What is the atomic mass of Sulfur?

Write the Henderson-Hasselbalch equation for a propanoic acid solution ( CH3CH2CO2H , pKa=4.874 ) using the symbols HA and A− ,

zysi [14]

Answer:

a) [A⁻]/[HA] = 0.227

b) [A⁻]/[HA] = 0.991

c) [A⁻]/[HA] = 2.667

Explanation:

In the Henderson-Hasselbalch equation, HA stands from an acid an A⁻ stands from its conjugate base, as follows:

CH₃CH₂CO₂H = HA

CH₃CH₂CO₂⁻ = A⁻

pH = pka + Log [A⁻]/[HA]

pH = 4.874 + Log[CH₃CH₂CO₂⁻]/[CH₃CH₂CO₂H]

(a)

4.23 = 4.874 + Log [A⁻]/[HA]

-0.644 = Log [A⁻]/[HA]

$10^{-0.644}$ = [A⁻]/[HA]

0.227 = [A⁻]/[HA]

(b)

4.87 = 4.874 + Log [A⁻]/[HA]

-0.004 = Log [A⁻]/[HA]

$10^{-0.004}$ = [A⁻]/[HA]

0.991 = [A⁻]/[HA]

(c)

5.30 = 4.874 + Log [A⁻]/[HA]

0.426 = Log [A⁻]/[HA]

$10^{0.426}$ = [A⁻]/[HA]

2.667 = [A⁻]/[HA]

6 0

3 years ago

A compound has the empirical formula . a 256-ml flask, at 373 k and 750. torr, contains 0.527 g of the gaseous compound. give th

CaHeK987 [17]

Answer: From the ideal gas law, MM=mRTPV; where MM = molecular mass; m = mass; P = pressure in atmospheres; V= volume in litres; R = gas constant with appropriate units. So, 0.800â‹…gĂ—0.0821â‹…Lâ‹…atmâ‹…Kâ’1â‹…molâ’1Ă—373â‹…K0.256â‹…LĂ—0.987â‹…atm = 97.0 gâ‹…molâ’1. nĂ—(12.01+1.01+2Ă—35.45)â‹…gâ‹…molâ’1 = 97.0â‹…gâ‹…molâ’1. Clearly, n = 1. And molecular formula = C2H2Cl2. I seem to recall (but can't be bothered to look up) that vinylidene chloride, H2C=C(Cl)2 is a low boiling point gas, whereas the 1,2 dichloro species is a volatile liquid. At any rate we have supplied the molecular formula as required.

4 0

3 years ago

The element lanthanum has two stable isotopes, lanthanum-138 with an atomic mass of 137.9 AMU and lanthanum-139 with an atomic m

dmitriy555 [2]

Answer:

A. there is an isotope of lanthanum with an atomic mass of 138.9

Explanation:

By knowing the different atomic masses of both Lanthanum atoms, we can not tell anything about their occurence in nature. Therefore, all the last three options are incorrect. Because, the atomic mass does not tell anything about the availability or natural abundance of an element.

Now, the isotopes of an element are those elements, which have same number of electrons and protons as the original element, but different number of neutrons. Therefore, they have same atomic number but, different atomic weight or atomic masses.

Hence, by looking at an elements having same atomic number, but different atomic masses, we can identify them as isotopes.

Thus, the correct option is:

A. there is an isotope of lanthanum with an atomic mass of 138.9.

7 0

2 years ago

The atomic mass is the sum of the number of ________ and ________ in the nucleus of an atom. Choose two answers from the options

iVinArrow [24]

Answer :

The atomic mass is the sum of the number of protons and neutrons in the nucleus of an atom.

Explanation :

Atomic number : it is defined as the number of protons or number of electrons.

Atomic number = number of protons = number of electrons

Mass number or atomic mass : It is defined as the sum of the number of protons and the number of neutrons.

Mass number or atomic mass = Number of protons + Number of neutrons

For example :

Number of protons in carbon = 6

Number of neutrons in carbon = 6

Mass number or atomic mass = Number of protons + Number of neutrons

Mass number or atomic mass = 6 + 6

Mass number or atomic mass = 12

Thus, the atomic mass is the sum of the number of protons and neutrons in the nucleus of an atom.

3 0

3 years ago

Read 2 more answers

Three 1.0-l flasks, maintained at 308 k, are connected to each other with stopcocks. initially the stopcocks are closed. one of

Licemer1 [7]

Answer:

0.6103 atm.

Explanation:

We need to calculate the vapor pressure of each component after the stopcocks are opened.

Volume after the stopcocks are opened = 3.0 L.

1) For N₂:

P₁V₁ = P₂V₂

P₁ = 1.5 atm & V₁ = 1.0 L & V₂ = 3.0 L.

P₂ of N₂ = P₁V₁ / V₂ = (1.5 atm) (1.0 L) / (3.0 L) = 0.5 atm.

2) For H₂O:

Pressure of water at 308 K is 42.0 mmHg.

we need to convert from mmHg to atm: (1.0 atm = 760.0 mmHg).

P of H₂O = (1.0 atm x 42.0 mmHg) / (760.0 mmHg) = 0.0553 atm.

We must check if more 2.2 g of water is evaporated,

n = PV/RT = (0.0553 atm) (3.0 L) / (0.082 L.atm/mol.K) (308 K) = 0.00656 mole.

m = n x cmolar mass = (0.00656 mole) (18.0 g/mole) = 0.118 g.

It is lower than the mass of water in the flask (2.2 g).

3) For C₂H₅OH:

Pressure of C₂H₅OH at 308 K is 102.0 mmHg.

we need to convert from mmHg to atm: (1.0 atm = 760.0 mmHg).

P of C₂H₅OH = (1.0 atm x 102.0 mmHg) / (760.0 mmHg) = 0.13421 atm.

We must check if more 0.3 g of C₂H₅OH is evaporated,

n = PV/RT = (0.13421 atm) (3.0 L) / (0.082 L.atm/mol.K) (308 K) = 0.01594 mole.

m = n x molar mass = (0.01594 mole) (46.07 g/mole) = 0.7344 g.

It is more than the amount in the flask (0.3 g), so the pressure should be less than 0.13421 atm.

We have n = mass / molar mass = (0.30 g) / (46.07 g/mole) = 0.00651 mole.

So, P of C₂H₅OH = nRT / V = (0.00651 mole) (0.082 L.atm/mole.K) (308.0 K) / (3.0 L) = 0.055 atm.

So, total pressure = P of N₂ + P of H₂O + P of C₂H₅OH = 0.5 atm + 0.0553 atm + 0.055 atm = 0.6103 atm.

3 0

2 years ago