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monitta
4 years ago
15

Consider the hydrate below. FeCl3•6H2O

Chemistry
1 answer:
fiasKO [112]4 years ago
5 0
So this is nomenclature.
The name would be Iron (III) Chloride Hexahydrate
Reason why it's Iron (III) is because of the 3 after Cl, if you take FeCl3 apart it's Fe^{3+} and Cl^-.
Hexahydrate because hexa = 6 and hydrate=water

The term hydrate would be wrong because that clearly doesn't come first

It would be the anhydrous compound that comes first. The "an" in "anhydrous" means "without". Hydrous=water, so anyhdrous= without water
The part without water is FeCl3. The water is 6H2O

The prefix for the coefficient of water is 6, or hexa, but that's not first in the name
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Why concentrated HNO3 is 69% by mass of nitric acid? Please explain what it means.
Lyrx [107]

Answer:

See the answer below

Explanation:

<u>The percentage by mass of a substance in a mixture is the ratio of the mass of the substance and the mass of the entire mixture.</u>

<em>Hence, when concentrated HNO3 is said to be 69% by mass of nitric acid, it means that the solution of the acid has 69% HNO3 and 31% water in terms of mass. More specifically, 100 g of the solution of the acid would contain 69 g of HNO3 and 31 g of water respectively,</em>

7 0
3 years ago
An excited ozone molecule, O3*, in the atmosphere can undergo one of the following reactions,O3* → O3 (1) fluorescenceO3* → O +
Maurinko [17]

Answer:

The simplified expression for the fraction  is  \text {X} =    \dfrac{  {k_3  \times cM} }{k_1 +k_2 + k_3 }

Explanation:

From the given information:

O3* → O3                   (1)    fluorescence

O + O2                      (2)    decomposition

O3* + M → O3 + M    (3)     deactivation

The rate of fluorescence = rate of constant (k₁) × Concentration of reactant (cO)

The rate of decomposition is = k₂ × cO

The rate of deactivation = k₃ × cO × cM

where cM is the concentration of the inert molecule

The fraction (X) of ozone molecules undergoing deactivation in terms of the rate constants can be expressed by using the formula:

\text {X} =    \dfrac{ \text {rate of deactivation} }{ \text {(rate of fluorescence) +(rate of decomposition) + (rate of deactivation) }  } }

\text {X} =    \dfrac{  {k_3 \times cO \times cM} }{  {(k_1 \times cO) +(k_2 \times cO) + (k_3 \times cO \times cM) }  }

\text {X} =    \dfrac{  {k_3 \times cO \times cM} }{cO (k_1 +k_2 + k_3  \times cM) }

\text {X} =    \dfrac{  {k_3  \times cM} }{k_1 +k_2 + k_3  }    since  cM is the concentration of the inert molecule

7 0
4 years ago
What is the density of 10cm^3 volume of water that has a mass of 10g?
prisoha [69]

Answer:

The answer is

<h2>1.0 g/cm³</h2>

Explanation:

The density of a substance can be found by using the formula

density =  \frac{mass}{volume} \\

From the question

mass = 10 g

volume = 10 cm³

It's density is

density =  \frac{10}{10}  = 1 \\

We have the final answer as

<h3>1.0 g/cm³</h3>

Hope this helps you

3 0
4 years ago
Read 2 more answers
What is the mass, in grams, of 2.50 * 10-3 mol of aluminum sulfate?
Nonamiya [84]

Answer:

The mass is 0.855 grams (option A)

Explanation:

Step 1: Data given

aluminium sulfate = Al2(SO4)3

Numer of moles Al2(SO4)3 = 2.50 * 10^-3 moles

atomic mass Al = 26.99 g/mol

atomic mass S = 32.065 g/mol

Atomic mass O = 16 g/mol

Step 2: Calculate molar mass Al2(SO4)3

Molar mass = 2* 26.99 + 3*32.065 + 12*16

Molar mas = 342.175 g/mol

Step 3: Calculate mass Al2(SO4)3

Mass Al2(SO4)3 = moles Al2(SO4)3 * molar mass Al2(SO4)3

Mass Al2(SO4)3 = 2.5 *10^-3 moles * 342.175 g/mol

Mass Al2(SO4)3 = 0.855 grams

The mass is 0.855 grams (option A)

5 0
3 years ago
2C3H6O + 8O2 --&gt; 6CO2 + 6H2O
grigory [225]

Answer: 0.45 moles of H_2O will be produced from 0.15 moles of propanol.

Explanation:

The balanced chemical reaction is:

2C_3H_6O+8O_2\rightarrow 6CO_2+6H_2O

C_3H_6O is the limiting reagent as it limits the formation of product and O_2 is the excess reagent.

According to stoichiometry :

2 moles of C_3H_6Oproduces = 6 moles of H_2O

Thus 0.15 moles of V will produce=\frac{6}{2}\times 0.15=0.45moles  of H_2O

Thus 0.45 moles of H_2O will be produced from 0.15 moles of propanol.

7 0
3 years ago
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