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3 years ago

How many traditional areas of study can chemistry be divided into

1 answer:

6 0

<span>There is five main area of study in Chemistry, these are:

Analytical, this focusses on experimental equipment and methods used in chemistry (e.g., NMR, Spectroscopic methods, etc.)

Biochemistry - focuses on the chemistry of compounds and processes in living things (e.g., amino acids, proteins, DNA, cellular respiration, Krebs cycle, etc.)

Organic - focuses on the chemistry on most carbon-based molecules found in living things (e.g., hydrocarbons, alcohols, carbolic acids. Amines, ester, etc.)

Inorganic - (focuses on all elements other than carbon (e.g., fluorine, silicon, xenon, etc.)

Physical - focuses on the basic structure and energetic son atoms and molecules (e.g., subatomic structure, is nice and covalent bonding, thermodynamics, reactions, etc.)</span>

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Mencionar 5 objetos o sustancias químicas que tengan Carbono.

RoseWind [281]

Answer:

agua

madera

plastico

ropa

bebida carbonatada

cáscara

Explanation:

6 0

3 years ago

A common antacid contains the acid-neutralizing agent sodium hydrogen carbonate. Write the chemical formula for sodium hydrogen

kaheart [24]

Sodium hydrogen carbonate has:
$sodium \: ion ( {Na}^{+}), hydrogen \\ ion ( {H}^{+}), and \: carbonate \\ ion ( {CO3}^{2-})$
The + from Na+ and + from H+ makes 2+, and therefore the 2- from CO3 evens the charges to neutral.

Not to be confused with Sodium Carbonate, (Na2CO3), which lack an H, Sodium Hydrogen Carbonate is sometimes shortened to just Sodium Bicarbonate. It is used in the kitchen (cooking) as "baking soda."

Formula: NaHCO3

3 0

3 years ago

You calculate the Wilson equation parameters for the ethanol (1) 1 1-propanol (2) system at 258C and find they are L12 5 0.7 and

Gala2k [10]

Here is the correct question.

You calculate the Wilson equation parameters for the ethanol (1) +1 - propanol (2) system at 25° C and find they are ∧₁₂ = 0.7 and ∧₂₁ = 1.1 . Estimate the value of parameters at 50° C

Answer:

the values of the parameters at 50° C are 0.766 and 1.047

Explanation:

From "critical point , enthalpy of phase change and liquid molar volume " the liquid molar volume (v) of ethanol and 1 - propanol is represented as follows:

Compound Liquid molar volume (cm³/mol)

Ethanol (1) 58.68

1 - propanol (2) 75.14

To calculate the temperature dependent parameters of the Wilson equation ∧₁₂.

∧₁₂ = $\frac{V_2}{V_1} \ exp \ (\frac{-a_{12}/R}{T} )$ ------------ equation (1)

where:

$a_{12}/R$ = Wilson parameter = ???

$V_2$ = liquid molar volume of component 2 = 75.14 cm³/mol

$V_1$ = liquid molar volume of component 1 = 58.68 cm³/mol

T = temperature = 25° C = ( 25 + 273.15) K = 298.15 K

Replacing our values in the above equation ; we have:

$0.7 = \frac{75.14 \ cm^3/mol}{58.68 \ cm^3/mol} \ exp \ (\frac{-a_{12}/R}{298.15 \ K} )$

$0.7 = 1.281 \ exp \ (\frac{-a_{12}/R}{298.15 \ K} )$

$In (0.547) = \ (\frac{-a_{12}/R}{298.15 \ K} )$

$-a_{12}/R= 0.60 * 298.15 \ K$

$-a_{12}/R= - 178.89 \ K$

$a_{12}/R= 178.89 \ K$

To calculate the temperature dependent parameters of the Wilson equation ∧₂₁

∧₂₁ = $\frac{V_1}{V_2} \ exp \ (\frac{-a_{12}/R}{T} )$ ---------- equation (2)

$1.1 = \frac{58.68 \ cm^3/mol}{75.14 \ cm^3/mol} \ exp \ (\frac{-a_{12}/R}{298.15 \ K} )$

$1.1 = 0.7809 \ exp \ (\frac{-a_{12}/R}{298.15 \ K} )$

$\frac{1.1}{0.7809}= exp \ (\frac{-a_{12}/R}{298.15 \ K} )$

$1n ( 1.4086)= (\frac{-a_{12}/R}{298.15 \ K} )$

$-a_{12}/R = 0.3426 * 298.15 \ K$

$-a_{12}/R =102.15 \ K$

$a_{12}/R = -102.15 \ K$

From equation (1) ; let replace 178.98 K for $a_{12}/R$

$V_2 =$ 75.14 cm³/mol

$V_1$ = 58.68 cm³/mol

T = 50° C = ( 50 + 273.15 ) K = 348.15 K

So;

∧₁₂ = $\frac{75.14 \ cm^3/mol}{58.68 \ cm^3/mol} \ exp \ (\frac{- 178,.89 \ K}{348.15 \ K} )$

∧₁₂ = 1.281 exp(-0.5138)

∧₁₂ = 1.281 × 0.5982

∧₁₂ =0.766

From equation 2; let replace 102.15 K for $a_{12}/R$

$V_2 =$ 75.14 cm³/mol

$V_1$ = 58.68 cm³/mol

T = 50° C = ( 50 + 273.15 ) K = 348.15 K

So;

∧₂₁ = $\frac{58.68 \ cm^3/mol}{75.14 \ cm^3/mol} \ exp \ (\frac{-(-102.15)\ K}{298.15 \ K} )$

∧₂₁ = 0.7809 exp (0.2934)

∧₂₁ = 0.7809 × 1.3410

∧₂₁ = 1.047

Thus, the values of the parameters at 50° C are 0.766 and 1.047

8 0

3 years ago

The same elements may combine in different ratios to form the same compound?

xeze [42]

Ans:

It is also known as law of multiple proportion. It states that if two elements form more than one compound between them then the ratios of the masses of the second element which combine with the fixed mass of first element will be ratios of small whole numbers.

Hope it helps!

5 0

2 years ago

Read 2 more answers

What is the molality of a solution of water and KCl if the freezing point of the solution is –3°C? (Kf = 1.86°C/m; molar mass of

Elina [12.6K]

Some of the solutions exhibit colligative properties. These properties depend on the amount of solute dissolved in a solvent. These properties include freezing point depression, boiling point elevation, osmotic pressure and vapor pressure lowering. Calculations are as follows:

<span> ΔT(freezing point) = (Kf)mi
3 = 1.86 °C kg / mol (m)(2)
3 =3.72m
m = 0.81 mol/kg</span>

4 0

3 years ago

Read 2 more answers