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3 years ago

Which of the following is a possible formula unit?

Chemistry

1 answer:

Black_prince [1.1K]3 years ago

5 0

<h3><u>Answer;</u></h3>

AlF₃

<h3><u>Explanation;</u></h3>

A formula unit is the empirical formula of any ionic or covalent compound.

In this case; the possible formula unit; AlF₃

Aluminium fluoride is a chemical compound. Its chemical formula is AlF₃.

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In the laboratory, a general chemistry student measured the pH of a 0.529 M aqueous solution of phenol (a weak acid), C6H5OH to

Artyom0805 [142]

Answer:

The dissociation constant of phenol from given information is $9.34\times 10^{-11}$ .

Explanation:

The measured pH of the solution = 5.153

$C_6H_5OH\rightarrow C_6H_5O^-+H^+$

Initially c

At eq'm c-x x x

The expression of dissociation constant is given as:

$K_a=\frac{[C_6H_5O^-][H^+]}{[C_6H_5OOH]}$

Concentration of phenoxide ions and hydrogen ions are equal to x.

$pH=-\log[x]$

$5.153=-\log[x]$

$x=7.03\times 10^{-6} M$

$K_a=\frac{x\times x}{(c-x)}=\frac{x^2}{(c-x)}=\frac{(7.03\times 10^{-6} M)^2}{ 0.529 M-7.03\times 10^{-6} M}$

$K_a=9.34\times 10^{-11}$

The dissociation constant of phenol from given information is $9.34\times 10^{-11}$ .

4 0

3 years ago

Fish breathe the dissolved air in water through their gills. Assuming the partial pressures of oxygen and nitrogen in air to be

amid [387]

Answer:

X(O₂) = 0.323

X(N₂) = 0.677

Explanation:

We have the partial pressures of oxygen (O₂) and nitrogen (N₂):

P(O₂) = 0.20 atm

P(N₂) = 0.80 atm

In order to solve the problem, you need the solubilities of each gas in water at 298 K. We can consider 1.3 x 10⁻³ mol/(L atm) for oxygen (O₂) and 6.8 x 10⁻⁴mol/(L atm) for nitrogen (N₂) from the bibliography.

s(O₂) = 1.3 x 10⁻³ mol/(L atm)

s(N₂) = 6.8 x 10⁻⁴mol/(L atm)

So, we calculate the concentration (C) of each gas as the product of its partial pressure (P) and the solubility (s):

C(O₂) = P(O₂) x s(O₂) = 0.20 atm x 1.3 x 10⁻³ mol/(L atm) = 2.6 x 10⁻⁴mol/L

C(N₂) = P(N₂) x s(N₂) = 0.80 atm x 6.8 x 10⁻⁴mol/(L atm) = 5.44 x 10⁻⁴ mol/L

In 1 liter of water, we have the following number of moles (n):

n(O₂) = 2.6 x 10⁻⁴ mol

n(N₂) = 5.44 x 10⁻⁴ mol

Thus, the total number of moles (nt) is calculated as the sum of the number of moles of the gases in the mixture:

nt = n(O₂) + n(N₂) = 2.6 x 10⁻⁴ mol + 5.44 x 10⁻⁴ mol = 8.04 x 10⁻⁴ mol

Finally, the mole fraction of each gas is calculated as the ratio between the number of moles of each gas and the total number of moles:

X(O₂) = n(O₂)/nt = 2.6 x 10⁻⁴ mol/(8.04 x 10⁻⁴ mol) = 0.323

X(N₂) = n(N₂)/nt = 5.44 x 10⁻⁴ mol/(8.04 x 10⁻⁴ mol) = 0.677

5 0

2 years ago

Determine the volume of an object that has a mass of 455.6 g and a density of 19.3 g/cm3. (1cm3 = 1ml) select one:

Dmitry [639]

The answer is B. The density, or more precisely, the volumetric mass density, of a substance is its mass per unit volume.

4 0

3 years ago

Tyrosine absorbs maximally at 290 nm. A 0.2162 g sample containing some amount of Tyrosine was placed into a 250.0 mL volumetric

mojhsa [17]

Answer:

B. 0.069 %

Explanation:

It should be initially noted in this answer in particular that, the amino acids tryptophan and tyrosine have a very precise 280 nm absorption rate, which allows a direct A280 size of protein concentration. The 280 nm UV absorbance rate is regularly utilized to approximate protein concentration in laboratories due to its simplistic nature, its affordability and also the ease of usage.

kindly check the attached image below for the solution to the above question.

7 0

3 years ago

Calculate the DH°rxn for the decomposition of calcium carbonate to calcium oxide and carbon dioxide. DH°f means delta or change

USPshnik [31]

Answer: +178.3 kJ

Explanation:

The chemical equation follows:

$CaCO_3(s)\rightarrow CaO(s)+CO_2(g)$

The equation for the enthalpy change of the above reaction is:

$\Delta H^o_{rxn}=[(1\times \Delta H^o_f_{(CaO(s))})+(1\times \Delta H^0f_{CO_2}]-[(1\times \Delta H^o_f_{(CaCO_3(s))})]$

We are given:

$\Delta H^o_f_{(CaO(s))}=-635.1kJ/mol\\\Delta H^o_f_{(CaCO_3(s))}=-1206.9kJ/mol\\\Delta H^o_f_{(CO_2(g))}=-393.5kJ/mol\\\Delta H^o_{rxn}=?$

Putting values in above equation, we get:

$\Delta H^o_{rxn}=[(1\times (-635.1))+(1\times (-393.5))]-[(1\times (-1206.9))]$

The DH°rxn for the decomposition of calcium carbonate to calcium oxide and carbon dioxide is +178.3 kJ

5 0

2 years ago