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2 years ago

Which pair of elements form an ionic bond with

Chemistry

2 answers:

Vesna [10]2 years ago

5 0

Answer:

It's only A

Explanation:

Potassium (K) is a metal and Chlorine (Cl) is a nonmetal, and ionic compounds are formed between metals and nonmetals. So KCl forms ionic bonds.

Hydrogen (H) and Chlorine (Cl) are both nonmetals, and only covalent bonds form between nonmetals. (Only exception is when its an acid, then it's ionic)

Also, the question indicated which "pair" and not "pairs" meaning that there is only one answer, and I'm more likely to go with A rather than D.

KATRIN_1 [288]2 years ago

3 0

Answer:

Yes Both A) KCL and D) HCL are correct...

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Ionic bonds form between which two types of elements?

Alla [95]

Answer:

Ionic bonds usually occur between metal and nonmetal ions. For example, sodium (Na), a metal, and chloride (Cl), a nonmetal, form an ionic bond to make NaCl.

Explanation:

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3 years ago

In the following equation, which element has been reduced?

Mila [183]

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2 years ago

Determine the maximum number of electrons in the 2f designation

sergejj [24]

Google said

How many electrons fit in each shell around an atom?

The maximum number of electrons that can occupy a specific energy level can be found using the following formula:

Electron Capacity = 2n2

The variable n represents the Principal Quantum Number, the number of the energy level in question.

Energy Level
(Principal Quantum Number) Shell Letter Electron Capacity
1 K 2
2 L 8
3 M 18
4 N 32
5 O 50
6 P 72
Keep in mind that an energy level need not be completely filled before electrons begin to fill the next level. You should always use the Periodic Table of Elements to check an element's electron configuration table if you need to know exactly how many electrons are in each level.

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3 years ago

Calculate the standard enthalpy change for the reaction at 25 ∘C. Standard enthalpy of formation values can be found in this lis

weqwewe [10]

Answer:

$\Delta H_{rxn}=-2043.999kJ$

Explanation:

$\Delta H_{rxn}^{0}=\sum [n_{i}\times \Delta H_{f}^{0}(product)_{i}]-\sum [n_{j}\times \Delta H_{f}^{0}(reactant_{j})]$

Where $n_{i}$ and $n_{j}$ are number of moles of product and reactant respectively (equal to their stoichiometric coefficient).

$\Delta H_{f}^{0}$ is standard heat of formation and $\Delta H_{rxn}^{0}$ is standard enthalpy change for reaction at $25^{0}\textrm{C}$

So, $\Delta H_{rxn}=[3mol\times \Delta H_{f}^{0}(CO_{2})_{g}]+[4mol\times \Delta H_{f}^{0}(H_{2}O)_{g}]-[1mol\times \Delta H_{f}^{0}(C_{3}H_{8})_{g}]-[5mol\times \Delta H_{f}^{0}(O_{2})_{g}]$

or, $\Delta H_{rxn}=[3mol\times -393.509kJ/mol]+[4mol\times -241.818kJ/mol]-[1mol\times -103.8kJ/mol]-[5mol\times 0kJ/mol]$

or, $\Delta H_{rxn}=-2043.999kJ$

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3 years ago