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3 years ago

Compare and contrast ionic and Covalent compounds

Chemistry

1 answer:

Elden [556K]3 years ago

3 0

Answer:

Ionic bonding occurs between atoms that have opposite needs for electrons (metals and nonmetals) and results in a transfer of electrons. Covalent bonding occurs between atoms that have similar needs for electrons (two nonmetals) and results in a sharing of electrons.

Explanation:

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Round off the following measurement to three significant figures. 1.296 g

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At standard temperature and pressure. 0.500 mole of xenon gas occupies

ANEK [815]

Answer:

0.500 mole of Xe (g) occupies 11.2 L at STP.

General Formulas and Concepts:

Gas Laws

STP (Standard Conditions for Temperature and Pressure) = 22.4 L per mole at 1 atm, 273 K

Stoichiometry

Mole ratio
Dimensional Analysis

Explanation:

Step 1: Define

Identify.

0.500 mole Xe (g)

Step 2: Convert

[DA] Set up: $\displaystyle 0.500 \ \text{mole Xe} \bigg( \frac{22.4 \ \text{L Xe}}{1 \ \text{mole Xe}} \bigg)$
[DA] Evaluate: $\displaystyle 0.500 \ \text{mole Xe} \bigg( \frac{22.4 \ \text{L Xe}}{1 \ \text{mole Xe}} \bigg) = 11.2 \ \text{L Xe}$

Topic: AP Chemistry

Unit: Stoichiometry

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2 years ago

How do I find the molar mass of Copper (II) Nitrate?

QveST [7]

Answer:

The molar mass of copper (II) nitrate is 187.5 g/mol.

Explanation:

The molar mass is the mass of all the atoms in a molecule in grams per mole. To calculate the molar mass of a molecule, we first obtain the atomic weights from the individual elements in a periodic table. We then count the number of atoms and multiply it by the individual atomic masses.

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3 years ago

The reactant concentration in a first-order reaction was 8.10×10−2 M M after 15.0 s s and 1.80×10−3 M M after 90.0 s s . What is

kipiarov [429]

Answer:

The answer to the question is

The rate constant for the reaction is 1.056×10⁻³ M/s

Explanation:

To solve the question, e note that

For a zero order reaction, the rate law is given by

[A] = -k×t + [A]₀

This can be represented by the linear equation y = mx + c

Such that y = [A], m which is the gradient is = -k, and the intercept c = [A]₀

Therefore the rate constant k which is the gradient is given by

Gradient = $\frac{[A]_{2} - [A]_{1} }{t_{2} - t_{1} }$ where [A]₁ = 8.10×10⁻² M and [A]₂ = 1.80×10⁻³ M

= $\frac{1.80*10^{-3} M- 8.10*10^{-2} M}{90 s - 15 s}$ = -0.001056 M/s = -1.056×10⁻³ M/s

Threfore k = 1.056×10⁻³ M/s

3 0

3 years ago

Compare and contrast ionic and Covalent compounds​

Compare and contrast ionic and Covalent compounds