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LenaWriter [7]
3 years ago
15

Is the number of total molecules on the left side of a balanced equation always equal to the number of total molecules on the ri

ght side of the equation
Chemistry
1 answer:
Nadya [2.5K]3 years ago
3 0

Answer:

No

Explanation:

No, but the total mass of reactants must equal the total mass of products to be a balanced equation.

Example: Consider the following reaction ...

3H₂ + N₂ => 2NH₃ and 'amu' is atomic mass units (formula weights from periodic table)

In terms of molecules, there are 4 molecules on the left (3 molecular hydrogens (H₂) and 1 molecular nitrogen (N₂) and 2 molecules of ammonia on the right side of equation arrow. ∑reactant molecules ≠ ∑product molecules.

In terms of mass of reactants & mass of products, the 3H₂ + N₂ => 6amu + 28amu = 34amu & mass of products (2NH₃) => 2(14amu) + 6(1amu) = 34amu for sum of product masses.

∑mass reactants = ∑mass products <=> 34amu = 34amu.

The expression '∑mass reactants = ∑mass products' as applied to chemical equations is generally known as 'The Law of Mass Balance'.

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The value of Zeff for a valence electron in an atom of silicon is?
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Zeff is the effective nuclear charge wherein, Z resembles the number of protons in the nucleus while S corresponds to the number of non-valence electrons.

Zeff = Z - S

Silicon has 14 protons; its electron configuration is [Ne] 3s2 3p2. Its non-valence electrons is in the n = 1 and n =2 shells. There are 2 electrons in n = 1 and 8 in n = 2, so there are a total of 10 non-valence electron.

<span><span>Z<span>eff</span></span>= 14−10= 4</span>

So, the answer is 4.
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3 years ago
What would happen to the rate of a reaction with rate law rate = k [NO]^2[H2] if
Ede4ka [16]

The rate of a reaction would be one-fourth.

<h3>Further explanation</h3>

Given

Rate law-r₁ = k [NO]²[H2]

Required

The rate of a reaction

Solution

The reaction rate (v) shows the change in the concentration of the substance (changes in addition to concentrations for reaction products or changes in concentration reduction for reactants) per unit time.  

Can be formulated:  

Reaction: aA ---> bB  

\large{\boxed{\boxed{\bold{v~=~-\frac{\Delta A}{\Delta t}}}}

or  

\large{\boxed{\boxed{\bold{v~=~+\frac{\Delta B}{\Delta t}}}}

The concentration of NO were halved, so the rate :

\tt r_2=k[\dfrac{1}{2}No]^2[H_2]\\\\r_2=\dfrac{1}{4}k.[No]^2[H_2]\\\\r_2=\dfrac{1}{4}r_1

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Answer:

A.

Explanation:

Common sense<3

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Answer:

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