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3 years ago

Which 3 of the following is an exception to the law of superposition

Chemistry

1 answer:

Aleksandr [31]3 years ago

3 0

the understanding of dehidatianismistering can be difficult in the modern world created for these specific terms

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How many grams of NaF should be added to 500 mL of a 0.100 M solution of HF to make a buffer with a pH of 3.2

Korolek [52]

Answer:

2.25g of NaF are needed to prepare the buffer of pH = 3.2

Explanation:

The mixture of a weak acid (HF) with its conjugate base (NaF), produce a buffer. To find the pH of a buffer we must use H-H equation:

pH = pKa + log [A-] / [HA]

<em>Where pH is the pH of the buffer that you want = 3.2, pKa is the pKa of HF = 3.17, and [] could be taken as the moles of A-, the conjugate base (NaF) and the weak acid, HA, (HF). </em>

The moles of HF are:

500mL = 0.500L * (0.100mol/L) = 0.0500 moles HF

Replacing:

3.2 = 3.17 + log [A-] / [0.0500moles]

0.03 = log [A-] / [0.0500moles]

1.017152 = [A-] / [0.0500moles]

[A-] = 0.0500mol * 1.017152

[A-] = 0.0536 moles NaF

The mass could be obtained using the molar mass of NaF (41.99g/mol):

0.0536 moles NaF * (41.99g/mol) =

<h3>2.25g of NaF are needed to prepare the buffer of pH = 3.2</h3>

4 0

3 years ago

The equilibrium constant Kp for the reaction (CH3),CCI (g) = (CH3),C=CH, (g) + HCl (g) is 3.45 at 500. K. (5.00 x 10K) Calculate

Karolina [17]

<u>Answer:</u> The value of $K_p$ for the reaction is 6.32 and concentrations of $(CH_3)_2C=CH,HCl\text{ and }(CH_3)_3CCl$ is 0.094 M, 0.094 M and 0.106 M respectively.

<u>Explanation:</u>

Relation of $K_p$ with $K_c$ is given by the formula:

$K_p=K_c(RT)^{\Delta ng}$

where,

$K_p$ = equilibrium constant in terms of partial pressure = 3.45

$K_c$ = equilibrium constant in terms of concentration = ?

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature = 500 K

$\Delta n_g$ = change in number of moles of gas particles = $n_{products}-n_{reactants}=2-1=1$

Putting values in above equation, we get:

$3.45=K_c\times (0.0821\times 500)^{1}\\\\K_c=\frac{3.45}{0.0821\times 500}=0.084$

The equation used to calculate concentration of a solution is:

$\text{Molarity}=\frac{\text{Moles}}{\text{Volume (in L)}}$

Initial moles of $(CH_3)_3CCl(g)$ = 1.00 mol

Volume of the flask = 5.00 L

So, $\text{Concentration of }(CH_3)_3CCl=\frac{1.00mol}{5.00L}=0.2M$

For the given chemical reaction:

$(CH_3)_3CCl(g)\rightarrow (CH_3)_2C=CH(g)+HCl(g)$

Initial: 0.2 - -

At Eqllm: 0.2 - x x x

The expression of $K_c$ for above reaction follows:

$K_c=\frac{[(CH_3)_2C=CH]\times [HCl]}{[(CH_3)_3CCl]}$

Putting values in above equation, we get:

$0.084=\frac{x\times x}{0.2-x}\\\\x^2+0.084x-0.0168=0\\\\x=0.094,-0.178$

Negative value of 'x' is neglected because initial concentration cannot be more than the given concentration

Calculating the concentration of reactants and products:

$[(CH_3)_2C=CH]=x=0.094M$

$[HCl]=x=0.094M$

$[(CH_3)_3CCl]=(0.2-x)=(0.2-0.094)=0.106M$

Hence, the value of $K_p$ for the reaction is 6.32 and concentrations of $(CH_3)_2C=CH,HCl\text{ and }(CH_3)_3CCl$ is 0.094 M, 0.094 M and 0.106 M respectively.

8 0

3 years ago

Rank the types of galaxies by their ability to form new stars, from most to least activity

ankoles [38]

wfegmgghmzgchmnj . bgmhmxbs

6 0

3 years ago

Calculate the standard enthalpy change of formation of CHA given that the standard enthalpies change of combustion of methane, g

Nostrana [21]

Answer : The standard enthalpy of formation of methane is, -74.8 kJ/mole

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The formation reaction of $CH_4$ will be,

$C(s)+2H_2(g)\rightarrow CH_4(g)$ $\Delta H_{formation}=?$

The intermediate balanced chemical reaction will be,

(1) $CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(l)$ $\Delta H_1=-890kJ/mole$

(2) $C(s)+O_2(g)\rightarrow CO_2(g)$ $\Delta H_2=-393.4kJ/mole$

(3) $H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l)$ $\Delta H_3=-285.7kJ/mole$

Now we will reverse the reaction 1, multiply reaction 3 by 2 then adding all the equations, we get :

(1) $CO_2(g)+2H_2O(l)\rightarrow CH_4(g)+2O_2(g)$ $\Delta H_1=890kJ/mole$

(2) $C(s)+O_2(g)\rightarrow CO_2(g)$ $\Delta H_2=-393.4kJ/mole$

(3) $2H_2(g)+2O_2(g)\rightarrow 2H_2O(l)$ $\Delta H_3=2\times (-285.7kJ)=-571.4kJ/mole$

The expression for enthalpy of formation of $CH_4$ will be,

$\Delta H_{formation}=\Delta H_1+\Delta H_2+\Delta H_3$

$\Delta H=(+890kJ/mole)+(-393.4kJ/mole)+(-571.4kJ/mole)$

$\Delta H=-74.8kJ/mole$

Therefore, the standard enthalpy of formation of methane is, -74.8 kJ/mole

3 0

3 years ago

How would you write and name the ions for tin?

Yakvenalex [24]

For Single Element Ions (e.g. K+ , Mg2+, P3-) To name positive (+) ions write the name as from the Periodic Table and add the word 'ion' afterwards. To name negative (-) ions write the name from the Periodic Table but replace the ending with 'ide'. Put the word 'ion' after the name.

3 0

4 years ago