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2 years ago

The symbol for xenon (xe) would be a part of the noble gas notation for the element

Chemistry

1 answer:

daser333 [38]2 years ago

6 0

Answer:

Explanation:

To write the electronic configuration of an element using the preceding noble gas configuration, we simply use the noble gas in the previous period of that particular element. We do not use the noble gas of that period in which the element belongs to but the one in the preceding group.

A is wrong

This is because the noble gas in the preceding period is krypton.

In fact, the electronic configuration is [Kr] 4d105s25p3

C is wrong

The noble gas in the last period here is Radon. The electronic configuration of Radium in fact is [Rn] 7s2

D is wrong

The noble gas in the last period before the period of uranium is Radon also. In fact, the electronic configuration of the element is [Rn] 5f36d17s2

B is correct, the actual electronic configuration of the element Cesium is [Xe] 6s1

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Industrial production of nitric acid, which is used in many products including fertilizers and explosives, approaches 10 billion

mylen [45]

Answer: $9.361\times 10^{4} kJ$

Explanation:

The balanced chemical equation :

$4NH_3(g)+5O_2(g)\rightarrow 4NO(g)+6H_2O(g)$ $\Delta H^0_{rxn}=-902.0kJ$

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}=\frac{7.056\times 10^3g}{17g/mol}=415.1moles$

According to stoichiometry:

4 moles of $NH_3$ produces = 902.0 kJ of energy

415.1 moles of $NH_3$ produces = $\frac{902.0}{4}\times 415.1=9.361\times 10^{4} kJ$ of energy

Thus the change in enthalpy is $9.361\times 10^{4} kJ$

5 0

2 years ago

Write the lewis structure for ch2clcoo−. assign a formal charge for any atom with a non-zero formal charge.

Murljashka [212]

The Lewis structure of Chloroacetate (H₂CClCO₂) is given below. In structure it is shown that carbon has a double bond with one oxygen atom and two single bonds with CH₂Cl and O⁻.

Formal Charge;
Formal charge is caculated as,

Formal charge = # of valence e⁻ - [# of lone pair of e⁻ + 1/2 # of bonded e⁻]

Formal charge on Oxygen (Highlighted Red);

Formal charge = 6 - [ 6 + 2/2]

Formal charge = 6 - [6 + 1]

Formal charge = 6 - [7]

Formal charge = -1

8 0

2 years ago

A 8.00g of a certain Compound X, known to be made of carbon, hydrogen and perhaps oxygen, and to have a molecular molar mass of

Ne4ueva [31]

Answer:

C3H6.

Explanation:

Data obtained from the question:

Mass of the compound = 8g

Mass of CO2 = 24.01g

Mass of H2O = 13.10g

Next, we shall determine the mass of C, H and O present in the compound. This is illustrated below:

Molar Mass of CO2 = 12 + (2x16) = 44g/mol

Molar Mass of H2O = (2x1) + 16 = 18g/mol

Mass of C in compound = Mass of C/Molar Mass of CO2 x 24.01

=> 12/44 x 24.01 = 6.5g

Mass of H in the compound = Mass of H/Molar Mass of H2O x 13.1

=> 2x1/18 x 13.1 = 1.5g

Mass of O in the compound = Mass of compound – (mass of C + Mass of H)

=> 8 – (6.5 + 1.5) = 0

Next, we shall determine the empirical formula of the compound. This is illustrated below:

C = 6.5g

H = 1.

Divide by their molar mass

C = 6.5/12 = 0.54

H = 1.4/1 = 1.

Divide by the smallest

C = 0.54/0.54 = 1

H = 1/0.54 = 2

Therefore, the empirical formula is CH2

Finally, we shall determine the molecular formula as follow:

The molecular formula of a compound is a multiple of the empirical formula.

Molecular formula = [CH2]n

[CH2]n = 44

[12 + (2x1)]n = 44

14n = 44

Divide both side by 14

n = 44/14

n = 3

Molecular formula = [CH2]n = [CH2]3 = C3H6

Therefore, the molecular formula of the compound is C3H6

4 0

2 years ago

penicillin. an important antibiotic (antibacterial agent), was discovered accidentally by the scottish bacteriologist alexander

dmitriy555 [2]

Answer:

mass percent of carbon = 57.78 %

mass percent of hydrogen = 6.40 %

mass percent of nitrogen = 8.96 %

mass percent of oxygen = 20.49 %

mass percent of sulfur = 10.24 %

Explanation:

Given data

Molecular formula = C₁₄H₂₀N₂O₄S

molecular mass (total mass) = 312.39 g/mol

Percentage of carbon = ?

Percentage of hydrogen = ?

Percentage of oxygen = ?

Percentage of nitrogen = ?

Percentage of sulfur = ?

Solution

1st we find out number of moles of each element from the molecular formula

Number of moles of carbon = 14 mol

Number of moles of hydrogen = 20 mol

Number of moles of nitrogen = 2 mol

Number of moles of oxygen = 4 mol

Number of moles of sulfur = 1 mol

Now we find out the mass of each element

as we know that

<em>mass = number of moles × molecular mass</em>

mass of carbon = 14 mol × 12 g/mol

mass of carbon = 168 g

mass of hydrogen = 20 mol × 1 g/mol

mass of hydrogen = 20 g

mass of nitrogen = 2 mol × 14 g/mol

mass of nitrogen = 28 g

mass of oxygen = 4 mol × 16 g/mol

mass of oxygen = 64 g

mass of sulfur = 1 mol × 32 g/mol

mass of sulfur = 32 g

now we find out the mass percent of each element

<em> mass percent = ( mass ÷ total mass ) × 100</em>

mass percent of carbon = ( 168 g ÷ 312.39 g/mol ) × 100

mass percent of carbon = 57.78 %

mass percent of hydrogen = ( 20 g ÷ 312.39 g/mol ) × 100

mass percent of hydrogen = 6.40 %

mass percent of nitrogen = ( 28 g ÷ 312.39 g/mol ) × 100

mass percent of nitrogen = 8.96 %

mass of oxygen =( 64 g ÷ 312.39 g/mol ) × 100

mass percent of oxygen = 20.49 %

mass percent of sulfur = ( 32 g ÷ 312.39 g/mol ) × 100

mass percent of sulfur = 10.24 %

7 0

2 years ago

How many molecules of glucose are in 1 l of a 100 mm glucose solution?

Karolina [17]

Answer is: 6.022·10²² molecules of glucose.

c(glucose) = 100 mM.

c(glucose) = 100 · 10⁻³ mol/L.

c(glucose) = 0.1 mol/L; concentration of glucose solution.

V(glucose) = 1 L; volume of glucose solution.

n(glucose) = c(glucose) · V(glucose).

n(glucose) = 0.1 mol/L · 1 L.

n(glucose) = 0.1 mol; amount of substance.

N(glucose) = n(glucose) · Na (Avogadro constant).

N(glucose) = 0.1 mol · 6.022·10²³ 1/mol.

N(glucose) = 6.022·10²².

6 0

2 years ago