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kozerog [31]
2 years ago
12

Looking at the same nonmetal group on the periodic table, how does the reactivity of an element in period 2 compare to the react

ivity of an element in period 4?
A. The period 2 element would be more reactive because the attractive force of protons is stronger when there are fewer neutrons interfering.

B. The period 2 element would be more reactive because the attractive force of protons is stronger when electrons are attracted to a closer electron shell.

C. The period 4 element would be more reactive because the attractive force of protons is stronger when there are more neutrons helping.

D. The period 4 element would be more reactive because the attractive force of protons is stronger when electrons are attracted to a farther electron shell.
Chemistry
2 answers:
Snezhnost [94]2 years ago
9 0

Answer is: B. The period 2 element would be more reactive because the attractive force of protons is stronger when electrons are attracted to a closer electron shell.

For example, fluorine (the period 2) is more reactive than bromine (the period 4).

Fluorine (F) is nonmetal with greatest electronegativity, which means it easily gain electrons.

Fluorine jas atomic number 9, which means it has 9 protons and 9 electrons. It gain one electron to form fluorine anion (F⁻) with stable electron configuration like closest noble gas neon (Ne) with 10 electrons.

Electron configuration of fluorine: ₉F 1s² 2s² 2p⁵.

Bas_tet [7]2 years ago
4 0

Answer : Option B) The period 2 element would be more reactive because the attractive force of protons is stronger when electrons are attracted to a closer electron shell.

Explanation : The reactivity of the Periods decreases as we go from left to right across a period. The farther to the left and down the periodic chart we go, the easier it is for electrons to be donated or taken away, resulting in higher reactivities of the elements. The attractive force of the protons is found to be stronger when electrons are found to be attracted to a closer electron shell.

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Determine the mass of oxygen in a 7.20 g sample of Al2(SO4)3.
Mekhanik [1.2K]

Given:

7.20 g sample of Al2(SO4)3

Required:

Mass of oxygen

Solution:

                Since you are not given a chemical reaction, just base your solution to the chemical formula given.

Molar mass of Al2(SO4)3 = 342.15 g/mol

7.20 g Al2(SO4)3 (1 mol/342.15g)(3mol O/2 mol Al)(1 mol O2/1/2 mol O2)(32g O2/1mol O2) = 4.04 g O2

5 0
2 years ago
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A 50.6 grams sample of magnesium hydroxide (Mg(OH)2) is reacted with 45.0 grams of hydrochloric acid (HCl). What is the theoreti
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            <span> Mg(OH)2(s) + 2HCl(aq) yield MgCl2(aq) + 2H2O(l)

grams HCl required = (50.6 grams Mg(OH)2) * (1 mol Mg(OH)2 / 58.3197 grams Mg(OH)2) * (2 mol HCl / 1 mol Mg(OH)2) * (36.453 grams HCl / 1 mol HCl) = 63.26 grams HCl required

Since there are only 45.0 grams HCl, then HCl is the limiting reactant.

theoretical yield MgCl2 = (45.0 grams HCl) * (1 mol HCl / 36.453 grams HCl) * (1 mol MgCl2 / 2 mol HCl) * (95.211 grams MgCl2 / 1 mol MgCl2) = 58.6 grams MgCl2 </span>
7 0
2 years ago
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One reactant with a mass of 10 grams is combined with another reactant with a mass of 8 grams in a sealed container. After the r
disa [49]
18g is the most reasonable mass after the reaction
3 0
2 years ago
The molar absorptivity of a compound at 500 nm wavelength is 252 M-1cm-1. Suppose one prepares a solution by dissolving 0.00140
adell [148]

Answer:

The absorbance of the solution is 0.21168.

Explanation:

Given that,

Wavelength = 500 nm

Molar absorptivity = 252 M⁻¹ cm⁻¹

Number of moles = 0.00140

Volume of solution = 500.0 mL

Length = 3.00 mm

We need to calculate the molar concentration

Using formula of the molar concentration

C=\dfrac{N}{V}

Where, N = number of moles

V = volume

Put the value into the formula

C=\dfrac{0.00140}{0.5000}

C=0.0028\ M

We need to calculate the absorbance of the solution

Using formula of absorbance

A=\epsilon C l

Put the value into the formula

A=252\times0.0028\times0.300

A=0.21168

Hence, The absorbance of the solution is 0.21168.

5 0
2 years ago
Upon combustion, a 0.8009 g sample of a compound containing only carbon, hydrogen, and oxygen produces 1.6004 g co2 and 0.6551 g
dlinn [17]
The compound contains Carbon, Hydroxide and Oxide
1 mole of carbon iv oxide contains 44 g, out of which 12 g are carbon.
Therefore, 1.6004 g of CO2 will contain;
    1.6004 ×12/44 = 0.4365 g of carbon
1 mole of water contains 18 g of which 2 g is hydrogen,
Therefore, 0.6551 g of H2O will hace ;
         0.6551 × 2/18 = 0.0728 g of hydrogen.
The total mass of the compound is 0.8009 g,
Thus the mass of oxygen = 0.8009 -(0.4365 +0.0728)
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To get the empirical formula we first get the number of moles of each element;\
 Carbon = 0.4365/12= 0.036375 moles
Hydrogen = 0.0728/1 = 0.0728 moles
Oxygen = 0.2916/16 = 0.018225 moles
Then, to get the smallest ratio we divide each with the smallest value;
Carbon :                        Hydrogen :                    Oxygen
= (0.036375/0.018225) : (0.0728/0.018225) : ( 0.018225/0.018225)
= 1.996                         :   3.995                    : 1
≈ 2 : 4 : 1
Therefore, the empirical formula is C2H4O


3 0
2 years ago
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