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3 years ago

Do you notice a pattern in the charge for elements in each group? Explain.

Chemistry

1 answer:

MrRissso [65]3 years ago

7 0

Answer:

Below are some patterns in the charge for elements in each group.

Group 1 elements : 1 valence electron and form ions with charge +1

Group 2 elements : 2 valence electrons and form ions with charge + 2

Group 3 elements : 3 valence electrons and form ions with charge + 3(there are some exceptions as well)

Elements in groups 4 and 5 are unpredictable also the D block elements consist of multiple oxidation states..

Group 6 elements : 6 valence electrons, form ions with charge -2

Group 7 elements: 7 valence electrons, form ions with charge -1

The octet rule is being followed, the elements form either ionic bond or covalent bond to fulfill it.

eg: when a K atom forms a K+ ion, the ion has the same electron configuration as the noble gas Ar (argon).

When an O atom gains 2 electrons to form the O²⁻ ion, the ion has the same electron configuration as the noble gas Ne (neon).

Note: Boron (B) is in Group 3 but doesn't form ions.

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Answer the following for the reaction: 3AgNO3(aq)+Na3PO4(aq)→Ag3PO4(s)+3NaNO3(aq)

Brums [2.3K]

Answer:1) Volume of $AgNO_3$ required is 55.98 mL.

2) 0.62577 grams of $Ag_3PO_4$ is produced.

Explanation:

$3AgNO_3(aq)+Na_3PO_4(aq)\rightarrow Ag_3PO_4(s)+3NaNO_3(aq)$

1) Molarity of $AgNO_3,M_1=0.225 M$

Volume of $AgNO_3.V_1=?$

Molarity of $Na_3PO_4,M_2=0.135 M$

Volume of $Na_3PO_4,V_2=31.1 mL=0.0311 L$

$Molarity=\frac{\text{number of moles}}{\text{volume of solution in liters}}$

$\text{number of moles }Na_3PO_4=M_2\times V_2=0.135 mol/L\times 0.0311 L=0.0041985 moles$

According to reaction, 1 mole of $Na_3PO_4$ reacts with 3 mole of $AgNO_3$ , then, 0.0041985 moles of $Na_3PO_4$ will react with:

$\frac{3}{1}\times 0.0041985$ moles of $AgNO_3$ that is 0.0125955 moles.

$M_1=0.225 M=\frac{\text{number of moles of }AgNO_3}{V_1}$

$V_1=\frac{0.0125955 moles}{0.225 M}=0.05598 L=55.98 mL$

Volume of $AgNO_3$ required is 55.98 mL.

$Molarity=0.195 M=\frac{\text{number of moles}}{\text{volume of solution in liters}}$

Number of moles of $AgNO_3=0.195\times 0.023 L=0.004485 moles$

According to reaction, 3 moles of $AgNO_3$ gives 1 mole of $Ag_3PO_4$ , then 0.004485 moles of $AgNO_3$ will give: $\frac{1}{3}\times 0.004485$ moles of $Ag_3PO_4$ that is 0.001495 moles.

Mass of $Ag_3PO_4$ =

Moles of $Ag_3PO_4$ × Molar Mass of $Ag_3PO_4$

= 0.001495 moles × 418.58 g/mol = 0.62577 g

0.62577 grams of $Ag_3PO_4$ is produced.

7 0

3 years ago

Hydrogen and Methanol have both been proposed as alternatives to hydrocarbon fuels. Write balanced reactions for the complete co

Arlecino [84]

Answer:

The order of energy released per mass is

CH₃OH (-2.268 × 10⁴ kJ/kg) < C₈H₁₈ (-4.826 × 10⁴ kJ/kg) < H₂ (-2.835 × 10⁵ kJ/kg)

Explanation:

In order to calculate the enthalpy of a reaction (ΔH°r) we can use the following expression.

ΔH°r = ∑np × ΔH°f(p) - ∑nr × ΔH°f(r)

where

ΔH°f(i) are the enthalpies of formation of reactants and products

ni are the moles of reactants and products

Combustion of hydrogen

H₂(g) + 1/2 O₂(g) ⇒ H₂O(l)

ΔH°r = 2 mol × ΔH°f(H₂O) - 1 mol × ΔH°f(H₂) - 1/2 mol × ΔH°f(O₂)

ΔH°r = 2 mol × (-285.8 kJ/mol) - 1 mol × 0 - 1/2 mol × 0

ΔH°r = -571.6 kJ

571.6 kJ are released when 1 mole of H₂ is burned. The amount of heat released per kilogram is:

$\frac{-571.6kJ}{1molH_{2}} .\frac{1molH_{2}}{2.016gH_{2}} .\frac{10^{3}gH_{2} }{1kgH_{2}} =-2.835 \times 10^{5} kJ/kgH_{2}$

Combustion of methanol

CH₃OH(l) + 3/2 O₂(g) ⇒ CO₂(g) + 2 H₂O(l)

ΔH°r = 1 mol × ΔH°f(CO₂) + 2 mol × ΔH°f(H₂O) - 1 mol × ΔH°f(CH₃OH) - 3/2 mol × ΔH°f(O₂)

ΔH°r = 1 mol × (-393.5 kJ/mol) + 2 mol × (-285.8 kJ/mol) - 1 mol × (-238.4 kJ/mol) - 3/2 mol × 0

ΔH°r = -726.7 kJ

726.7 kJ are released when 1 mole of CH₃OH is burned. The amount of heat released per kilogram is:

$\frac{-726.7kJ}{1molCH_{3}OH} .\frac{1molCH_{3}OH}{32.04gCH_{3}OH} .\frac{10^{3}gCH_{3}OH }{1kgCH_{3}OH} =-2.268 \times 10^{4} kJ/kgCH_{3}OH$

Combustion of octane

C₈H₁₈(l) + 12.5 O₂(g) ⇒ 8 CO₂(g) + 9 H₂O(l)

ΔH°r = 8 mol × ΔH°f(CO₂) + 9 mol × ΔH°f(H₂O) - 1 mol × ΔH°f(C₈H₁₈) - 12.5 mol × ΔH°f(O₂)

ΔH°r = 8 mol × (-393.5 kJ/mol) + 9 mol × (-285.8 kJ/mol) - 1 mol × (-208.4 kJ/mol) - 12.5 mol × 0

ΔH°r = -5511.8 kJ

5511.8 kJ are released when 1 mole of C₈H₁₈ is burned. The amount of heat released per kilogram is:

$\frac{-5511.8kJ}{1molC_{8}H_{18}} .\frac{1molC_{8}H_{18}}{114.2gC_{8}H_{18}} .\frac{10^{3}gC_{8}H_{18} }{1kgC_{8}H_{18}} =-4.826 \times 10^{4} kJ/kgC_{8}H_{18}$

5 0

3 years ago