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3 years ago

An elements’ neutral atoms have 5 valance electrons and 3 electron shells. What element is it?

Chemistry

1 answer:

Elodia [21]3 years ago

3 0

Answer:

Phosphorus

Explanation:

Group 5A has 5 valence electrons.

3 electron shells equals 15 electrons on Phosphorus. Furthermore, 2 on shell 1, 8 on shell 2, and 5 on shell 3.

- Hope that helps! Please let me know if you need further explanation.

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Assuming equal concentrations and complete dissociation, rank these aqueous solutions by their freezing points from highest to l

muminat

Answer:

NH4Cl > Li2SO4 > CoCl3

Explanation:

Let us recall that the freezing point depression depends on the molality of the solution and the number of particles present.

Let us also recall that freezing point depression is a colligative property. It depends on the number of particles present in solution.

Usually, the more the number of particles present, the lower the freezing point. Hence, NH4Cl which has only two particles will have the highest freezing point while CoCl3 which has four particles will have the lowest freezing point.

4 0

3 years ago

An unknown compound contains only carbon, hydrogen, and oxygen (CxHyOz). Combustion of 5.00g of this compound produced 7.33g of

Olegator [25]

A combustion reaction involves an organic compound reacted with oxygen. The general chemical equation is as follows:

Organic Compound + Oxygen = CO2 + H2O

To calculate the amount of C present in the original sample, we use the values given and assume that there is complete combustion that is happening.

7.33 g CO2 ( 1 mol CO2 / 44.01 g CO2)(1 mol C / 1 mol CO2) = 0.167 mol C

Therefore, 0.167 mol of C was originally in the sample.

8 0

3 years ago

Read 2 more answers

25.00 mL of a H2SO4 solution with an unknown concentration was titrated to a phenolphthalein endpoint with 28.11 mL of a 0.1311

LuckyWell [14K]

Answer:

Concentration of the H₂SO₄ solution is 0.0737 M

Explanation:

Equation of the neutralization reaction between the acid, H₂SO₄, and the base, NaOH, is given below:

H₂SO₄ + 2NaOH -----> Na₂SO₄ + 2H₂O

From the above equation, one mole of acid requires 2 moles of base for complete neutralization which occurs at phenolphthalein endpoint.

mole ratio of acid to base, nA/nB = 1:2

Concentration of the base, Cb = 0.1311 M

Volume of base, Vb, = 28.11 mL

Concentration of acid, Ca = ?

Volume of acid, Va + 25.0 mL

Using the formula, CaVa/CbVb = nA/nB

making Ca subject of the formula, Ca = Cb*Vb*nA/Va*nB

substituting the values into the equation

Ca = (0.1311 * 28.11 * 1) / 25.0 * 2 = 0.0737 M

Therefore, concentration of the H₂SO₄ solution is 0.0737 M

5 0

4 years ago

What is the potential energy of the ball when it gets to its maximum height just before falling back to the ground?J

erastovalidia [21]

Answer:

9.8 joules

Explanation:

due to acceleration due to gravity

4 0

4 years ago

Menthol, the substance we can smell in mentholated cough drops, is composed of c, h, and o. a 9.045×10−2 −mg sample of menthol i

Ket [755]

Answer:

Empirical Formula = C₁₀H₂₀O

Solution:

Data Given:

Mass of Menthol = 9.045 × 10⁻² mg = 9.045 × 10⁻⁵ g

Mass of CO₂ = 0.2546 mg = 0.0002546 g

Mass of H₂O = 0.1043 mg = 0.0001043 g

Step 1: Calculate %age of Elements as;

%C = (mass of CO₂ ÷ Mass of sample) × (12 ÷ 44) × 100

%C = (0.0002546 ÷ 9.045 × 10⁻⁵) × (12 ÷ 44) × 100

%C = (2.814) × (12 ÷ 44) × 100

%C = 2.814 × 0.2727 × 100

%C = 76.73 %

%H = (mass of H₂O ÷ Mass of sample) × (2.02 ÷ 18.02) × 100

%H = (0.0001043 ÷ 9.045 × 10⁻⁵) × (2.02 ÷ 18.02) × 100

%H = (1.153) × (2.02 ÷ 18.02) × 100

%H = 1.153 × 0.1120 × 100

%H = 12.91 %

%O = 100% - (%C + %H)

%O = 100% - (76.73% + 12.91%)

%O = 100% - 89.64%

%O = 10.36 %

Step 2: Calculate Moles of each Element;

Moles of C = %C ÷ At.Mass of C

Moles of C = 76.73 ÷ 12.01

Moles of C = 6.3888 mol

Moles of H = %H ÷ At.Mass of H

Moles of H = 12.91 ÷ 1.01

Moles of H = 12.7821 mol

Moles of O = %O ÷ At.Mass of O

Moles of O = 10.36 ÷ 16.0

Moles of O = 0.6475 mol

Step 3: Find out mole ratio and simplify it;

C H O

6.3888 12.7821 0.6475

6.3888/0.6475 12.7821/0.6475 0.6475/0.6475

9.86 19.74 1

≈ 10 ≈ 20 1

Result:

Empirical Formula = C₁₀H₂₀O₁

8 0

4 years ago