction). Also the enthalpy of formation is the heat required to form the compound from pure elements, then products are just atoms of reagents organized in a different form.

In this case:

ΔH rxn = [(2 * -1675.7) - (3 * -520.0)] kJ/mol = -1815.4 kJ/mol

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2 years ago

The volume of blood plasma in adults is 3.1 L. it's density is 1.03 g/cm3. Approximately how many pounds of blood plasma are the

e-lub [12.9K]

The important thing in this question is the unit. The mass equals density * volume. 3.1 L = 3.1 * 10^3 cm3. So the mass is 3.193*10^3 g. 1 pound = 453.95 g. So the answer is 7.04 pounds.

3 0

2 years ago

Read 2 more answers

A voltaic cell is constructed with two silver-silver chloride electrodes, where the half-reaction is AgCl (s) + e− → Ag (s) + Cl

ANTONII [103]

Answer : The cell emf for this cell is 0.118 V

Solution :

The half-cell reaction is:

$AgCl(s)+e^\rightarrow Ag(s)+Cl^-(aq)$

In this case, the cathode and anode both are same. So, $E^o_{cell}$ is equal to zero.

Now we have to calculate the cell emf.

Using Nernest equation :

$E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Cl^{-}{diluted}]}{[Cl^{-}{concentrated}]}$

where,

n = number of electrons in oxidation-reduction reaction = 1

$E_{cell}$ = ?

$[Cl^{-}{diluted}]$ = 0.0222 M

$[Cl^{-}{concentrated}]$ = 2.22 M

Now put all the given values in the above equation, we get:

$E_{cell}=0-\frac{0.0592}{1}\log \frac{0.0222M}{2.22M}$

$E_{cell}=0.118V$

Therefore, the cell emf for this cell is 0.118 V

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2 years ago

A student mixed together aqueous solutions of Y and Z. A white precipitate(solid)formed. which could not be Y and Z

maks197457 [2]

(A)hydrochloric acid + silver nitrate
HCl(aq) + AgNO3(aq) -----> AgCl(s) +HNO3(aq)

(B)hydrochloric acid + sodium hydroxide
HCl(aq) + NaOH(aq) -----> NaCl(aq) + H2O(l)

(C)calcium chloride + silver nitrate
CaCl2(aq) + AgNO3(aq) ----> AgCl(s) +Ca(NO3)2(aq)

(D)sodium chloride + silver nitrate
NaCl(aq) + AgNO3(aq) ----> AgCl(s) +NaNO32(aq)

AgCl is a white precipitate.
In (B) no precipitate was formed, so answer is B.

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2 years ago

An atom of lead has a radius of and the average orbital speed of the electrons in it is about . Calculate the least possible unc

victus00 [196]

The question is incomplete. Here is the complete question.

An atom of lead has a radius of 154 pm and the average orbitalspeed of the electron in it is about 1.8x $10^{8}$ m/s. Calculate the least possible uncertainty in a measurement of the speed of an electron in an atom of lead. Write your answer as a percentage of the average speed, and round it to significant 2 digits.