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BaLLatris [955]
3 years ago
13

Which type of bond is salt

Chemistry
2 answers:
tigry1 [53]3 years ago
8 0

Answer:

An Ionic Bond

Explanation:

Salt compounds are formed by the "donating" of electrons.

Alex17521 [72]3 years ago
5 0

Answer:

NaCl, or Salt, is an ionic bond.

Explanation:

Ionic bonds are formed from one metal and one nonmetal, Sodium being the metal and Chlorine being the nonmetal.

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defon

Hello there!

How is density different from weight and mass you say? It's simple!

1. Unlike weight, density is composed of mass and volume.

2. By dividing the mass over the volume, you can determine an object's density, while weight is the product of mass and gravity.

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3 years ago
What would be the Name of a Chemical made of 14 g Lithium, 32 g Sulfur, and 64 g Oxygen?
bulgar [2K]

The name of the chemical made of 14 g Lithium or 2 moles Lithium, 32 g Sulfur or 1 mol sulfur and 64 g oxygen or 4 moles of oxygen is Lithium sulfate. From the chemical reation:

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What are the units used to measure specific heat capacity?
Andrews [41]

The units used to measure specific heat capacity is Joules per kilogram per Kelvin.

<h3>What is specific heat capacity?</h3>

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brainly.com/question/1747943

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7 0
2 years ago
Consider the following reaction:
adell [148]

Answer:

1. d[H₂O₂]/dt = -6.6 × 10⁻³ mol·L⁻¹s⁻¹; d[H₂O]/dt = 6.6 × 10⁻³ mol·L⁻¹s⁻¹

2. 0.58 mol

Explanation:

1.Given ΔO₂/Δt…

    2H₂O₂     ⟶      2H₂O     +     O₂

-½d[H₂O₂]/dt = +½d[H₂O]/dt = d[O₂]/dt  

d[H₂O₂]/dt = -2d[O₂]/dt = -2 × 3.3 × 10⁻³ mol·L⁻¹s⁻¹ = -6.6 × 10⁻³mol·L⁻¹s⁻¹

 d[H₂O]/dt =  2d[O₂]/dt =  2 × 3.3 × 10⁻³ mol·L⁻¹s⁻¹ =  6.6 × 10⁻³mol·L⁻¹s⁻¹

2. Moles of O₂  

(a) Initial moles of H₂O₂

\text{Moles} = \text{1.5 L} \times \dfrac{\text{1.0 mol}}{\text{1 L}} = \text{1.5 mol }

(b) Final moles of H₂O₂

The concentration of H₂O₂ has dropped to 0.22 mol·L⁻¹.

\text{Moles} = \text{1.5 L} \times \dfrac{\text{0.22 mol}}{\text{1 L}} = \text{0.33 mol }

(c) Moles of H₂O₂ reacted

Moles reacted = 1.5 mol - 0.33 mol = 1.17 mol

(d) Moles of O₂ formed

\text{Moles of O}_{2} = \text{1.33 mol H$_{2}$O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{2 mol H$_{2}$O}_{2}} = \textbf{0.58 mol O}_{2}\\\\\text{The amount of oxygen formed is $\large \boxed{\textbf{0.58 mol}}$}

8 0
3 years ago
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