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Marrrta [24]
2 years ago
6

Initially, 0.740 mol of A is present in a 5.00 L solution. 2 A ( aq ) − ⇀ ↽ − 2 B ( aq ) + C ( aq ) At equilibrium, 0.200 mol of

C is present. Calculate K.
Chemistry
1 answer:
drek231 [11]2 years ago
4 0

Answer:

POINTZZZ

Explanation:

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Which type of bonding involves the complete transfer of a valence electron from a less electronegative atom to a more electroneg
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Answer:

Ionic or electrovalent bonds

Explanation:

Ionic or electrovalent bonds are interatomic or intramolecular bonds which are formed between two kinds of atoms having a large electronegativity difference usually 2.1.

Electronegativity is the property that combines the ability of an atom to gain or lose electrons. It is expressed as the tendency with which atoms of elements attracts valence electrons in a chemical bond.

In this bond type, a metal transfers its electrons to a more electronegative atom which is a non-metal.

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Balance the equation and state the limiting reagent in the following reaction:
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Which of the following has a melting point greater than room temperature?
nikdorinn [45]
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Why do you think different liquids have different freezing points
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5 0
3 years ago
A chemical reaction produces 3.81g potassium chloride (KCl). How many formula units of potassium chloride are there? *Find Molar
Aleks04 [339]

Answer:

The number of formula units in 3.81 g of potassium chloride (KCl) is approximately 3.08 × 10²²

Explanation:

The given parameters is as follows;

The mass of potassium chloride produced in the chemical reaction (KCl) = 3.81 g

The required information = The number of formula units of potassium chloride (KCl)

The Molar Mass of KCl = 74.5513 g/mol

The \ number \ of \ moles \ of \ a \ substance, n = \dfrac{The  \  mass  \  of  \  the substance}{The   \  Molar  \   Mass   \  of  \   the   \  substance}

Therefore, we have;

The \ number \ of \ moles \ of \ KCl= \dfrac{3.81 \ g}{74.5513 \ g/mol} \approx 0.051106 \ moles

1 mole of a substance, contains Avogadro's number (6.022 × 10²³) of  formula units

Therefore;

0.051106 moles of KCl contains 0.051106 × 6.022 × 10²³  ≈ 3.077588 × 10²² formula units

From which we have, the number of formula units in 3.81 g of potassium chloride (KCl) ≈ 3.08 × 10²² formula units.

8 0
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