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3 years ago

What is the formula for manganese (ii) fluoride decahydrate? (you may use a * to represent the dot in the formula.)?

2 answers:

4 0

These types of molecules are called hydrates. They have a certain number of moles attached to the salt. Their characteristic is being hygroscopic. That means that when they are exposed to air, they readily solvate.

The formula for Manganese Fluoride Decahydrate will involve the formula Mn, F and H₂O. In ionic form, Manganese is Mn⁺² while fluoride is in F⁻. When they are brought together, their superscripts are 'cross-multiplied' and becomes their respective subscripts. The compound becomes MnF₂. Then, we add the decahydrate which means 10 moles of H₂O. Hence, the formula for Manganese Fluoride Decahydrate is MnF₂*10H₂O.

goblinko [34]3 years ago

3 0

Answer:

MnF₂*10H₂O.

Explanation:

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Dennis_Churaev [7]

The neutral grain spirit is essentially known as rectified spirits or ethyl alcohol.

<h3>Meaning of neutral grain spirit</h3>

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2 years ago

Select all that apply. Which of the following are characteristics of bases. contain hydroxide ion or produce it in a solution ta

Blababa [14]

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4 years ago

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State the definition of the partial molar Gibbs energy.

balu736 [363]

Explanation :

As we know that the Gibbs free energy is not only function of temperature and pressure but also amount of each substance in the system.

$G=G(T,P,n_1,n_2)$

where,

$n_1\text{ and }n_2$ is the amount of component 1 and 2 in the system.

Partial molar Gibbs free energy : The partial derivative of Gibbs free energy with respect to amount of component (i) of a mixture when other variable $(T,P,n_j)$ are kept constant are known as partial molar Gibbs free energy of $i^{th}$ component.

For a substance in a mixture, the chemical potential $(\mu)$ is defined as the partial molar Gibbs free energy.

The expression will be:

$\bar{G_i}=\mu_i=\frac{\partial G}{\partial n_i}_{(T,P,n_j)}$

where,

T = temperature

P = pressure

$n_i\text{ and }n_j$ is the amount of component 'i' and 'j' in the system.

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3 years ago

Write a balanced chemical equation, including states of matter, for the combustion of gaseous benzene, c6h6.

snow_tiger [21]

Almost all hydrocarbon 'burn' reactions involve oxygen; it's by far the most reactive substance in air.

Hydrocarbon combustions always involve 
[some hydrocarbon] + oxygen --> carbon dioxide + steam. 

C6H6(l) + O2 (g)--> CO2 (g)+ H2O (g)

Balance carbon, six on each side: 
C6H6(l) + O2 (g)--> 6CO2 (g)+ H2O (g)

Balance hydrogen, six on each side: 
C6H6(l) + O2 (g)--> 6CO2(g) + 3H2O (g)

Now, we have fifteen oxygens on the right and O2 on the left. 
Two ways to deal with that. We can use a fraction: 
C6H6 (l)+ (15/2)O2 (g)--> 6CO2 (g)+ 3H2O (g)

Or, if you prefer to have whole number coefficients, double everything 
to get rid of the fraction: 
2C6H6 (l)+ 15O2 (g)--> 12CO2 (g)+ 6H2O (g)

With the SATP states thrown in... 
C6H6(l) + (15/2)O2(g) --> 6CO2(g) + 3H2O(g)

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3 years ago

Using phenol, dimethyl sulfate, and naoh, show how you would synthesize methyl phenyl ether.

Jlenok [28]

The synthesis of Methyl Phenyl Ether is shown below,

The synthesis takes place in two steps,

Step 1: Formation of Sodium Phenoxide:
Phenol being more acidic than Alcohols with pKa ≈ 10 can loose its proton attached to oxygen atom, as the resulting phenoxide ion is stabilized by resonance. Therefore, when phenol is treated with NaOH (bBase) it looses proton and forms Sodium Phenoxide.

Step 2: Formation of Methyl Phenyl Ether:
The Phenoxide ion formed in first step when treated with Dimethyl Sulfate produces Methyl Phenyl Ether through a Williamson's Ether synthesis Reaction. Dimethyl sulfate is a well known alkylating agent when treated with phenols, thiols and amines. Dimethyl sulfate readily transfers the methyl group and forms NaSO₄CH₃.

The reaction is as follow,

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3 years ago