Addition of an excess of lead (II) nitrate to a 50.0mL solution of magnesium chloride caused a formation of 7.35g of lead (II) c

Response:

FALSE

Rationale:

Sedimentary rocks are defined as rocks formed through the processes of compaction and cementation. Initially, sediments derived from various locations must accumulate. Over time, these deposited sediments undergo substantial compaction due to the weight of the layers above. This process converts loose sediments into solid rock. This is the process through which sedimentary rocks, comprised of sand-sized particles, are formed. For instance, examples include Shale, Sandstone, and Mudstone.

Thus, both compaction and cementation are crucial in the formation of sedimentary rocks.

Therefore, the statement above is False.

Answer:

Both CaCl2 and CaBr2 consist of elements (bromine and chlorine) from the same group (group 7).

Explanation:

In the periodic table, elements are arranged into groups based on their valence electron count in the outermost shell. Elements in the same group, which possess a similar number of valence electrons, typically exhibit similar chemical behaviors.

Chlorine and Bromine in CaCl2 and CaBr2 belong to group 7, known as HALOGENS, characterized by having 7 valence electrons in their outer shell.

The similarity in properties between CaCl2 and CaBr2 arises because both contain Chlorine and Bromine, leading to analogous reactions and behaviors when interacting with other compounds.

Answer:

The rate law for the decomposition reaction is:

The unit for the rate constant will be

Explanation:

The rate law can be expressed as:

..[1]

When the drug concentration is tripled, the decomposition rate rises by a factor of nine.

...[2]

[1] ÷ [2]

Solving for x results in:

x = 2.

This indicates a second-order reaction.

The decomposition reaction's rate law is:

The unit for the rate constant will be:

The unit for the rate constant will be .

Answer:

The empirical formula is =

The Valproic acid formula is =

Explanation:

Mass of the produced water = 0.166 g

Molar mass of water = 18 g/mol

The moles of are calculated as 0.166 g /18 g/mol = 0.00922 moles.

In 1 mole of water, there are 2 moles of hydrogen atoms.

Thus,

Moles of H = 2 x 0.00922 = 0.01844 moles

Each hydrogen atom's molar mass is 1.008 g/mol.

Hydrogen mass in the molecule = 0.01844 x 1.008 = 0.018588 g

Mass of produced carbon dioxide = 0.403 g

Molar mass of carbon dioxide = 44.01 g/mol

The moles of are calculated as 0.403 g  /44.01 g/mol = 0.009157 moles.

Each carbon atom's presence is 1 mole in 1 mole of carbon dioxide.

So,

Moles of C = 0.009157 moles

The molar mass of carbon is 12.0107 g/mol.

Carbon mass in molecule = 0.009157 x 12.0107 = 0.11 g

Since Valproic acid comprises only hydrogen, oxygen, and carbon, the oxygen mass in the sample = Total mass - Carbon mass  - Hydrogen mass.

The sample's overall mass = 0.165 g.

Oxygen mass in the sample = 0.165 - 0.11 - 0.018588 = 0.036412 g  

The molar mass of oxygen is 15.999 g/mol.

Moles of O  = 0.036412  / 15.999  = 0.002276 moles

Taking the simplest ratio for H, O, and C yields:

0.01844: 0.002276: 0.009157

= 8: 1: 4

The empirical formula becomes

While molecular formulas detail the precise count of atoms for each element, empirical formulas represent the simplest form or reduced ratio of these elements in the compound.

Consequently,  

The molecular mass equals n × Empirical mass.

Here, n is a positive integer from 1, 2, 3...

Empirical mass = 4×12 + 8×1 + 16 = 72 g/mol.

Molar mass = 144 g/mol.

Thus,  

The molecular mass = n × Empirical mass.

144 = n × 72

⇒ n = 2

The Valproic acid formula is =