5. What is the pH of an aqueous solution of the strong acid, Perchloric Acid (HClO4), with a concentration of 0.007 M?

An electron can be added to halogen atom to force a halide ion with

Lilit [14]

An electron can be added to halogen atom to force a halide ion with 8 valence electrons

An atom can be defined as the smallest part of an element which can take part in a chemical reaction.

However whenever, an electron is added to halogen atom to force a halide ion with 8 different valence electrons

So therefore; an electron can be added to halogen atom to force a halide ion with 8 valence electrons

Learn more about halogens:

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4 0

2 years ago

How do you solve this ??

kotegsom [21]

Answer : Option A) 2.00 eV

Explanation : The conversion of J to eV is done with the following formula;

$E_{eV} = E_{J} X (6.241 X 10^{18})$

Here, we have the value of particle in terms of Joules which is 3.2 X $10^{-19}$

So, on substituting we get,
$E_{eV}$ = 3.2 X $10^{-19}$ X $(6.241 X 10^{18} )$

$E_{eV}$ = 1.99 eV so, it can be rounded off to 2.00 eV.

3 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

A 0.1 gram sample of an unknown liquid is vaporized completely at 70 degrees C to fill a 750mL flask. The pressure is 0.05951atm

Deffense [45]

Answer:

The molar mass of the liquid 62.89 g/mol

Explanation:

Step 1: Data given

Mass of the sample = 0.1 grams

Temperature = 70°C

Volume = 750 mL

Pressure = 0.05951 atm

Step 2: Calculate the number of moles

p*V = n*R*T

n = (p*V)/(R*T)

⇒ with n = the number of moles gas = TO BE DETERMINED

⇒ with p = The pressure = 0.05951 atm

⇒ with V = The volume of the flask = 750 mL = 0.750 L

⇒ with R = The gasconstant = 0.08206 L*atm/K*mol

⇒with T = the temperature = 70 °C = 343 Kelvin

n = (0.05951 *0.750)/(0.08206*343)

n = 0.00159 moles

Step 3: Calculate molar mass

Molar mass = mass / moles

Molar mass =0.1 gram / 0.00159 moles

Molar mass = 62.89 g/mol

The molar mass of the liquid 62.89 g/mol

6 0

3 years ago

Cellular respiration uses glucose and oxygen, which have high levels of free energy, and releases carbon dioxide and water, whic

Alja [10]

Answer: Cellular respiration is spontaneous and exergonic. The energy released from the glucose is stored in ATP molelcules.

Explanation:

Spontaneous reactions have an increase in entropy (level of disorder) and a decrease in enthalpy (total energy). Cellular respiration goes from a more ordered state (one molecule of glucose) to a more disordered state (several molecules of CO2), and goes from a state with a lot of free energy to one with much less free energy. As a result, respiration is a spontaneous process.

As free energy from the glucose is released as ATP molecules during oxidation, the reaction is exergonic.

4 0

2 years ago

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